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Test: Quantitative Reasoning (Level 700) - 2 - GMAT MCQ


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21 Questions MCQ Test Practice Questions for GMAT - Test: Quantitative Reasoning (Level 700) - 2

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Test: Quantitative Reasoning (Level 700) - 2 - Question 1

Find the value of 

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 1


Now, since 

Answer: B.

Test: Quantitative Reasoning (Level 700) - 2 - Question 2

A beaker contains 100 milligrams of a solution of salt and water that is x% salt by weight such that x < 90. If the water evaporates at a rate of y milligrams per hour, how many hours will it take for the concentration of salt to reach (x+10)%, in terms of x and y?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 2

Amount of solution after 'k' hours = 100 - ky

100x = 100x+1000-kyx -10ky
k(xy+10y) = 1000

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Test: Quantitative Reasoning (Level 700) - 2 - Question 3

In the multiplication above, each letter stands for a different non-zero digit, with A x B < 10. What is the two-digit number AB?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 3

Given that A x B < 10, the possible values of AB are 23, 32, 24 and 42.
It is also given that A * B = C and units digit of (A* A + C * B) = B
Since, A = 4 and B =2 only satisfies above condition, hence AB =42.
Answer: (E)

Test: Quantitative Reasoning (Level 700) - 2 - Question 4

Find the numbers of ways in which 4 boys and 4 girls can be seated in a row of 8 seats if they sit alternately and if there is a boy named John and a girl named Susan amongst this group who cannot be put in adjacent seats ?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 4

Total Arrangements of 4 boys and 4 girls can be calculated in these two ways
Case 1: BGBGBGBG i.e. total ways = 4!*4! = 24*24 = 576
Case 2: GBGBGBGB i.e. total ways = 4!*4! = 24*24 = 576
Total ways = 576+576 = 1152
Unfavorable ways = cases in which John and Susan are together = 14*3!*3! = 504 (here number 14 comes as pairs of BG or GB who are John and Susan and remaining 3!*3! are ways in which remaining 3 boys can sit alternately and 3 girls can sit alternately)
Favorable cases = 1152 - 504 = 648
Answer: Option D

Test: Quantitative Reasoning (Level 700) - 2 - Question 5

The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a rate of 3 feet per second. Once Bill reaches the group of people, he stops walking and stands with them until the walkway ends. What is Bill’s average rate of movement for his trip along the moving walkway?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 5

This is how I broke it down:
Total Distance = 300 feet
default rate is 3fps (feet per second)
walking rate is 6fps (default + walking)
catch up rate: is walking rate - default rate = 6-3 = 3fps

Part 1: Time for Catch up
As he gets on the walk way he instantly starts "catching up" to the group that is 120 feet ahead of him. How long does this take?
D/R = Time; so 120feet/3fps [ note I'm using the catch up rate here ] = 40 seconds of catch up
After 40 seconds, how much of the 300 feet of the moonwalk has been used? well if he was moving for 40 seconds at 6 feet per second (he was walking) then he covered 240 feet of actual walkway.
300-240 feet = 60 feet.
There are only 60 feet left of walkway.

Part 2: Standing with the crowd
Now bill just idles with the crowd for the last 60 feet. 60ft/3fps [ note this is the default rate ] = 20 seconds.

Part 3: Average rate
6 fps for 40 seconds = 240 feet
3 fps for 20 seconds = 60 feet
X fps for 60 seconds = 300 feet

300/60 = 5fps average rate. Answer is E.

Test: Quantitative Reasoning (Level 700) - 2 - Question 6

Five integers in a set are written ascending order. The median of this set is 17, and the average of the smallest and largest integers is 16. When the smallest and largest numbers are removed from the set, the average of the new smallest and largest integers is 15. What can be the minimum value of the largest of the original five integers?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 6

Here we have N1, N2, N3, N4, N5 where the median N3 = 17.

We are given that (N1+N5)/2 = 16 and (N2+N4)/2 = 30. Simplify these to be N1+N5=32 and N2 + N4 = 30.

The difference in the values of N1+N5 and N2+N4 is 2. This means that from the lower and upper bounds, the highest value number and/or the lowest value number has to be at least 2 from the median (N4 cannot be valued below 17 and N2 cannot be above 17 - that would break the rule). We are asked what can be the minimum value of the largest of original five inetegers AKA what is the value of N5? I think we found the answer in the prior sentence (N5 is at least 2 higher than N3), but let's test cases to be sure.
N5 + N1 = 32
19 + N1 = 32 => N1 = 13
Now we must address N2 + N4 = 30
Our bounds are now 13 ≤ N2 ≤ 17 and 17 ≤ N4 ≤ 19 but we have to make the overall value drop by 2. We obviously cannot have N2 < N1 so we must have N2 ≥ N1 and manipulate the upper bound.
13+N4=30 => N4 = 17.
If we were to increase N2 to 14, then N4 would have to equal 16 but we cannot have N4 cross the value of the median so this does not work.
As such you can see 19 is the correct answer for the lowest value for N5. IMO answer is C.

Test: Quantitative Reasoning (Level 700) - 2 - Question 7

The sum of the digits of a positive integer N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 7

N = 11a+7
i.e. Possible values of N may be 18, 29, 30, 41, 52, 63, 74, 85, 96, 107, 118, 129, 140, 151, 162, 173, 184, 195 and so on
11a may be a 3 digit number such as 121, 242, 363, 979
N = 979+7 = 986
9+8+6 = 23 Satisfied!
986 when divided by 33 leaves remainder = 29
Answer: Option E

Test: Quantitative Reasoning (Level 700) - 2 - Question 8

100 people are attending a newspaper conference. 45 of them are writers and more than 38 are editors. Of the people at the conference, x are both writers and editors and 2x are neither. What is the largest possible number of people who are both writers and editors?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 8

{Total} = {Writers} + {Editors} - {Both} + {Neither}.
{Total} = 100;
{Writers} = 45;
{Editors} > 38;
{Both} = x;
{Neither} = 2x;100 = 45 + {Editors} - x + 2x
x = 55 - {Editors}.
We want to maximize x, thus we should minimize {Editors}, minimum possible value of {Editors} is 39, thus:
x = {Both} = 55 - 39 = 16.
Answer: B.

Test: Quantitative Reasoning (Level 700) - 2 - Question 9

Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 9

Diana and her two brothers can be assigned to one of the 3 groups, so each has 3 choices, so total # of different assignments of Diana and her 2 brothers to 3 groups is 3∗3∗3 = 3 = 27.
Now, "Diana leaves at the same time as AT LEAST one her brothers" means that Diana is in the same group as at least one her brothers.
Let's find the opposite probability of such event and subtract it from 1. Opposite probability would be the probability that Diana is not in the group with any of her brothers.
In how many ways we can assign Diana and her two brothers to 3 groups so that Diana is not in the group with any of her brothers? If Diana is in the first group, then each of her two brothers will have 2 choices (either the second group or the third) and thus can be assigned in 2 ∗ 2 = 2= 4 ways to other two groups. As there are 3 groups, then total # of ways to assign Diana and her 2 brothers to these groups so that Diana is not in the group with any her bothers is 3 ∗ 4 = 12 (For each of Diana's choices her brother can be assigned in 4 ways, as Diana has 3 choices: first, second or the third group, then total 3 ∗ 4 = 12). So probability of this event is 12/27.
Probability that Diana is in the same group as at least one her brothers would be 
Answer: E.

Test: Quantitative Reasoning (Level 700) - 2 - Question 10

K = 1*2+2*3+________+21*22

What is the remainder when K is divided by 23?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 10

Given:

  • K = 1 * 2 + 2 * 3 + 3 * 4 + … + 21 * 22


To find:

  • The remainder when K is divided by 23

Approach and Working:
If we observe the terms of the series K, we can generalise it as follows:

  • 1st term = 1 * 2
  • 2nd term = 2 * 3
  • 3rd term = 3 * 4
  • Hence, nth term = n * (n + 1) = n2+n

Now, sum of all the terms of 

  • If n = 21, 

Therefore, it will be always divisible by 23, giving a 0 remainder

Hence, the correct answer is option E.

Test: Quantitative Reasoning (Level 700) - 2 - Question 11

Rank the following quantities in order, from smallest to biggest.

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 11

First of all, clearly 

So, II is bigger than I. Now, what about III? When we take higher order roots, the values move closer to one. If the number starts larger than one, then higher and higher roots make it smaller, closer to one. If the number starts between 0 and 1, then higher and higher roots make it larger, closer to one. Therefore, III is larger than II. From smallest to biggest, I, II, III.
Answer = (A).

Test: Quantitative Reasoning (Level 700) - 2 - Question 12

A manufacturer produces a certain men's athletic shoe in integer sizes from 8 to 17. For this particular shoe, each unit increase in size corresponds to a 1/4-inch increase in the length of the shoe. If the largest size of this shoe is 20% longer than the smallest size, how long, in inches, is the shoe in size 15?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 12

From size 8 to size 17, there are 9 increments of .25 each making the size 17 shoe 9*.25 = 2.25 inches more than the size 8 shoe.
This 2.25 accounts for 20% of the length of size 8 shoe so length of size 8 shoe is 2.25*5 = 11.25 inches
Length of size 15 shoe must be 11.25 + .25*7 = 13 inches

Test: Quantitative Reasoning (Level 700) - 2 - Question 13

A certain thermometer measured the temperature in City X in degreed Fahrenheit every day for one month, and found that the range of temperatures was 21 degrees. If, subsequently, it was discovered that the thermometer always overstated the temperature by 15%, what would the approximate range of measurements have been, in degrees, if the temperatures had been measured accurately?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 13

Original range of temperatures: 21 degrees Fahrenheit
The thermometer always overstated by 15%
Overstatement of 15% on 21 degrees = 0.15 * 21 = 3.15 degrees
Adjusted range with accurate measurements: 21 - 3.15 = 17.85 degrees
Approximate range after accurate measurements: 18 degrees
Therefore, the approximate range of measurements with accurate readings would be 18 degrees, which corresponds to option C.

Test: Quantitative Reasoning (Level 700) - 2 - Question 14

Two candles of the same height are lighted at the same time. The first candle is consumed in 4 hours and the second one in 3 hours. Assume that each candle burns at the same rate. In how many hours will the first candle measure twice the height of the second candle?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 14

We can let the height of both candles be 12 inches. So the first candle burns at a rate of 3 inches per hour and the second candle at a rate of 4 inches per hour. We can let x = the number of hours it takes the first candle to be twice the height of the second candle.
12 - 3x = 2(12 - 4x)
12 - 3x = 24 - 8x
5x = 12
x = 2.4 hours = 2 hours 24 minutes

Test: Quantitative Reasoning (Level 700) - 2 - Question 15

40 men and 20 women together can complete a work in 12 days and ratio of efficiency of a man to a woman is 2:3, find how many men are required to complete half of the work in 7 days?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 15

work rate of men =2x
Work rate of women = 3x
Total work done in 12 days = (40*2x(men's Work) + 20*3x(women's Work))*12
Half work to be done by men in 7 days will be equal to =(140*12x/7)*1/2x(men's Work rate)*1/2(Half work)
gives 60 Men
Answer D.

Test: Quantitative Reasoning (Level 700) - 2 - Question 16

Note: Total stopping distance is the sum of the distance traveled during reaction time and the distance traveled after brakes have been applied.

Of the following, which is the greatest speed, in miles per hour, at which the car can travel and stop with a total stopping distance of less than 200 feet?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 16

Should be A. 50mph as the distance comes out to be 55 ft + 137 ft = 192 ft
Rest all go more than 200 ft easily

Test: Quantitative Reasoning (Level 700) - 2 - Question 17

C individuals pledged to pay equal contributions so that a charity’s goal of $x could be reached. If d of the contributors failed to pay their share, which of the following represents the additional number of dollars that each of the remaining individuals would have to pay in order to allow the charity to reach its goal?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 17

Number of individuals = C
Amount paid by each individual = n
Total expected amount = Charity's goal = nC = x
n = x/C
Number of individuals who fail to pay = d
Contribution from individuals who would fail to pay = dx/C --> Additional amount
Number of individuals who are paying = C - d
Additional amount has to be divided among the (C - d) individuals --> dx/C(C - d)
Answer: E

Test: Quantitative Reasoning (Level 700) - 2 - Question 18

Which one of the following diets or the combination of diets supplies more protein for every dollar?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 18


From the above table clearly, only for food B, the protein content per dollar is more compared to that of any other foods or combinations.

Test: Quantitative Reasoning (Level 700) - 2 - Question 19

For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 19

Given set in ascending order is {n, n+1, n+2, n+4, n+8}.

Test: Quantitative Reasoning (Level 700) - 2 - Question 20

A bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely. What is the value of

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 20

What is the value of (10− 102) ∗ 0.00(12) ? (You can indicate repeated part of the decimal by putting it in brackets) You can solve it as Pinali suggests above, just open the brackets and multiply: (10− 102) ∗ 0.00(12) = 10,000∗0.00(12) − 100∗0.00(12) = 12.(12) − 0.(12) = 12, you can see that 0.(12) part is subtracted from 12.(12) which gives 12. But you can do this problem in another way too: 0.00(12) can be written as fraction 12/9,900 (as many 9's as numbers in repeated pattern and as many zeros after as zeros after decimal point). For more on how to convert a recurring decimal to fraction see Number Theory chapter of Math Book (link in my signature).

Answer: E.

Test: Quantitative Reasoning (Level 700) - 2 - Question 21

For any integer P greater than 1, P! denotes the product of all integers from 1 to P, inclusive. A number is chosen at random from the list of factors of 8!. What is the probability that the chosen number is a multiple of 28?

Detailed Solution for Test: Quantitative Reasoning (Level 700) - 2 - Question 21

8! = 27∗32∗5∗7
Any factor of 8! can be written as 2a∗3b∗5c∗7d , where a, b, c and d are integers
a can take 8 values (0 to7)
b can take 3 values (0 to 2)
c can take 2 values (0 to 1)
d can take 2 values (0 to 1)

Number of factors = 8*3*2*2

If the factor of 8! is multiple of 28, in that case-
a can take 6 values (2 to 7)
b can take 3 values (0 to 2)
c can take 2 values (0 to 1)
d can take 1 value (only 1)

Number of multiples of 28 = 6*3*2*1

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