# Ged Math practice quiz 7 - online test

One third of the candidates for a job were 25 years old or younger. Two sevenths of the candidates were at least 50 years old.
If 84 people applied for the job, how many were between 25 and 50?

32

28

24

33

One third of 84 is 84 × 1/3 = 84÷3 = 28. 28 people 25 years old or younger applied for the job. Two seventh of 84 is 84 × 2/7= 2×84÷7 = 24. 24 people who were at least 50 years old applied for the job. 84 - 28 - 24 = 32. 32 people applied for the job who were between 25 and 50.

Which of the following is a factor of 450 and multiple of 25?

30

75

90

100

It can be confirmed by division that (a), (b) and (c) are factors of 450, while 100 is not. The multiples of 25 are 25, 50, 75,.... 75 is both a factor of 450 and multiple of 25

The formula h - 15 = 3.2t gives the height h in inches of a plant t weeks after planting.
Which is the rate at which the plant's height is increasing?

15 inches per week

18.2 inches per week

3.2 inches per week

48 inches per week

h =15 + 3.2t gives the height of a shrub in inches after t weeks. When planting occurs the height is 15 + 3.2× 0 = 15 inches After one week the height is 15 + 3.2× 1 = 18.2 inches The rate at which the height increases is 18.2 - 15 = 3.2 inches

How high was the initial height of the shrub?

15 inches

3.2 inches

18.2 inches

11.8 inches

Initial means at the beginning. At the beginning t is 0. h =15 + 3.2t gives the height of a shrub in inches after t weeks. When planting occurs the height is 15 + 3.2× 0 = 15 inches

The formula h -15 = 3.2t gives the height h in inches of a plant t weeks after planting.
When was the height 31 inches?

Some time between 11 and 18 weeks

Some time between 7 and 12 weeks

Some time between 6 and 10 weeks

Some time between 3 and 7 weeks

h =15 + 3.2t gives the height of a shrub in inches after t weeks. When the height is 31 inches 31 = 15 + 3.2t
3.2t = 16
t = 16÷3.2
t = 5
The height is 31 inches after 5 weeks The next three questions refer to the following diagram:
A graph of the temperature T (degrees Fahrenheit) in a refrigerator t hours after midday.
What is the equation of the axis of symmetry of this graph?

T = 9

t = 3

T = 0

t = 0

It is clear from the diagram that the T -axis is the mirror line or axis of symmetry of this diagram. The equation of the T- axis is t = 0. The equation of the axis of symmetry of this graph is t = 0 Refer to the following diagram:
It is necessary to find the times when the temperature is 8o F. Which of the following is the correct equation to do this.

9 - t = 8

9 +t2 = 8

9 - t2 = 8

9t - t2 = 8

It is apparent from the graph that it is not linear hence the answer cannot be (A) If an attempt is made to solve (B) it will be found there is no real solution. One of the solutions to (D) is 8, which gives a negative value. (C) is the only possible solution. Which of the following is the equation of the straight line joining the points on the graph corresponding to t = 2 and t = 1?

T + 3t = 11

T - 3t = 11

T + 11t = 3

T + 3t = 1

The point on the graph corresponding to t = 1 is (1,8).
The point on the graph corresponding to t = 2 is (2,5).
The slope of the straight line joining them is (5 - 8)÷(2 - 1) = -3.
The equation is T = -3t + c
As (2,5) lies the line we can substitute for t and T.
5 = -3×2 + c
c = 11
Equation is T = -3t + 11 which is the same as (a)

The following two questions relate to the sphere shown below:
The formula for the volume V of the sphere = 4(pi)r3 / 3.
Which of the following is the closest approximation to the volume when r, the radius, is 11 inches?

6000 cubic inches

5590 cubic inches

5573 cubic inches

5000 cubic inches

It is necessary to substitute 11 for r in the formula. The calculation is V = 4/3 x (pi=3.14) x 113 = 5572.45....... Clearly the closest approximation to this is (c) If the radius of the sphere were doubled by what factor would the surface area be enlarged?
Note: The formula for the surface area S of the sphere = 4 x (pi) x r2

2

4

8

16

If r = 1 then the surface area of the smaller sphere is 12.56...... If the radius is doubled the surface area is 50.26...... The factor is 50.26... ÷ 12.56.... = 4. The surface area is enlarged by a factor of 4.