Problem solving Practice Test 2 - GRE Maths Practice

n and p are integers greater than 1 5n is the square of a number 75np is the cube of a number.

The smallest value for n + p is

A. 14
B. 18
C. 20
D. 30
E. 50

Solution

The smallest value for n such that 5n is a square is 5.
75np can now be written as 75 x 5 x p.
This gives prime factors.... 3 x 5 x 5 x 5 x p
To make the expression a perfect cube, p will have to have factors 3 x 3 , and hence p =9
n + p = 5 + 9 = 14

After being dropped a certain ball always bounces back to 2/5 of the height of its previous bounce. After the first bounce it reaches a height of 125 inches. How high (in inches) will it reach after its fourth bounce?

A. 20
B. 15
C. 8
D. 5
E. 3.2

Solution

If after each bounce it reaches 2/5 of the previous height, then after the second bounce it will reach 2/5 x 125. After the third it will reach 2/5 x 2/5 x 125. After the fourth it will reach 2/5 x 2/5 x 2/5 x 125. This cancels down to 2 x 2 x 2 = 8

If n ≠ 0, which of the following expressions could have a value less than n?

Indicate ALL such expressions.

A. 2n
B. n²
C. 2 – n

Solution

Correct Answer: A,B,C

Explanation:

Remember that n could be positive negative or a fraction. Try out a few cases:
In case A, if n is -1, then 2n is less than n.
In case B, if n is a fraction such as ½ then n2 will be less than n.
In case C, if n is 2, then 2-n = 0, which is less than n.
Therefore, all of the choices could have a value less than n

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ABCD is a square of side 3, and E and F are the mid points of sides AB and BC respectively. What is the area of the quadrilateral EBFD ?

A. 2.25
B. 3
C. 4
D. 4.5
E. 6

Solution

(Total area of square - sum of the areas of triangles ADE and DCF) will give the area of the quadrilateral
9 - (2 x ½ x 3 x 1.5) = 4.5

A circular logo is enlarged to fit the lid of a jar. The new diameter is 50 per cent larger than the original. By what percentage has the area of the logo increased?

A. 50
B. 80
C. 100
D. 125
E. 250

Solution

If the diameter of the old lid is 100, the diameter of the new lid will be 150.
This gives a ratio of 2 : 3
The areas of the lids will be in the ratio (2)² : (3)² . or 4 : 9
This represents an increase in area of 5units for every 4 units
5/4 is equivalent to 125 percent.

Two sets of 4 consecutive positive integers have exactly one integer in common. The sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set?

A. 4
B. 7
C. 8
D. 12
E. it cannot be determined from the information given.

Solution

If two sets of four consecutive integers have one integer in common, the total in the combined set is 7., and we can write the sets as
n + (n + 1) + (n + 2) + (n + 3 ) and
(n + 3) + (n + 4) + (n + 5) + (n + 6)
Note that each term in the second set is 3 more than the equivalent term in the first set. Since there are four terms the total of the differences will be 4 x 3 = 12

A 3 by 4 rectangle is inscribed in circle. What is the circumference of the circle?

A. 2.5π
B. 3π
C. 5π
D. 4π
E. 10π

Solution

Draw the diagram. The diagonal of the rectangle is the diameter of the circle. The diagonal is the hypotenuse of a 3,4,5 triangle, and is therefore, 5.
Circumference = π.diameter = 5π

Jo’s collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to British stamps is 5 to 1, what is the ratio of US to British stamps?

A. 5 : 1
B. 10 : 5
C. 15 : 2
D. 20 : 2
E. 25 : 2

Solution

Indian stamps are common to both ratios. Multiply both ratios by factors such that the Indian stamps are represented by the same number.
US : Indian = 5 : 2, and Indian : British = 5 : 1. Multiply the first by 5, and the second by 2.
Now US : Indian = 25 : 10, and Indian : British = 10 : 2
Hence the two ratios can be combined and US : British = 25 : 2

Helpers are needed to prepare for the fete. Each helper can make either 2 large cakes or 35 small cakes per hour. The kitchen is available for 3 hours and 20 large cakes and 700 small cakes are needed. How many helpers are required?

A. 10
B. 15
C. 20
D. 25
E. 30

Solution

20 large cakes will require the equivalent of 10 helpers working for one hour. 700 small cakes will require the equivalent of 20 helpers working for one hour. This means if only one hour were available we would need 30 helpers. But since three hours are available we can use 10 helpers.

Solution

Correct Answer: B,C

Explanation:

The marked angle, ABC must be more than 90 degrees because it is the external angle of triangle BDC, and must be equal to the sum of angles BDC (90) and DCB.
Also ABC is not a straight line and must be less than 180.
Therefore 90 < 5x < 180
The only values for x which satisfy this relation are 20 and 30.