# How to Go From a 48 to 51 in GMAT Quant – Part II This post is continuation of last week’s post which you can check here.

Another method of saving time on simple questions – use data given in one statement to examine the other!

Now you might think we have lost it! After all, you know very well that in Data Sufficiency questions of GMAT, you must examine each statement independently. You CANNOT use data from one to analyze the other – absolutely correct. So you should ignore the other statement completely while examining one – hmm, not necessarily!

Sometimes, one statement could give us ideas about the next one such that we could save time while examining it. Needless to say, we need to be very careful but it certainly is a useful strategy. Also, it could help us verify that our calculations are correct. Here is why…

When we say DS question, think of a puzzle. The question stem gives you the statement of a puzzle ending with something like “What is the value of x?” or “Is x 7?” etc. You have to answer the question asked in the puzzle. Think of the two statements that come with the question as clues to the puzzle. So the puzzlemaster gives you the first clue (statement 1) and asks you: can you answer the question now? If you are able to, your answer is either (A) or (D).

Then he tells you to ignore the first clue and gives you another clue (statement 2). Again he asks you: can you answer the question now? Again, you may or may not able to. If you are able to, your answer will be (B) or (D) depending on how you fared in statement 1. If you are unable to answer the question, he tells you to consider both statements together and then try to answer. If you are able to, your answer is (C).

The point to note here is that both clues lead you to answer the same puzzle. Say if the puzzle is: What is x? If clue 1 tells you that x is 6, clue 2 cannot tell you that x is 9. They both must lead you to the same value of x. Clue 1 could tell you that x is either 6 or 8 and clue 2 could tell you that x is either 8 or 9. In this case, when we use both clues together, we find that x must be 8 to satisfy both. Hence the statements never contradict each other. This means, if we get possible values of x from statement 1, we know that statement 2 will also give us at least one of those values.

This is how one statement could give us a starting point for the next one. Now that you understand the “why”, let’s go on to “how”, using a question.

Question: If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

Statement 1: N is divisible by 3

Statement 2: N is divisible by 7

Solution:

Given: N = 4321 + K

1 <= K < 10

So N could range from 4322 (when K = 1) to 4330 (when K = 9). To find the value of K, we need to find the unique value of N.

Statement 1 tells us that N is divisible by 3.

4321 is not divisible by 3 since the sum of its digits is 4+3+2+1 = 10. It is 1 more than a multiple of 3. So the next multiple of 3 will be 4323. Hence N could be 4323. But there are some other multiples of 3 which could be the value of N. After 4323, 4326 and 4329 could also be the values of N since they are multiples of 3 too. We know this because if A is a multiple of 3, A+3, A+6, A+9, A-3, A-6 etc are also multiples of 3. So since 4323 is a multiple of 3, 4326 and 4329 will also be multiples of 3. We did not get a unique value for N so statement 1 alone is not sufficient.

Now let’s go on to statement 2. This tells us that N must be a multiple of 7. In 10 consecutive numbers, there will be either one multiple of 7 or two multiples of 7. If there is only one multiple of 7 in the range 4322 to 4331, statement 2 alone will be sufficient to give us the value of N. If there are two multiples of 7 in this range, then statement 2 alone will not be sufficient.

Recall that from statement 1, we already know that N will take one of three values: 4323, 4326 or 4329.

Let’s check for 4326 because it is in the middle. If 4326 is divisible by 7, there will be no other multiple of 7 in the range 4322 to 4331 because the closest multiples of 7 to 4326 will be 4326 – 7 and 4326 + 7. When we divide 4326 by 7, we find that it is divisible. This means that statement 2 gives us a single value of N. Hence statement 2 alone is sufficient.

Hypothetically, what if we had found that 4326 is not divisible by 7? Then we would have known that either 4323 or 4329 must be a multiple of 7. In both cases, statement 2 would have given us 2 multiples of 7 because both 4330 (7 more than 4323) and 4322 (7 less than 4329) are in the possible range. Then we would have known that the answer will be (C) i.e. we will need both statements to answer the question since the possible values from the two statements will have only one overlap in either case.

Note that what we gleaned from statement 1 helped us quickly examine statement 2 and get to the answer right away. But this is an advanced technique and you should use it only if you understand it very well. Else, it is best to stick to completely ignoring one statement while working on the other.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# How to Go From a 48 to 51 in GMAT Quant – Part I People often ask – how do we go from 48 to 51 in Quant? This question is very hard to answer since we don’t have a step by step plan – do theory from here – do questions from there – take a test from here – read posts from there etc. Today and in the next few weeks, we will discuss how to go from 48 to 51 in Quant.

Above Q48, the waters are pretty choppy! Questions are hard less because of the content and more because they look so unique – even though they’re testing the same concepts.  Training yourself to see familiarity in the obscure is difficult, and that happens from seeing a lot of problems. There is barely any scope for making silly mistakes – you must run through all simple questions quickly and neatly, leaving you plenty of time to think through the tougher ones. It’s important to have enough time and keep your cool, which is easier to do if you have more time.

The question for today is: how do you handle simple questions quickly? We have mentioned many times that most GMAT Quant questions do not need Algebra. We can easily solve them by just analyzing while reading the question stem!

Here is how we can do that:

Question: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

Solution:  This is a pretty simple non-tricky PS question. To solve it, most people use an algebraic method which looks something like this.

Girls in school A : Girls in school B = 4 : 3
So number of girls in school A = 4n and number of girls in school B = 3n
Since in school A, 40% students are girls and 60% are boys, number of boys is 6n.
Since in school B, 60% students are girls and 40% students are boys, number of boys is 2n.

If we transfer 20 boys from school A, we are left with 6n – 20 and when 20 boys are added to school B, we get 2n + 20 boys in school B.
(6n – 20)/(2n + 20) = 5/3
You get n = 20

Boys at school A after transfer = 6*20 – 20 = 100
Boys at school B after transfer = 2*20 + 20 = 60
Difference = 40

This method gives you the correct answer, obviously, but it does take quite a bit of time. On the other hand, this is what should go through your mind while reading the question if you are focused on using logic:

“School A is 40% girls and school B is 60% girls.”

School A – 40% girls
School B – 60% girls

“The ratio of the number of girls at school A to the number of girls at school B is 4:3”
When we read this line, we should take a step back to the previous line with the % figures. We see that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100 (use easy numbers). So school A has 80 girls while school B has 60 girls. This gives us a ratio of 4:3. (If you do not get 4:3 on your first try, you should tweak the assumed numbers a bit but you should stick to simple numbers.) Then verify the rest of the data against these numbers and get your answer.

School A has 120 boys and school B has 40 boys. Transfer 20 boys from school A to school B to get 100 boys in school A and 60 boys in school B giving us a difference of 40 boys.

This takes lesser time but requires some ingenuity. That could be the difference between Q48 and Q51.

Hope this gave you some ideas. Try the reasoning approach on other simple questions. With practice, you can save a ton of time!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Find the Correct Answer for Diagonals of a Polygon in This GMAT Question In today’s post, let us discuss n sided polygons and the number of diagonals they have.

We will discuss the following:

1. How do we find the number of diagonals an n sided polygon has?
2. How many diagonals are subtended by each vertex?
3. What happens when one or more vertices do not make diagonals?

Given an n sided polygon, how many diagonals will it have?

An n sided polygon has n vertices. If we join every distinct pair of vertices we will get nC2 lines. These nC2 lines include the n sides of the polygon as well as its diagonals.
So the number of diagonals is given by the expression nC2 – n.
nC2 – n = n(n-1)/2 – n = n(n – 3)/2

Alternatively, you can think of it this way – every vertex makes a diagonal with (n – 3) vertices. It does not make a diagonal with itself, and the two vertices next to it on either side (since it forms sides with these two). So we get n*(n – 3) diagonals. But here, each diagonal is double counted, once for each of its two vertices. Hence we divide n*(n – 3) by 2 to get the actual number of diagonals. That is another way of arriving at n(n – 3)/2

Let’s take some examples to solidify this concept:

Example 1: How many diagonals does a polygon with 25 sides have?
No. of diagonals = nC2 – n = n(n – 3)/2 = 25*(25 – 3)/2 = 275

Example 2: How many diagonals does a polygon with 20 sides have, if one of its vertices does not send any diagonal?

The number of diagonals of a 20 sided figure = 20*(20 – 3)/2 = 170

But one vertex does not send any diagonals. Each vertex makes a diagonal with (n-3) other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right. With all other vertices, it makes a diagonal. So, we need to remove 20 – 3 = 17 diagonals from the total.

Total number of diagonals if one vertex does not make any diagonals = 170 – 17 = 153 diagonals.

We hope everything done till now makes sense so that we can build on the concept a bit. Now let’s we will give we a question with two solutions and two different answers. We have to find out the correct answer and explain why the other is wrong.

Question: How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

Answer: We will use two different methods to solve this question:

Method 1: Using the formula discussed above

Number of diagonals in a polygon of 18 sides = 18*(18 – 3)/2 = 135 diagonals

Each vertex makes a diagonal with n-3 other vertices – as discussed before.

So, each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any diagonals = 135 – 15*3 = 135 – 45 = 90 diagonals.

Method 2:
The polygon has a total of 18 vertices. 3 vertices do not participate so we need to make all diagonals that we can with 15 vertices.

Number of lines we can make with 15 vertices = 15C2 = 15*14/2 = 105

But this 105 includes the sides as well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices do not participate, 4 sides will not be formed. 15 vertices will have 14 sides which will be a part of the 105 we calculated before.

Total number of diagonals if 3 vertices do not send any diagonals = 105 – 14 = 91

Note that the two answers do not match. Method 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct and the correct one is method 2. So then, what is the problem with method 1?

When we subtract 45 from 135 (for each of the 15 diagonals made by the 3 vertices which we need to ignore), we are double counting 1 diagonal in this figure of 45. We actually need to subtract only 44 diagonals. Which diagonal are we double counting? The one which connects 2 of the three ignored vertices. Try to make a polygon with a few vertices. Make a few diagonals. Remove 3 vertices next to each other. 2 of the three vertices which have a vertex between them will be joined by a diagonal. When we remove 15 diagonals for each vertex, we are removing that diagonal twice.

Hence, what we need to do is 135 – 44 = 91.

The correct answer is 91 diagonals.

Mind you, we assumed that the vertices which were removed were next to each other. If they are not, the answer would be different since there would be more double counting. Let’s take an example of that.

Question: How many diagonals does a polygon with 18 sides have if three of its non-adjacent vertices, do not send any diagonals? (No two of the three vertices are next to each other)

Number of diagonals in a polygon of 18 sides = 18*(18 – 3)/2 = 135 diagonals

Each vertex makes a diagonal with n-3 other vertices – as discussed before.

So, each vertex will make 15 diagonals.

Since 3 vertices are not counted, they will not send out each of these 15 diagonals i.e. 45 diagonals. But in this figure of 45, we have double counted 3 diagonals. To understand this, say we number the vertices from 1 to 18. Say, we leave out vertices 1, 4 and 16.

Vertex 1 makes diagonals with vertices 3, 4, 5 … 16, 17 (a total of 15 diagonals)

Vertex 4 makes diagonals with vertices 6, 7, 8, … 16, 17, 18, 1, 2 (a total of 15 diagonals)

Vertex 16 makes diagonals with vertices 18, 1, 2, 3, 4, … 15, 16 (a total of 15 diagonals)

In our total of 45, we have counted the diagonal of vertices 1 and 4 twice. We have also counted the diagonal of vertices 1 and 16 twice and diagonal of vertices 4 and 16 twice. So, we have double counted 3 diagonals in our figure of 45. We need to subtract only 42 diagonals from 135.
Total number of diagonals if 3 vertices do not send any diagonals = 135 – 42 = 93 diagonals.

The answer here is 93 diagonals.

Note that in the previous example, only 1 diagonal was double counted while here, 3 diagonals were double counted. Understand why – in the previous example, since the vertices were adjacent, they made sides of the polygon. So, assuming vertices 1, 2 and 3 were ignored, 1 and 2 joined to make a side of the polygon and 2 and 3 joined to make a side of the polygon. Only vertices 1 and 3 joined to make a diagonal and hence this diagonal was double counted. In our formula n(n – 3)/2, we have already gotten rid of the sides so they don’t come in the picture at all. The formula only gives us the number of diagonals. So, in the previous example, we had already ignored the two sides formed by adjacent vertices (when we used the formula n(n-3)/2) and had to take care of double counting of only one diagonal. In this example, each ignored vertex made a diagonal with the other ignored vertex and hence we had to handle the double counting of 3 diagonals.

We hope all this is clear to you and you will be able to effortlessly handle any question regarding diagonals of polygons now.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Want more practice with diagonals-of-polygons and other geometric concepts? Check out the Veritas Prep Question Bank for hundreds of free practice problems.

# The Reason Behind Absolute Value Questions on the GMAT Even after working extensively on absolute value questions, sometimes students come up with “why?” i.e. why do we have to take positive and negative values? Why do we have to consider ranges etc. They know the process but they do not understand the reason they need to follow the process. So here today, in this post, we will try to explain the reason.

You know how to solve an equation such as x + 2x = 4. Simple enough, right? Just add x with 2x to get 3x and separate out the x on one side. But what do you do when you have an equation with absolute values? How will you handle that equation? Say, you have |x| + 2x = 4. Is this your regular equation? No! You CANNOT say that x + 2x = 4 => 3x = 4 => x = 4/3. You have an absolute value and that complicates matters. You need to get rid of it to get a solution for x. How do you get rid of absolute values? The definition of absolute value helps us here:

|x| = x if x >= 0

|x| = -x if x < 0

So you can substitute x for |x| to make it a regular equation but only if x is non negative. If x is negative, then you put -x instead of |x| to convert it into a simple equation. And that is the reason you need to take positive and negative values of what is inside the absolute value sign.

Similarly,

|x-5| = (x-5) if (x-5) >= 0 i.e. if x >= 5

|x-5| = -(x-5) if (x-5) < 0 i.e. if x < 5

Let’s go back to the previous example and see how we can get rid of the absolute value to make it a regular equation:

Question 1: What is the value of x given |x| + 2x = 4?

We don’t know whether x is positive or negative so we will look at what happens in both cases:

Case 1: x is positive or 0

If x >= 0 then equation becomes x + 2x = 4 => x = 4/3

Our initial condition is that x is non negative. We get a positive solution on solving it and hence 4/3 is a valid solution.

Case 2: x is negative

If x < 0 then equation becomes -x + 2x = 4 => x = 4

Our initial condition is that x is negative. We get a positive solution on solving it and hence x = 4 is not a valid solution. Had we obtained a negative solution, it would have been valid.

So there is only one solution x = 4/3.

We hope the entire process makes more sense now. Let’s follow it up with a complex question from our algebra book.

Question 2:  If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

Statement 1: x < 4

Statement 2: x > -3

Solution: Now what do you do when you have y = |x+3| + |4-x|? How do you convert this into a regular equation? You don’t know whether whatever is in the absolute value sign is positive or negative. How will you get rid of the sign then? You will work on all the cases (messy algebra coming up!).

Now, we see the same logic in this question:

y = |x+3| + |4-x|

|x+3| = (x+3) if (x+3) >= 0. In other words, if x >= -3

|x+3| = -(x+3) if (x+3) < 0. In other words, if x < -3

|4-x| = (4-x) if (4-x) >= 0. In other words, if x <= 4

|4-x| = -(4-x) if (4-x) < 0. In other words, if x > 4

So our absolute values behave differently when x < -3, between -3 and 4 and when x > 4. We say that -3 and 4 are our transition points.

Case 1:

When x < -3, |x+3| = -(x+3) and |4-x| = (4-x).

So the equation becomes y = -(x+3) + (4-x)

y = 1 – 2x

For different values of x, y will take different values. Recall that x must be less than -3. Say x = -4, then y = 9. If x = -5, y = 11.

Case 2:

When -3 <= x <= 4, |x+3| = (x+3) and |4-x| = (4-x).

So the equation becomes y = (x+3) + (4-x)

y = 7

In this range, y will always be 7.

Case 3:

When x > 4, |x+3| = (x+3) and |4-x| = -(4-x)

So the equation becomes y = (x+3) – (4-x)

y = 2x – 1

For different values of x, y will take different values. Recall that x must be more than 4. Say x = 5, then y is 9. If x = 6, then y is 11.

Note that y equals 7 only when x is between -3 and 4. Both statements together tell us that x is between -3 and 4. No statement alone gives us this information. Hence, using both statements, we know that y must be 7.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Easy (A)/(B) Trap in Data Sufficiency Questions on the GMAT We know that ‘Easy C’ is a common trap of DS questions – have you wondered whether there could be trap called ‘Easy A/B’ such that the answer would actually be (C)? Such questions also exist! The point is that whenever you feel that the question was way too simple, you might want to take a step back and review. GMAT will try every trick in the trade to delineate you. Let us show you a question which looks like an easy (A) but isn’t:

Question: 25 integers are written on a board. Are there at least two consecutive integers among them?

Statement 1: For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

Statement 2: At least one value occurs more than once in the list.

Solution: Let’s first review the information given to us here:

25 integers are written on the board – we don’t know whether they are all distinct. We want to know if there is any pair of consecutive integers among them.

Let’s look at the statements:

Statement 1: For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

It is easy to fall for statement 1 and think that it is sufficient alone. Say, if any single value is increased by 1 and it doesn’t match any other value already there in the list, it means that there are no consecutive integers, doesn’t it? Well, no! But we will talk about that in a minute. Let’s first look at why we might think that statement 1 is sufficient.

Say, the numbers are: 1, 5, 8, 10, 35, 76 …

If you increase 1 by 1, you get 2 and the list looks like this:

Now the numbers are 2, 5, 8, 10, 35, 76 …

Note that the number of distinct integers is the same.

Had there been two consecutive integers such as 1, 2, 8, 10, 35, 76 …

If we increase 1 by 1, the list would have become 2, 2, 8, 10, 35, 76 … – this would have decreased the number of distinct integers.

You might be tempted to say here that statement 1 alone is sufficient. What you might forget is that when you increase a number by 1, one distinct integer could be getting wiped out and another taking its place! It may not occur to you that the case might be different when one value occurs more than once, but statement 2 should give you a hint. Obviously, statement 2 alone is not sufficient but let’s analyze what happens when we take both statements together.

Since statement 1 doesn’t tell you that all values are distinct, statement 2 should make you think that you need to consider the case where one value occurs more than once in the list. In that case, is it possible that number of different values in the list does not change even though there is a pair of consecutive integers?

Say the numbers are 1, 1, 2, 8, 10, 35, 76 …
Now if you increase 1 by 1, the list would look like 1, 2, 2, 8, 10, 35, 76 …

Here, the number of distinct integers stays the same even when you increase a number by 1 and you have consecutive integers! In this case, if there were no consecutive integers, the number of distinct integers would have increased. Hence if the numbers are not all distinct and the number of distinct numbers needs to stay the same, there must be a pair of consecutive integers.

This tells you that statement 1 is not sufficient alone but both statements together answer the question with a ‘Yes’.

Takeaway – Just as when you get an easy (C), you must check whether the answer could be (A) or (B), when you feel that the answer is an easy (A) or (B), you might want to check whether the other statement gives some relevant data and is necessary.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Understanding Conjunctions on the GMAT We would like to discuss a bit about conjunctions today – just whatever is relevant for GMAT. We will start by defining the kinds of conjunctions, then move on to the different ways in which they are used, and finally, we will see how they can be tested in a question.

Conjunction is a word that connects or joins together words, phrases, clauses, or sentences. There are two kinds of conjunctions:

1. Coordinating conjunctions – Connect two equal parts of a sentence

Further, coordinating conjunctions are of two types:

Pure Conjunctions – and, but, or, for, nor, yet, so (the first letters of these make the acronym FANBOYS) – try to keep these in mind.

Conjunctive Adverbs – These words sometimes act as conjunctions and at other times, as adverbs – accordingly, in fact, again, instead, also, likewise, besides, moreover, consequently, namely, finally, nevertheless, for example, otherwise, further, still, furthermore, that is, hence, then, however, therefore, indeed, thus

2. Subordinating conjunctions – Connect two unequal parts of a sentence e.g. independent and dependent clauses – after, since, when, although, so that, whenever, as, supposing, where, because, than, whereas, before, that, wherever, but that, though, whether, if, though, which, in order that, till, while, lest, unless, who, no matter, until, why, how, what, even though 1. Two independent clauses can be joined by a comma and a pure conjunction. However, a comma by itself will not work to join together two sentences and will create a comma splice!

Examples:

The rain slashed the town, and the people scurried for shelter.

The policeman dodged the bullets, but a bystander was shot.

If you omit the conjunctions ’or’ and ‘but’ above, you will create a comma splice.

2. When two independent clauses are joined by a conjunctive adverb we need to insert a semicolon between the two clauses. Note that conjunctive adverbs are not really full conjunctions, and they can’t do that job by themselves. It is the semicolon that does the real job of joining the two independent clauses.

Examples:

The rain slashed the town; furthermore, the people scurried for shelter.

The policeman dodged the bullets; however, a bystander was shot.

Note that if we use a comma instead of a semicolon in the examples above, we will create a comma splice.

3. A dependent clause at the beginning of a sentence is introductory, and it is usually followed by a comma.

Examples:

While the rain slashed the town, the people scurried for shelter.

Although the policeman dodged the bullets, a bystander was shot.

On the other hand, no punctuation is necessary for the dependent clause following the main clause.

Let’s take one of our own questions to understand the application of these concepts:

Question: Unlike the previous year’s bidding, the contract was awarded not simply to the firm offering to complete the work on time for the least cost; the thoroughness of the design submission was also factored into the decision.

(A) Unlike the previous year’s bidding, the contract this year was awarded not simply to the firm offering to complete the work on time for the least cost;

(B)   This year, unlike last year, the contract was awarded not simply to the firm offering to complete the work on time for the least cost;

(C)   Unlike the previous year’s bidding, this year the contract was awarded not simply to the firm offering to complete the work on time for the least cost;

(D)   Unlike the previous year’s bidding, the bidding for the contract this year was awarded not simply to the firm offering to complete the work on time for the least cost, instead

(E)    Unlike the previous year’s bidding, the contract’s bidding this year were awarded not simply to the firm offering to complete the work on time for the least cost;

Solution: Other than the comparison errors contained in (A) – compares bidding with contract – and (C) – compares bidding with year – we have sentence structure errors.

There are two independent clauses here:

–          the contract was awarded not simply to the firm offering to complete the work on time for the least cost.

–          the thoroughness of the design submission was also factored into the decision.

There are two ways to join them – we can use a conjunction or a semi colon. Options (A), (B), (C) and (E) use a semi colon.

Option (D) tries to use a conjunction with a comma but note that “instead” is a conjunctive adverb. It needs a semi colon before it. The use of instead with a comma has created a comma splice. Options (D) and (E) also have meaning errors since they award ‘bidding’ to the firm instead of awarding the ‘contract’ to the firm. (E) is also incorrect in its use of ‘were awarded’. The contract is singular and hence, ‘was awarded’ should be used.

Option (B) rectifies all these errors and is the answer!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Rounding Up Some Official GMAT Questions! Last week we looked at some rounding rules. Today, let’s go over some official questions on rounding. They are quite simple and if we just keep the “Slip to the side and look for a 5” rule in mind, they can be easily solved.

Question 1: If n = 2.0453 and n* is the decimal obtained by rounding n to the nearest hundredth, what is the value of n* – n?

(A) -0.0053
(B) -0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153

Solution: A quick note on place value nomenclature:

Given a decimal 345.789, we know that 5 represents the units digit, 4 the tens digit and 3 the hundreds digit. Also, 7 represents the tenths digit, 8 the hundredths digit and 9 the thousandths digit and so on…

Now let’s go back to this question:

n = 2.0453

We need to round n to the nearest hundredth which means we will retain 2 digits after the decimal. The third digit after the decimal is 5 so 2.0453 rounded to the nearest hundredth is 2.05.

Thus n* – n = 2.05 – 2.0453 = 0.0047

Question 2: If digit h is the hundredths digit in the decimal n = 0.2h6, what is the value of n, rounded to the nearest tenth?

Statement 1: n < 1/4

Statement 2: h < 5

Solution: Given that n = 0.2h6

We need to find the value of n rounded to the nearest tenth i.e. we need to keep only one digit after the decimal.

Statement 1: n < 1/4

In decimal form, it means n < 0.25

If h were 5 or greater, n would become 0.256 or 0.266 or higher. All these values would be more than 0.25 so h must be less than 5 such as 0.246 or 0.236 etc. In all such cases, n would be rounded to 0.2

This statement alone is sufficient.

Statement 2: h < 5

This is even simpler. Since we have been given that h is less than 5, when we round n to the tenths digit, we will get 0.2

This statement alone is also sufficient.

Question 3: If d denotes a decimal number, is d >= 0.5?

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

Statement 2: When d is rounded to the nearest integer, the result is 1.

Solution: Again, a simple question!

We need to find whether d is greater than or equal to 0.5 or not.

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

This means that whatever d is, when we round it to the nearest tenth, we get 0.5. What are the possible values of d? If d is anywhere from 0.450 to 0.5499999…, it will be rounded to 0.5

Some of these numbers are less than 0.5 and others are greater than 0.5 so this statement alone is not sufficient.

Statement 2: When d is rounded to the nearest integer, the result is 1.

In this case d must be at least 0.5; only then can it be rounded to 1.

d can be anything from 0.50 to 1.499999… In any case, d will be greater than or equal to 0.5.

This statement alone is sufficient to answer the question.

We hope you see that if we just remember the rules, we can solve most rounding questions very quickly and efficiently.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Rounding Rules on the GMAT: Slip to the Side and Look for a Five! The famous rounding song by Joe Crone is pretty much all you need to solve the trickiest of rounding questions on GMAT:

You just slip to the side, and you look for a five.

Well if the number that you see is a five or more,

You gotta round up now, that’s for sure.

If the number that you see is a four or less,

You gotta round down to avoid a mess.

To put it in our own words, when we round a decimal, we drop the extra decimal places and apply certain rules:

–          If the first dropped digit is 5 or greater, we round up the last digit that we keep.

–          If the first dropped digit is 4 or smaller, we keep the last digit that we keep, the same.

For Example, we need to round the following decimals to two digits after decimal:

(a) 3.857

We drop 7. Since 7 is ‘5 or greater’, we are left with 3.86

(b) 12.983

We drop 3. Since 3 is ‘4 or smaller’, we are left with 12.98

(c) 26.75463

We drop 463. Since 4 is ‘4 or smaller’, we are left with 26.75

(d) 8.9675

We drop 75. Since 7 is ‘5 or greater’, we are left with 8.97

Note example (c) carefully:

When we round 26.75463 to two decimal places, we do not start rounding from the rightmost digit i.e. this is incorrect: 26.75463 becomes 26.7546 which becomes 26.755 which further becomes 26.76 – this is not correct. .00463 is less than .005 and hence should be ignored. You only need to worry about the digit right next to the digit you are keeping. Just slip to the side, and look for a five!

A logical question arises: what happens when we have, say, 2.5 and we need to round it to the nearest integer? 2.5 is midway between 2 and 3. In that case, why do we round the number up, as the rule suggests? Note that a 2.5 is a tie and we have many tie breaking rules that can be used. They are ‘Round half to odd’, ‘Round half to even’, ‘Round up’, ‘Round down’, ‘Round towards 0’, ‘Round away from 0’ etc. We don’t need to worry about all these since GMAT uses only Round up i.e. 2.5 will be rounded up to 3.

Let’s take a look at a question now which uses these fundamentals.

Question: The exact cost price to make each unit of a widget is \$7.6xy7, where x and y represent single digits. What is the value of y?

Statement 1: When the cost is rounded to the nearest cent, it becomes \$7.65.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes \$7.65.

Solution: The question is based on rounding. We need to figure out the value of y given some rounding scenarios. Let’s look at them one by one.

Statement 1: When the cost is rounded to the nearest cent, it becomes \$7.65.

When rounded to the nearest cent, the cost becomes 7 dollars and 65 cents. 6xy7 cents got rounded to 65 cents. When will .6xy7 get rounded to .65? When .6xy7 lies anywhere in the range .6457 to .6547. Note that in all these cases, when you round the number to 2 digits, it will become .65.

Say price is 7.6468. We need to drop 68 but since 6 is ‘5 or greater’, 4 gets rounded up to 5.

Similarly, say the price is 7.6543. We need to drop 43. Since 4 is ‘4 or smaller’, 5 stays as it is.

So x and y can take various different values. This statement alone is not sufficient.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes \$7.65

Now the cost is rounded to the tenth of a cent which means 3 places after the decimal. But the cost is given to us as \$7.65. Since we need 3 places, the cost must be \$7.650 (which will be written as \$7.65)

When will 7.6xy7 get rounded to 7.650? Now this is the tricky part of the question – from 7.6xy7, you need to drop the 7 and round up y. When you do that, you get 7.650. This means 7.6xy7 must have been 7.6497. Only in this case, when we drop the 7, we round up the 9 to make 10, carry the 1 over to 4 and make it 5. This is the only way to get 7.650 on rounding 7.6xy7 to the tenth of a cent. Hence x must be 4 and y must be 9. This statement alone is sufficient to answer the question.

Hope you see that a few simple rules can make rounding questions quite easy.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# When Permutations & Combinations and Data Sufficiency Come Together on the GMAT! While discussing Permutations and Combinations many months back, we worked through several examples of arranging people in seats. Today we bring you an interesting question based on those concepts. It brings to the fore the tricky nature of both Data Sufficiency and Combinatorics – so much so that when the two get together, it is unlimited fun!

In some circumstances, we suggest you to travel the whole nine yards – i.e. solve for the answer in Data Sufficiency questions too even if you feel that sufficiency has already been established. This is especially true for quadratic equations which we assume will give us two values of x but might actually give just a single unique value (such that both roots are the same). In Combinatorics too, sometimes what may look like two distinct cases could actually give the same answer. Let’s jump on to the question.

Question 1: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

Solution:

There are x children and y chairs.

x and y are prime numbers.

Statement 1: x + y = 12

Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:

Case 1: x=5 and y=7

There are 5 children and 7 chairs.

Case 2: x=7 and y=5

There are 7 children and 5 chairs

At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient  alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.

Case 1: x=5 and y=7

If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Case 2: x = 7 and y = 5

If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.

So actually this statement alone is sufficient! Most people would not have seen that coming!

Statement 2: There are more chairs than people.

We don’t know how many children or chairs there are. This statement alone is not sufficient.

We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!

Now, what if we alter the question slightly and make it:

Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Medians, Altitudes and Angle Bisectors in Special Triangles on the GMAT No concept group on GMAT geometry questions is more important than triangles, and no discussion of triangles on the GMAT would be fully comprehensive without covering medians, altitudes, and angle bisectors. So in today’s post, let’s discuss these interesting triangle concepts as a group and show how they can be applicable to challenging GMAT geometry questions. To begin, we are assuming you know the terms median, angle bisector and altitude but still, just to be sure, we will start our discussion today by defining them:

Median – A line segment joining a vertex of a triangle with the mid-point of the opposite side.

Angle Bisector – A line segment joining a vertex of a triangle with the opposite side such that the angle at the vertex is split into two equal parts.

Altitude – A line segment joining a vertex of a triangle with the opposite side such that the segment is perpendicular to the opposite side.

Usually, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. But, importantly, in special triangles such as isosceles and equilateral triangles, they can overlap. And, as always, any time you can identify a triangle as a special triangle, you have even more rules you can apply to better understand it. We will now give you some of these properties which can be very useful.

I.

In an isosceles triangle (where base is the side which is not equal to any other side):

– the altitude drawn to the base is the median and the angle bisector;

– the median drawn to the base is the altitude and the angle bisector;

– the bisector of the angle opposite to the base is the altitude and the median.

II.

The reverse of what we just learned is also true. Consider a triangle ABC:

– If angle bisector of vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this angle bisector is also the altitude.

– If altitude drawn from vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this altitude is also the angle bisector.

– If median drawn from vertex A is also the angle bisector, the triangle is isosceles such that AB = AC and BC is the base. Hence this median is also the altitude.

and so on…

III.

In an equilateral triangle, each altitude, median and angle bisector drawn from the same vertex, overlap.

Try to prove all these properties on your own. That way, you will not forget them.

A few things this implies:

-Should an angle bisector in a triangle which is also a median be perpendicular to the opposite side? Yes.

-Can we have an angle bisector which is also a median which is not perpendicular? No. Angle bisector which is also a median implies isosceles triangle which implies it is also the altitude.

-Can we have a median from vertex A which is perpendicular to BC but does not bisect the angle A? No. A median which is an altitude implies the triangle is isosceles which implies it is also the angle bisector.

and so on…

Let’s take a look at a sample Data Sufficiency question to demonstrate the applicability of these concepts:

Question: What is the measure of angle A in triangle ABC?

Statement 1: The bisector of angle A is a median in triangle ABC.
Statement 2: The altitude of B to AC is a median in triangle ABC.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.

(C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Solution: We are given a triangle ABC but we don’t know what kind of a triangle it is.

So let’s move on to work on the statements directly.

Statement 1: The bisector of angle A is a median in triangle ABC.

The angle bisector is also a median. This means triangle ABC must be an isosceles triangle such that AB = AC. But we have no idea about the measure of angle A. This statement alone is not sufficient.

Statement 2: The altitude of angle B to AC is a median in triangle ABC.

The altitude is also a median. This means triangle ABC must be an isosceles triangle such that AB = BC (Note that the altitude is drawn from vertex B here). But we have no idea about the measure of angle A. This statement alone is not sufficient.

Using both statements together, we see that AB = AC = BC. So the triangle is equilateral! So angle A must be 60 degrees. Sufficient! Therefore the correct answer is (C).

And the ever-important takeaway from this problem: here we were able to use our knowledge of how medians, altitudes, and angle bisectors appear in special types of triangles to prove that we were dealing with a special, equilateral triangle. This is an important lesson both for the concepts of medians and altitudes and for GMAT geometry as a whole, particularly in Data Sufficiency. You have to be proactively on the lookout for special triangles, because special triangles allow you to determine quite a few facts from a limited set of information. They set up perfectly for Data Sufficiency, where the entire question type depends on “cleverly-hidden” information that can be unlocked by recognition of key concepts. So it is very important that you:

1) Embrace any GMAT geometry rules that help you identify or leverage special triangles (equilateral, isosceles, 30-60-90, etc.)

2) Look proactively to find special triangle relationships so that you can apply the rules that accompany them. This is particularly important with Data Sufficiency but also in Problem Solving problems where if you don’t see a relationship you can be doomed to staring in frustration, but if you do you can quickly apply all that you know to reach an answer.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

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# A Remainders Shortcut for the GMAT We firmly believe that teaching someone is a most productive learning for oneself and every now and then, something happens that strengthens this belief of ours. It’s the questions people ask – knowingly or unknowingly – that connect strings in our mind such that we feel we have gained more from the discussion than even our students!

The other day, we came across this common GMAT question on remainders and many people had solved it the way we would expect them to solve. One person, perhaps erroneously, used a shortcut which upon reflection made perfect sense. Let me give you that question and the shortcut and the problem with the shortcut. We would like you to reflect on why the shortcut actually does make sense and is worth noting down in your log book.

Question: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

Solution: We are assuming you know how people do the question usually:

The logic it uses is discussed here and the solution is given below as Method I.

Method I:

Positive integer n leaves a remainder of 4 after division by 6. So n = 6a + 4

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 4, 10, 16, 22, 28, …

Positive integer n leaves a remainder of 3 after division by 5. So n = 5b + 3

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 3, 8, 13, 18, 23, 28, …

The first common value is 28. So n = 30k + 28

Hence remainder when positive integer n is divided by 30 is 28.

Perfect! But one fine gentleman came up with the following solution wondering whether he had made a mistake since it seemed to be “super simple Math”.

Method II:

Given in question: “n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5.”

Divide the options by 6 and 5. The one that gives a remainder of 4 and 3 respectively will be correct.

(A) 3 / 6 gives Remainder = 3 -> INCORRECT
(B) 12 / 6 gives Remainder = 0 -> INCORRECT
(C) 18 / 6 gives Remainder = 0 -> INCORRECT
(D) 22 / 6 gives Remainder = 4 but 22 / 5 gives  Remainder = 2 -> INCORRECT
(E) 28 / 6 gives Remainder = 4 and 28 / 5 gives Remainder = 3 ->  CORRECT

Now let us point out that the options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. The question says that n must give a remainder of 4 upon division by 6 and a remainder of 3 upon division by 5. This solution divided the options (which are not the values of n) by 6 and 5 and got the remainder as 4 and 3 respectively. So the premise that when you divide the correct option by 6 and 5, you should get a remainder of 4 and 3 respectively is faulty, right?

This is where we want you to take a moment and think: Is this premise actually faulty?

The fun part is that method II is perfectly correct too. Method I seems a little complicated when compared with Method II, doesn’t it? Let us give you the logic of why method II is correct:

Recall that division is nothing but grouping. When you divide n by 30, you make complete groups of 30 each. The number of groups you get is called the quotient (not relevant here) and the leftover is called the remainder. If this is not clear, check this post first.

When n is divided by 30, groups of 30 are made. Whatever is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. Now, whatever is leftover (given in the options) after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6 (i.e. divide the option by 6), we must have remainder 4 since n leaves remainder 4. When we split it into groups of 5 (i.e. divide the option by 5), we must have remainder 3 since n leaves remainder 3. And, that is the reason we can divide the options by 6 and 5, check their remainders and get the answer!

Now, isn’t that neat!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# A Take on GMAT Takeaways Once you have covered your fundamentals, we suggest you to practice advanced questions and jot down your takeaways from them. Sometimes students wonder how to find that all important “takeaway”. Today, let’s discuss how to elicit a takeaway from a question which seems to have none.

What is a takeaway? It is a small note to yourself which you would do well to remember while going for the exam. Even if you don’t remember the exact property you jotted down, knowing that such a property exists is enough. You can always try it on a couple of numbers in the test to recall the exact content.

The question we will discuss today serves another purpose – it discusses properties of squares of odd and even integers so in a sense is a continuation of our advanced number properties discussion.

Question: Given x and y are positive integers such that y is odd, is x divisible by 4?

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.
Statement 2: x – y = 3

Solution: As of now, we don’t know any specific properties of squares of odd and even integers. However, we do have a good (presumably!) understanding of divisibility. To recap quickly, divisibility is nothing but grouping. To take an example, if we divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is the quotient and 1 is the remainder. For more on these concepts, check out our previous posts on divisibility.

Coming back to our question,

First thing that comes to mind is that if y is odd, y = (2k + 1).

We have no information on x so let’s proceed to the two statements.

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

The statement tell us something about y^2 so let’s get that.

If y = (2k + 1)

y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1

Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), k(k+1) will be even. So 4k(k+1) will be divisible by 4*2 i.e. by 8. So when y^2 is divided by 8, it will leave a remainder 1.

When y^2 is divided by 8, remainder is 1. To get a remainder of 5 when x^2 + y^2 is divided by 8, we should get a remainder of 4 when x^2 is divided by 8. So x must be even. If x were odd, the remainder when x^2 were divided by 8 would have been 1. So we know that x is divisible by 2 but we don’t know whether it is divisible by 4 yet.

x^2 = 8a + 4 (when x^2 is divided by 8, it leaves remainder 4)

x^2 = 4(2a + 1)

So x = 2*?Odd Number

Square root of an odd number will be an odd number so we can see that x is even but not divisible by 4. This statement alone is sufficient to say that x is NOT divisible by 4.

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (Since Even – Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say. This statement alone is not sufficient.

So could you point out the takeaway from this question?

Note that when we were analyzing y, we used no information other than that it is odd. We found out that the square of any odd number when divided by 8 will always yield a remainder of 1.

Now what can you say about the square of an even number? Say you have an even number x.

x = 2a

x^2 = 4a^2

This tells us that x^2 will be divisible by 4 i.e. we can make groups of 4 with nothing leftover. What happens when we try to make groups of 8? We join two groups of 4 each to make groups of 8. If the number of groups of 4 is even, we will have no remainder leftover. If the number of groups of 4 is odd, we will have 1 group leftover i.e. 4 leftover. So when the square of an even number is divided by 8, the remainder is either 0 or 4.

Looking at it in another way, we can say that if a is odd, x^2 will be divisible by 4 and will leave a remainder of 4 when divided by 8. If a is even, x^2 will be divisible by 16 and will leave a remainder of 0 when divided by 8.

Takeaways

– The square of any odd number when divided by 8 will always yield a remainder of 1.

– The square of any even number will be either divisible by 4 but not by 8 or it will be divisible by 16 (obvious from the fact that squares have even powers of prime factors so 2 will have a power of 2 or 4 or 6 etc). In the first case, the remainder when it is divided by 8 will be 4; in the second case the remainder will be 0.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Advanced Number Properties on the GMAT – Part III Continuing our discussion on number properties, today we will discuss how factorials affect the behavior of odd and even integers. Since we are going to deal with factorials, positive integers will be our concern. Using a question, we will see how factorials are divided.

Question: If x and y are positive integers, is y odd?

Statement 1: (y+2)!/x! = odd
Statement 2: (y+2)!/x! is greater than 2

Solution: The question stem doesn’t give us much information – just that x and y are positive integers.

Question: Is y odd?

Statement 1: (y+2)!/x! = odd

Note that odd and even are identified only for integers. Since (y+2)!/x! is odd, it must be a positive integer. This means that x! must be equal to or less than (y+2)!

Now think, how are y and y+2 related? If y+2 is odd, y+1 is even and hence y is odd. If y+2 is even, by the same logic, y is even.

(y+2)! = 1*2*3*4*…*y*(y+1)*(y+2)

x! = 1*2*3*4*…*x

Note that (y+2)! and x! have common factors starting from 1. Since x! is less than or equal to (y+2)!, x will be less than or equal to (y+2). So all factors in the denominator, from 1 to x will be there in the numerator too and will get canceled leaving us with the last few factors of (y+2)!

To explain this, let us take a few examples:

Example 1: Say, y+2 = 6, x = 6

(y+2)!/x! = 6!/6! = 1

Example 2: Say, y+2 = 7, x = 6

(y+2)!/x! = 7!/6! = (1*2*3*4*5*6*7)/(1*2*3*4*5*6) = 7 (only one leftover factor)

Example 3: Say, y+2 = 6, x = 4

(y+2)!/x! = 6!/4! = (1*2*3*4*5*6)/(1*2*3*4) = 5*6 (two leftover factors)

If the division of two factorials is an integer, the factorial in the numerator must be larger than or equal to the factorial in the denominator.

So what does (y+2)!/x! is odd imply? It means that the leftover factors must be all odd. But the leftover factors will be consecutive integers. So after one odd factor, there will be an even factor. If we want (y+2)!/x! to be odd, we must have either no leftover factors (such that (y+2)!/x! = 1) or only one leftover factor and that too odd.

If we have no leftover factor, it doesn’t matter what y+2 is as long as it is equal to x. It could be odd or even. If there is one leftover factor, then y+2 must be odd and hence y must be odd. Hence y could be odd or even. This statement alone is not sufficient.

Statement 2: (y+2)!/x! is greater than 2

This tells us that y+2 is not equal to x since (y+2)!/x! is not 1. But all we know is that it is greater than 2. It could be anything as seen in examples 2 and 3 above. This statement alone is not sufficient.

Both statements together tell us that y+2 is greater than x such that (y+2)!/x! is odd. So there must be only one leftover factor and it must be odd. The leftover factor will be the last factor i.e. y+2. This tells us that y+2 must be odd. Hence y must be odd too.

Takeaways: Assuming a and b are positive integers,

–          a!/b! will be an integer only if a >= b

–          a!/b! will be an odd integer whenever a = b or a is odd and a = b+1

–          a!/b! will be an even integer whenever a is even and a = b+1 or a > b+1

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Advanced Number Properties on the GMAT – Part II Before we get started, be sure to take a look at Part I of this article. Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly.

To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem.

Question 1: Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Solution: What do we know about sum of two prime numbers?

e.g. 3 + 5 = 8

5 + 11 = 16

5 + 17 = 22

23 + 41 = 64

Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum.

We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd.

2 + 3 = 5

2 + 7 = 9

2 + 17 = 19

Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number.

Let’s check the odd options first:

(A)   19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.)
(B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.)
(D) 79 = 2 + 77 (77 is not prime.)

79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers.

81 = 2 + 79

Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2.

Question 2: If m, n and p are positive integers such that m < n < p, is m a factor of the odd integer p?

Statement 1: m and n are prime numbers such that (m + n) is a factor of 119.

Statement 2: p is a factor of 119.

Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on.

The question stem tells us that p is an odd integer. Also, m < n < p.

Question: Is m a factor of p?

There isn’t much information in the question stem for us to process so let’s jump on to the statements directly.

Statement 1: m and n are prime numbers such that (m + n) is a factor of 119.

Write down the factors of 119 first to get the feasible range of values.

119 = 1, 7, 17, 119

All factors of 119 are odd numbers. So (m + n), a sum of two primes must be odd. This means one of m and n is 2. There are many possible values of m and n e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119).

Note that we also have m < n. This means that in each case, m must be 2 and n must be the other number of the pair.

So now we know that m is 2. We also know that p is an odd integer. Is m a factor of p? No. Odd integers are those which do not have 2 as a factor. Since m is 2, p does not have m as a factor.

This statement alone is sufficient to answer the question!

Statement 2: p is a factor of 119

This tells us that p is one of 7, 17 and 119. p cannot be 1 because m < n < p where all are positive integers.

But it tells us nothing about m. All we know is that it is less than p. For example, if p is 7, m could be 1 and hence a factor of p or it could be 5 and not a factor of p. Hence this statement alone is not sufficient.

Something to think about: In this question, if you are given that m is not 1, does it change our answer?

Key Takeaways:

– When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2.

– Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2.

– Remember the special position 2 occupies among prime numbers – it is the only even prime number.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Advanced Number Properties on the GMAT – Part I

Don’t worry, we are not going to discuss (Even + Even = Even) and (Odd + Odd = Even) type of basic number properties in this post. What we have in mind for today is something based on this but far more advanced. Often, people complain that they thoroughly understand the theory but have difficulties applying it and hence are stuck at a score of 600. They look for practice questions and tend to ignore concepts since they already “know” them. We often ask them to go back to concepts since we believe that a strong foundation of concepts is necessary for ‘score increase’. Mind you, when we do that, we don’t mean to ask them to review the basic concepts again, we mean to ask them to deduce and work on advanced concepts. Let’s show you with the help of a question.

Question: If two integers are chosen at random out of first 5 positive integers, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

– Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.

If the product is odd, it can always be written as product of two odd numbers.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Takeaway:

When can we write a number as difference of squares?

– When the number is odd

or

– When the number has 4 as a factor

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# A GMAT Formula to Remember: Profit on One, Loss on Another I am no fan of formulas, especially the un-intuitive ones but the one we are going to discuss today has proved quite useful. It is for a concept tested on GMAT Prep so it might be worth your while to remember this little formula.

When two items are sold at the same selling price, one at a profit of x% and the other at a loss of x%, there is an overall loss. The loss% = (x^2/100)%

We will see how this formula is derived but the algebra involved is tedious. You can skip it if you wish.

Say two items are sold at \$S each. On one, a profit of x% is made and on the other a loss of x% is made.

Say, cost price of the article on which profit was made = Ct

Ct (1 + x/100) = S

Ct = S/(1 + x/100)

Cost Price of the article on which loss was made = Cs

Cs (1 – x/100) = S

Cs = S/(1 – x/100)

Total Cost Price of both articles together = Ct + Cs = S/(1 + x/100) + S/(1 – x/100)

Ct + Cs = S[1/(1 + x/100) + 1/(1 – x/100)]

Ct + Cs = 2S/(1 – (x/100)^2)

Total Selling Price of both articles together = 2S

Overall Profit/Loss = 2S – (Ct + Cs)

Overall Profit/Loss % = [2S – (Ct + Cs)]/[Ct + Cs] * 100

= [2S/(Ct + Cs) – 1] * 100

= [2S/[2S/(1 – (x/100)^2)] – 1] * 100

= (x/100)^2 * 100

= x^2/100

Overall there is a loss of (x^2/100)%.

Let’ see how this formula works on a GMAT Prep question.

Question: John bought 2 shares and sold them for \$96 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had

(A) a profit of \$10
(B) a profit of \$8
(C) a loss of \$8
(D) a loss of \$10
(E) neither a profit nor a loss

Solution:

Note that the question would have been straight forward had the COST price been the same, say \$100. A 20% profit would mean a gain of \$20 and a 20% loss would mean a loss of \$20. Overall, there would have been no profit no loss.

Here the two shares are sold at the same SALE price. One at a profit of 20% on cost price which must be lower than the sale price (to get a profit) and the other at a loss of 20% on cost price which must be higher than the sale price (to get a loss). 20% of a lower amount will be less in dollar terms and hence overall, there will be a loss.

The loss % = (20)^2/100 % = 4%.
But we need the amount of loss, not the percentage of loss.

Total Sale price of the two shares = 2*96 = \$192
Since there is a loss of 4%, the 96% of the total cost price must be the total sale price

(96/100)*Cost Price = Sale Price
Cost Price = \$200
Loss = \$200 – \$192 = \$8

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Determining the Area of Similar Triangles on the GMAT Recall the important property that we discussed about the relation between the areas of the two similar triangles last week – if the ratio of their sides is ‘k’, the ratio of their areas will be k^2. As mentioned last week, it’s an important property and helps you easily solve otherwise difficult questions. The question I have in mind today also brings in focus the Pythagorean triplets.

There are some triplets that you should know out cold: (3, 4, 5), (5, 12, 13) and (8, 15, 17). Usually you will find one of these three or their multiples on GMAT. Given a right triangle and two sides, say the two legs, of length 20 and 48, we need to immediately bring them down to the lowest form 20:48 = 5:12. So we know that we are talking about the 5, 12, 13 triplet and the hypotenuse will be 13*4 = 52. These little things help us save a lot of time. Why is it that some people get done with the Quant section in less than an hour while others fall short on time? It is these little things that an adept test taker has mastered which make all the difference.

Anyway, let us go on to the question we have in mind.

Question: In the figure given below, the length of PQ is 12 and the length of PR is 15. The area of right triangle STU is equal to the area of the shaded region. If the ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR, what is the length of TU? (A) (9?2)/4

(B) 9/2

(C) (9?2)/2

(D) 6?2

(E) 12

Solution: The information given in the question seems to overwhelm us but let’s take it a bit at a time.

“length of PQ is 12 and the length of PR is 15”

PQR is a right triangle such that PQ = 12 and PR = 15. So PQ:PR = 4:5. Recall the 3-4-5 triplet. A multiple triplet of 3-4-5 is 9-12-15. This means QR = 9.

“ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR”
ST/TU = PQ/QR

The ratio of two sides of PQR is equal to the ratio of two sides of STU and the included angle between the sides is same ( = 90). Using SAS, triangles PQR and STU are similar.

“The area of right triangle STU is equal to the area of the shaded region”

Area of triangle PQR = Area of triangle STU + Area of Shaded Region

Since area of triangle STU = area of shaded region, (area of triangle PQR) = 2*(area of triangle STU)

In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2. If the ratio of the areas is given as 2 (i.e. k^2 is 2), the sides must be in the ratio ?2 (i.e. k must be ?2).

Since QR = 9, TU must be 9/?2. But there is no 9/?2 in the options – in the options the denominators are rationalized. TU = 9/?2 = (9*?2)/(?2*?2) = (9*?2)/2.

The question could take a long time if we do not remember the Pythagorean triplets and the area of similar triangles property.

Takeaways:

1. Pythagorean triplets you should know: (3, 4, 5), (5, 12, 13) and (8, 15, 17) and their multiples.
2. In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Looking for Similar Triangles on the GMAT Our Geometry book discusses the various rules we use to recognize similar triangles such as SSS, AA, SAS and RHS so we are assuming that we needn’t take those up here.

We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines.

This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context!

What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often.

We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k

Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’.

Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2.

So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4.

Now we are all ready to tackle the question we have in mind.

Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area? (A)   3:8

(B)   3:5

(C)   5:8

(D)   8:5

(E)    5:3

Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!).

Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below) Now look at the original figure.

HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.

Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.

The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.

Hence, area of shaded region : Area of unshaded region = 5:3

Try to think of other ways in which you can solve this question.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# A Tricky Question on Negative Remainders Today, we will discuss the question we left you with last week. It involves a lot of different concepts – remainder on division by 5, cyclicity and negative remainders. Since we did not get any replies with the solution, we are assuming that it turned out to be a little hard.

It actually is a little harder than your standard GMAT questions but the point is that it can be easily solved using all concepts relevant to GMAT. Hence it certainly makes sense to understand how to solve it.

Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Solution: As we said last week, this question can easily be solved using cyclicity and negative remainders. What is the remainder when a number is divided by 5? Say, what is the remainder when 2387646 is divided by 5? Are you going to do this division to find the remainder? No! Note that every number ending in 5 or 0 is divisible by 5.

2387646 = 2387645 + 1

i.e. the given number is 1 more than a multiple of 5. Obviously then, when the number is divided by 5, the remainder will be 1. Hence the last digit of a number decides what the remainder is when the number is divided by 5.

On the same lines,

What is the remainder when 36793 is divided by 5? It is 3 (since it is 3 more than 36790 – a multiple of 5).

What is the remainder when 46^8 is divided by 5? It is 1. Why? Because 46 to any power will always end with 6 so it will always be 1 more than a multiple of 5.

On the same lines, if we can find the last digit of 3^(7^11), we will be able to find the remainder when it is divided by 5.

Recall from the discussion in your books, 3 has a cyclicity of 4 i.e. the last digit of 3 to any power takes one of 4 values in succession.

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

and so on… The last digits of powers of 3 are 3, 9, 7, 1, 3, 9, 7, 1 … Every time the power is a multiple of 4, the last digit is 1. If it is 1 more than a multiple of 4, the last digit is 3. If it is 2 more than a multiple of 4, the last digit is 9 and if it 3 more than a multiple of 4, the last digit is 7.

What about the power here 7^(11)? Is it a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4 or 3 more than a multiple of 4? We need to find the remainder when 7^(11) is divided by 4 to know that.

Do you remember the binomial theorem concept we discussed many weeks back? If no, check it out here.

7^(11) = (8 – 1)^(11)

When this is divided by 4, the remainder will be the last term of this expansion which will be (-1)^11. A remainder of -1 means a positive remainder of 3 (if you are not sure why this is so, check last week’s post here). Mind you, you are not to mark the answer as (D) here and move on! The solution is not complete yet. 3 is just the remainder when 7^(11) is divided by 4.

So 7^(11) is 3 more than a multiple of 4.

Review what we just discussed above: If the power of 3 is 3 more than a multiple of 4, the last digit of 3^(power) will be 7.

So the last digit of 3^(7^11) is 7.

If the last digit of a number is 7, when it is divided by 5, the remainder will be 2. Now we got the answer!

Interesting question, isn’t it?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# All About Negative Remainders on the GMAT I could have sworn that I had discussed negative remainders on my blog but the other day I was looking for a post discussing it and much as I would try, I could not find one. I am a little surprised since this concept is quite useful and I should have taken it in detail while discussing divisibility. Though we did have a fleeting discussion of it here.

Since we did miss it, we will discuss it in detail today but you must review the link given above before we proceed.

Consider this: When n is divided by 3, it leaves a remainder 1.

This means that when we divide n balls in groups of 3 balls each, we are left with 1 ball. This also means that n is 1 MORE than a multiple of 3. Or, it also means that n is 2 less than the next multiple of 3, doesn’t it?

Say, n is 16. When you split 16 balls into groups of 3 balls each, you get 5 groups of 3 balls each and there is one ball leftover. n is 1 more than a multiple of 3 (the multiple being 15). But we can also say that it is 2 LESS than the next multiple of 3 (which is 18). Hence, the negative remainder in this case is -2 which is equivalent to a positive remainder of 1.

Generally speaking, if n is divided by m and it leaves a remainder r, the negative remainder in this case is -(m – r).

When n is divided by 7, it leaves a remainder of 4. This is equivalent to a remainder of -3.

n is 3 more than a multiple of m. It is also 2 less than the next multiple of m. This means m is 5.

This concept is very useful to us sometimes, especially when the divisor and the remainder are big numbers.

Let’s take a question to see how.

Question 1: What is the remainder when 1555 * 1557 * 1559 is divided by 13?

(A)   0

(B)   2

(C)   4

(D)   9

(E)    11

Solution: Since it is a GMAT question (a question for which we will have no calculator), multiplying the 3 numbers and then dividing by 13 is absolutely out of question! There has to be another method.

Say n = 1555 * 1557 * 1559

When we divide 1555 by 13, we get a quotient of 119 (irrelevant to our question) and remainder of 8. So the remainder when we divide 1557 by 13 will be 8+2 = 10 (since 1557 is 2 more than 1555) and when we divide 1559 by 13, the remainder will be 10+2 = 12 (since 1559 is 2 more than 1557).

So n = (13*119 + 8)*(13*119 + 10)*(13*119 + 12) (you can choose to ignore the quotient and just write it as ‘a’ since it is irrelevant to our discussion)

So we need to find the remainder when n is divided by 13.

Note that when we multiply these factors, all terms we obtain will have 13 in them except the last term which is obtained by multiplying the constants together i.e. 8*10*12.

Since all other terms are multiples of 13, we can say that n is 8*10*12 (= 960) more than a multiple of 13. There are many more groups of 13 balls that we can form out of 960.

960 divided by 13 gives a remainder of 11.

Hence n is actually 11 more than a multiple of 13.

We did not use the negative remainders concept here. Let’s see how using negative remainders makes our calculations easier here. The remainder of 8, 10 and 12 imply that the negative remainders are -5, -3 and -1 respectively.

Now n = (13a – 5) * (13a – 3) * (13a – 1)

The last term in this case is -5*-3*-1 = -15

This means that n is 15 less than a multiple of 13 i.e. actually 2 less than a multiple of 13 because when you go back 13 steps, you get another multiple of 13. This gives us a negative remainder of -2 which means the positive remainder in this case will be 11.

Here we avoided some big calculations.

I will leave you now with a question which you should try to solve using negative remainders.

Question 2: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Hint: I solved this question orally in a few secs using cyclicity and negative remainders. Don’t get lost in calculations!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Is This GMAT Question Suspect? I came across a discussion on one of our questions where the respondents were split over whether it is a strengthen question or weaken! Mind you, both sides had many supporters but the majority was with the incorrect side. You must have read the write up on ‘support’ in your Veritas Prep CR book where we discuss how question stems having the word ‘support’ could indicate either strengthen or inference questions. I realized that we need a write up on the word ‘suspect’ too so here goes.

First let me give you the question stem of that question:

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

So do we have to find a reason which indicates that the new course WILL increase interest in the academy or do we have to find a reason that indicates that the new course WILL NOT increase interest in the academy? That is, is this a type of ‘strengthen’ question or the opposite – a ‘weaken’ question?

I would have expected most people to tag it as a strengthen question i.e. we are looking for a reason which indicates that the new course WILL increase interest in the academy but that is not the case. Many people get waylaid by the word ‘suspect’ and incorrectly tag it as a weaken question. Yes, suspect could mean ‘doubtful’ but just because the question stem has it, it doesn’t mean you have a weaken question at hand. Similar to the situation where the word ‘support’ in the question stem doesn’t necessarily imply that you have been given a strengthen question to deal with.

Let’s discuss various meanings of the word ‘suspect’ and how they are used (using merriam-webster.com):

Suspect can be used as a verb, noun or adjective. In our question stem, it is used as a verb and that’s what we will focus most on but let’s take up the other two briefly first.

SUSPECT (- NOUN) – one that is suspected; especially a person suspected of a crime

1. One suspect has been arrested.
2. She is a possible suspect in connection with the kidnapping.

SUSPECT (- ADJECTIVE) regarded or deserving to be regarded with suspicion

doubtful, questionable

1. The room had a suspect odor.
2. Since she was carrying no cash or credit cards, her claim to the store’s detectives that she had intended to pay for the items was suspect.

SUSPECT ( – VERB) – The verb suspect can be used in 3 different ways:

1. to imagine (one) to be guilty or culpable on slight evidence or without proof

For Example: He’s suspected in four burglaries.

2.  to have doubts of: distrust

For Example: The fire chief suspects arson. I suspect his intentions.

3. to imagine to exist or be true, likely, or probable

For Example: I suspect it will rain.

Given a construction “I suspect A will happen”, which meaning will it have? In this case, it has the meaning of ‘imagine to be true’ or ‘think to be true’. There is absolutely no ambiguity here. When I ask “Which option supplies the best reason to suspect that the new course will increase interest in the academy?” you are definitely looking for the option that indicates that the new course WILL increase interest. Let me give you the whole question now. I am sure you will be able to solve it easily.

Question: The metropolitan cooking academy surveyed prospective students and found that students wanted a curriculum that focused on today’s healthy dining trends. In order to reverse the trend of declining interest in the school’s programs, administrators propose a series of new courses focused on cooking exotic species of fish, alternative grains such as quinoa, and organically produced vegetables.

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

(A) Cooking fish, grains, and vegetables relies on same culinary fundamentals as does the preparation of other ingredients.
(B) In the food and beverage industry, many employers no longer have time to train apprentices and therefore demand basic culinary skills from their new hires.
(C) Local producers in the area near the Metropolitan cooking academy are excellent sources of exotic fish and organic vegetables.
(D) Many other cooking schools have found a decline in the level of interest in their program.
(E) Many advocates of healthy dining stress the importance of including fish, grains and organically produced vegetables in one’s diet.

Solution: Let’s break down the argument:

Premises:

A survey found that students wanted a curriculum that focused on today’s healthy dining trends.

Administrators propose a series of new courses focused on cooking fish, alternative grains and organically produced vegetables

What will indicate that the new course will increase interest in the academy? If fish, grains and organic vegetables are considered ‘today’s healthy dining trends’, then probably the course will become popular. That is what option (E) says. Hence the answer is (E).

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# An Official Question on Absolute Values Now that we have discussed some important absolute value properties, let’s look at how they can help us in solving official questions.

Knowing these basic properties can help us quickly analyze the question and arrive at the answer without getting stuck in analyzing different ranges, a cumbersome procedure.

First we will look at a GMAT Prep question.

Question 1: Is |m – n| > |m| – |n|?

Statement 1: n < m
Statement 2: mn < 0

Solution 1:

Recall the property number 2

Property  2: For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y| when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

So if m and n have the same sign with |m| >= |n|, equality will hold.

Also, if n is 0, equality will hold.

If we can prove that both these conditions are not met, then we can say that |m – n| is definitely greater than |m| – |n|.

Statement 1: n < m

We have no idea about the signs of m and n. Are they same? Are they opposite? We don’t know. Also n may or may not be 0. Hence we don’t know whether the equality will hold or the inequality. Statement 1 alone is not sufficient to answer the question.

Statement 2: mn < 0

Since mn is negative, it means one of m and n is positive and the other is negative. This also implies that n is definitely not 0. So we know that m and n do not have the same sign and n is not 0. So under no condition will the equality hold.  Hence |m – n| is definitely greater than |m| – |n|. Statement 2 alone is sufficient to answer the question.

Let’s look at one more question now.

Question 2: If xyz ? 0, is x(y + z) >= 0?

Statement 1: |y + z| = |y| + |z|
Statement 2: |x + y| = |x| + |y|

Solution 2:  xyz ? 0 implies that all, x, y and z, are non zero numbers.

Question: Is x(y + z) >= 0?

If we can prove that x(y + z) is not negative that is x and (y+z) do not have opposite signs, we can say that x(y + z) is positive or 0.

Looking at the statements given, let’s review our property number 1:

Property 1: For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

The two statements give us equalities which means that the relevant part of the property is this:
|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

We are also given in the question stem that x, y and z are not 0. Hence, given |x + y| = |x| + |y|, we can infer that x and y have the same sign.

Statement 1: |y + z| = |y| + |z|

This implies that y and z have the same signs. But we have no information about the sign of x hence this statement alone is not sufficient.

Statement 2: |x + y| = |x| + |y|

This implies that x and y have the same signs. But we have no information about the sign of z hence this statement alone is not sufficient.

Using both statements together, we know that x, y and z have the same sign. Whatever is the sign of y and z, the same will be the sign of (y+z). Hence x and (y+z) have the same sign. This implies that x(y + z) cannot be negative.

Hence we can answer our question with a definite ‘yes’.

Mind you, both these questions can get time consuming (even though they aren’t really tough) if you don’t understand these properties well. You can certainly start your thinking from the scratch, arrive at the properties and then proceed or resort to more desperate measures such as number plugging but that is best avoided in DS questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Properties of Absolute Values on the GMAT – Part II We pick up this post from where we left the post of last week in which we looked at a few properties of absolute values in two variables. There is one more property that we would like to talk about today. Thereafter, we will look at a question based on some of these properties.

(III)     |x – y| = 0 implies x = y

x and y could be positive/negative integer/fraction; if the absolute value of their difference is 0, it means x = y. They cannot have opposite signs while having the same absolute value. They must be equal. This also means that if and only if x = y, the absolute value of their difference will be 0.

Mind you, this is different from ‘difference of their absolute values’

|x| – |y| = 0 implies that the absolute value of x is equal to the absolute value of y. So x and y could be equal or they could have opposite signs while having the same absolute value.

Let’s now take up the question we were talking about.

Question: Is |x + y| < |x| + |y|?

Statement 1: | x | is not equal to | y |
Statement 2: | x – y | > | x + y |

Solution: One of the properties we discussed last week was

“For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs”

We discussed in detail the reason absolute values behave this way.

So our question “Is |x + y| < |x| + |y|?” now becomes:

Question: Do x and y have opposite signs?

We do not care which one is greater – the one with the positive sign or the one with the negative sign. All we want to know is whether they have opposite signs (opposite sign also implies that neither one of x and y can be 0)? If we can answer this question definitively with a ‘Yes’ or a ‘No’, the statement will be sufficient to answer the question. Let’s go on to the statements now.

Statement 1: | x | is not equal to | y |
This statement tells us that absolute value of x is not equal to absolute value of y. It doesn’t tell us anything about the signs of x and y and whether they are same or opposite. So this statement alone is not sufficient.

Statement 2:| x – y | > | x + y |
Let’s think along the same lines as last week – when will | x – y | be greater than | x + y |? When will the absolute value of subtraction of two numbers be greater than the absolute value of their addition? This will happen only when x and y have opposite signs. In that case, while subtracting, we would actually be adding the absolute values of the two and while adding, we would actually be subtracting the absolute values of the two. That is when the absolute value of the subtraction will be more than the absolute value of the addition.

For Example: x = 3, y = -2

| x – y | = |3 – (-2)| = 5

| x + y | = |3 – 2| = 1

or

x = -3, y = 2

| x – y | = |-3 – 2| = 5

| x + y | = |-3 + 2| = 1

If instead, x and y have the same sign, | x + y | will be greater than| x – y |.

If at least one of x and y is 0, | x + y | will be equal to| x – y |.

Since this statement tells us that | x – y | > | x + y |, it implies that x and y have opposite signs. So this statement alone is sufficient to answer the question with a ‘Yes’.

Takeaway from this question:

If x and y have the same signs, | x + y | >| x – y |.

If x and y have opposite signs, | x + y | <| x – y |.

If at least one of x and y is 0, | x + y | =| x – y |.

You don’t need to ‘learn this up’. Understand the logic here. You can easily recreate it in the exam if need be.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Properties of Absolute Values on the GMAT We have talked about quite a few concepts involving absolute value of x in our previous posts. But some absolute value questions involve two variables. Then do we need to consider the positive and negative values of both x and y? Certainly! But there are some properties of absolute value that could come in handy in such questions. Let’s take a look at them:

(I)    For all real x and y, |x + y| <= |x| + |y|

(II)   For all real x and y, |x – y| >= |x| – |y|

We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one.

(I)     For all real x and y, |x + y| <= |x| + |y|

The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side.

For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it.

The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs.

For example, x = 4, y = 8:

|4 + 8| = |4| + |8| = 12

OR x = -3, y = -4:

|-3 -4| = |-3| + |-4| = 7

Also, when at least one of x and y is 0, the equality will hold.

For example, x = 0, y = 8:

|0 + 8| = |0| + |8| = 8

OR x = -3, y = 0:

|-3 + 0| = |-3| + |0| = 3

What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y|

For example, x = -4, y = 8:

|-4 + 8| < |-4| + |8|

4 < 12

OR x = 3, y = -4:

|3 -4| < |3| + |-4|

1 < 7

Let’s look at our second property now:

(II) For all real x and y, |x – y| >= |x| – |y|

Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side.

When will the equality hold?

When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values

For example, x = 9, y = 2;

|9 – 2| = |9| – |2| = 7

OR x = -7, y = -3

|-7 – (-3)| = |-7| – |-3| = 4

Also when y is 0, the equality will hold.

For example, x = 8, y = 0:

|8 – 0| = |8| – |0| = 8

OR x = -3, y = 0:

|-3 – 0| = |-3| – |0| = 3

What happens when x and y have the same sign but absolute value of y is greater than that of x?

It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative.

For example, x = -4, y = -9

|x – y| = |-4 – (-9)| = 5

|x| – |y| = |-4| – |-9| = -5

So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again.

Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal.

For example, x = 0, y = -5

|x – y| = |0 – (-5)| = 5

|x| – |y| = |0| – |5| = -5

Taking further absolute value, |(|x| – |y|)| = |-5| = 5

So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0.

Now let’s look at the inequality part of property 2.

Whenever x and y have opposite signs, |x – y| > |x| – |y|

On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs

For example, x = -4, y = 8:

|-4 – 8| > |-4| – |8|

12 > -4

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4

Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)|

OR x = 3, y = -4:

|3 –(-4)| > |3| – |-4|

7 > -1

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1

Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)|

Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not.

To sum it all neatly,

(I) For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

(II) For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

(III) For all real x and y, |x – y| >= |(|x| – |y|)|

|x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x – y| > |(|x| – |y|)| when (1) x and y have opposite signs

Note that property (III) matches property (I).

There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to use.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# How Well Do You Know Your Factors? In the last three weeks, we discussed a couple of strategies we can use to solve max-min questions: ‘Establishing Base Case’ and ‘Focus on Extremes’. Now try to use those to solve this question:

Question: A carpenter has to build 71 wooden boxes in one week. He can build as many per day as he wants but he has decided that the number of boxes he builds on any one day should be within 4 off the number he builds on any other day.
(A) What is the least number of boxes that he could have build on Saturday?
(B) What is the greatest number of boxes that he could have build on Saturday?

Meanwhile, let’s move on to something else today. What we will discuss today is a very simple concept but it seems odd to us when we first confront it even if we are very comfortable with factors and divisibility. If we tell you the concept right away, you will probably not believe us when we say that many people are unable to come up with it on their own. Hence, we will first give you a question which you need to answer in 30 seconds. If you are unable to do so, then we will discuss the concept with you!

Question: A, B, C and D are positive integers such that A/B = C/D. Is C divisible by 5?

Statement 1: A is divisible by 210

Statement 2: B = 7^x, where x is a positive integer

Solution: Let’s discuss the solution till the point I assume you will be quite comfortable.

We need to find whether C is divisible by 5. So let’s separate the C out of the variables.

Since C is an integer, AD will be divisible by B but what we don’t know is that after the division, is the quotient divisible by 5?

Statement 1: A is divisible by 210

We still have no idea what B is so this statement alone is not sufficient. Let’s take an example of how the value of B could change our answer. Assume A is 210.

If B is 3, AD/B will be divisible by 5.

If B is 10, AD/B may not be divisible by 5 (depending on the value of D).

Statement 2: B = 7^x, where x is a positive integer

We have no idea what A and D are hence this statement alone is not sufficient.

Using both together: Now, this is where the trick comes in. Using both statements together, we see that C = (210*a*D)/(7^x)

Now we can say for sure that C will be divisible by 5. If you are not sure why, read on.

The Concept:

As you know, factors (also called divisors) of a number N are those positive integers which completely divide number N i.e. they do not leave a remainder on dividing N. If F is a factor of N, N/F leaves no remainder. This also means that N can be written as F*m where m is an integer. Sure you feel this is elementary but this concept is not as internalized in your conscience as you believe. To prove it, let me give you a question.

Example 1: Is 3^5 * 5^9 * 7 divisible by 18?

Did you take more than 2 seconds to say ‘No’ confidently?

For N to be divisible by F, you should be able to write N as F*m i.e. N must have F as a factor. F here is 18 (= 2*3^2) but we have no 2 in N (which is 3^5 * 5^9 * 7) though we do have a couple of 3s. Hence this huge product is not divisible by 18.

This helps us deduce that odd numbers are never divisible by even numbers.

Example 2: Is 3^5*7^6*11^3 divisible by 13?

For the numerator to be divisible by the denominator, the denominator MUST BE a factor of the numerator. In the entire numerator, there is no 13 so the numerator is not divisible by 13.

Example 3: On the other hand, is 3^5*7^6*11^3*13 divisible by 13?

Yes, it is. 13 gets cancelled and the quotient will be 3^5*7^6*11^3.

Example 4: Is 2^X divisible by 3?

No. No matter what X is, you will only have X number of 2s in the numerator and will never have a 3. So this will not be divisible by 3.

Let’s come back to the original question now:

Given that C = (210*a*D)/(7^x)

Whatever x is, 7^x will get cancelled out by the numerator and we will be left with something. That something will include 5 (obtained from 210) since only 7s will be cancelled out from the numerator. Hence C is divisible by 5.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Max-Min Strategies: Focus on Extremes In the last two weeks, we discussed some max min strategies. Today, let’s look at another max-min question in which we apply the strategy of focusing on the extremes. The largest or the smallest values are often found at the extremes of a given range.

Question: If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16

(B) -14

(C) 0

(D) 14

(E) 16

Solution: To get the sum of the maximum and minimum possible values of xy, we need to know the maximum and minimum values of xy. For those, we need to find the values that x and y can take. So first, we should review the information given:

(x + 1)^2 <= 36

(y – 1)^2 < 64

We need to find the values that x and y can take. There are many ways of doing that. We can solve the inequality using the wave method discussed in this post or using the concept of absolute values. Let’s discuss both the methods.

Wave method to solve inequalities:

Solve for x: (x + 1)^2 <= 36

(x + 1)^2 – 6^2 <= 0

(x + 1 + 6)(x + 1 – 6) <= 0

(x + 7)(x – 5) <= 0

-7 <= x <= 5 (Using the wave method)

Solve for y: (y – 1)^2 < 64

(y – 1)^2 – 8^2 < 0

(y – 1 + 8)(y – 1 – 8) < 0

(y + 7)(y – 9) < 0

-7 < y < 9 (Using the wave method)

Or you can solve taking the square root on both sides

Solve for x: (x + 1)^2 <= 36

|x + 1| <= 6

-6 <= x + 1 <= 6 (discussed in your Veritas Algebra book)

-7 <= x <= 5

So x can take values: -7, -6, -5, -4, … 3, 4, 5

Solve for y: (y – 1)^2 < 64

|y – 1| < 8

-8 < y – 1 < 8 (discussed in your Veritas Algebra book)

-7 < y < 9

So y can take values: -6, -5, -4, -3, … 6, 7, 8.

Now that we have the values of x and y, we should try to find the minimum and maximum values of xy.

Note that the values of xy can be positive as well as negative. The minimum value will be the negative value with largest absolute value (largest negative) and the maximum value will be the positive value with the largest absolute value.

Minimum value – For the value to be negative, one and only one of x and y should be negative. Focus on the extreme values: if x is -7 and y is 8, we get xy = -56. This is the negative value with largest absolute value.

Maximum value – For the value to be positive, both x and y should have the same signs. If x = -7 and y = -6, we get xy = 42. This is the largest positive value.

The sum of the maximum value of xy and minimum value of xy is -56 + 42 = -14

Try to think of it in terms of a number line. x lies in the range -7 to 5 and y lies in the  range -6 to 8. The range is linear so the end points give us the maximum/minimum values. Think of what happens when you plot a quadratic – the minimum/maximum could lie anywhere.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Max-Min Strategies: Establishing Base Case Continuing our discussion on maximizing/minimizing strategies, let’s look at another question today. Today we discuss the strategy of establishing a base case, a strategy which often comes in handy in DS questions. The base case gives us a starting point and direction to our thoughts. Otherwise, with the number of possible cases in any given scenario, we may find our mind wandering from one direction to another without reaching any conclusions. That is a huge waste of time, a precious commodity.

Question: Four friends go to Macy’s for shopping and buy a top each. Three of them buy a pillow case each too. The prices of the seven items were all different integers, and every top cost more than every pillow case. What was the price, in dollars, of the most expensive pillow case if the total price of the seven items was \$89?

Statement 1: The most expensive top cost \$16.

Statement 2: The least expensive pillow case cost \$9.

Solution: The first problem here is figuring out the starting point. There must be many ways in which you can price the seven items such that the total cost is \$89. So we need to establish a base case (which conforms to all the conditions given in the question stem) first and then we will tweak it around according to the additional information obtained from our statements.

‘Seven items for \$89’ means the average price for each item is approximately \$12. But 12 is not the exact average. 12*7 = 84 which means another \$5 were spent.

A sequence with an average of 12 and different integers is \$9, \$10, \$11, \$12, \$13, \$14, \$15.

But actually another \$5 were spent so the prices could be any one of the following variations (and many others):

\$9, \$10, \$11, \$12, \$13, \$14, \$20 (Add \$5 to the highest price)

\$9, \$10, \$11, \$12, \$13, \$16, \$18 (Split \$5 into two and add to the two highest prices)

\$9, \$10, \$12, \$13, \$14, \$15, \$16 (Split \$5 into five parts of \$1 each and add to the top 5 prices)

\$7, \$9, \$13, \$14, \$15, \$16, \$17 (Take away some dollars from the lower prices and add them to the higher prices along with the \$5)

etc

Let’s focus on another piece of information given in the question stem: “every top cost more than every pillow case.”

This means that when we arrange all the prices in the increasing order (as done above), the last four are the prices of the four tops and the first three are the prices of the three pillow cases. The most expensive pillow case is the third one.

Now that we have accounted for all the information given in the question stem, let’s focus on the statements.

Statement 1: The most expensive top cost \$16.

We have already seen a case above where the maximum price was \$16. Is this the only case possible? Let’s look at our base case again:

\$9, \$10, \$11, \$12, \$13, \$14, \$15

(a further \$5 needs to be added to bring the total price up to \$89)

Since the prices need to be all unique, if we add 1 to any one price, we also need to add at least \$1 to each subsequent price. E.g. if we increase the price of the least expensive pillow case by \$1 and make it \$10, we will need to increase the price of every subsequent item by \$1 too. But we have only \$5 more to give.

If the maximum price is \$16, it means the rightmost price can increase by only \$1. So all prices before it can also only increase by \$1 only and except the first two prices, they must increase by \$1 to adjust the extra \$5.

Hence the only possible case is \$9, \$10, \$12, \$13, \$14, \$15, \$16.

So the cost of the most expensive pillow case must have been \$12.

Statement 1 is sufficient alone.

Statement 2: The least expensive pillow case cost \$9.

A restriction on the lowest price is much less restrictive. Starting from our base case

\$9, \$10, \$11, \$12, \$13, \$14, \$15,

we can distribute the extra \$5 in various ways. We can do what we did above in statement 1 i.e. give \$1 to each of the 5 highest prices: \$9, \$10, \$12, \$13, \$14, \$15, \$16

We can also give the entire \$5 to the highest price: \$9, \$10, \$11, \$12, \$13, \$14, \$20

So the price of the most expensive pillow case could take various values. Hence, statement 2 alone is not sufficient.

Note that the answer is a little unexpected, isn’t it? If we were to read the question and guess within 20 secs, we would probably guess that the answer is (C), (D) or (E). The two statements give similar but complementary information. It would be hard to guess that one will be sufficient alone while other will not be. This is what makes this question interesting and hard too.

Our strategy here was to establish a base case and tweak it according to the information given in the statements. This strategy is often useful in DS – not just in max-min questions but others too.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog series!

# How to Deal with Maximizing/Minimizing Strategies on the GMAT We haven’t dealt with maximizing/minimizing strategies in our QWQW series yet (except in sets). The reason for this is that the strategy to be used varies from question to question. What works in one question may not work in another. You might have to think up on what to do in a question from scratch and you have only 2 mins to do it in. The saving grace is that once you know what you have to do, the actual work involved to arrive at the answer is very little.

Let’s look at some maximizing minimizing strategies in the next few weeks. We start with an OG question today with a convoluted question stem.

Question: List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S?

I. -16
II. 6
III. 10

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Solution:

There is a lot of information in the question stem and a lot of variables are explained. Let’s review the given data in our own words first.

T has 30 decimals. The sum of all the decimals is S.

10 decimals have even tenths digit. They will be rounded up.

20 decimals have odd tenths digit. They will be rounded down.

The sum of rounded numbers is E.

E – S can take many values so how do we figure which ones it cannot take? We need to find the minimum value E – S can take and the maximum value it can take. That will help us figure out the values that E – S cannot take. Note that E could be greater than S and it could be less than S. So E – S could be positive or negative.

Step 1: Getting Minimum Value of E – S

Let’s try to make E as small as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very small. The estimate should add a very small number to round it up so that E is not much greater than S. Say the numbers are something similar to 3.8999999 (the tenths digit is the largest even digit) and they will be rounded up to 4 i.e. the estimate gains about 0.1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately .1*10 = 1 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be as large as possible. Say the numbers are something similar to 3.999999 (tenths digit is the largest odd digit) and they will be rounded down to 3 i.e. the estimate loses approximately 1 per number. Since there are 20 such numbers, the estimate is 1*20 = 20 less than actual.

Overall, the estimate will be approximately 20 – 1 = 19 less than actual.

Minimum value of E – S = -19

Step 2: Getting Maximum Value of E – S

Now let’s try to make E as large as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very high. Say the numbers are something similar to 3.000001 (tenths digit is the smallest even digit) and they will be rounded up to 4 i.e. the estimate gains 1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately 1*10 = 10 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be very little. Say the numbers are something similar to 3.1 (tenths digit is the smallest odd digit). They will be rounded down to 3 i.e. the estimate loses approximately 0.1 per number. Since there are 20 such numbers, the estimate is approximately 0.1*20 = 2 less than actual.

Overall, the estimate will be approximately 10 – 2 = 8 more than actual.

Maximum value of E – S = 8.

The minimum value of E – S is -19 and the maximum value of E – S is 8.

So E – S can take the values -16 and 6 but cannot take the value 10.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog series!

# Converting Non-Terminating Repeating Decimals to Fractions Last week we discussed the properties of terminating decimals. We also discussed that non-terminating but repeating decimals are rational numbers.

For GMAT, we must know how to convert these non-terminating repeating decimals into rational numbers. We know how to do vice versa i.e. given a rational number, we can divide the numerator by the denominator to find its decimal equivalent.

For example:

1/3 = 0.333333333… (infinite number of 3s)

But given 0.555555555…, how will you convert it to its exact fraction equivalent?

Take a look at this file: Non-terminating Decimals

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Terminating Decimals in Data Sufficiency on the GMAT Last week, we discussed the basics of terminating decimals. Let me review the important points here:

–  To figure out whether the fraction is terminating, bring it down to its lowest form.

–  Focus on the denominator – if it is of the form 2^a * 5^b, the fraction is terminating, else it is not.

Keeping this in mind, let’s look at a couple of DS questions on terminating decimals.

Question 1: If a, b, c, d and e are integers and m = 2^a*3^b and n = 2^c*3^d*5^e, is m/n a terminating decimal?

Statement 1: a > c
Statement 2: b > d

Solution:

Given: a, b, c, d and e are integers

Question: Is m/n a terminating decimal?

Or Is (2^a*3^b)/(2^c*3^d*5^e)?

We know that powers of 2 and 5 in the denominator are acceptable for the decimal to be terminating. If there is a power of 3 in the denominator after reducing the fraction, then the decimal in non- terminating. So our question is basically whether the power of 3 in the denominator gets canceled by the power of 3 in the numerator. If b is greater than (or equal to) d, after reducing the fraction to lowest terms, it will have no 3 in the denominator which will make it a terminating decimal. If b is less than d, even after reducing the fraction to its lowest terms, it will have some powers of 3 in the dominator which will make it a non-terminating decimal.

Question: Is b >= d?

Statement 1: a > c

This statement doesn’t tell us anything about the relation between b and d. Hence this statement alone is not sufficient.
Statement 2: b > d

This statement tells us that b is greater than d. This means that after we reduce the fraction to its lowest form, there will be no 3 in the denominator and it will be of the form 2^c * 5^e only. Hence it will be a terminating decimal. This statement alone is sufficient.

Now onto another DS question.

Question 2: If 0 < x < 1, is it possible to write x as a terminating decimal?

Statement 1: 24x is an integer.

Statement 2: 28x is an integer.

Solution:

Given: 0 < x < 1

Question: Is x a  terminating decimal?

Again, x will be a terminating decimal if it is of the form m/(2^a * 5^b)

Statement 1: 24x is an integer.

24x = 2^3 * 3 * x = m (an integer)

x = m/(2^3 * 3)

Is x a terminating decimal? We don’t know. If m has 3 as a factor, x will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Statement 2: 28x is an integer.

28x = 2^2 * 7 * x = n (an integer)

x = n/(2^2 * 7)

Is x a terminating decimal? We don’t know. If n has 7 as a factor, x will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Taking both together,

m/24 = n/28

m/n = 6/7

Since m and n are integers, m will be a multiple of 6 (and thereby of 3 too) and n will be a multiple of 7. So x will be a terminating decimal.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# How to Identify Terminating Decimals on the GMAT If you’ve been studying for the GMAT for any length of time, you are most likely pretty solid on the basic concepts of decimals and rational numbers.

Decimals can be either rational or irrational.

Decimals which terminate and those which are non-terminating but repeating are rational. These decimals can be written as fractions in the form a/b . As an example, .375=3/8

Decimals which are non-terminating and non-repeating are irrational, like roots such as root 2, root 3 etc. The digits after the decimal point will continue on forever.

The problem comes when we get a question based on these rather straightforwards concepts. That’s the point at which we realize that our basic knowledge and understanding are not as strong or as sufficient as we assumed them to be. As an example, let’s take a look at the following question:

Question: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189

(B) 15/196

(C) 16/225

(D) 25/144

(E) 39/128

If your first thought is that the best strategy is to simply divide the numerator by the denominator in each answer choice and thus figure out which  one terminates and which ones do not, congratulations! You have a ton of computation work ahead of you, and it will be extremely time-consuming and tedious, and committing to doing that many calculations dramatically increases your chances of making a careless mistake somewhere. Assuming that the process goes well, you would eventually arrive at the right answer, but we don’t have that kind of time to burn, and we know that the GMAT does not really care if we can crank out a lot of calculations in order to brute force our way to the correct answer. Because we know that the exam is trying to evaluate our ability to reason our way to the correct answer, there simply has to be another, more logical (and significantly less exhausting and tedious!) approach to this problem. Well, let’s walk through it:

A fraction, in its most reduced form, can be expressed as a terminating decimal if and only if the denominator has powers of only 2 and/or 5. Take a look at a simple example to try to understand why that is the case:

Let’s say that the variables a and b represent two integers. We would also know that the following relationship is true:

a/b=a*1/b

For ab to be a terminating decimal, 1b must be a terminating decimal. Well, what happens when you start dividing 1 by b? You add a decimal point and start adding 0s. You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate.

Taken that way, it’s a bit more obvious than it seemed, isn’t it?

1/3 = .333333333333333333…

1/7 = .142857142857142857…

1/11 = .09090909090909090…

Now that we understand how to identify if a fraction will yield a terminating decimal, let’s revisit the question we posed above. With this new understanding in mind, it should be a little more straightforward now.

Question 1: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189

(B) 15/196

(C) 16/225

(D) 25/144

(E) 39/128

Instead of having to crunch through each answer choice by hand, we can simply look at the denominators of each fraction to determine which one will give us a termination decimal. Let’s take a look at the prime factors of each denominator:

189=3^2 * 7 (will not terminate)

196=2^2 * 7^2 (will not terminate)

225=3^2 * 5^2 (will not terminate)

144=2^4 * 3^2 (will not terminate)

128=2^7 (bingo!)

Because 128 is a power of 2, our fraction will terminate, and E is thus the correct answer. All of the denominators in the other four answer choices contain other prime numbers as well and as a result will not terminate.

Using the same concepts as the previous question, let’s look at the following problem:

Question 2: If 1211*517is expressed as a terminating decimal, how many non-zero digits will the decimal have?

(A) 1

(B) 2

(C) 4

(D) 6

(E) 11

Solution:

A good first step is to realize that we can combine 2s and 5s to make 10s, which will make our job of figuring out the number of decimal places much simpler:

2^11 * 5^17 = 2^11 * 5^11 * 5^6 = 10^11 * 5^6

As a result, we know that  1/(10^11 * 5^6) is really just .00000…001 / 5^6.

From this point, we should try to figure out the answer intuitively:

What do you get when you divide .01 by 5? You get .002. You write 0s till you get 10 and then you get a non-zero digit.

What do you get when you divide .01 by 125 (which is 5^3)? You get .00008.

Do you notice something interesting there? The non zero term is 8, or 2^3.

The reason is this: You have 1 followed by as many 0s as you require in the dividend. 125 is 5^3, so you will need 2^3 i.e. you will need 10^3 as the dividend and then 125 will be able to divide out completely (in other words, the decimal will terminate).

Now, trying to apply the same logic, what will be the nonzero digits if you divide .00001 by 625?

Look at the exponents: 625 = 5^4, which tells us that, to get to a power of 10 ,we will need 2^4, or 16, to get 10^4, thus allowing the decimal to terminate. As a result, we know that the result will exactly have two nonzero digits: the 16 from the 2^4.

Going back to where we left off with the original problem, what will you get when you divide .000…0001 by 5^6? Applying the same reasoning as before, we know the the nonzero digits will be 64 (since we’re dividing by 5^6, we need 2^6, which is 64, to get to 10^6 and terminate the decimal).

As usual with GMAT questions, there is another way to approach it using similar logic but a different process. If we can manipulate the fraction to give us only a power of 10 in the denominator, we can then look at the numerator to know exactly what nonzero digits will appear in the resulting terminating decimal (the power of 10 in the denominator will simply tell us how many zeroes appear between the decimal place and the nonzero digits the numerator yields). Here, we need to turn the 56 in the denominator into 10s, so let’s multiply the entire fraction by 2^6 / 2^6:

1/ (10^11 * 5^6) * 2^6 / 2^6 = 2^6 /(10^11 * 10^6) = 2^6 / 10^17

The numerator turns into 64, and since the denominator is a power of 10, we can look at the exponent to know how many zeroes will appear between the decimal point and the nonzero digits. Because the denominator has an exponent of 17, the 64 from the numerator will be shifted 17 places to the right of the decimal place. The nonzero digits are the 64 that the numerator gives us, so we end up with two nonzero digits. The correct answer is thus B.

Let’s conclude our discussion of identifying terminating decimals on GMAT problems by taking a look at one more example, this time in the from of Data Suffiency, everyone’s favorite kind of quant problem:

Question 3: Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

90 < r < 100

(2) s = 4

Right off the bat, notice how much easier this question is if we rephrase it as “if rs is in its most simplified form, does the prime factorization of the denominator consist entirely of 2’s or 5’s?”

Now that we are experts on the properties and identification of terminating decimals, Statement 1 clearly can’t be sufficient on its own because it simply tells us nothing at all about the denominator. 91/2 and 91/10 are terminating decimals, for example, but 91/3 and 91/12 are not.

Statement 2 tells us that the denominator is 4, or 2^2. If we’ve mastered the rule for identifying terminating decimals, we should see right away that this must be sufficient because we know anything divided by 4 will result in a terminating decimal. The answer is B, Statement 2 alone is sufficient to answer the question. Beware of the C trap here: some GMAT takers may feel uncomfortable choosing an answer that involves no knowledge of the numerator. As demonstrated on the previous examples, however, we know that the numerator of the fraction is irrelevant in the process of identifying a terminating decimal.  Only the prime factors of the denominator (in its most reduced form) matter.

Our discussion of terminating decimals here also has a broader takeaway for GMAT preparation in general: many problems that appear to come down to tedious and/or time-consuming calculations are built around a concept, rule, or logical principle that will not be apparent to the unprepared GMAT student. Make sure your study plan includes time for understanding how concepts and rules work, not simply what they mean, and you will find yourself solving conceptually more often and using brute force less often, leading to great improvements in your time efficiency on the quant section!

# Can You Find the Correct Answer to This Tricky GMAT Question? This is hard to confess publicly but I must because it is a prime example of how GMAT takes advantage of our weaknesses – A couple of days back, I answered a 650 level question of weighted averages incorrectly. Those of you who have been following my blog would understand that it was an unpleasant surprise – to say the least. I know my weighted averages quite well, thank you! For this comedown, I blame the treachery of GMAT because it knows how to get you when you become too complacent. The takeaway here is – no matter how easy and conventional the question seems, you MUST read it carefully.

Let me share that particular question with you. I will also share two solutions which give you two different answers. It is an exercise for you to figure out which one is the correct solution (that is, if one of them is the correct solution). Needless to say, the error in the solution(s) is conceptual and very easy to see (not some sly calculation mistake). It’s just that in your haste, it’s very easy to miss this important point. I hope to see some comments with some good explanations.

Question: The price of each hair clip is ¢ 40 and the price of each hair band is ¢ 60. Rashi selects a total of 10 clips and bands from the store, and the average (arithmetic mean) price of the 10 items is ¢ 56. How many bands must Rashi put back so that the average price of the items that she keeps is ¢ 52?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Solution 1:

Price of each clip (Pc) = 40

Price of each band (Pb) = 60

Average price of each item (Pavg) = 56

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 56)/(56 – 40) = 1/4 (our weighted average formula)

Since the total number of items is 10, number of clips = 1*2 = 2 and number of bands = 4*2 = 8

If the average price is changed to 52,

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 52)/(52 – 40) = 2/3

Now the ratio has changed to 2:3. This gives us number of clips as 4 and number of bands as 6.

Since previously she had 8 bands and now she has 6 bands, she must have put back 2 bands.

Solution 2:

Say the number of hair clips is C and the number of hair bands is 10 – C.

(40C + 60(10 – C))/10 = 56 (Using the formula: Average = Sum/Number of items)

On solving, you get C = 2

Number of clips is 2 and number of bands is (C – 2) = 8.

Now, let’s consider the scenario when she puts back some bands, say x.

(2*40 + (8 – x)*60)/(10 – x) = 52

On solving, you get x = 5

So she puts back 5 bands so that the average price is 52.

Obviously, there is only one correct answer. It’s your job to figure out whether it is (B) or (E) or some third option. Also what’s wrong with one or both of these solutions?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Elementary, My Dear Watson! While eagerly awaiting the kick off of season 3 of BBC’s Sherlock, let’s put our time to good use. Though we have already spent a lot of it speculating over what really happened to Sherlock (HOW did he come back?!), perhaps we can take a leaf out of his book and learn to notice little things in whatever is leftover. There is a good reason to do that – there are little clues in some questions that the test maker unwittingly leaves to bring clarity to the question. If we understand those clues, a seemingly mysterious problem could be easily unraveled. Let us show you with an example.

Question: Peter and Jacob are at the northwest corner of a field, which is a rectangle 300 ft long and 160 ft wide. Peter walks in a straight line directly to the southeast corner of the field. If Jacob walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, what is the difference in the distance traveled by the two?

(A) 20
(B) 40
(C) 80
(D) 120
(E) 140

Solution: The first thing we do in these “direction” questions is draw the diagram. But there is a problem here: how do we decide the orientation of the rectangle? It could be either of these two. A few things help us decide this. There are two definitions of length:
1. Length is the longest side of the rectangle.
2. Width is from side to side and length is whatever width isn’t (i.e. the side from up to down in a rectangle) (this definition is less embraced than the first one)

If the side from up to down is the longest side, then there is no conflict.
Keeping this in mind, when drawing the figure, given that length is the longer of the two, one could make the rectangle on the left and there will be no conflict. But the question maker may not want to take for granted that you know this.

So he/she leaves a clue – the question mentions that ‘Jacob walks 180 ft down the west side of the field’. There needs to be at least 180 ft on the west side of the field for him to travel that much. So the orientation on the left makes sense. This is something the question maker would have put to try to give you a hint of the orientation. Now that we know what our diagram should look like, we can proceed to solve this question. If you just remember some of your pythagorean triplets, this question can be solved in moments (and that’s why we suggest you to remember them!) If not, it would involve some calculations.
QR = 160, RS = 300
So QR:RS = 8:15
Remember 8-15-17 pythagorean triplet? (the third triplet after 3-4-5 and 5-12-13)
Since the two sides are in the ratio 8:15, the hypotenuse must be 17. The common multiplier is 20 so QS  should be 17*20 = 340
Therefore, Peter traveled 340 feet.

TP = 120, PS = 160
TP:PS = 3:4
Does it remind you of 3-4-5 triplet?
120 is 3*40 and 160 is 4*40 so TS will be 5*40 = 200
So Jacob traveled a total distance of 180 + 200 = 380 feet.

Difference between the distance traveled = 380 – 340 = 40 feet

Note: The following triplets come in handy: (3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) and (9, 40, 41)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! When faced with an unusual or hard quadratic equation problem, some people waste a lot of time while trying to ‘split the middle term’. The common refrain is ‘I am just not good at it.’ Actually, difficult quadratic problems have little to do with intuition and a lot to do with understanding how numbers work. If I am looking at a tough quadratic equation and am unable to find the required factors, I will go back to check my quadratic to see if it is correct rather than try to use the esoteric quadratic formula.

To solve a quadratic equation, you need to find the two factors that add up to give the middle term and multiply to give the product of the constant term and co-efficient of x^2. What you may have problems with is ‘splitting the middle term’. Even though solving a quadratic is a basic skill one must possess to crack the GMAT, we have seen people struggle with it especially if the coefficient of x^2 is something other than 1. Let’s discuss how we can split the middle term quickly in such cases.

Tough Quadratic Equation Question 1: Solve for x: 5x^2 – 34x + 24 = 0

To factorize, we need to find two numbers a and b such that:
a + b = -34
a*b = 5*24

Step 1: Prime factorize the product.
a*b = 5*24 = 2*2*2*3*5

Step 2: Check the signs and decide what you need. Here the sum of a and b is negative (-34) while product is positive. This means a and b both are negative. Both will add to give a negative number and multiply to give a positive number (more on this at the end*). It also means that both a and b are smaller than 34 (since both are negative, their absolute values will be added to give 34).

Step 3: Try to split the prime numbers into two groups such that their sum is 34. Try the most obvious group first i.e.
2*2*2 and 3*5 -> 8 and 15

If a = 8 and b  = 15, we get a + b = 23
But the sum needs to be 34, i.e. a number greater than 23.

Before we discuss the next step, let’s talk about how adding numbers works:
Let’s say the prime factorization we have is 2*2*5*5.

We split it into 2 groups -> 2*5 and 2*5 (10 and 10). The sum of 10 and 10 is 20.
We split it in another way -> 2*2 and 5*5 (4 and 25). Their sum is 29. The sum increased.
We split it in yet another way -> 2 and 2*5*5 (2 and 50). Their sum is 52. The sum increased further.

Notice that further apart the numbers are, the greater is their sum. We get the least sum (20) when the numbers are equal. If we need a higher sum, we increase the distance between the numbers.

Going back to the original question, the prime factorization is 2*2*2*3*5 and we split it as 2*2*2 and 3*5. This gave us a sum of 8 + 15 = 23. We need 34 so we need to get the numbers farther from each other but not too far either. Let’s say, we pick a 2 from 8 and give it to 15. We get two numbers 4 and 30. They are farther apart and their sum is 34. So the numbers we are looking for are -4 and -30 (to get -34 as sum).

Now the quadratic is simply: 5*(x –  4/5)*(x – 30/5) = 0 (a shortcut to the usual ‘5x^2 – 4x – 30x + 24 = 0 and proceed’ method)

x = 4/5 or 6

Tough Quadratic Equation Question 2: Solve for x: 8x^2 – 47x – 63 = 0

To factorize, we need to find two numbers a and b such that:

a + b = -47
a*b = 8*(-63) = – 2*2*2*3*3*7

The product of a and b is negative which means one of a and b is negative. The sum of a and b is also negative therefore, the number with higher absolute value is negative and the other is positive. When these two numbers will be added, the difference of their absolute values will be the sum and the sign of the sum will be negative. So at least one of a and b is greater than 47. The greater one will be negative and the smaller one will be positive.

Let’s try to split the factors now. To start, we split the primes into two easy groups: 2*2*2 and 3*3*7 to get 8 and 63 but -63 + 8 = -55. We need the sum to have lower absolute value than 55 so we need to get the numbers closer together.

Take off a 3 from 63 and give it to 8 to get 2*2*2*3 and 3*7. Now a and b are -24 and 21; the numbers are too close.

Instead, take off 7 from 63 and give it to 8 to get 2*2*2*7 and 3*3. Now a and b are -56 and 9.

-56 + 9 = -47 -> the required sum.

Now the quadratic is 8*(x + 9/8)*(x – 56/8) = 0

x = -9/8 or 7

With a little bit of practice, the hardest quadratic equation questions can be quickly solved.

*How do you decide the sign of a and b:

Product is positive – This means both a and b have the same sign. If sum is negative, both a and b are negative; if sum is positive, they both are positive.

Product is negative – This means a and b have opposite signs; one is negative, the other is positive. If sum is positive, the number which is positive has a higher absolute value. If sum is negative, the number which is negative has a higher absolute value.

Keep in mind, also, that on the GMAT these difficult quadratic equation problems will always be accompanied by answer choices. Consequently, you generally have the option to test the answer choices to see which would solve the tough quadratic as opposed to slogging through some difficult quadratic formula math. If we revisit Tough Quadratic Equation Question #2, let’s see it with some answer choices:

Solve for x: 8x^2 – 47x – 63 = 0

(A) -12
(B) -7
(C) 7
(D) 12
(E) 21

Here a first step to lighten your load would be to consider negative vs. positive answer choices. Since the first term of the quadratic will be squared, that means that 8x^2 will be a large positive number regardless of whether x is positive or negative. The second term is subtracted, meaning that if x is positive the -47x term will be negative, but if x is negative the -47x term will be another big positive number. For this reason, you logically don’t have to try choices (A) and (B) – negative answer choices would leave too large a positive value for -63, the third term, to counteract.

From there you can choose which value to try next. Remember: tough quadratic equation problems on the GMAT are often solved via number properties and factors/multiples, so you can think of how the answer choice relate. Two are odd (7 and 21) and one is even (12); two are multiples of 7; and two are multiples of 3. The even/odd relationship can help you eliminate (D) without much extra work: if x were even, then the first two terms would each be even and the last, -63, would be odd. Even – Even – Odd would leave an odd result, but you need the even number 0 to round out this quadratic equation, so your choice is now between (C) and (E).

If you plug in 7, the easier number, note that again you can rely on factoring to solve this difficult quadratic. If x = 7 then the quadratic is:

8(7^2) – 47(7) – 63 = 0

Every term in this tough quadratic equation is a multiple of 7, so you can factor out your 7s to simplify and check that the quadratic does equal 0:

7(8 * 7) – 7(47) – 7(9) = 0

Divide both sides by 7 and you have:

56 – 47 – 9 = 0

This checks out, so (C) is the correct answer to this tough-looking (but not necessarily that tough) quadratic equation problem.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Integrated Reasoning – Cumulative Graphs Coming back to Integrated Reasoning question types, let’s discuss a cumulative graph today. They are usually a little trickier than your usual line/pie/bar graphs since you have to focus on not the data points but ‘the change’ from one data point to another.  Every subsequent data point will be either above or at the same level as the previous data point.

Let’s try to understand what a cumulative graph is before we look at questions on one. Here, the x axis gives the hourly wage and the y axis gives us the cumulative number of people. This means that the point giving the number of people with hourly wage of \$30 actually gives the number of people whose hourly wage is UP TO \$30. The point giving the number of people with hourly wage of \$40 gives the number of people whose hourly wage is up to \$40 and hence includes the number of people whose hourly wage is up to \$30. That is the reason the graph will always be flat or will have a positive slope.

If we want to focus on the people whose hourly wage varies from \$30 to \$40, we need to look at the slope of the graph in between these two points. The difference between these two points gives us the number of people with hourly wage in the range of \$30 – \$40. This implies that if the slope is steep, many people lie in this range.

Set 1: 200 people were surveyed to find out their hourly wage. 100 people had college degrees while other 100 were those who had not completed high school. The following graph gives the cumulative number of people and their hourly wages. Question 1: True/False: As per the given graph, the average hourly wages of people who did not complete high school is higher than the average hourly wages of people who have college degrees.

Question 2: The median wage of the people with college degrees lies between ___________ while the median wage of people who did not complete high school lies between ___________ (select two options, one for each blank)

(A)   \$10–\$20 per hr

(B)   \$20–\$30 per hr

(C)   \$30–\$40 per hr

(D)   \$40–\$50 per hr

(E)    \$50–\$60 per hr

Question 3: Approximately what percentage of ‘people with college degrees’ have hourly wage in the range \$50 – \$60?

(A) 25%

(B) 35%

(C) 45%

Solutions:

Solution 1: False

Just because the graph of ‘people who did not complete high school’ lies above the graph of ‘people who have college degrees’, it doesn’t mean that average hourly wage of people who did not complete high school is higher. Since the y axis gives the cumulative number of people, having a higher graph early on implies that many people have lower salaries. For example, about 45 people who did not complete high school have hourly wage up to \$10. On the other hand, only 18 people with college degrees have hourly wage up to \$10. Looking at the data, we can say that the average hourly wage of people with college degree will be much higher than the average hourly wage of people who did not complete high school. Solution 2: The median wage will be the average of the wage of the 50th person and 51st person. The thick black line shows the range in which these two lie.

Let’s first look at people who have college degrees. The 50th and 51st people will have wages lying in the range \$50 – \$60. Answer (E)

What about people who did not complete high school? The 50th person and the 51st person lie in the range \$10 – \$20. Answer (A)

Solution 3: Number of people with hourly wage up to \$50 is 45. Number of people with hourly wage up to \$60 is 80. Hence number of people whose hourly wage lies in the range \$50 – \$60 is about 80 – 45 = 35. Since 100 people were surveyed, the required percentage is about 35%. Answer (B)

Ensure you understand these graphs well.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# ‘Which’ vs ‘That’ Debate I know I promised that I will bring you some tricky Integrated Reasoning questions this week, but I am really irked by the ‘which’ vs ‘that’ debate and would like to put it to rest once and for all. Hence, in this post I would like to talk about restrictive and non-restrictive clauses, about ‘which’ and ‘that’, about when to use a comma and some other such things.

First of all, it is NOT necessary that ‘which’ has to be preceded by a comma.  Just because you see a ‘which’ clause without commas, it does not mean the option is wrong.

To understand the uses of ‘which’ and ‘that’, we need to understand defining and non-defining relative clauses.

What is a Relative Clause? It is the clause that begins with a relative pronoun (who, which, that, whom, whose)! We use relative clauses to clarify which person or thing we are talking about or to add extra information about a noun.

For example:

My father, who is 70, goes running every day.

My youngest son, whose work takes him all over the world, is coming home tomorrow.

My son who works for a consultancy is coming home tomorrow.

I’m going to wear the shirt that I bought in Paris.

The relative clauses have been underlined. Note that some are surrounded by commas and some are not.

The ones that are not surrounded by commas clarify which person or thing we are talking about. These are defining relative clauses.

The ones that are surrounded by commas provide extra information about a noun. They are called non-defining relative clauses.

Defining relative clauses: They define the noun. What do we mean by that? Let’s see.

Example: My son who works for a consultancy is coming home tomorrow.

‘My son who works for a consultancy’ implies that I probably have more than one son and one of them works for a consultancy. He is the one who is coming.

Example: I’m going to wear the shirt that I bought in Paris.

‘the shirt that I bought in Paris’ defines the shirt. I have many shirts but I am going to wear the one I bought in Paris.

• Defining relative clauses can begin with ‘who’, ‘which’ or ‘that’. You use ‘who’ or ‘that’ for people and ‘which’ or ‘that’ for things.

For example: All the sentences given below are correct.

My son who works for a consultancy is coming home tomorrow.

I’m going to wear the shirt which I bought in Paris.

I’m going to wear the shirt that I bought in Paris.

• Also, sometimes you can omit the relative pronoun of defining relative clauses. When the relative pronoun acts as the object of the relative clause, you can omit the relative pronoun.

Example: I’m going to wear the shirt I bought in Paris.– Correct

‘shirt’ here is the object of the verb ‘bought’. The relative clause is ‘I bought the shirt in Paris.’ The relative pronoun replaces ‘the shirt’ which is the object of this clause.

• When the relative pronoun is the subject of the relative clause, you cannot omit it.

My son who works for a consultancy is coming home tomorrow. – Incorrect

‘who’ is the subject of the verb ‘works’. You cannot omit the relative pronoun here.

Non-defining Relative Clauses: They provide extra information about the noun. In these cases, we already know the person/thing we are talking about.

Example: My father, who is 70, goes running every day.

‘My father’ clearly talks about my father (who we assume is unique). ‘who is 70’ only gives us more information about my father.

Example: My youngest son, whose work takes him all over the world, is coming home tomorrow.

My youngest son’ already clarifies that we are talking about my youngest son. ‘whose work…’ only tells us more about him.

• Non defining relative clauses can use most relative pronouns but they cannot use ‘that’. Also, you cannot omit the pronoun.

My father, that  is 70, goes running every day. – Incorrect. Cannot use ‘that’

My father,  is 70, goes running every day. – Incorrect. Cannot work without the pronoun

We hope this clarifies the use of ‘which’ and ‘that’ and also when to use commas.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# IR Questions: Multi Source Reasoning Now that we have seen some basic Integrated Reasoning question types, let’s start working on tricky Integrated Reasoning questions. The first set we would like to discuss is from GMAT Prep Software’s practice questions. This question has elements of RC, CR, PS and DS, all combined in one!

It was one of the first IR questions I had come across and I thought to myself – GMAT just got more complicated! Not because the given information was hard to understand but because there was a lot of it which needed to be analyzed together to arrive at a conclusion – not unlike our real life situations.

Anyway, let’s take a close look at this set to get a feel of what official IR questions are like.   From our discussion of the previous few weeks, it must be apparent that this is a multi source reasoning set. There are three tabs containing data about the techniques, artifacts and budget.

A quick read of the three tabs gives us the following general idea:

Techniques – Discusses techniques of dating different types of material

Artifacts – Discusses the collection of artifacts found and the accuracy of dating them

Budget – Discusses the cost of the techniques

In this way, we have got the lay of the land – so to speak. Now we can come back to the relevant information after reading the question.

Question 1: Which one of the following pieces of information would, on its own, provide the strongest evidence that the given artifact was actually produced on Kaxna?

(A)   A radiocarbon date of 1050 BC for a wooden bowl

(B)   IRMS analysis of a necklace made from animal bones and teeth.

(C)   A TL date for a fired-clay brick that places it definitively in the period of the Kaxna Kingdom

(D)   ICP-MS analysis of a metal tool that reveals element ratios unique to a mine on Kaxna

(E)    Determination that a stone statue was found near a quarry known to produce stone statues during the Kaxna Kingdom

Solution 1: The options give us various artifacts and ask us to find the one which has the strongest evidence of being produced on Kaxna. We would probably need the data given in the first two tabs – Techniques (to tell us which technique gives is what information about the artifact) and Artifacts (to tell us the accuracy of the techniques)

Let’s look at each option:

(A)   A radiocarbon date of 1050 BC for a wooden bowl

From tab 1, we know that radiocarbon dating gives us the approximate date of the plant’s death (and hence the approximate date of the creation of the object – within two years) but it doesn’t tell us the location of the creation of the object. We don’t know whether the bowl was made on the Kaxna islands.

(B)   IRMS analysis of a necklace made from animal bones and teeth.

IRMS analysis gives us clues about the animal’s diet and mineral content in water. Knowing these ratios for Kaxna island can give us an idea of whether the artifact was created on Kaxna. But this ratio of mineral content may not be unique to Kaxna and hence we cannot say whether the artifact was in fact created on Kaxna.

(C)   A TL date for a fired-clay brick that places it definitively in the period of the Kaxna Kingdom

TL dating may place the brick in the period of the Kaxna Kingdom but it doesn’t tell us the location where it was created. It may not have been created on the Kaxna island.

(D)   ICP-MS analysis of a metal tool that reveals element ratios unique to a mine on Kaxna.

ICP-MS analysis reveals ratios unique to a mine on Kaxna. This means the tool must have come from that mine on Kaxna. We can say with fair bit of certainty that the artifact came from Kaxna. Hence this seems to be the correct option. Let’s still take a look at option (E) too.

(E)    Determination that a stone statue was found near a quarry known to produce stone statues during the Kaxna Kingdom

Just because the stone statue was found near the quarry, it doesn’t mean that it was produced in the quarry. Hence we cannot say that the statue belongs to the Kaxna period.

We need to understand the given data really well to be able to answer this question. Let’s look at another question which uses the third Tab now.

Question 2: For each of the following combinations of Kaxna artifacts, select Yes if, based on the information provided, the cost of all pertinent techniques described can be shown to be within the museum’s first-year Kaxna budget. Otherwise, select No.

Yes                 No

O                     O             2 bone implements and 5 fired-clay cups decorated with gold

O                     O             7 wooden statues and 20 metal implements

O                     O             15 wooden statues decorated with bone

Solution 2: This question is similar to the Two-Part Analysis questions we have seen before. You have to answer three parts correctly here to get the right answer.

To answer this one, we need the first and the third tabs – Techniques to match the artifacts with the technique and Budget to get the cost of the technique. Let’s find out.

Total Budget: Unlimited IRMS + \$7000 (4TL + 15RC OR 40 ICP-MS)

Note that we have no idea about the costs of TL, RC, ICP-MS tests. We don’t know how their costs are related either.

1. 2 bone implements and 5 fired-clay cups decorated with gold

Bone implements need IRMS (unlimited available).

Fired clay objects need TL technique tests (5).

We know that \$7000 is enough for 4 TL tests but we don’t know whether it is enough for 5 TL tests. E.g. each TL may cost \$1500 or it may cost \$150 – we don’t know. Hence, we cannot say that the cost is within the budget.

2. 7 wooden statues and 20 metal implements

Plant matter needs RC dating tests (7).

Metals need ICP-MS  tests (20).

We know that \$7000 is enough for 40 ICP-MS tests so we know that 20 ICP-MS tests will cost less than or equal to \$3500.

We also know that \$7000 is enough for 4TL + 15RC tests which means \$3500 is enough for 2TL + 7.5 RC tests. So we can be certain that \$3500 is enough for 7 RC tests.

Therefore, the 7 RC tests and the 20 ICP-MC tests can be done within the \$7000 budget.

3. 15 wooden statues decorated with bone

Plant matter needs RC dating tests (15)

Animal bones need IRMS tests which happen in-house.

We know that \$7000 is enough for 4TL + 15RC tests which means it is certainly enough for 15RC tests.

Therefore, the 15 RC tests and 15 IRMS tests can be done within the \$7000 budget.

So we will click on ‘No’ in the first row, ‘Yes’ in the second row and ‘Yes’ in the second row.

We hope you realize that IR questions may seem hard but when you get down to it, they are pretty much like the other question types with which we are already very familiar.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Critical Reasoning: Some Common Mistakes Now that we have seen some basic Integrated Reasoning question types, we will look at some tricky questions but not this week. This week, we would like to discuss a Critical Reasoning question. This question is simple and straight forward but still many people falter in it. The reasons for this are not hard to find. Let’s analyze this question in detail.

Question:

People often criticize their local government for not providing enough funds to the public libraries in their district. They complain of lacking infrastructure and out-of-date and worn out reading material. Surprisingly, the most frequent and vociferous complaints come from those who live in districts where the libraries are most well maintained and kept current.

All of the following, considered individually, help to explain the apparent paradox EXCEPT:

(A) People from districts of well maintained libraries are more likely to use the public libraries.
(B) People have no knowledge of the facilities and infrastructure provided by the other libraries in their district.
(C) Good facilities cause people’s expectations to rise leading them to demand even more.
(D) The people in districts with well maintained libraries are likely to complain when the library they use is not as well maintained as the other libraries in that district.
(E) Most complaints about libraries come from political activists, most of who live in districts with well maintained libraries.

Solution:

The first thing to note here is that it is an ‘explain the paradox’ question but with an ‘EXCEPT’. This means that four of the five options will explain the paradox and our answer will be the one which will NOT. Test takers are often used to looking for the option that does explain the paradox and hence select an option which does this well. They often forget that they were actually required to do the opposite.  This is the first reason why test takers answer this question incorrectly.

Let’s try to understand the argument now:

People blame their local government for not maintaining their public libraries. The surprising thing is that most of these people come from districts with most well maintained libraries. You would expect that people living in districts with better libraries will have less to complain about and would be happier! Hence, here is the paradox.

Let’s do some pre-thinking now. How can you explain this paradox?

Some reasons come to mind: People who visit well maintained libraries have even higher expectations. They visit the library more and expect more out of it. If they find out that their friends in the same district have a library with great facilities, they become unhappy with their own library etc.

Let’s look at the options now:

(A)   People from districts of well maintained libraries are more likely to use the public libraries.

People who visit a place more are more aware of its follies. They get used to the amenities and worry about facilities that are not available. Also, if a library is frequented by many people, its books, videos and infrastructure in general will wear out faster. Another thing that could explain the paradox is that if more people visit the library, the probability of some people complaining about it is higher. Hence this option definitely explains why these people may complain more.

(B) People living in districts with good libraries have no knowledge of the facilities and infrastructure provided by the other libraries in their district.

This does not help explain our paradox. It doesn’t matter whether they know about the facilities provided by other libraries. If they do know the status of other libraries in their district, their behavior might be different but if they do not know, it has no effect on their behavior. Knowing may make them complain less or even more.Not knowing doesn’t make them complain more. In fact, if they do know about the facilities provided by other libraries in their district, they might start complaining even more if they like the facilities of other libraries more. It doesn’t help explain why they complain a lot right now.

(C) Good facilities cause people’s expectations to rise leading them to demand even more.

This option tells us that people get used to amenities and start expecting even more. This explains the paradox.

(D) The people in districts with well maintained libraries are likely to complain when the library they use is not as well maintained as the other libraries in that district.

This helps explain the paradox too. If many people in their district are getting better facilities than them, they might complain about the not equally good or better facilities available at their own library.

(E) Most complaints about libraries come from political activists, most of who live in districts with well maintained libraries.

This option is the most frequently chosen incorrect answer. Test takers reason that it is out of scope since it doesn’t matter who lives where. Actually, this explains the paradox too. Think about it – the option doesn’t talk about doctors or hair dressers; it talks about political activists. Political activists are usually very vocal and critical of things around them. These are the people who complain about everything and will complain about their library too. They live in districts with well maintained libraries and visit them. Since they visit well maintained libraries, many complaints will come from districts with well maintained libraries.

Also note that most does not necessarily imply just a little more than 50%. It could imply 80%, 90% etc too. Some test takers feel that 51% complaints come from political activists and more than half of them (i.e. approx. 26% of total) live in districts with well maintained libraries. This accounts for only 26% complaints and hence doesn’t explain the paradox.  But ‘most’ could just as easily imply 90% and hence this option certainly helps explain the paradox.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Integrated Reasoning – Two Part Analysis Questions Let’s continue our series and look at another Integrated Reasoning question type today – two part analysis. As complicated as it sounds, it’s actually the simplest of the IR question types in my opinion. The reason for this is that it tests no new skills; it checks your ability to handle the same old PS and CR questions.

The only reason it is new is that it reduces the probability of guessing correctly and it puts more time pressure on you! Your probability of guessing correctly is 20% in PS/CR questions; it goes down to 4% in two part analysis because you have to guess correctly twice. As for time pressure, you get about 2 mins for every PS and about 1.5 mins for every CR question. For each part of two part analysis, you have only 1.25 mins.

Anyway, let’s look at a sample question to get familiar with this question type.

Question: A grocery store sells fruits in pre packed closed bags such that individual pieces of fruit are not sold. Mangoes are sold at the rate of \$5 per bag (each bag contains two mangoes) and apples at the rate of \$8 per bag (each bag contains five apples). During a particular day, the store started with some mangoes and apples and sold them all by the end of the day. The revenue at the end of the day from selling mangoes and apples that day was \$128. Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?

Choose only one from each column: Solution:

Note that it is a PS question. Only the format of the question is different. Also, the use of the word ‘could’ in the question stem suggests that there could be multiple solutions to this problem. Let’s take a closer look at how to solve it.

Say, number of bags of mangoes is ‘m’ and number of bags of apples is ‘p’.

Then 5m + 8p = \$128 (total revenue)

Each bag of mangoes has 2 mangoes and each bag of apples has 5 apples.

So number of mangoes sold = 2m (to be selected in the first column)

Number of apples sold = 5p (to be selected in the second column)

We need to solve for this equation: 5m + 8p = 128

It is easy to see that one solution to this equation is m = 0, p = 16. The next solution will be m = 8, p = 11. Another will be m = 16, p = 6. Yet another will be m = 24, p = 1. If you are wondering how we are landing on one solution after another so effortlessly, you need to check out a previous post of QWQW – Integral Solutions to Equations in Two Variables.

So there are three solutions possible to our question: Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?

There are three different cases possible:

Case 1: Number of mangoes sold could be 16 (= 2m when m is 8). In that case number of apples sold will be 55 ( = 5p when m is 8, p is 11)

Case 2: Number of mangoes sold could be 32 (= 2m when m is 16). In that case number of apples sold will be 30 ( = 5p when m is 16, p is 6)

Case 3: Number of mangoes sold could be 48 ( = 2m when m is 24). In that case number of apples sold will be 5 (= 5p when m is 24, p is 1)

The case we select should be that of which both numbers are included in the options. Case 3 satisfies this condition. So we select 48 in the first column and 5 in the second column.

This is the only ‘exotic’ step of the two part analysis. The rest of the question is just like any other PS question.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Integrated Reasoning – Multi Source Reasoning For the past couple of weeks, we have been talking about integrated reasoning. Today we will continue with that and take up a multi-source reasoning question. These questions often include substantial data and require you to make inferences based on it. They test your logical and reasoning aptitude so don’t get lost in the data. Review the given information and then jump on to the questions. Then come back to the relevant part of the given information and peruse it in detail.

Given Data:

Minerals possess a number of properties that are used as an aid in their identification. These are listed below with a brief description:

Color – The color of the specimen as it appears to the naked eye under normal lighting conditions. Some minerals such as gold will only appear as one color, but due to impurities and crystal light distortion, many minerals can appear in multiple colors. Therefore, observable specimen color is the least effective property in identification.

Streak – The color of a mineral in powdered form. A streak test is performed by dragging a freshly cleaved mineral surface across an unglazed porcelain (Mohs hardness 7) surface. If the mineral is less hard than the porcelain, it will leave a stripe of color (the mineral in a powdered state). This is the true color of a mineral specimen as it lessens the impurity impact and eliminates the light distortion from the crystal. Although a mineral may have multiple observable specimen colors, it will only have one streak color.

Hardness – Minerals are identified roughly by their hardness based on the Mohs scale of mineral hardness, a list of ten minerals from #1 (softest) to #10 (hardest). All minerals will fall somewhere along the scale, based on their ability to scratch any mineral with a number lower than theirs and their inability to scratch any mineral with a number higher than theirs.

Mohs Scale of Mineral Hardness

1         Talc

2         Gypsum

3         Calcite

4         Fluorite

5         Apatite

6         Orthoclase

7         Quartz

8         Topaz

9         Corundum

10     Diamond

Specific Gravity – It is the relative weight of a mineral as compared to the weight of an equal volume of water. The specific gravity is also referred to as density and is expressed normally as an average of a small range of densities.

Common Minerals and Their Specific Gravity

Halite – 2.1

Diamond – 2.26

Gypsum – 2.3

Quartz – 2.7

Talc – 2.8

Muscovite Mica – 2.8

Corundum – 4.0

Cinnabar – 8.0

Gold – 19.3

Optical Properties – Used mainly by scientists, X-rays are sent through thin slices of mineral, producing identifying patterns of light which measure their index of refraction which is distinct for each mineral.

Properties of 3 unidentified minerals:

1. Mineral A was not able to scratch any of the top 9 minerals on Mohs scale. Its specific gravity is 2.3 (rounded to one decimal place).
2. The streak color of Mineral B is white. Its specific gravity is 2.3 (rounded to one decimal place)
3. Mineral C scratches Calcite and Topaz. It is pink in color and its index of refraction is 2.417.

Questions:

Question 1: For each of the following, select ‘Yes’ if the mineral can be uniquely identified based on the information provided. Otherwise, select No.

Question 2. State True/False: It is possible that mineral A and mineral B are the same mineral.
Solutions:

Solution 1. Mineral A – No

We know that the mineral lies between 9  and 10 (excluding 9 but including 10) on the Mohs scale. Also its specific gravity could be anything from 2.25 to 2.35 (excluding 2.35). There could be many minerals with these two properties. From the given data, we see that diamond is one such mineral but it may not be the only one.

Mineral B – No

Again, there can be many minerals with streak color white. Note that every mineral has a single streak color but every streak color may not belong to a single mineral. But we can say that its hardness must be less than 7 (hardness of unglazed porcelain) on Mohs scale since it has a streak color. Also its specific gravity could be anything from 2.25 to 2.35 (excluding 2.35).

Mineral C – Yes

Index of refraction is distinct for each mineral hence given the index of refraction, we can uniquely identify the mineral.

Solution 2. We need to compare mineral A with mineral B. The known properties of the two should not clash if they are to be the same mineral. Note from above that mineral A has a hardness of more than 9 on Mohs scale while mineral B has a hardness of less than 7. So it is not possible that mineral A and B are the same.

Hope the example gave you some idea about the multi source reasoning questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!