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)
    Simplify expression inside radical
    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

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