The exciting thing is that pre-thinking is useful in Quant too. If you take a step back to review what the question asks and think about what you are going to do and what you expect to get, it is highly likely that you will not get distracted mid-way during your solution. Let’s show you with the help of an example:

**Question**: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

**Statement I**: Train B arrived at Newcastle before train A arrived at Birmingham.

**Statement II**: The distance between Newcastle and Birmingham is greater than 140 km.

Following are the things that would ideally constitute pre-thinking on this question:

- Quite a bit of data is given in the question stem with some speed and time taken.

- Distance traveled by both the trains is the same since they travel along the same route.

- We could possibly make an equation by equating the two distances and come up with multiple answers for the time at which train B arrived at Newcastle.

- The statements do not provide any concrete data. We cannot make any equation using them but they might help us choose one of the answers we get from the equation of the question stem.

Mind you, the thinking about the statements helping us to arrive at the answer is just speculation. The answer may well be (E). But all we wanted to do at this point was find a direction.

The diagram given above incorporates the data given in the question stem. Train A starts from Newcastle toward Birmingham at 3:00 and meets train B at 4:00. Train B starts from Birmingham toward Newcastle at 3:50 and meets train A at 4:00. Let x be the distance from Birmingham to the meeting point.

Speed of train A = 100 km/hr

Speed of train B = Distance/Time = x/(10 min) = x/(1/6) km/hr = 6x km/hr (converted min to hour)

If we get the value of x, we get the value of speed of train B and that tells us the time it takes to travel from the meeting point to Newcastle (a distance of 100 km). So all we need to figure out is whether the statements can give us a unique value of x.

By 4:00, train A has already travelled for 1 hour and train B has already travelled for 10 mins i.e. 1/6 hour. Total time taken by both is 2 hrs. The remaining (5/6) hrs is the time needed by both together to reach their respective destinations.

Time taken by train A to reach Birmingham + Time taken by train B to reach Newcastle = 5/6

Distance(x)/Speed of train A + 100/Speed of train B = 5/6

x/100 + 100/6x = 5/6

3x^2 – 250x + 5000 = 0

3x^2 – 150x – 100x + 5000 = 0

3x(x – 50) – 100(x – 50) = 0

(3x – 100)(x – 50) = 0

x = 100/3 or 50

So speed of train B = 6x = 200 km/hr or 300 km/hr

Statement 1: Train B arrived at Newcastle before Train A arrived at Birmingham.

If x = 50, time taken by train A to reach Birmingham = 50/100 = 1/2 hour and time taken by train B to reach Newcastle = 100/300 = 1/3 hour. Train B takes lesser time so it arrives first.

If x = 33.33, time taken by train A to reach Birmingham = (100/3)/100 = 1/3 hour and time taken by train B to reach Newcastle = 100/200 = 1/2 hour. Here, train A takes lesser time so it arrives first at its destination.

Since train B arrived first, x must be 50 and train B must have taken 1/3 hour i.e. 20 mins to arrive at Newcastle. So train B must have arrived at 4:20.

This statement is sufficient alone.

Statement 2: The distance between Newcastle and Birmingham is greater than 140 km.

Total distance between Newcastle and Birmingham = (100 + x) km. x must be 50 to make total distance more than 140.

Time taken by train B must be 1/3 hr (as calculated above) and it must have arrived at 4:20.

This statement is sufficient alone.

Answer (D)

So our speculation was right. Each of the statements provided us relevant information to choose one of the two values that the quadratic gave us.

*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 few months back, we had discussed a 700 level ‘Races’ question.

**Question 1**: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately.

(A) 8, 10

(B) 4, 5

(C) 5, 9

(D) 6, 9

(E) 7, 10

Check out its complete solution here.

Now, what if we had only 30 seconds to guess on it and move on? Then we could have easily guessed (B) here and moved on. Actually, the question implies that the only possible options are those in which the time taken by B is somewhere between 3 mins and 6 mins (excluding) – we would guess 4 mins or 5 mins. Since only option (B) has time taken by (B) as 5 mins, that must be the answer – no chances of error here – perfect! Had there been 2 options with 4 mins/5 mins, we would have increased the probability of getting the correct answer to 50% from a mere 20% within 30 seconds.

Now you are probably curious as to how we got the 3 min to 6 min range. Here is the logic:

Read one sentence of the question at a time -

“*A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds.*”

So first, A gives B a head start of 1/10th of the race but still beats him. This means B is certainly quite a bit slower than A. This should run through your mind on reading this sentence.

“*Next, A gives B a head start of 3 mins and is beaten by 1000 m.*”

Next, A gives B a head start of 3 mins and B beats him by 1000 m i.e. half of the race. What does this imply? It implies that B ran more than half the race in 3 mins. To understand this, say B covers x meters in 3 mins. Once A, who is faster, starts running, he starts reducing the distance between them since he is covering more distance than B every second. At the end, the distance between them is still 1000 m. This means the initial distance that B created between them by running for 3 mins was certainly more than 1000 m (This was intuitively shown in the diagram in this post). Since B covered more than 1000 m in 3 mins, he would have taken less than 6 mins to cover the length of the race i.e. 2000 m. A must be even faster and hence would take even lesser time.

Only option (B) has time taken by B as 5 mins (less than 6 mins) and hence satisfies our range! So the answer has to be (B).

Let’s try the same technique on another question.

**Question 2**: If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?

(A) 4.2 days

(B) 6.8 days

(C) 8.3 days

(D) 9.8 days

(E) 10.2 days

Solution: If we try to use algebra here, the calculations involved will be quite complicated. The options are not very close together so we can try to get a ballpark value and move forward. Let’s take each sentence at a time:

“*If 12 men and 16 women can do a piece of work in 5 days*”

Say rate of work of each man is M and that of each woman is W. This statement gives us that

12M + 16W = 1/5 (Combined rate done per day)

In lowest terms, it is 3M + 4W = 1/20

“*13 men and 24 women can do it in 4 days*,”

This gives us 13M + 24W = 1/4

“*how long will 7 men and 10 women take to do it?”*

Required: 7M + 10W = ?

Solving the two equations above will be tedious so let’s estimate:

(3M + 4W = 1/20) * 6 gives 6M + 8W = 1/10

So 6 men and 8 women working together will take 10 days. Hence, 7 men and 10 women will certainly take fewer than 10 days.

(13M + 24W = 1/4) / 2 gives 6.5M + 12W = 1/8

So 7 men and 10 women might take about 8 or perhaps a little bit more than 8 days to complete the work. There is only 0.5 additional man (hypothetically) but 2 fewer women to complete it. So we would guess that the number of days would lie between 8 to 10 and closer to 8 days.

Answer (C) fits.

Note that it seems like there are many equations here but all you have actually done is made two equations. Once you write them down, you don’t even need to actually multiply them with some integer to get them close to the required equation. Just looking at the first one, you can say that 6 men and 8 women will take 10 days. It takes but a couple of seconds to arrive at these conclusions.

*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!*

In your haste to complete the test on time, don’t overlook the important details. Getting too many easy questions wrong is certainly disastrous. Take a step back and ensure that what they asked is what you have found and that your logic is solid. To illustrate the problem, let’s give you a question – people gloss over it, consider it an easy remainders problem, answer it incorrectly and move on. But guess what, it isn’t as easy as it looks!

**Question**: If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

**Statement 1**: The remainder when (m + n) is divided by 7 is 1.

**Statement 2**: The remainder when (m – n) is divided by 3 is 1.

First let’s give you the **incorrect** solution provided by many.

Question: What is the remainder when (m^2 – n^2) is divided by 21?

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

Therefore, remainder of product (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) when it is divided by 21 is 1.

Answer (C)

This would have been correct had the statements been:

Statement 1: The remainder when (m + n) is divided by 21 is 1.

Statement 2: The remainder when (m – n) is divided by 21 is 1.

Statement 1: (m + n) = 21a + 1

Statement 2: (m – n) = 21b + 1

(m^2 – n^2) = (m + n)*(m – n) = (21a + 1)*(21b + 1) = 21*21ab + 21a + 21b + 1

Here, every term is divisible by 21 except the last term 1. So when we divide (m^2 – n^2) by 21, the remainder will be 1.

But let’s go back to our original question. If you solved it the way given above and got the answer as (C), you are not the only one who jumped the gun. Many people end up doing just that. But here is the correct solution:

The statements given are:

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

This gives us (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) = 21ab + 7a + 3b + 1

Here only the first term is divisible by 21. We have no clue about the other terms. We cannot say that 7a is divisible by 21. It may or may not be depending on the value of a. Similarly, 3b may or may not be divisible by 21 depending on the value of b. So how can we say here that the remainder must be 1? We cannot. We do not know what the remainder will be in this case even with both statements together.

Say, if a = 1 and b = 1,

m^2 – n^2 = 21*1*1 + 7*1 + 3*1 + 1 = 21 + 11

The remainder when you divide m^2 – n^2 by 21 will be 11.

Say, if a = 2 and b = 1,

m^2 – n^2 = 21*2*1 + 7*2 + 3*1 + 1 = 21*2 + 18

The remainder when you divide m^2 – n^2 by 21 will be 18.

Hence, both statements together are not sufficient to answer the question.

Answer (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!*

Today, let’s bring back the beloved lazy genius through a question. Try to solve it lazily i.e. try to do minimum work on paper. This means making equations and solving them is a big no-no and doing too many calculations is cumbersome.

**Question**: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)

(B) yz/(yz + xz – xy)

(C) yz/(yz + xz + xy)

(D) xyz/(yz + xz – xy)

(E) (yz + xz – xy)/yz

Note that you have variables in the question and the options. Since we are looking for a lazy solution, making equations out of the variables is not acceptable. So then, should we plug in numbers? With three variables to take care of, that might involve a lot of calculations too. Then what else?

Here is our minimum-work-solution to this problem; try to think one of your own and don’t forget to share it with us.

Plugging in numbers for the variables can be troublesome but you can give some very convenient values to the variables so that the effect of a pump and a drain will cancel off.

There are no constraints on the values of x, y and z except z > x (drain Gamma empties slower than pipe Alpha fills)

Let’s say, x = 2 hrs, y = 4 hrs, z = 4 hrs

What did we do here? We made the rate of Beta same as the rate of Gamma i.e. 1/4 of the tank each. This means, whenever both of them are working together, drain Gamma cancels out the work of pump Beta. Every hour, pump Beta fills 1/4^{th} of the tank and every hour drain Gamma empties 1/4^{th} of the tank. So the entire tank will be filled by pump Alpha alone. Hence, if y = z, pump Alpha fills the entire tank i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1.

In the options, put y = z and see which option gives you 1. Note that you don’t have to put in the values of 2, 4 and 4. We gave those values only for illustration purpose.

If y = z, xy = xz.

So in option (B), xz cancels xy in the denominator giving yz/yz = 1

Again, in option (E), xz cancels xy in the numerator giving yz/yz = 1

The other options will not simplify to 1 even though when we put y = z, the answer should be 1 irrespective of the value of x, y and z. The other options will depend on the values of x and/or y. Hence the only possible options are (B) and (E). But we still need to pick one out of these two.

Now let’s say, x = 4, y = 2, z = 4.00001 ( z should be greater than x but let’s assume it is infinitesimally greater than x such that we can approximate it to 4 only)

Rate of work of Gamma (1/4^{th} of the tank per hour) is half the rate of work of Beta (1/2 the tank per hour). Rate of work of Gamma is same as rate of work of Alpha. Half the work done by pump Beta is removed by drain Gamma. So if pump Beta fills the tank, drain Gamma empties half of it in that time – the other half would be filled by pump Alpha i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1/2.

Put x = z in the options (B) and (E). The one that gives you 1/2 with these values should be the answer. Again, you don’t need to plug in the actual values till the end.

If x = z, yx = yz

(B) yz/(yz + xz – xy)

yz cancels xy in the denominator giving us yz/xz = y/x = 2/4 = 1/2

(E) (yz + xz – xy)/yz

yz cancels xy in the numerator giving us xz/yz = x/y = 4/2 = 2

Only option (B) gives 1/2. Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. If you do, you are on your way to becoming a lazy genius yourself!

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

Let’s recap the concepts before we see the question:

I. When you divide a number by 100, the remainder is formed by the last two digits of the number.

II. When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x.

III. We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

These three were discussed with examples last week.

IV. When m is divided by n and a negative remainder (–r) is obtained, we can find the actual remainder simply as (n – r).

This is discussed with examples in this post.

Now, let’s solve a question involving all these concepts.

**Question**: What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96

(B) 76

(C) 56

(D) 36

(E) 16

**Solution**: We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.

Find remainder of (301*402*503*604*646*547*448*349)^2/100

Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.

We need to find the remainder of:

(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100

This implies we need to find the remainder of:

(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50

We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.

We need the remainder of:

(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50

Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:

We need:

the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50

the remainder of (12)*(24)*(24)*(24) / 50

Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.

We need:

the remainder of (6)*(24)*(24)*(24) / 25

the remainder of (6)*(25-1)*(25-1)*(25-1) / 25

Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:

the remainder of 6*(-1)*(-1)*(-1) / 25

The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76

Answer (B)

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

Actually, we can find the last two digits quite easily in most such cases by using the concepts of remainders.

There are two concepts you need to understand before we go on to see how to solve such questions:

I. **When you divide a number by 100, the remainder is formed by the last two digits of the number.** Say, you divide 138 by 100, the remainder will be 38 (last two digits). Take another example – divide 1275 by 100, the remainder will be 75 and so on.

II. **When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x**. This should remind you of the binomial theorem we discussed many weeks ago. When we multiply all these terms together (px + a), (qx + b) etc, each term obtained will have at least one x except the last term which is obtained by multiplying the remainders together. To get a better idea, let’s take some numbers:

Let’s say we need to find the remainder when we divide 12*23*52*81 by 10.

K = 12*23*52*81 = (10 + 2)*(20 + 3)*(50 +2)*(80 + 1)

When you multiply these four terms together, you will get many terms such as 10*20*50*80, 10*20*50*1, 10*20*2*80 etc. All these will have a multiple of 10 except the last one. The last one will be 2*3*2*1 = 12. That doesn’t have a multiple of 10. Now divide 12 by 10 to get the remainder 2. So when you divide K by 10, the remainder will be 2.

Now, let’s look at a question:

Question 1: What are the last two digits of 63*35*37*82*71*41?

(A) 10

(B) 30

(C) 40

(D) 70

(E) 80

Solution: Using concept 1, we know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

*Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.*

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)

Next week, we will see some more complicated questions using these and other fundamentals.

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

Last Digit of Base:

0 – Last digit of expression with any power will be 0.

1 – Last digit of expression with any power will be 1.

2 – 2, 4, 8, 6, 2, 4, 8, 6… Cyclicity is 4.

3 – 3, 9, 7, 1, 3, 9, 7, 1… Cyclicity is 4.

4 – 4, 6, 4, 6, 4, 6, 4, 6… Cyclicity is 2.

5 - Last digit of expression with any power will be 5.

6 – Last digit of expression with any power will be 6.

7 – 7, 9, 3, 1, 7, 9, 3, 1… Cyclicity is 4.

8 – 8, 4, 2, 6, 8, 4, 2, 6… Cyclicity is 4.

9 – 9, 1, 9, 1, 9, 1, 9, 1… Cyclicity is 2.

Cyclicity is nothing but pattern recognition. You see that when you multiply 2 by itself, there is a pattern of last digit which goes 2, 4, 8, 6, 2, 4, 8, 6 and so on. We can use the same principle for when a question asks us for the last two digits of the expression. Let me remind you first that here at QWQW, we sometimes flirt with the lines that define GMAT scope. Obviously, we do point out whenever we are indulging and that’s exactly what we are going to do in this post. We are carrying on for the love of Math and the Q51 score.

The last two digits of the base decide the last two digits of the expression. For example,

**Example 1: **Let’s look at powers of 11.

11^1 = 11

11^2 = 121

11^3 = 1331

11^4 = …41 (we should just multiply the last two digits together and ignore the rest)

11^5 = …51

11^6 = …61

11^7 = …71

Note that the last two digits are displaying a pattern depending on the power. So we expect the cyclicity here to be 10.

11^8 = …81

11^9 = …91

11^(10) = …01

11^(11) = …11

11^(12) = …21

and so on. So the last two digits should go from 11, 21 to 91, 01 and then go to 11 again. The cycle of 10 starts from power of 1, 11, 21 etc. This means that 11^(46) should have last two digits as 61, 11^(92) should have last two digits as 21 and 11^(168) should have last two digits as 81.

Let’s look at some other numbers now:

**Example 2:** Say, we need the last two digits of 6^{58}

6^1 = 6 (No second last digit)

6^2 = 36

6^3 = 216

6^4 = …96 (Just multiply the last two digits)

6^5 = …76

6^6 = …56

6^7 = …36

and hence starts the cycle again:

3, 1, 9, 7, 5, 3, 1, 9, 7, 5 and so on.

The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit.

**Example 3:** How about the last two digits of 7^102?

7^1 = 7 (No second last digit)

7^2 = 49

7^3 = 343

7^4 = …01

7^5 = …07

7^6 = …49

7^7 = …43

We see a cyclicity of 4 here: 49, 43, 01, 07, 49, 43, 01, 07 … and so on. The new cycle begins at 2, 6, 10, 14 i.e. even powers which are not multiples of 4. So a new cycle will begin at 102 too. So the last two digits of 7^(102) will be 49.

Now there can be many variations in the questions asking us to find the last two digits. We will use different concepts for different question types. Today we saw how to use pattern recognition. We will look at some other methods next week.

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

Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark’s remaining chips are $20 chips, how much money did Mark bet?

You can view this as a word problem where you assume the number of chips and then go splitting them up or you can view this as a mixtures problem even though it doesn’t use words such as ‘mixture’, ‘solution’, ‘combined’ etc. As we have seen enough number of times, our mixture problems are solved in seconds using the weighted average concept.

The question discussed here also belongs to the same category – looks super tricky but can be easily solved with weighted averages formula. But we have seen plenty and more of such questions in our blog posts. Today we will take a look at a different type of sinister question and I suggest you to think about the concept being tested in that before trying to solve it.

**Question**: Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?

(A)10:00

(B)10:34

(C)11:02

(D)11:48

(E)12:20

Before we look at the solution, think about the concept being tested here – clocks? Circular motion?

Neither!

**Solution**: Note that when giving data about watch1, you are told how it varies with the actual time. Data about all other watches tells us about the time they show relative to the incorrect watches. The concept being tested here is Relative Speed!

What do we mean by “gains 15 mins” or “loses 20 mins” etc? When a watch gains 15 mins every hour, it means that even though it should show that one hour has passed, it shows that 1 hr 15 mins have passed. So the watch runs faster than it should. Hence the speed of the watch is more than the speed of a correct watch. Now the question is how much more? The minute hand of the correct watch travels one full circle in one hour. The minute hand of the incorrect watch travels one full circle and then a quarter circle in one hour (to show that 1 hour 15 mins have passed even when only an hour has passed). So it is 5/4 times the speed of a correct watch. On the same lines, let’s analyze each watch.

Say the speed of a correct watch is s.

- “Watch1 loses 15 minutes every hour. “

Watch1 covers only three quarters of the circle in an hour.

Speed of watch1 = (3/4)*s

- “Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15).”

Now we have the speed of watch2 relative to speed of watch1. Speed of watch2 is (5/4) times the speed of watch1.

Speed of watch2 = (5/4)*(3/4)s = (15/16)*s

- “Watch3 loses 20 minutes every hour relative to watch2.”

Watch3 loses 20 mins every hour means its speed is (2/3)rd the speed of watch2

Speed of watch3 = (2/3)*(15/16)*s = (5/8)*s

- “Finally, watch4 gains 20 minutes every hour relative to watch3.”

Speed of watch4 = (4/3)*Speed of watch3 = (4/3)*(5/8)*s = (5/6)*s

So the speed of watch4 is (5/6)th the speed of a correct watch. So if a correct watch shows that 6 hours have passed, watch4 will show that 5 hours have passed. If a correct watch shows that 12 hours have passed, watch4 will show that 10 hours have passed. From 12 noon to 12 midnight, a correct watch would have covered 12 hours. Watch4 will cover 10 hours and will show the time as 10:00.

Answer (A)

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

*While traveling from Detroit to Novi, a car averaged 10 miles per gallon, and while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?*

We have two figures for mileage given here – 10 miles per gallon and 18 miles per gallon. We need to find the average mileage. So we can use the weighted average formula but what will the weights be? Will they be 1:2 since the distance between the two cities is given to be in the ratio 1:2? If you think that taking the distance to be the weights in this problem is correct, then you fell for the trap in this question.

To explain the concept, let us use a simpler example first:

When talking about average speed, what are the weights? We know that the weight given to each speed is the time for which that speed was maintained, right? Yes! But why?

Let’s review our weighted average formula:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)

Average Speed = (Distance1 + Distance2)/(Time 1 + Time2)

Average Speed = Total Distance/Total Time

This is an accurate representation of average speed.

Now see what happens when you use distance as the weights.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Speed*Distance doesn’t represent any physical quantity. So this doesn’t make sense. The units of the quantities will help you see the relation clearly.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)

Average Speed = (miles/hour * hour + miles/hour * hour)/(hour + hour)

Average Speed = (miles + miles)/(hour + hour)

Average Speed = Total miles/Total hours

What happens when you take distance as the weights?

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Average Speed = (miles/hour * miles + miles/hour * miles)/(miles + miles)

miles^2/hour doesn’t represent a physical quantity and hence doesn’t make sense here. Therefore, whenever you are confused what the weights should be, look at the units.

Let’s go back to the original question now. Average required is miles per gallon. So you are trying to find the weighted average of two quantities whose units must be miles/gallon.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

The unit of Cavg, C1 and C2 is miles/gallon so w1 and w2 should be in gallons to get

miles/gallon = (miles/gallon * gallon + miles/gallon * gallon)/(gallon + gallon)

miles/gallon = Total miles/Total gallons

So how will we actually solve this question?

**Question**: While traveling from Detroit to Novi, a car averaged 10 miles per gallon while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

**Solution**:

Let the distance between Detroit and Novi be D. So the distance between Novi and Lapeer must be 2D.

Amount of fuel used to cover distance D = D/10

Amount of fuel used to cover distance 2D = 2D/18 = D/9

So the two weights used must be D/10 and D/9

Average miles/gallon = (10*D/10 + 18*D/9)/(D/10 + D/9) = 3D*90/19D = 270/19 = 14.2 miles/gallon

Or simply, Average miles/gallon = Total miles/Total gallons = 3D/(D/10 + D/9) = 14.2 miles/gallon

Food for thought: Which one of the following can you solve?

- If a vendor sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was his overall profit%?

- If a vendor sold apples at a profit of 10% and oranges at a profit of 20%, what was his overall profit% if cost price of each apple was $0.20 and the cost price of each orange was $.06?

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

Myth 3: Use of ‘being’ is always wrong on GMAT!

Often, the way we use ‘being’ in our day-to-day communication, is incorrect. For example,

Being a doctor, he is very well respected.

But there are correct ways of using ‘being’. Since most students believe that ‘being’ is wrong, don’t trust the GMAC to not use this nugget of information to misdirect the test takers. The correct answers of questions at higher ability are worded in such a way that they make the test takers uncomfortable!

So how is ‘being’ used correctly?

‘Being’ is used to express a temporary state.

The little boy started screaming when he saw his dog being impounded.

‘Being impounded’ is a temporary state and would be over – unlike being a doctor. So the use of being is correct here.

Let’s look at one of our own sentence correction questions now:

**Question**: The data being collected in the current geological survey are providing a strong warning for engineers as they consider the new dam project, but their greatest importance might lie in how they influence the upcoming decision by those same engineers on whether to retrofit 75 bridges in the survey zone.

A. The data being collected in the current geological survey are providing a strong warning for engineers as they consider the new dam project, but their greatest importance

B. The data being collected in the current geological survey provide a strong warning for engineers as they consider the new dam project, but its greatest importance

C. The data collected in the current geological survey is providing a strong warning for engineers as they consider the new dam project, but their greatest importance

D. The data collected in the current geological survey provides a strong warning for engineers in consideration of the new dam project, but its greatest importance

E. The data collected in the current geological survey provide a strong warning for engineers in consideration for the new dam project, but the greatest importance

**Solution: **Let’s find the decision points:

First decision point: being collected vs collected

‘The data being collected’ is a temporary state here. Data won’t always be collected, but are being collected for a short time right now, so ‘being’ is used properly here. And the sentence makes it very clear that this is temporary; look at the word ‘current’ before ‘geological survey.’ If you were skeptical of the word ‘being’ before, that is understandable, but the word ‘current’ should serve as a clear warning that this is a temporary, ongoing event. (And furthermore, the non-underlined portion talks about an upcoming decision, even further cementing the idea that this is a temporary survey with data ‘being’ collected for a short period of time). Alas, even with that temporary state, there isn’t really anything wrong with ‘The data collected in …’ either so we retain all answer options.

It’s worth noting that they put the ‘being collected’ vs. ‘collected’ decision early in the sentence/answer-choices as a great place to put a ‘false decision point.’ The authors of these questions know that they want to reward emphases on logical meaning and core grammar rules, and they also know that students like to study quick tips and tricks, so they leave that bait there for the tips/tricks folks while they hold off the bigger reward for those willing to prioritize decision points from most important to least and not just from left to right in order of appearance.

Second decision point: are/is

Technically, data is plural of datum. In academic writing it is almost always treated as plural. It is treated as singular in informal writing but GMAT favors treating it as plural.

Even if you do not know this, the use of “*they* influence the upcoming” – in the portion of the sentence that is not underlined – should tell you that ‘data’ is used in plural form here.

Hence the use of ‘are’ is appropriate. Hence, options (C) and (D) are eliminated.

Third decision point: Pronouns

There are many pronouns used here. Antecedent of each pronoun is present in the sentence. The usage clarifies which pronoun refers to data and which refers to engineers.

Original Sentence: The data being collected in the current geological survey are providing a strong warning for engineers as **they** consider the new dam project, but **their** greatest importance might lie in how they influence the upcoming decision by those same engineers on whether to retrofit 75 bridges in the survey zone.

**they** – only engineers can consider the new dam project so ‘they’ refers to engineers

**their/its** - greatest importance will be of data, which is plural, so ‘their’ would be the correct usage. Eliminate option (D)

There is no ambiguity in the use of pronouns. The nouns are present and the usage clarifies the antecedent.

Now we are left with options (A) and (E). In option (E), “in consideration for the new dam project” is bad diction. Also, it doesn’t tell us ‘whose greatest importance?’.

Answer is (A)

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