Every solution below shows the working and ends with the final answer in bold. Compare each step with your own work, especially the subtraction used for repeating decimals and the squaring checks used for square roots.
Problem 1: Terminating decimal
Divide 3 by 8: 8 goes into 30 three times (remainder 6), into 60 seven times (remainder 4), and into 40 five times (remainder 0). So \(\frac{3}{8}=\)\(0.375\), and the decimal terminates.
Problem 2: Repeating decimal
Divide 5 by 9: every step gives digit 5 with remainder 5, so the digit 5 repeats. \(\frac{5}{9}=\)\(0.\overline{5}\).
Problem 3: Two-digit repeat
70 divided by 11 is 6 with remainder 4. 40 divided by 11 is 3 with remainder 7. The remainder 7 returns, so 63 repeats. Check: \(0.\overline{63}=\frac{63}{99}=\frac{7}{11}\). Answer: \(0.\overline{63}\).
Problem 4: Sort the numbers
\(\sqrt{9}=3\) is an integer: rational. \(\sqrt{10}\): 10 is not a perfect square, so irrational. \(-\frac{4}{5}\) is a fraction: rational. \(0.\overline{12}\) repeats: rational. \(\pi\) never repeats: irrational.
Rational: \(\sqrt{9},\ -\frac{4}{5},\ 0.\overline{12}\). Irrational: \(\sqrt{10},\ \pi\).
Problem 5: One repeating digit
Let \(x=0.444\)…. Then \(10x=4.444\)…. Subtract: \(9x=4\), so \(x=\frac{4}{9}\).
Problem 6: Two repeating digits
Let \(x=0.4545\)…. Then \(100x=45.4545\)…. Subtract: \(99x=45\). So \(x=\frac{45}{99}=\)\(\frac{5}{11}\) after dividing by 9.
Problem 7: Three repeating digits
Three digits repeat, so multiply by 1,000: \(1000x=135.135\)…. Subtract \(x\): \(999x=135\). Then \(x=\frac{135}{999}\). Both numbers divide by 27, so \(x=\frac{5}{37}\).
Problem 8: Delayed repeat
Let \(x=0.1666\)…. Then \(100x=16.666\)… and \(10x=1.666\)…. Subtract: \(90x=15\). So \(x=\frac{15}{90}=\)\(\frac{1}{6}\).
Problem 9: Between two integers
\(5^{2}=25\) and \(6^{2}=36\). Since \(25< 30< 36\), \(5<\sqrt{30}< 6\). Answer: between 5 and 6.
Problem 10: Nearest tenth
\(7^{2}=49\) and \(8^{2}=64\), so the root is just above 7. Test \(7.1^{2}=50.41\), which is above 50, while \(7.0^{2}=49\) is below. Since 50 is closer to 50.41 (difference 0.41) than to 49 (difference 1), \(\sqrt{50}\approx\)\(7.1\).
Problem 11: Square root of 2
\(1.41^{2}=1.9881\) and \(1.42^{2}=2.0164\). Because \(1.9881< 2< 2.0164\), \(1.41<\sqrt{2}< 1.42\).
Problem 12: Plot the root
\(2.2^{2}=4.84\) and \(2.3^{2}=5.29\), so \(\sqrt{5}\) is between 2.2 and 2.3. Since \(2.24^{2}=5.0176\) is just above 5, \(\sqrt{5}\approx\)\(2.24\). Plot a point a little left of the 2.25 mark.
Problem 13: Compare with a decimal
\(2.6^{2}=6.76\), and \(6.76< 7\). So \(2.6<\sqrt{7}\). Answer: \(\sqrt{7}>2.6\). (Indeed \(\sqrt{7}\approx2.646\).)
Problem 14: Compare roots
\(3.5^{2}=12.25\), and \(12< 12.25\). So \(\sqrt{12}< 3.5\). Answer: \(\sqrt{12}< 3.5\).
Problem 15: Order four numbers
Decimals: \(3.14\), \(\pi\approx3.1416\), \(\frac{22}{7}\approx3.1429\), \(\sqrt{10}\approx3.1623\). Order: \(3.14<\pi<\frac{22}{7}<\sqrt{10}\).
Problem 16: Rational in between
\(\sqrt{2}\approx1.414\) and \(\sqrt{3}\approx1.732\). The decimal 1.5 is between them. Check: \(1.5^{2}=2.25\), and \(2< 2.25< 3\). Answer: \(1.5\) (many answers work, such as 1.6).
Problem 17: Square garden
The side is \(\sqrt{40}\). Since \(6^{2}=36\) and \(7^{2}=49\), it is between 6 and 7. Test \(6.3^{2}=39.69\) and \(6.4^{2}=40.96\). 40 is closer to 39.69, so the side is about \(6.3\text{ ft}\).
Problem 18: Tile border
\(9^{2}=81\) and \(10^{2}=100\). Test \(9.4^{2}=88.36\) and \(9.5^{2}=90.25\). The side is about \(9.5\text{ ft}\). The border is \(4\times 9.5=38\). Answer: side \(\approx9.5\text{ ft}\), border \(\approx38\text{ ft}\).
Problem 19: Two lots
Priya’s area is \(6.7^{2}=44.89\) square yards. Diego’s area is 45 square yards, and \(45>44.89\). Equivalently, Diego’s side is \(\sqrt{45}>6.7\). Answer: Diego’s lot is larger (by 0.11 square yard).
Problem 20: Thirds and nines
\(0.\overline{3}=\frac{3}{9}=\frac{1}{3}\) and \(0.\overline{6}=\frac{6}{9}=\frac{2}{3}\). The sum is \(\frac{1}{3}+\frac{2}{3}=1\). The decimals add to \(0.\overline{3}+0.\overline{6}=0.\overline{9}\). So \(0.\overline{9}=1\).
Problem 21: Square root claim
\(49=7^{2}\), so \(\sqrt{49}=7\), which is the fraction \(\frac{7}{1}\). Answer: Ethan is wrong; 49 is a perfect square, so the root is rational.
Problem 22: Is 22/7 equal to pi?
\(\frac{22}{7}=3.\overline{142857}\) repeats, so it is rational. Pi starts \(3.14159265\) and never repeats, so it is irrational. A rational number cannot equal an irrational number, and the difference is about \(0.001\). Answer: Mei is not correct; 22/7 is only an approximation.
See every chapter on the Grade 8 math page.