# (Paper) GRE Math Problems Practice Question with solutions

**GRE Math Problems Practice
Question with solutions**

**Q1. If both x and y are prime numbers, which of the following CANNOT
be the difference of x and y?**

(A) 1

(B) 3

(C) 9

(D) 15

(E) 23

**Answer: **Choice E is correct. This problem is solved fastest by
process of elimination. Both 2 and 3 are prime and their difference is one
(Eliminate Choice A). Both 5 and 2 are prime and their difference is 3
(Eliminate Choice B). Both 11 and 2 are prime and their difference is 9
(Eliminate C). Both 17 and 2 are prime and their difference is 15 (Eliminate
D).

**Q2. Car X and Car Y are five miles apart and are on a collision course.
Car X is driving directly north and Car Y is driving directly east. If the
point of impact is one mile closer to the current position of Car X than to
the current position of Car Y, how many miles away from the point of impact
is Car Y at this time?**

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

**Answer: **Choice D is correct. This problem can be solved by using the
Pythagorean theorem as Cars X and Y are 5 miles apart, which is the
hypotenuse of a right triangle. Let d be the distance Car Y is from the
point of collision. Then the distance Car X is from the collision is d-1.
Solving for D: dd + (d-1)(d-1) = 25, d=4, -3. Since d denotes distance, we
reject -3 as a valid answer.

**Q3. In the diagram above, AD = BE = 6 and CD = 3(BC). If AE = 8, then BC
= ?**

(A) 6

(B) 4

(C) 3

(D) 2

(E) 1

**Answer:** Choice E is correct. Since AE is a line segment and all the
lengths are additive, AE = AD + DE. We know that AD = 6 and AE = 8. So DE =
AE - AD = 8 - 6 = 2. We also know that BE = 6. So BD = BE - DE = 6 - 2 = 4.
We know BD is 4, but need to find BC.

Since CD = 3(BC), we can solve for BC: x + 3x = 4. x = 1.

**Q4. If the length of rectangle A is one-half the length of rectangle B,
and the width of rectangle A is one-half the width of rectangle B, what is
the ratio of the area of rectangle A to the area of rectangle B?**

(A) 1/4

(B) 1/2

(C) 1/1

(D) 2/1

(E) 4/1

**Answer:** Choice A is correct. This problem includes a common mistake:
the ratio of areas is NOT the same as the ratio of lengths. Instead, the
ratio of areas for similar polygons is equal to the square of the lengths of
the lengths. If we use 4 and 2 as the length and width for rectangle A, its
area is 8. Rectangle B would have an area of (8)(4) = 32, four times that of
A.

**Q5. If the length of rectangle A is one-half the length of rectangle B,
and the width of rectangle A is one-half the width of rectangle B, what is
the ratio of the area of rectangle A to the area of rectangle B?**

(A) 1/4

(B) 1/2

(C) 1/1

(D) 2/1

(E) 4/1

**Answer: **Choice A is correct. This problem includes a common mistake:
the ratio of areas is NOT the same as the ratio of lengths. Instead, the
ratio of areas for similar polygons is equal to the square of the lengths of
the lengths. If we use 4 and 2 as the length and width for rectangle A, its
area is 8. Rectangle B would have an area of (8)(4) = 32, four times that of
A.

**Q6. A cube and a rectangular solid are equal in volume. If the length of
the edges of the rectangular solid are 4, 8, and 16, what is the length of
an edge of the cube?**

(A) 4

(B) 8

(C) 12

(D) 16

(E) 64

**Answer:** Choice B is correct. We have all of the dimensions to
calculate the volume of the rectangular solid, which is 16 x 8 x 4. This is
also the volume of the cube. So, the length of an edge of the cube is the
cubic root of (16 x 8 x 4), or 8.