Detailed solutions to pre-algebra placement test practice questions with step-by-step explanations. Essential for high school and college math preparation.
Solution:
Use order of operations to evaluate multiplication and division first from left to right:
\[9 \div 3 \cdot 2 = 6\]Insert the result in the whole expression:
\[72 - 6 + 2\]Evaluate addition and subtraction from left to right:
\[72 - 6 + 2 = 68\]Solution:
Rewrite \(0.0000022\) using \(10^{-5}\):
\[0.0000022 = 0.22 \times 10^{-5}\]Add the expressions:
\[3.0 \times 10^{-5} + 0.22 \times 10^{-5} = 10^{-5}(3.0 + 0.22)\] \[3.22 \times 10^{-5}\]Solution:
Reduce all fractions to the least common denominator 12:
\[\frac{2}{3} + \left(\frac{1}{2} - \frac{1}{6}\right) + \left(\frac{1}{3} - \frac{3}{4}\right)\] \[= \frac{8}{12} + \left(\frac{6}{12} - \frac{2}{12}\right) + \left(\frac{4}{12} - \frac{9}{12}\right)\] \[= \frac{8 + 6 - 2 + 4 - 9}{12} = \frac{7}{12}\]Solution:
Evaluate expressions inside brackets:
\[\frac{4}{3} \times \frac{3}{5} = \frac{4}{5}\] \[\frac{1}{4} \div \frac{4}{5} = \frac{1}{4} \times \frac{5}{4} = \frac{5}{16}\]Substitute expressions:
\[\frac{1}{4} + \frac{4}{5} - \frac{5}{16}\]Rewrite with denominator 80:
\[= \frac{20}{80} + \frac{64}{80} - \frac{25}{80} = \frac{59}{80}\]Solution:
Original price: \( \$20\), Selling price: \( \$26\)
Absolute change:
\[26 - 20 = \$6\]Relative change:
\[\frac{6}{20} = \frac{6 \times 5}{20 \times 5} = \frac{30}{100} = 30\%\]Solution:
Convert all terms to decimals:
\[\frac{3}{4} = 0.75,\quad 20\% = 0.2\] \[0.75 + 0.85 + 0.2 = 1.8\]Solution:
Earnings: \(6 \times 5.50 = \$33\)
Magazines cost: \(2 \times 9.50 = \$19\)
Total spent: \(19 + 8.25 = \$27.25\)
Money left: \(33 - 27.25 = \$5.75\)
Solution:
Convert mixed numbers to decimals:
\[1\frac{3}{4} = 1.75,\quad 4\frac{1}{2} = 4.5\] \[1.75 + 3.75 + 4.5 = 10 \text{ pounds}\]Solution:
Let \(x\) be total students. \(60\%\) are older than 8:
\[60\% x = 120\] \[x = \frac{120}{0.6} = 200\]Students 8 or younger: \(40\% \times 200 = 80\)
Solution:
Let \(x\) be total students. Students with siblings: \(\frac{4}{5}x\)
Students with more than one sibling: \(60\%\) of \(\frac{4}{5}x\):
\[0.6 \times \frac{4}{5}x = 0.48x = 48\%\]Solution:
Let \(x\) be kilometers driven:
\[\text{Plan A: } 20 + 0.05x\] \[\text{Plan B: } 15 + 0.07x\]Set equal:
\[20 + 0.05x = 15 + 0.07x\] \[5 = 0.02x\] \[x = 250 \text{ km}\]Solution:
Hourly rate: \(\frac{y}{x}\) dollars/hour
Earnings in \(z\) hours: \(\frac{y}{x} \times z = \frac{yz}{x}\)
Solution:
Total marks for class: \(30 \times 80 = 2400\)
Total for girls: \(20 \times 85 = 1700\)
Total for boys: \(2400 - 1700 = 700\)
Boys' average: \(\frac{700}{10} = 70\)
Solution:
Given: \(\frac{y}{5} = \frac{10}{25}\)
Cross multiply: \(25y = 50\)
Solution: \(y = 2\)
Solution:
Simplify inside radical:
\[\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]Solution:
Males: \(24 - 15 = 9\)
Ratio males to total: \(9:24\)
Solution:
Convert to mixed number:
\[2.05 = 2 + \frac{5}{100} = 2 + \frac{1}{20} = 2\frac{1}{20}\]Solution:
Divisibility rule for 3: sum of digits divisible by 3
Only \(934566\) has digit sum \(9+3+4+5+6+6=33\) (divisible by 3)
Solution:
\(27\) has divisors \(1, 3, 9, 27\) → not prime
Solution:
Slope through \((2,0)\) and \((-1,3)\):
\[m = \frac{3-0}{-1-2} = \frac{3}{-3} = -1\]Parallel lines have equal slope: \(m = -1\)
Solution:
Factors of 32: \(1, 2, 4, 8, 16, 32\)
Factors of 48: \(1, 2, 3, 4, 6, 8, 12, 16, 24, 48\)
Greatest common factor: \(16\)
Solution:
Lowest common multiples:
Only pair (6,8) has LCM = 24