# GRE QUANT QUESTIONS WITH SOLUTIONS PDF

Are you prepared for the GRE Quantitative Reasoning section? Try some practice questions with solutions and tips from Jamboree faculty.

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## GRE Quant Practice Question #1

A magician marks off on a stick of length 1 yard in thirds and fifths and breaks the stick at the marked points. What is the maximum number of pieces which are equal in length?

• 2

• 3

• 4

• 5

• 6

Solution: Thirds means = 13, 23, 33, 43,…..and so on.
Fifth means = 15, 25, 35, 45,….. and so on.
A stick of length 1 yard is marked in thirds and fifths.
Stick is marked at these points. Hence, we need to find out the length of each point.
AB = 15 - 0
BC = 13 -15, and so on.
But, this method would be time-consuming, as it involves 7 fractions subtraction.
Therefore, to reduce calculations, let's multiply the entire line by product of denominators (i.e., 3*5=15).
Multiply all values by 15.
Now, length of AB=3, BC=2, CD=1, DE=3, EF=1, FG=2, GH=3
There are a maximum of 3 pieces which have the same length (AB, DE, GH)
Jamboree Tip: working with integers will be more convenient than working with fractions.

## GRE Quant Practice Question #2

Ron will walk from intersection A to intersection B along a route that is confined to the square grid of three streets and four avenues shown in the map above. How many routes from A to B can Ron take that have a minimum possible length?

• 8

• 5

• 15

• 10

• 11

Solution: To go from A to B, let's represent
1 right move? r
1 up move u
On analysis, we can infer that going from
intersection A to intersection b along the
grid section, Ron will need to take 5 steps
3 right and 2 up
So, some of the different routes can be:
RRRUU
UURRR
URURR
RURRU, and so on.
But, we won't prefer writing all the routes and rather, we need a method.
Every time when we write a different route, it has 2 u’s and the 3 r’s but in different arrangements. Therefore, we treat it like a question of P and C.
[ N things can be arranged in a line in n! Ways. If there are repeating alphabets, we divide by their factorial.]
It’s a 5 alphabetic word, with 2 u’s and the 3 r’s.
Therefore, uuurr= 5!3!*2!
(U represents thrice, so 3!
R represents twice, so 2!)
= 5*4*3*2*13*2*1*2*1
=10

## GRE Quant Practice Question #3

Investment of \$4000 at R% per annum compounded annually will become \$16000 in 8 years. If \$2000 is invested at R% per annum compounded annually, in how many years will the investment become \$16000?

• 8

• 12

• 24

• 13

• 10

Solution:
Principal, P1 = \$4000
Rate, R1 = R%
Amount, A1 = \$16000
Time, T1 = 8 years.
Using the formula of compound amount: A = P (1 + R100)T
16000 = 4000 (1 + R100)8 (1 + R100)8 = 4
Applying the property of exponent: xa = y
x = (y)1/a
So, (1 + R100) = (4)1/8 …………………………………………….. (1)
Now,
Principal, P2 = \$2000
Rate, R2 = R%
Amount, A2 = \$16000
Time, T2 = T years.
Using the formula of compound amount: A = P (1 + R100)T
16000 = 2000 (1 + R100)T
(1 + R100)T = 8
So, (1 + R100) = (8)1/T …………………………………………….. (2)
Equating values of (1 + R100) from equation (1) & (2);
(4)1/8 = (8)1/T
(22)1/8 = (23)1/T
(2)1/4 = (2)3/T { (xa)b = (x)a b }
14 = 3T {Xa = Xb; and x ?1 and-1 then a = b.}
T = 12 years.
Alternatively:
As the investment of \$4000 is becoming \$16000 in 8 years so in other words we could say that the investment is becoming 4 times in 8 years or is becoming 2 times in 4 years.
So the investment of \$2000 will become \$4000 in 4 years
And \$4000 will become \$8000 in the next 4 years
And \$8000 will become \$16000 in the next 4 years
or investment of \$2000 will become \$16000 in 12 years.

## GRE Quant Practice Question #4

For any integer n greater than 1, n! denotes the product of all integers from 1 to n, inclusive. How many prime numbers are there between 15! + 5 and 15! + 13?

• 5

• 10

• 1

• 0

• 20

Solution: Prime numbers between 15! +5 and 15! + 13.

As we know t

hat,

n! = product of all integers from 1 to n.

→5! = 1×2×3×4×5

Or

15! = 1×2×3×4×5×6×……×13×14×15

Firstly, Let’s take 15! + 5 from the list.

15! + 5, 15! + 6, 15! + 7, 15! + 8, 15! + 9, ……15! + 13

15! + 5 = 5(1×2×3×4×6×7×8×……×13×14×15 + 1),

As we can take 5 common, hence 15! + 5 is non -prime.

15! + 6 = 6(1×2×3×4×5×7×8×……×13×14×15 + 1),

As we can take 6 common, hence 15! + 6 is non -prime.

15! + 7 = 7(1×2×3×4×5×6×8×……×13×14×15 + 1),

As we can take 7 common, hence 15! + 7 is non -prime.

15! + 8 = 8(1×2×3×4×6×7×9×……×13×14×15 + 1),

As we can take 8 common, hence 15! + 8 is non -prime.

.

.

.

15! + 12 = 12(1×2×3×4×6×7×…×13×14×15 + 1),

As we can take 12 common, hence 15! + 12 is non -prime.

And

15! + 13 = 13(1×2×3×4×6×7×…×12×14×15 + 1).

As we can take 13 common, hence 15! + 13 is non -prime.

Such that, all the numbers from 15! + 5, 15! + 6, 15! + 7, 15! + 8, 15! + 9, …. To 15! + 13 are

product of two or more prime numbers.

Therefore, there is no prime number in the list

15! + 5, 15! + 6, 15! + 7, 15! + 8, 15! + 9, ……,15! + 13

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