Olympic College Math 94 Test on Topics 1 to 5
Math 94 – Test on Topics 1 to 5
1. Evaluate each expression for the given value of the variable or variables:
(a)  x 2 , for x  4
(b) y 2  x 2 , for x  4, y  1
2.
Combine like terms where possible and simplify the result.
(a)
4x – x + 5x – x
=
(b)
2x – 5 + 5x – 1
=
(c)
2(2x – 3y + 5) + 5y – 4x – 10 =
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Olympic College Math 94 Test on Topics 1 to 5
3.
Solve the following equations.
(a)
x – 7= – 3
(b)
x+
(c)
5z = – 80
(d)
4.
1
2
=
3
3
2
x  – 30
3
Solve the following equations. (Give any decimal answers to 2 decimal places)
(a)
(c)
(b) 4 – 2w = 4w – 16
5x + 7 = 32
4 – 0.3z
=
(e) 5(2  x)  2( x  6)  x
3+z
(d)
2x – 3 = 10x + 13
(f) 5(x – 5) – 5 =
2x + 3( x – 2)
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Olympic College Math 94 Test on Topics 1 to 5
5.
Write the following ratios, in lowest terms:
(a) the ratio of 54 pounds to 24 pounds.
(b) the ratio of 3 hours to 40 minutes
6.
Solve the following proportion equations.
(a)
5 x

15 3
(b)
12 2

30 x
7.
One mile equals approximately 1.6 kilometers. Find the distance in kilometers from Seattle to
Ketchikan – a distance of 600 miles.
8.
An airplane ascends 150 feet as it flies a horizontal distance of 1,000 feet. How much altitude will it
gain as it flies a horizontal distance of 2 mile? [Use 5,280 feet = 1 mile.]
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Olympic College Math 94 Test on Topics 1 to 5
9.
In a scale drawing, a building 100 feet tall is drawn 4 inches high. The building next to it is 120 feet
tall. How high will it be drawn in the scale drawing?
10. The formula for calculating the voltage V in a wire is
V
=
IR
where R is the resistance and I is the current.
(a) What is the Voltage when the resistance is 10 and the Resistance is 12?
(b) What is the resistance in a wire that has a Voltage of 240 and a current of 30.
11. The formula for getting the area of a circle is:A
=
 r2
where  = 3.14 and r is the radius of the circle.
(a) What is the area of a circular swimming pool with a radius of 15 feet.
(b) A circular table has an area of 20 square feet what is the radius of this circle?
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Olympic College Math 94 Test on Topics 1 to 5
12. In the following formula change the subject to the given variable.
(a)
T
=a+b – c
Solve for b
(b)
T
=a+b – c
Solve for c
(c)
M 
1
dm 2
4
Solve for D
(d). V = LWH
Solve for H.
(e) 6y – 5x = 14
Solve for y.
(f). K = M + 2MD
Solve for D.
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Olympic College Math 94 Test on Topics 1 to 5
13. Last week, Pat earned $40 less than Tony.
If their combined earnings for the week were $400, how much did each earn?
14.
Mark weighs 50 pounds less than twice as much as Laura. The sum of their weights is 250 pounds.
How much do they each weigh?
15.
The length of a rectangular swimming pool is 20 meters more than twice its width.
If its perimeter is 180 meters, find its dimensions.
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Olympic College Math 94 Test on Topics 1 to 5
16.
17.
The Smiths invested $2100, part at 9% simple interest and part at 6% simple interest.
If they received a total annual interest of $168, how much did they invest at each rate?
Two boats the Norton and the Scotia are 200 miles apart, They are traveling towards each other.
One boat the Norton can travel at 5 mph faster than the other boat the Scotia. The two boats meet
after 6 hours. What are the speeds of the two boats?
Bonus Question.
Solve the equation 15 – 4(x – 3) = 5(3 – 2x) – 3(x + 2)
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Olympic College Math 94 Test on Topics 1 to 5
Test on Topics 1 to 5 Solutions
1. Evaluate each expression for the given value of the variable or variables:
(a)  x 2 , for x  4
x2
=
– (– 4)2
=
– (16) =
– (16)
(b) y 2  x 2 , for x  4, y  1
y2x2=
2.
(1) 2  (4) 2 =
1 – 16
=
Combine like terms where possible and simplify the result.
(a)
4x – x + 5x – x
=
7x
(b)
2x – 5 + 5x – 1
=
2x + 5x – 5 – 1 =
2(2x – 3y + 5) + 5y – 4x – 10 =
=
=
=
Solve the following equations.
(c)
3.
– 15
(a)
(b)
(c)
(d)
x–7
x
x
=
=
=
1
3
=
x
=
x
=
x+
5z =
z =
2
x
3
7x – 6
4x – 6y + 10 + 5y – 4x – 10
4x – 4x – 6y + 5y + 10 – 10
0 – y+0
–y
–3
– 3+7
4
add 7 from both sides
2
3
2 1
–
3 3
1
3
subtract
1
from both sides
3
– 80
– 16
divide both sides by 5
– 45
divide both sides by
– 30
x =
2
3
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Olympic College Math 94 Test on Topics 1 to 5
4.
Solve the following equations. (Give any decimal answers to 2 decimal places)
(a)
(c)
5x + 7
5x
x
4 – 0.3z
4
1
0.77
(e) 5(2 – x)
10 – 5x
10 – 5x
10 – 5x
10
–2
– 0.33
5.
=
=
=
=
=
=
=
=
=
=
32
25
5
=
=
=
=
3+z
3 + 1.3z
1.3z
z
2(x + 6) – x
2x + 12 – x
2x – x + 12
x + 12
6x + 12
6x
x
4w – 16
6w – 16
6w
w
add 2w to both sides
add 16 to both sides
divide both sides by 6
(d) 2x – 3
–3
– 16
–2
=
=
=
=
10x + 13
8x + 13
8x
x
subtract 2x from both sd
subtract 13 from both sd
divide both sides by 8
(f) 5(x – 5) – 5 =
2x + 3(x – 2)
5x – 25 – 5 = 2x + 3x – 6
5x – 30
= 5x – 6
5x
= 5x + 24
0
= 24
No Solution possible
54 : 24 =
180 : 40 =
9:4
9:2
divide by 6
divide by 20
Solve the following proportion equations.
(a)
7.
=
=
=
=
Write the following ratios, in lowest terms:
(a) the ratio of 54 pounds to 24 pounds.
(b) the ratio of 3 hours to 40 minutes
6.
(b) 4 – 2w
4
20
3.33
5 x

15 3
15x =
x
=
(b)
15
1
12 2

30 x
12x = 60
x = 5
One mile equals approximately 1.6 kilometers. Find the distance in kilometers from Seattle to
Ketchikan – a distance of 600 miles.
1
=
1.6
x =
600
x
960 km
8. An airplane ascends 150 feet as it flies a horizontal distance of 1,000 feet. How much altitude will it
gain as it flies a horizontal distance of 2 mile? [Use 5,280 feet = 1 mile.]
150
1000
1000x
x
=
=
=
x
10560
1584000
1584 feet
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Olympic College Math 94 Test on Topics 1 to 5
9.
In a scale drawing, a building 100 feet tall is drawn 4 inches high. The building next to it is 120 feet
tall. How high will it be drawn in the scale drawing?
100
4
100x
x
=
=
=
120
x
480
4.8 inches
10. The formula for calculating the voltage V in a wire is
V = IR
where R is the resistance and I is the current.
(a) What is the Voltage when the resistance is 10 and the Current is 12?
V
=
IR =
10 x 12 =
120
(b) What is the resistance in a wire that has a Voltage of 240 and a current of 30.
V =
240 =
240
=
30
IR
30R
R
11. The formula for getting the area of a circle is:A
=
 r2
where  = 3.14 and r is the radius of the circle.
(a) What is the area of a circular swimming pool with a radius of 15 feet.
A
=
 r2
=
3.14x 152
=
3.14 x 225
=
706.5
(b) A circular table has an area of 20 square feet what is the radius of this circle?
A =
20 =
6.36
=
6.36 =
2.52
=
 r2
3.14 x r2
r2
r
r
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Olympic College Math 94 Test on Topics 1 to 5
12. In the following formula change the subject to the given variable.
(a)
T
=a+b – c
Solve for b
b =T + c – a
(b)
T
=a+b – c
Solve for c
c
(c)
M 
1
dm 2
4
Solve for d
(d).
V = LWH
4M
m2
V
H
LW
14  5x
y
6
KM
D
2M
d
Solve for H.
(e) 6y – 5x = 14
Solve for y.
(f). K = M + 2MD
Solve for D.
=a+b – T
13. Last week, Pat earned $40 less than Tony.
If their combined earnings for the week were $400, how much did each earn?
Tony
Pat
=
=
x
$40 less than Tony
=
x – 40
combined earnings for the week
x + x – 40
2x – 40
2x
x
=
=
=
=
=
$400
400
400
440
220
So Tony =
14.
x
=
$220
and
Pat =
x – 40
=
220 – 40
=
$180
Mark weighs 50 pounds less than twice as much as Laura. The sum of their weights is 250 pounds.
How much do they each weigh?
Laura
=
x
and
Sum of their weights
x + 2x – 50
2x – 50
2x
x
So Laura
=
x
=
Mark
=
=
=
=
=
=
$50 less than twice Laura =
2x – 50
250
250
250
300
150
150 pounds and
Mark
=
2x – 50 =
300 – 50 = 250 pounds
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Olympic College Math 94 Test on Topics 1 to 5
15.
The length of a rectangular swimming pool is 20 meters more than twice its width.
If its perimeter is 180 meters, find its dimensions.
2x +20
Width = x
Length =
Length =
20 meters more than twice its width
x
2x + 20
x
Perimeter
= 180
x + 2x + 20 + x + 2x + 20 = 180
2x +20
6x + 40
= 180
6x
= 140
x
= 23.3
So Width = x = 23.3 metres and Length = 2x + 20 = 2(23.3) + 20 = 66.6 metres
16.
The Smiths invested $2100, part at 9% simple interest and part at 6% simple interest.
If they received a total annual interest of $168, how much did they invest at each rate?
$2100
9%
2100 – x
x
Interest =
6%
9% of x + 6% of (2100 – x)
0.09x + 0.06(2100 – x)
=
0.09x + 126 – 0.06x
0.09x – 0.06x + 126
0.03x + 126
0.03x
=
168
=
=
=
=
x
=
x
=
168
168
168
168
42
42
0.03
1400
So $1400 is invested at 9% and the rest $700 is invested at 6%
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Olympic College Math 94 Test on Topics 1 to 5
17.
Two boats the Norton and the Scotia are 200 miles apart, They are traveling towards each other.
One boat the Norton can travel at 5 mph faster than the other boat the Scotia. The two boats meet
after 6 hours. What are the speeds of the two boats?
x+5
x
Norton
Scotia
200 miles
Distance traveled by the Norton
Distance traveled by the Scottia
Total distance traveled
6x + 30 + 6x
12x + 60
12x
=
=
=
=
x
=
x
=
=
=
6(x + 5) =
6x
=
6x + 30
6x
500
500
500
440
440
12
36.67
So the speed of the Scotia was 36.67 mph and the speed of the Norton was 41.67 mph
Bonus Question.
Solve the equation
15 – 4(x – 3)
15 – 4x +12
15 +12 – 4x
27 – 4x
27
18
–2
=
=
=
=
=
=
=
5(3 – 2x) – 3(x + 2)
15 – 10x – 3 x – 6
15 – 6 – 10x – 3 x
9 – 13x
9 – 9x
– 9x
x
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