Quantitative Practice Questions - 1 | Practice Questions for GMAT PDF Download

The four numbers n, n + 1,n + 3,and n + 8 appear in ascending order, so their median M must be the arithmetic mean of the middle two numbers. Therefore, 

n + 2 = 72

n = 70

• n = 40
• p = .625(success probability)
• q = .375(Failure probability)
• standard deviation = 
• z = (.75 − .625)/.0765 = 1.63 
• z = 1.63
• Percentile =  .9484
• 1 − .9484 = .0516

In the first three hours, he travels 180 miles.

3 × 60 = 180 

In the next two hours, he travels 144 miles.

2 × 72 = 144 

for a total of 324 miles.

180 + 144 = 324 

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

324 / 5 = 64.8 mph

Mary makes $13/hour and works 40 hours.  So she makes

$13⋅40 hours = $520 . 

However, we need to subtract the cost of the items that she bought.  If the sweater retails for $50, Mary buys it for .75.50 = $37.50 because of her 25% discount.  Similarly, she buys the jacket for .75.144 = $108 .75⋅144 = $108 .  So her net profit is

520 − 37.50 − 108 = $374.50 .

1/3 of Audrey’s age is N/3.

2 years more than 13 of Audrey’s age is

N/3 + 2.

However, be careful because the question is asking for Matt’s age in 5 years, so you need to add 5 years to Audrey’s current age:

N/3 + 2 + 5 = N/3 + 7

I = p x r   t 

I = (76000)(0.10)(5) 

I = 38000

Let x be the price before discount of the merchandise Carolyn can purchase now using her 10% discount. Then Carolyn will pay 90% of that price, or 0.9x. Set up and solve the equation:

450 = 0.9x

x = 450 / 0.9 =  500

So Carolyn can use that gift card to purchase $500 worth of merchandise now.

Now let y be the price before discount of the merchandise Carolyn can purchase during Employee Appreciation Day using a 25% discount. Then Carolyn will pay 75% of that price, or 0.75y. Set up and solve the equation:

450 = 0.75x

y = 450 / 0.75 = 600

So Carolyn can use that gift card to purchase $600 worth of merchandise on Employee Appreciation Day.

Carolyn can save $100 by holding out.

Since there are 300 people, 300⋅.21 = 63 people play baseball and 300⋅.09 = 27 of those people play both baseball and soccer. Therefore, there are 63 − 27=36 people who play baseball but not soccer.

Probability: 36/300 = 325

By definition, a prime number is any number that is greater than  1  and is only divisible by  1  and itself. Therefore, by definition  1  is not a prime number.

We need to flip a coin until we get two heads in a row. The smallest number of possible flips is 2, which would occur if our first two flips are both heads. This eliminates three of our answer choices, because we know the sample space must start at 2. 

This leaves us with {x : x = 2, 3, 4 . . .} and {x : x = 2, 3, 4, 5, 6}. Let's think about {x : x = 2, 3, 4, 5, 6}. What if I flip a coin 6 times and get 6 tails? Then I have to keep flipping beyond 6 flips until I get two heads in a row; therefore the answer must be {x : x = 2, 3, 4 . . .}, because we don't have an upper limit on the number of flips it will take to produce two successive heads.

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 x 3 = 12 ) and 5 students are required to paint the small signs ( 12 × 5 = 60 ). In total, 8 students are required.

Mark spends $20 * 1,000 shares = $20,000. When the stock price increases to $25/share, he makes ($25 – $20) * 1,000 shares = $5,000. 

$5,000 / $20,000 = 1/4 = 25%

He does NOT increase his money by 125%, which would mean an additional $25,000, not $5,000.

We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d. There are seven possibilities:
a) {1, 2, 3}, sum = 6
b) {1, 2, 4}, sum = 7
c) {1, 2, 5}, sum = 8
d) {1, 3, 4}, sum = 8
e) {1, 2, 6}, sum = 9
f) {1, 3, 5}, sum = 9
g) {2, 3, 4}, sum = 9
For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits.  Thus, for each item on that list, there are six different possible four-digit numbers.  The total number of possible four-digit numbers would be 7*6 = 42. 

Total number of cars = 500

2D cars total = 165, so

4D cars total = 335

120 4D cars have BUC

“Eighteen percent of all the cars with back-up cameras have standard transmission.”

18% = 18/100 = 9/50

This means that the number of cars with BUC must be a multiple of 50.

How many 2D cars can we add to 120 4D cars to get a multiple of 50?  We could add 30, or 80, or 130, but after that, we would run out of 2D cars.  These leaves three possibilities for the total number with BUC:

If a total of 150 have BUC, then 18% or 27 of them also have ST.

If a total of 200 have BUC, then 18% or 36 of them also have ST.

If a total of 250 have BUC, then 18% or 45 of them also have ST.

Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”

40% = 40/100 = 2/5

This means that number of cars with both back-up cameras and standard transmission must be divisible by 5.  Of the three possibilities we have, only the third words.

Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)

18% or 45 of these also have ST.

40% of that is 18, the number of 2D cars with both BUC and ST.

Thus, the number of 4D cars with both BUC and ST would be

45 – 18 = 27

700 student total

4G = total number of fourth graders

5G = total number of fifth graders

We are told 4G = 320, so 5G = 700 – 320 = 380

5GM, 5GF = fifth grade boys and girls, respectively

We are told 5GF = 210, so 5GM = 380 – 210 = 170

4GC, 5GC = total number of 4th or 5th graders, respectively taking Chinese

We are told

5GC = 0.5(5G) = 0.5(380) = 190

4GC = 0.4(4G) = 0.4(320) = 128

4GFM, 4GMC, 5GFC, 5GMC = 4th/5th grade boys & girls taking Chinese

We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do.  Thus 5GMC = 80.

5GMC + 5GFC = 5GC

80 + 5GFC = 190

5GFC = 110

We are told:

4GFM < (0.5)(5GFC)

4GFM < (0.5)(100)

4GFM < 55

Thus, 4GFM could be as low as zero or as high as 54.

4GMC = 4GC – 4GFM

If 4GFM = 0, then 4GMC = 128 – 0 = 128

If 4GFM = 54, then 4GMC = 128 – 54 = 74

Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128.  Of the answer choices listed, the only one that works is 100.

There are three cases: AABC, ABBC, and ABCC.

In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C.  9*9*8 = 81*8 = 648.

In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C.  Again, 648.

48*3 = (50 – 2)*3 = 150 – 6 = 144

3*648 = 3(600 + 48) = 1800 + 144 = 1948

One point is (50, 70) and one is (100, 89): the line has to pass above both of those.  Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!

What is the slope from (50, 70) to (100, 90)?  Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5.  A line with a slope of 2/5 could pass just above these points.

Now, what about the third point?  For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points.  Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83).  This means it would pass under the third point, (80, 84).  A slope of 2/5 works for all three points.

We don’t have to do all the calculations, but none of the other slope values works.

The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.

Let’s say that the A = $100 at the beginning of the year.

End of January, 60% increase.  New price = $160

End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.

10% of $160 = $16

40% of $160 = 4(16) = $64

That’s the price at the end of February.

End of March, a 60% increase: that’s a increase of 60% of $64.

10% of $64 = $6.40

60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40

Add that to the starting amount, $64:

New price = $64 + $38.40 = $102.40

End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.

At this point, we are going to approximate a bit.  Approximate $102.40 as $100, so 40% of that would be $40.  The final price will be slightly more than $40.

Well, what is slightly more than $40, as a percent of the beginning of the year price of $100?  That would be slightly more than 40%.

The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan.  The German company must pay this amount.

Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan.  That’s the amount that the Germans pay to the Chinese.

That is the German company’s outlay, in euros.  Now, they make N metal chairs, and sell them, making a gross profit of P euros.

That must be the total revenue of the German company, in euros.  This comes from the sale to the American company.  Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.

That must be the total dollar amount that leaves the American company and goes to the German company.  This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.

Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch.  The problem tells us that:

P(A) > 0.6

P(A and B) < 0.5

P(A or B) = 0.7

First, let’s establish the minimum value.  If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7.  All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.

Now, the maximum value.  Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region.  See the conceptual diagram.

 

The top line, 1, is the entire probability space.  The second line, P(A or B) = 0.7, fixes the boundaries for A and B.  P(A) is the purple arrow, extending from the right.  P(B) is the green arrow extending from the left.  The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.

Let’s say that P(A) is just slightly more than 0.6.  That means the region outside of P(A), but inside of P(A or B) is slightly less than 1.  That’s the part of P(B) that doesn’t overlap with P(A).    Then, the overlap has to be less than 0.5.  If we add something less than 1 to something less than 5, we get something less than 6.  P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.

Thus, 0 ≤ P(B) < 0.6.

Terms are {4, 5, 24, x, y, z}
Mean = 10 i.e. Sum = 10*6 = 60
i.e.4+5+24+x+y+z = 60
i.e. x+y+z = 27

Media = between 7 and 8
But since number of terms in the set is even so median is the average of two middle terms which can only be 7.5
Hence, Median = 7.5
Media 7.5 is possible when two of the terms in the sets are {7,8} or {6,9} or {5,10} or {4,11} or {3,12} or {2,13} or {1,14}

Lets check options
(A) 13 {x, y, z} may be {2, 13, 12} hence Possible

(B) 12 {x, y, z} may be {5, 10, 12} hence Possible

(C) 11 {x, y, z}hence NOT Possible as the pair needed is {4, 11} but 4 can't be third term ever as sum of other two of x, y, z is 16 which can't be each smaller than 4. CORRECT ANSWER

(D) 10

(E) 5

Answer: Option C