Bunuel's Word Problems PS Diagnostic Test

Question

Bunuel's Word Problems Diagnostic Test

Distance/Rate Problems, Mixture Problems, Overlapping Sets, Work/Rate Problems, Word problems.

https://gmatclub.com/forum/download/file.php?id=69958

Please note that the questions are ranked in ascending level of difficulty according to the timer stats.

Grading & How to use this test:  
   Please time yourself and allocate 2 mins per question 
  17 - 18 or more correct, your Word Problems score is Q50+ 
  15 - 16 correct, your Word Problems score is Q49 
  13 - 14 correct, your Word Problems score is Q48 
  11 - 12 correct, your Word Problems score is Q47 
  9 - 10 correct, your Word Problems score is Q46 
  6 - 8 correct, your Word Problems score is Q45

Note that levels 7 and 8 are very hard questions. These questions are well within the GMAT parameters but their difficulty is driven by the amount of time you will need to spend to solve each of the questions. Do not be discouraged if you are not able to solve any of the Level 7 or 8 questions.

------------------- EASY -------------------

LEVEL 1:    
 A book store charges $21 for the first book ordered in the store, and $16 for each additional book. If Larry bought x books in the store, how much did he pay in terms of x ?

A. 16x - 5
B. 16x + 5
C. 16x + 16
D. 21x + 5
E. 21x + 16

B

In an x-meter race competition, Tom gives Jerry a 240-meters head start. If Tom runs 13/5 times as fast as Jerry and if the race ends in a tie, what is the value of x ?

(A) 330 meters
(B) 390 meters
(C) 600 meters
(D) 720 meters
(E) 730 meters

B

LEVEL 2:

A cyclist, after covering 2/5th of the race, noticed that to reach the midpoint of the race, she needs to cover 15 kilometers less than she had already covered. How long was the race in kilometers ?

A. 25
B. 30
C. 50
D. 100
E. 150

C

A predator is chasing its prey. The predator takes 4 leaps for every 6 leaps of the prey and the predator covers as much distance in 2 leaps as 3 leaps of the prey. What is the ratio of the speed of the predator to that of the prey ?

(A) 11 : 9
(B) 10 : 9
(C) 1 : 1
(D) 9 : 10
(E) 9 : 11

C

------------------- MEDIUM -------------------

LEVEL 3:

A bicycle wheel, x meters in circumference, is rolling at a constant rate in a straight line without slipping. If the wheel covers y meters in t hours, then what is the number of 360-degree rotations that the wheel make while rolling y meters for 1 hour in terms of x, y, and t?

A. xy/t
B. tx/y
C. x/(ty)
D. y/(tx)
E. xyt

D

Two cars, each moving at their respective constant speeds, leave simultaneously from points A and B towards each other. After they pass each other, car from A reaches B in 25 minutes, while car from B reaches A in 36 minutes. How many minutes it took the faster car to cover the whole distance between the two points ?

A. 44 minutes
B. 45 minutes
C. 50minutes
D. 55 minutes
E. 66 minutes

D

LEVEL 4:

Mary Poppins is choosing presents at a certain book shop. She wants books on poetry to comprise more than 45% but less than 50% of all the books she picks in the shop. What is the least number of books on poetry she can possibly get in the shop ?

A. 1
B. 5
C. 6
D. 10
E. 11

B

Athos, Porthos, Aramis and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get ?

A. 81
B. 189
C. 324
D. 405
E. 1024

B

------------------- HARD -------------------

LEVEL 5:

A shoe manufacturer started offering a standard 25% discount on the list price of a pair of shoes. However, a large retailer requested that in addition to the standard discount, they should additionally get 25 free pairs of shoes for every 100 pairs they purchase. The manufacturer calculated that even in this case, he would still make 20% profit on the cost. If the cost price per pair of shoes is $c, then what is the list price of the pair of shoes in terms of c ?

A. 2c
B. 3/2*c
C. 4/3*c
D. 5/4*c
E. 6/5*c

A

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes ?

A. 1440
B. 1600
C. 2460
D. 3920
E. 4000

A

LEVEL 6:

Hungry chipmunk notices that, along a straight path, there are walnuts placed at intervals of 1 meter. The chipmunk wants to gather all the walnuts in his burrow, which is exactly where the middle walnut is. The chipmunk starts gathering the walnuts from the leftmost walnut and carries only one walnut at a time. If after completing the task, the chipmunk covered a distance of 300 meters and collected odd number of walnuts, then how many walnuts did the chipmunk collect ?

A. 11
B. 13
C. 15
D. 23
E. 25

E

A crate contains 103 defective cellphones. If you distribute defective cell phones to x technicians such that each technician gets the same number of defective cell phones, the number of cellphones left in the crate is 2 less than the number of technicians. Which of the following cannot be the value of x ?

A. 3
B. 5
C. 7
D. 14
E. 15

D

------------------- VERY HARD -------------------

LEVEL 7:

A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?

A. 43
B. 44
C. 45
D. 46
E. 47

A

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the range of the shoe prices, which would allow the store to maintain or increase its current weekly revenue from the shoes ?

A. 12
B. 13
C. 23
D. 24
E. 25

D

LEVEL 8:

In a 1000 miles race Flash gives Superman a start of 40 miles and beats him by 19 milliseconds. If, in a 1000 miles race, Flash gives Superman a start of 30 milliseconds then Superman beats Flash by 40 miles. What is the ratio of speed of Flash to that of Superman ?

A. 4:5
B. 5:6
C. 6:5
D. 5:4
E. 4:3

C

Mbappe withdraws  \frac{100}{x}\%  of his money each time he visits the bank, where  x > 1 . If after 5 visits, he has less than  \frac{1}{x}^{th}  of the initial amount in the bank, what is the range of all possible value of  x  ? (Assume x is an integer)

A. 1
B. 2
C. 3
D. 4
E. 5

B

Ronaldo and Messi race each other in an x-meter race. They start simultaneously from the same point. If Ronaldo, running at a content rate, covers the distance in 8 seconds, while Messi, running at a content rate, covers the distance in 10 seconds, in how many seconds the distance left to cover by Ronaldo is half the distance left to cover by Messi ?

A. 3 seconds
B. 10/3 seconds
C. 5 seconds
D. 20/3 seconds
E. Cannot be determined

D

At a certain library, 12 students borrowed a total of 71 books. If the range of the number of books borrowed per student was 5, which of the following CANNOT be the number of students who borrowed 6 books?

I. 9
II. 10
III. 11

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

I. 9
II. 10
III. 11

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

B

OTHER DIAGNOSTIC TESTS:
 
   Bunuel's Word Problems PS Diagnostic Test 
   Bunuel's Algebra PS Diagnostic Test  
   Bunuel's Algebra DS Diagnostic Test

Bunuel · Accepted Answer

Official Solution:

Two cars, each moving at their respective constant speeds, depart simultaneously from points A and B towards each other. After passing each other, the car from point A reaches point B in 25 minutes, while the car from point B reaches point A in 36 minutes. How many minutes did it take for the faster car to cover the entire distance between the two points?

A. 44 minutes
B. 45 minutes
C. 50 minutes
D. 55 minutes
E. 60 minutes

Since the cars departed simultaneously and after meeting, the car from point A reached point B in less time than the car from point B took to reach point A, it's evident that the car from point A is faster.

https://gmatclub.com/gmat-practice-tests/resources/import/img/M40-78++.png

Assuming they met after  t  minutes, then the car from A spent 25 minutes covering the same distance the car from B covered in  t  minutes. Since they travel at constant speeds, the ratio of their speeds would be equal to the reciprocal of the ratio of their times: the ratio of speeds of A to that of B =  \frac{t}{25} . For instance, if one car took  t  minutes to cover 10 kilometers, and another took 25 minutes to cover the same distance, then the ratio of the speed of the first car to that of the second car would be  \frac{(\frac{10}{t})}{(\frac{10}{25})}=\frac{25}{t} .

Similarly, since the car from A spent  t  minutes covering the same distance the car from B covered in 36 minutes, we'd have the ratio of speeds of A to that of B =  \frac{36}{t} .

Equating these two gives:  \frac{t}{25} = \frac{36}{t} , which yields  t^2 = 25 * 36 , and finally,  t = 5 * 6 = 30  minutes.

Therefore, the faster car, the one from point A, took  t + 25 = 30 + 25 = 55  minutes to cover the entire distance between the two points.

Answer: D

mysterymanrog · Answer

At a certain library, 12 students borrowed a total of 71 books. If the range of the number of books borrowed per student was 5, which of the following CANNOT be the number of students who borrowed 6 books?

I. 9
II. 10
III. 11

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

I. 9
II. 10
III. 11

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

Lovely Q here!

Lets start by eliminating choices:

III) If this is the case, then there would be 1 students who has to borrow 5 books. The range here is not 5, but 1. So this is not a possible value of 6. Eliminate A

I) If this is the case, that means 54 books are borrowed among the 9 students, leaving 17 books for the remaining 3. Since 17 is not a multiple of 3, one student has to borrow 9 books, and the other two students will borrow 4 books. The range is (9-4=5), so this is possible. Eliminate E, A and D.

II) If this is the case, this means that 60 books are borrowed by 10 students, leaving 11 for the last 2. If one student borrows 8 and the other student borrows 3, the range condition is satisfied. Therefore, discard D.

B is the correct answer.

Bunuel  you should check the scoring guide, since it says 27+ for Q50, but there are only about 18 problems.

chanandler_bong · Answer

Would it be possible to get solutions  Bunuel  ? thank you

Bunuel · Answer

Official Solution:

In an  x -meter race competition, Tom gives Jerry a 240-meters head start. If Tom runs  \frac{13}{5}  times as fast as Jerry and if the race ends in a tie, what is the value of  x ?

A. 330 meters
B. 390 meters
C. 600 meters
D. 720 meters
E. 730 meters

The main idea in this question is that since they start simultaneously and the race ends in a tie, they run for the same amount of time. Tom runs  x  meters, while Jerry runs  x - 240  meters. Now, we equate the times:  time = \frac{distance}{rate} = \frac{x}{(\frac{13}{5}*r)} = \frac{x - 240}{r} , where  r  is the rate of Jerry. When we cancel out  r , we get  \frac{x}{(\frac{13}{5})} = x - 240 , which simplifies to  5x = 13x - 240*13 . Finally, this gives us  x = 390  meters.

We can also plug-in answer options. If we plug B, it would mean that Tom covered 390 meters and Jerry covered 390 - 240 = 150 meters. Since they run for the same amount of time, the ratio of distance covered should match the ratio of their rates. This means that 390/150 should equal 13/5, which it does. Thus, B is the correct answer.

Answer: B

Bunuel · Answer

Official Solution:

A bicycle wheel,  x  meters in circumference, is rolling at a constant rate in a straight line without slipping. If the wheel covers  y  meters in  t  hours, then what is the number of 360-degree rotations that the wheel makes while rolling  y  meters for 1 hour in terms of  x ,  y , and  t ?

A.  \frac{xy}{t} 
B.  \frac{tx}{y} 
C.  \frac{x}{ty} 
D.  \frac{y}{tx} 
E.  xyt

Rolling  y  meters in  t  hours, means that in 1 hour the bicycle moves  \frac{y}{t}  meters, and since the circumference of the wheel is  x  meters, then the wheel makes  \frac{(\frac{y}{t})}{x} = \frac{y}{tx}  360-degree rotations in that time.

Answer: D

Bunuel · Answer

Official Solution:

A predator is chasing its prey. The predator takes 4 leaps for every 6 leaps of the prey, and the predator covers as much distance in 2 leaps as 3 leaps of the prey. What is the ratio of the speed of the predator to that of the prey?

A. 11 : 9
B. 10 : 9
C. 1 : 1
D. 9 : 10
E. 9 : 11

Assuming one leap of the predator is  x  units and one leap of the prey is  y  units, then:

In 4 leaps, the predator covers the distance of  4x  units.

In 6 leaps, the prey covers the distance of  6y  units.

Note that these distances take the predator and the prey the same amount of time.

Since it is also given that the predator covers as much distance in 2 leaps as the prey does in 3 leaps, we get  2x = 3y . Substituting  2x = 3y  in the above, we get  6y = 2(3y) = 4x .

Thus, in the same time period, both the predator and the prey cover the same distance of  4x  units, indicating an equal speed.

Answer: C

Bunuel · Answer

Official Solution:

A bookstore charges $21 for the first book ordered in the store, and $16 for each additional book. If Larry bought  x  books in the store, how much did he pay in terms of  x ?

A.  16x - 5 
B.  16x + 5 
C.  16x + 16 
D.  21x + 5 
E.  21x + 16

According to the stem, Larry would be charged $21 for the first book and $16 for each of the additional  x-1  books. Thus, he'd be charged a total of  21*1 + 16(x - 1) = 16x + 5  dollars.

Answer: B

Bunuel · Answer

Official Solution:

A cyclist, after covering  \frac{2}{5}^{th}  of the race, noticed that to reach the midpoint of the race, she needs to cover 15 kilometers less than she had already covered. How long was the race in kilometers?

A. 25
B. 30
C. 50
D. 100
E. 150

Let's assume the race is  d  kilometers long. Then half of the race would be  \frac{d}{2}  kilometers long. We are told that to reach the midpoint of the race, so to cover additional  \frac{d}{2} - \frac{2}{5}d  kilometers, she needs to cover 15 kilometers less than she had already covered, so she needs to cover  \frac{2}{5}d -15 . Hence, we'd have:
 
 \frac{d}{2} - \frac{2}{5}d = \frac{2}{5}d-15 
 
Multiplying by 10 to get rid of the fractions yields:
 
 5d - 4d = 4d-150

3d = 150

d = 50 
 
Answer: C

Bunuel · Answer

Official Solution:

Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?

A. 81
B. 189
C. 324
D. 405
E. 1,024

After each distribution,  \frac{3}{4} th of the coins remain to be distributed to the next person. For instance:

Athos receives  \frac{1}{4}n  coins, leaving  \frac{3}{4}n  coins for distribution.

Then, Aramis receives  \frac{1}{4} th of that remaining amount, leaving  \frac{3}{4}(\frac{3}{4}n)=(\frac{3}{4})^2n  coins to distribute.

Next, Porthos receives  \frac{1}{4} th of the remaining amount, leaving  \frac{3}{4}((\frac{3}{4})^2n)=(\frac{3}{4})^3n  coins to distribute.

Subsequently, D'Artagnan receives  \frac{1}{4} th of the remaining amount, which is  \frac{1}{4}(\frac{3}{4})^3n  coins, leaving  \frac{3}{4}((\frac{3}{4})^3n)=(\frac{3}{4})^4n  coins to distribute.

In the final distribution, D'Artagnan also receives  \frac{1}{4} th of the remaining amount, an additional  \frac{1}{4}(\frac{3}{4})^4n  coins, making his total receipt  \frac{1}{4}(\frac{3}{4})^3n + \frac{1}{4}(\frac{3}{4})^4n=(\frac{3^3}{4^4} + \frac{3^4}{4^5})n=\frac{189}{1,024}n .

Given that  \frac{189}{1,024}n  must be an integer, the minimum value of  n  is 1,024. Therefore, the least number of coins D'Artagnan could have received is 189.

Answer: B

Bunuel · Answer

Official Solution:

A shoe manufacturer started offering a standard 25% discount on the list price of a pair of shoes. However, a large retailer requested that in addition to the standard discount, they should additionally receive 25 free pairs of shoes for every 100 pairs they purchase. The manufacturer calculated that even in this case, he would still make a 20% profit on the cost. If the cost price per pair of shoes is  $c , then what is the list price of the pair of shoes in terms of  c ?

A.  2c 
B.  \frac{3}{2}c 
C.  \frac{4}{3}c 
D.  \frac{5}{4}c 
E.  \frac{6}{5}c

Observe that the manufacturer essentially offers the retailer 125 pairs of shoes at the list price of 75 pairs — the price of 100 pairs after the 25% discount equals the list price of 75 pairs, and the retailer receives an additional 25 pairs as a gift. The cost price of 125 pairs of shoes is  $125c . Given that the manufacturer still maintains a 20% profit margin, the selling price becomes  $125c 	imes 1.2 = $150c .

Therefore, we deduce that the list price of 75 pairs equals  $150c , making the list price of one pair equal to  $2c .

Answer: A

Bunuel · Answer

Official Solution:

A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?

A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000

Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2  x  times. Then the price of a pair of shoes would be  32 - 2x  dollars, and a total of  80 + 20x  pairs of shoes would be sold, generating revenue equal to  (32 - 2x)(80 + 20x)  dollars.

We want to maximize  (32 - 2x)(80 + 20x) . Let's work with the expression:
 
 (32 - 2x)(80 + 20x)=

=-40(x - 16)(4 + x)=

=-40(x^2 - 12x - 64) 
 
Complete the square for  x^2 - 12 x - 64 :
 
 =-40(x^2 - 12x + 36 -36 - 64)

=-40

=-40(x - 6)^2 + 4,000 
 
Since the value of  -40(x - 6)^2  is 0 or negative, the maximum value of  -40(x - 6)^2 + 4,000  is  4,000 , that is when  -40(x - 6)^2  is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is  4,000 - 32*80 = 1,440  dollars.

Answer: A

Bunuel · Answer

Official Solution:

A shoe store sells a new model of running shoe for $32. At this price, the store sells 80 pairs each week. However, the manager estimated that for every $2 reduction in the price, the store will sell an additional 20 pairs of shoes per week. According to these estimations, what is the range of shoe prices that would allow the store to maintain or increase its current weekly revenue from the shoes?

A. 12
B. 13
C. 23
D. 24
E. 25

The current weekly revenue from the shoes is  32 * 80  dollars.

Assuming  x  is the number of times the price was reduced by 2 dollars, the price of a pair of shoes would be  32 - 2x  dollars, and a total of  80 + 20x  pairs of shoes would be sold, generating revenue equal to  (32 - 2x)(80 + 20x)  dollars.

The question asks to find the range of shoe prices such that  (32 - 2x)(80 + 20x) \geq 32 * 80 . Reduce by  40 :
 
 (16 - x)(4 + x) \geq 32*2

-x^2 + 12 x + 64 \geq 32*2

x^2 - 12x \leq 0

x(x - 12) \leq 0 
 
The above inequality holds for  0 \leq x \leq 12 . Thus, the shoe prices for which the store will maintain or increase its current weekly revenue start from  32 - 2 * 0 = 32  dollars and end with  32 - 2 *12 = 8  dollars. Therefore, the range of prices is  32 - 8 = 24 .

Answer: D

Bunuel · Answer

Official Solution:

Hungry chipmunk notices that, along a straight path, there are walnuts placed at intervals of 1 meter. The chipmunk wants to gather all the walnuts in its burrow, which is exactly where the middle walnut is. The chipmunk starts gathering the walnuts from the leftmost walnut and carries only one walnut at a time. If after completing the task, the chipmunk covered a distance of 300 meters and collected an odd number of walnuts, then how many walnuts did the chipmunk collect?

A. 11
B. 13
C. 15
D. 23
E. 25

Check the image below:

https://gmatclub.com/gmat-practice-tests/resources/import/img/M40-100.png

There are  n  walnuts to the left of the burrow,  n  walnuts to the right of the burrow, and 1 walnut exactly in the middle, at the burrow. Hence, the total number of walnuts is  n + n + 1=2n + 1 .

The chipmunk starts where the leftmost walnut is, so the first leg of the travel covers  n  meters. After that, the chipmunk reaches the burrow. To collect the remaining  n-1  walnuts, placed to the left of the burrow, the chipmunk should travel  n-1  meters twice: to the walnut and back to the burrow;  n-2  meters twice, to the walnut and back to the burrow; and so on until the first walnut to the left, where it travels 1 meter to the walnut and 1 meter back to the burrow. Thus, the chipmunk covers a total of:

2*1 + 2*2 + 2*3 + ... + 2(n-2) + 2(n-1)=2(1+2+3+...+(n-2)+(n-1))  meters.

The sum within the parentheses represents the sum of consecutive integers from 1 to  n-1 , which equals to  \frac{first + last}{2} * (number \ of \ terms) . Hence, we get:

2(1+2+3+...+(n-2)+(n-1))= 2(\frac{1 + (n-1)}{2}*(n-1))=n(n-1)

Don't forget the  n  meters the chipmunk covered to collect the first walnut to get the total distance covered to collect all walnuts to the left:

n(n-1) + n = n^2  meters.

Similarly, we can calculate the distance the chipmunk will cover to collect walnuts placed to the right of the burrow, with an expectation that the chipmunk has have to cover  n  meters twice, not once: to the rightmost walnut and back, making the distance covered to collect walnuts placed to the right equal to  n^2 + n  meters.

Thus, we have that  n^2 + (n^2 + n) = 300 , which gives  2n^2 +n- 300=0 . Solving gives  n=12 . Therefore, the number of walnuts is  2n+1=25 .

Answer: E

Bunuel · Answer

Official Solution:

A crate contains 103 defective cellphones. If you distribute defective cellphones to  x  technicians such that each technician gets the same number of defective cellphones, the number of cellphones left in the crate is 2 less than the number of technicians. Which of the following cannot be the value of  x ?

A. 3
B. 5
C. 7
D. 14
E. 15

Assuming each technician gets  n  cellphones, we get  xn + (x - 2) = 103 . This simplifies to  x(n + 1) = 105 , which implies that  x , the number of technicians, must be a factor of 105, such that  x - 2 \geq 0  (after all, there cannot be a negative number of cellphones left in the crate after the distribution of the cellphones). From the options, only option D, 14, is not a factor of 105, making D the correct answer.

Answer: D

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