GRE Numeric Entry Sample Questions

---

---

1. Question :

The height and radius of the base of a right circular cone are in the ratio 3:4. The ratio of the total surface
area to the curved surface area is 9:5. The sum of the total surface area and the curved surface area of the
cone is 224*pi sq.cm. Find the radius of the base.

Correct Answer:

8cm

Explanation:

Let the height and radius of the base be 3x and 4x respectively.

Let the total surface area and curved surface area be 9y and 5y respectively.

9y+5y=224*pi

14y=224pi

y=224pi/12=16pi

Curved surface area=5y=5*16pi=80pi …(1)

Curved surface area=pi*r*l, where r is the radius of the base and l is the slant height

l=sqrt[(3x)^2+(4x)^2]

=sqrt(9x^2+16x^2)

=sqrt(25x)=5x

Hence, curved surface area=pi*r*l=pi*4x*5x

=20pi*x^2

Equating with (1), we get

20pi*x^2=80pi

x^2=80/20=4

x=2

Radius=2*4=8cm

[x^2=x*x]

2. Question :

The distance between the points (x,-1) and (3,2) is 5. Find the value of x if the point lies in the fourth quadrant.

Correct Answer:

7

The distance between the points is given by

Sqrt[(x-3)^2+(-1-2)^2]=5

Squaring both sides, we get

(x-3)^2+(-3)^2=25

x^2-6x+9+9=25

x^2-6x-7=0

x^2-7x+x-7=0

x(x-7)+1(x-7)=0

x=7,-1

x=7 since (x,-1) lies in the fourth quadrant.

3. Question :

The sum of the squares of two positive numbers is 340. The sum of the numbers is 22. Find the difference between
the numbers.

Correct Answer:

14

Explanation:

Let the numbers be x and y

x+y=22 …(1)

x^2+y^2=340 …(2)

Put x=22-y from (1) in (2)

(22-y)^2+y^2=340

484-44y+y^2+y^2=340

2y^2-44y+484-340

y^2-22y+72=0

y^2-18y-4y+72=0

y(y-18)-4(y-18)=0

y=4, 18

When y=4, x=18

When y=18, x=4

Difference of the numbers=18-4=14

[x^2=x*x]

GRE Numeric Entry Sample Questions

---

---