Word Problems - GMAT Quantitative

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

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Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

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Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

Tap to reveal answer

Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

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Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

Tap to reveal answer

Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

Tap to reveal answer

Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

Tap to reveal answer

Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Define the universal set $U = \left \{ 1,2,3,4,...50\right \}$ .

Define to be the set of prime numbers, to be the set of integers that end in 2, 5, or 8, and $C = \left \{ 21,22,23,24,...40\right \}$ . If each number in the universal set were to be placed in its correct region in the above Venn diagram, how many integers would lie in the gray region?

Tap to reveal answer

Answer

The integers that go into the gray region are those that do not fall into any of the three sets , , or . We can eliminate the integers by taking the universal set and eliminating the elements that fall in any one of the three.

Out of the 50 integers from 1 to 50, we can eliminate the 20 that are in . This leaves 30 so far:

$\left \{ 1,2,3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,18, 19, 20, 41,42,43,44, 45, 46, 47, 48, 49, 50\right \}$

Now we can eliminate nine integers from - the ones that end in 2, 5, or 8. The 21 remaining numbers are

$\left \{ 1, 3,4, 6, 7, 9, 10, 11, 13, 14,16, 17, 19, 20, 41, 43,44, 46, 47, 49, 50\right \}$ .

We now eliminate the primes from - 3, 7, 11, 13, 17, 19, 41, 43, 47. This leaves us with

$\left \{ 1,4, 6,9, 10,14, 16, 20, 44, 46,49, 50\right \}$ ,

a set with 12 elements.

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Question

Mark will hire 5 of the 8 job applicants he interviews. In how many different ways can he do this?

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Answer

Since order doesn't matter here, set this up as a combination:

$\frac{8!}{5!3!}=\frac{8\times 7\times 6\times 5!}{5!(3\times 2)} = 8 \times7=56$

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Question

Choose the statement that is the logical opposite of:

"John is a Toastmaster but not an Elk."

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Answer

Let and be the set of all Toastmasters and Elks, respectively, and let be the set of all people. $\textup{John} \in T$ and $\textup{John} \notin E$ , so the set to which John belongs is the shaded set in this Venn diagram:

the logical opposite of this is that John belongs to the shaded set in the diagram:

A way of saying this is $\textup{John} \in E$ or $\textup{John} \notin T$ , or, equivalently, if $\textup{John} \notin E$ , then $\textup{John} \notin T$ .

In plain English, if John is not an Elk, then John is not a Toastmaster.

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Question

Two inlet pipes lead into a large water tank. One pipe can fill the tank in 45 minutes; the second pipe can fill it in 40 minutes. At 8:00 AM, the first pipe is opened; at 8:10 AM, the second one is opened. To the nearest minute, at what time is the tank full?

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Answer

Look at the work rates as "tanks per minute", not "minutes per tank".

The two pipes can fill the tank up at $\frac{1}{45}$ tanks per minute and $\frac{1}{40}$ tanks per minute.

Let be the time it took, in minutes, to fill the tank up. Then this is the amount of time that the first pipe had to let in water; the amount of time that the second pipe had, in minutes, is .

Since rate multiplied by time is equal to work, then the two pipes fill up $\frac{1}{45} t$ and $\frac{1}{40} \left ( t-10\right )$ tanks; together, they filled up $\frac{1}{45} t + \frac{1}{40} \left ( t-10\right )$ tank - one tank. This sets up the equation to be solved:

$\frac{1}{45} t + \frac{1}{40} \left ( t-10\right ) =1$

$360 \cdot \left [ \frac{1}{45} t + \frac{1}{40} \left ( t-10\right ) \right ] =360 \cdot 1$

$\frac{360}{45} t + \frac{360}{40} \left ( t-10\right ) =360$

$8 t + 9 \left ( t-10\right ) =360$

$t = 450 \div17 \approx 26.47$

This rounds to 26 minutes after the first pipe is opened, or 8:26 AM.

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Question

Andy and his wife Donna both work at the same building.

One morning, Andy left home at 8:00; Donna left 5 minutes later. Each arrived at their common destination at 8:50. Andy drove at an average speed of 45 miles per hour; what was Donna's average speed, to the nearest mile per hour?

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Answer

Andy traveled for 50 minutes, or $\frac{5}{6}$ of one hour, at 45 miles per hour, so the office building is $\frac{5}{6} \times 45 = 37 \frac{1}{2}$ miles from Andy and Donna's house. Donna traveled this distance for 45 minutes, or $\frac{3}{4}$ of one hour, so her speed was:

$37 \frac{1}{2} \div \frac{3}{4} = \frac{75}{2} \cdot \frac{4} {3} = 50$ miles per hour.

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Question

In order to qualify for the next heat, the race-car driver needs to average 60 miles per hour for two laps of a one mile race-track. The driver only averages 40 miles per hour on the first lap. What must be the driver's average speed for the second lap in order to average 60 miles per hour for both laps?

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Answer

If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:

$\frac{X+Y}{2}=60$

In our case we know lap one was driven at miles per hour. We substitute this value in for and solve for .

$\frac{40+Y}{2}=60$

Thus to average miles per hour for two laps with lap one being miles per hour, lap two would have to have a rate of miles per hour.

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Question

Together, Mary and I, can trade stocks at a rate of stocks every minutes. I alone on the other hand can only trade stock every minutes. How fast can Mary trade, alone?

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Answer

To solve this problem, we need to set up an equation as follows

$M+\frac{1}{2}=\frac{15}{4}$ ,.

is Mary's rate.

By simply manipulating the terms, we end up with

$M=\frac{13}{4}$ , which is the final answer.

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Question

Fill in the blank:

$3\frac{1}{4} \textup{ pounds } = \underline{ : : : : : : : : : : : : }\textup{ounces}$

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Answer

One pound is equal to 16 ounces.

Multiply:

$3 \frac{1}{4} \times 16 = 52$

52 ounces is the correct choice.

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Question

If it takes Sally 3 hours to drive $\dpi{100} \small q$ miles, how many hours will it take her to drive $\dpi{100} \small r$ miles at the same rate?

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Answer

If Sally drives q miles in 3 hours, her rate is 3/q miles per hour. Plug this rate into the distance equation and solve for the time:

$\dpi{100} \small Distance = rate\times time$

$\dpi{100} \small r=\frac{q}{3}\times t$

$\dpi{100} \small t=\frac{3r}{q}$

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Question

A cat runs at a rate of 12 miles per hour. How far does he run in 10 minutes?

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Answer

We need to convert hours into minutes and multiply this by the 10 minute time interval:

$\small \frac{12\ miles}{1\ hour}x\frac{1\ hour}{60\ min}x\frac{10\ min}{1}=\frac{120\ miles}{60}=2\ miles$

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Question

A group of students are making posters to advertise for a bake sale. 12 large signs and 60 small signs are needed. It takes 10 minutes to paint a small sign and 30 minutes to paint a large sign. How many students will be needed to paint all of the signs in 2 hours or less?

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Answer

In 2 hours, 1 student can paint 4 large signs or 12 small signs. Therefore, 3 students are required to paint the large signs ( $4\times 3=12$ ) and 5 students are required to paint the small signs ( $12\times 5=60$ ). In total, 8 students are required.

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Question

Jason is driving across the country. For the first 3 hours, he travels 60 mph. For the next 2 hours he travels 72 mph. Assuming that he has not stopped, what is his average traveling speed in miles per hour?

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Answer

In the first three hours, he travels 180 miles.

$3\times60=180$

In the next two hours, he travels 144 miles.

$2\times 72=144$

for a total of 324 miles.

Divide by the total number of hours to obtain the average traveling speed.

$\frac{324}{5}=64.8\ mph$

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Question

Randall traveled 75 kilometers in 600 minutes. What was Randall's per hour rate?

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Answer

We need to pay close attention to some details here.

We are given time in minutes, but asked for an answer in hours.

A rate can be defined as distance over time.

Taking the first detail, we convert 600 minutes to 10 hours, since there are 60 minutes in one hour.

Taking the second detail, we divide 75 kilometers by 10 hours. This gives us an answer of 7.5 kilometers per hour.

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