# Solutions to Pre-Algebra Placement Test Practice

Detailed solutions to the pre-algebra placement test practice are presented with detailed explanations.
1. Solution
Use order of operation to evaluate multiplication and division first from left to right
9 ÷ 3 � 2 = 6
Insert the result in the whole expression
72 - 6 + 2
Evaluate addition and subtraction from left to right to obtain final value of expression
72 - 6 + 2 = 68

2. Solution
We first rewrite 0.0000022 using 10-5 as follows
0.0000022 = 0.22 �—10-5
We now add 3.0 x 10-5 and  0.22�—10-5  as follows
3.0 x 10-5 +  0.22 �—10-5
= 10-5 (3.0 + 0.22)
= 3.22 x 10-5

3. Solution
Reduce all fractions to the least common denominator 12, add, subtract and reduce the final result.
2 / 3 + (1 / 2 - 1 / 6) + (1 / 3 - 3 / 4)
= 8 / 12 + (6 / 12 - 2 / 12) + (4 / 12 - 9 / 12)
= (8 + 6 - 2 + 4 - 9) / 12
= 7 / 12
The numerator after reduction is 7

4. Solution
Evaluate expressions inside brackets
(4 / 3) �— (3 / 5) = 4 / 5
(1 / 4) � (4 / 5) = (1 / 4) � (5 / 4) = 5 / 16
Substitute expressions inside brackets by their values found above
1 / 4 + (4 / 3 � 3 / 5) - (1 / 4 �4 / 5) = 1 / 4 + 4 / 5 - 5 / 16
Rewrite the fractions using to the least common denominator of 4, 5 and 16 which is 80 and evaluate from left to right to obtain the final answer.
= 20 / 80 + 64 / 80 - 25 / 80 = 59 / 80

5. Solution
The original price was $20 The selling price was$26
The absolute change was
26 - 20 = $6 The relative change was change / original price =$6 / $20 Change fraction into percent by multiplying numerator and denominator by 5 6 / 20 = (6 � 5 / (20 � 5) = 30 / 100 = 30 % 6. Solution Given expression 3 / 4 + 0.85 + 20 % Change fraction into decimal number (divide) 3 / 4 = 0.75 Change 20% to decimal 20% = 20 / 100 = 0.2 Evaluate the given express using decimal 3 / 4 + 0.85 + 20 % = 0.75 + 0.85 + 0.2 = 1.8 7. Solution Tom worked 6 hours at$5.50 per hour; he earned
6 �— $5.50 =$33
Buying 2 magazines at $9.50 each costs 2 �—$9.50 = $19 Shopping a pen at$8.25 costs
$8.25 Total spent on shopping$19 + $8.25 =$27.25
Tom has left = what he earned - what he spent
$33 -$27.25 = \$5.75

8. Solution
Change the mixed number 1 (3/4) into decimal number
1 (3/4) = 1 + 3/4 = 1 + 0.75 = 1.75
Change the mixed number 4 (1/2) into decimal number
4 (1/2) = 4 + 1/2 = 4 + 0.5 = 4.5
Total quantity of cheese bought
1 (3/4) + 3.75 + 4 (1/2) = 1.75 + 3.75 + 4.5 = 10 pounds

9. Solution
Let x be the total number of all students
Number of students that are 8 years old or younger is 40% of all students is written as
40% x
The number of the remaining students is 120
or
(100% - 40%) of total = 60% of total = 60% x
Hence
60% x = 120
Solve for x
x = 120 / 60% = 120 / (60 / 100) = 120 � 100 / 60 = 200
Number of students that are 8 years old or younger is given by
40% x = 40% 200 = 80

10. Solution
Let x be the total number of students Number of students with no brothers or sisters is 1/5 of the total and may be written as 1/5 of x = x (1/5)
Number of students with brothers or sisters
(1 - 1/5) = (4/5) x
40% of those with brothers or sisters have only one brother or sister
40% of (4 / 5) x = 40% (4/5) x
Students with more than one brother or sister
(100% - 40%) of (4 / 5) x = 60% (4 / 5) x = 48 % x
Percent of all students with more than one brother or sister is
48 %

11. Solution
Let x be the number of km. The cost CA for plan A is given by
CA = 20 + 0.05 x
The cost CB for plan B is given by
CB = 15 + 0.07 x
Same amount paid means CA = CB
20 + 0.05 x = 15 + 0.07 x
Solve the above equation to find x which is the number of km so that a customer would pay the same amount.
20 - 15 = 0.07 x - 0.05 x
Solutions: x = 250 km
For a distance of 250 km, a customer would pay the same amount in both plans A and B.

12. Solution
Joe earns y dollars in x hours, hence Joe earns per hour
y / x
Joe earns in z hours
z (y / x) = z y / x

13. Solution
Let T be the total of the marks for the 30 students. If the average of the class is 80, then
T / 30 = 80
T = 30 * 80 = 2400
Let TG be the total marks for the 20 girls. If the average for the girls is is 85, then
TG / 20 = 85
TG = 20 * 85 = 1700
Let TB be the total marks for the 10 boys, then
TB = 2400 - 1700 = 700
Average for boys is given by
700 / 10 = 70

14. Solution
Given equation
y / 5 = 10 / 25
Cross product gives
25 y = 5 * 10
Solve for y
y = 2

15. Solution
Given expression
√(32 + 42)
32 +  42 = 9 + 16 = 25
Simplify given expression by taking square root
√(32 + 42) = √25 = 5

16. Solution
The number of males is
24 - 15 = 9
The ratio of males to the total number of members of the club is
9:24

17. Solution
2.05 may be written as
2.05 = 2 + 0.05
Write 0.05 as a fraction
= 2 + 5/100
Reduce the fraction 5/100 by dividing the numerator and denominator by 5
= 2 + 1/20
Write as a mixed number
= 2(1/20)

18. Solution
An integer is divisible by 3 if the sum of its digits is divisible by 3. We first find the sum of the digits in each of the given numbers.
The sum of the digits in 101267 is equal to: 1+0+1+2+6+7 = 17.
The sum of the digits in 22345 is equal to: 2+2+3+4+5 = 16.
The sum of the digits in 934567 is equal to: 9+3+4+5+6+7 = 34.
The sum of the digits in 934566 is equal to: 9+3+4+5+6+6 = 33.
33 is divisible by 3 and therefore 934566 is divisible by 3

19. Solution
27 is divisible by 1, 3, 9 and 27 and hence is not a prime number.

20. Solution
The slope m of the line through thr points (2 , 0) and (-1 , 3) is given by
m = (3 - 0) /(-1 - 2) = 3 / (-3) = - 1
Two lines are parallel if their slopes are equal. Hence the slope of the line parallel to the line through (2 , 0) and (-1 , 3) is also equal to - 1

21. Solution
List all factors of 32 and 48 in ascending order
Factors of 32: 1, 2, 4, 8, 16, 32
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
List the common factors to 32 and 48
The common fators are: 1 , 2 , 4 , 8 , 16
The greatest common factor to 32 and 48 is
16

22. Solution
List of multiples of each pair stopping at the first (lowest) common multiple)
6: 6,12,18,24
8: 8,16,24
lowest common multiple of (6,8) = 24

6: 6,12
12: 12
lowest common multiple of (6,12) = 12

4: 4,8,12
6: 6,12
lowest common multiple of (4,6) = 12

4: 4,8,12
3: 3,6,9,12
lowest common multiple of (4,3) = 12

4: 4,8,12,16
8: 8,16
lowest common multiple of (4,8) = 16

6 and 8, the pair in a) has an LCM equal to 24