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 4331 (when K = 10). 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!*

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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

Answer (B)

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!*

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

An n sided polygon has n vertices. If you join every distinct pair of vertices you will get nC2 lines. These nC2 lines account for the n sides of the polygon as well as for the diagonals.

So the number of diagonals is given by nC2 – n.

nC2 – n = n(n-1)/2 – n = **n(n – 3)/2**

Taking quick examples:

**Example 1**: How many diagonals does a polygon with 25 sides have?

No. of diagonals = 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. Now let’s go on to the part which seems to make no sense at all!

**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 = 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 you 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. Your job is to tell us which method is correct and why the other method is incorrect.

*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!*

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.

Answer (C)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

**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’.

Answer (C)

**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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

A **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

Things to note about conjunctions:

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!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

**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

Answer (D)

**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.

Answer (D)

**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.

Answer (B)

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

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

*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.

Answer (B)

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

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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.

Answer: A

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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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 in special triangles such as isosceles and equilateral, they can overlap. We will now give you some 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 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 quick question on these concepts:

**Question**: What is ∠A in triangle ABC?

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

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

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

Jump on to the statements directly.

Statement 1: The bisector of ∠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 ∠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!

Answer (C)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*