Advanced Math 6 — every item

$\frac{1}{2} = \frac{4}{8}$, and there are $4$ eighths in four-eighths. By the rule: $\frac{1}{2} \times \frac{8}{1} = \mathbf{4}$. ($\frac{1}{16}$ comes from multiplying instead of dividing.)

$3\frac{1}{2} = \frac{7}{2}$. Then $\frac{7}{2} \div \frac{3}{4} = \frac{7}{2} \times \frac{4}{3} = \frac{28}{6} = \frac{14}{3} = \mathbf{4\tfrac{2}{3}}$. Sentence check: four three-quarters make $3$, and two-thirds of another three-quarter makes the extra half. ✓

$6 \div \frac{3}{4} = 6 \times \frac{4}{3} = \frac{24}{3} = \mathbf{8}$ servings. Check: $8 \times \frac{3}{4} = 6$. ✓

$\frac{8}{3} \times \frac{3}{2} = \frac{24}{6} = \mathbf{4}$. (Multiplying parts separately, $2 \times 1 + \frac{2}{3} \times \frac{1}{2} = 2\frac{1}{3}$, is the classic error.)

After the book, $\frac{3}{5}$ remains. Lunch uses $\frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$ of the original, leaving $\frac{3}{5} - \frac{1}{5} = \frac{2}{5}$. That $\frac{2}{5}$ is $24, so $\frac{1}{5}$ is $12 and the whole is $5 \times 12 = \mathbf{60}$ dollars.

$\frac{7}{8} = \frac{875}{1000} = \mathbf{0.875}$. It terminates because $8 = 2^3$ has no primes other than $2$ and $5$.

$8 = 2^3$, so $\frac{1}{8} = 0.125$ terminates. The others have a factor of $3$ in the denominator: $\frac{1}{3} = 0.\overline{3}$, $\frac{1}{6} = 0.1\overline{6}$, $\frac{1}{9} = 0.\overline{1}$. Answer: $\mathbf{\frac{1}{8}}$.

$42 \times 35 = 1470$. Three decimal places: $1.470 = \mathbf{1.47}$. Estimate check: $4 \times 0.35 = 1.4$. ✓

$7.2 \div 0.08 = 720 \div 8 = \mathbf{90}$. Sense check: $0.08$ is small, so the quotient should be much bigger than $7.2$. ✓

$3 \div 11 = 0.2727\ldots = 0.\overline{27}$. The repeating block is $\mathbf{27}$. (Every $\frac{k}{11}$ has a two-digit block equal to $9k$: $\frac{1}{11} = 0.\overline{09}$, $\frac{2}{11} = 0.\overline{18}$, $\frac{3}{11} = 0.\overline{27}$.)

$\lvert -250 \rvert = \mathbf{250}$ metres below the surface. The sign says below; the absolute value says how far.

Least to greatest: $-2.5 < -2 < -1.5 < 0 < 0.5$, so $\mathbf{-2.5,\ -2,\ -\tfrac{3}{2},\ 0,\ \tfrac{1}{2}}$. The first choice lists the negatives by absolute value, smallest first — the most common mistake.

$-3$ is to the right of $-7$, so $-3 > -7$: noon was warmer. The first choice compares the absolute values $7 > 3$ and gets the direction backwards.

$4 + 3 - 2 = \mathbf{5}$. A student who writes $\lvert -4 \rvert = -4$ gets $-4 + 3 + 2 = 1$; absolute value is never negative.

From $-6$ to $-1$ is $5$ steps to the right, so the distance is $\mathbf{5}$. (Equivalently $\lvert -1 - (-6) \rvert = \lvert 5 \rvert = 5$; subtraction of negatives is formalized in Math 7, but counting on the line works now.)

$5:2 = 35:14$ (both parts $\times 7$). You need $\mathbf{14}$ cups of sugar.

$2 + 5 = 7$ parts; one part $= 84 \div 7 = 12$. Larger share $= 5 \times 12 = \mathbf{60}$ dollars (the other gets $24; $60 + 24 = 84$ ✓).

$84 \div 12 = 7$ per part. Smallest share $= 2 \times 7 = \mathbf{14}$. (Shares: $14, 21, 49$; sum $84$ ✓.)

One unit $= 28 \div 4 = 7$; red $= 3 \times 7 = \mathbf{21}$. ($12$ treats $28$ as the total; $84$ multiplies instead of dividing.)

$10k + 20 = 12k \Rightarrow k = 10$. Originally $20$ boys and $30$ girls. Now $24$ boys and $30$ girls: $24 : 30 = 4 : 5$ ✓. Total now $= \mathbf{54}$.

$180 \div 2.5 = \mathbf{72}$ km/h. Check: $72 \times 2.5 = 180$. ✓

$7.50 \div 5 = \mathbf{1.50}$ dollars per pound.

$20 \times 3600 = 72{,}000$ m per hour $= \mathbf{72}$ km/h. Shortcut: multiply m/s by $3.6$.

Total distance $140$ km, total time $3$ h: $\frac{140}{3} = \mathbf{46\tfrac{2}{3}}$ km/h. The answer is closer to $40$ than to $60$ because more time was spent at $40$. ($50$ is the trap: averaging the speeds.)

Total time $= \frac{3}{4} + \frac{1}{4} = 1$ hour for $6$ km, so the average speed is $\mathbf{6}$ km/h — not $8$, the mean of $4$ and $12$. Jo spends three times as long walking as riding, so the slow speed dominates.

$25\%$ of $80$ is $20$; $80 + 20 = \mathbf{100}$. Or in one step: $80 \times 1.25 = 100$.

One percent is $36 \div 15 = 2.4$, so $100\%$ is $2.4 \times 100 = \mathbf{240}$. Check: $0.15 \times 240 = 36$. ✓

$\frac{10}{40} = \frac{1}{4} = \mathbf{25}\%$. (Measuring against the new price, $\frac{10}{50} = 20\%$, is the wrong base.)

$45 is three quarters of the original, so one quarter is $15 and the original is $4 \times 15 = \mathbf{60}$. ($56.25 adds $25\%$ of $45$ — the wrong base. $33.75 discounts again.)

Start at $100$: $\to 80 \to 96$. The final value is $\mathbf{96}\%$ of the original — a net $4\%$ loss, because the increase was $20\%$ of the smaller number. In one line: $0.8 \times 1.2 = 0.96$.

$3(4) + 2 = 12 + 2 = \mathbf{14}$. (Writing $34 + 2$ is the classic slip.)

$3(2x+4) + x = 6x + 12 + x = \mathbf{7x + 12}$. ($6x + 4$ forgets to distribute to the $4$; $6x + 12$ forgets the extra $x$; $7x^2$ multiplies $x$-terms instead of adding them.)

$7x + 12$ at $x = \frac{1}{2}$: $3.5 + 12 = \mathbf{15.5}$. Check without simplifying: $3(1 + 4) + 0.5 = 15 + 0.5 = 15.5$. ✓

$5y - 2y = 3y$ and $3 + 7 = 10$, so $\mathbf{3y + 10}$. ($13y$ mixes numbers into the $y$-term; $7y$ adds instead of subtracting.)

$12x + 18 = \mathbf{6(2x + 3)}$. Check by distributing: $6 \cdot 2x + 6 \cdot 3 = 12x + 18$ ✓. $2(6x+9)$ and $3(4x+6)$ are true but not fully factored; $6(2x + 18)$ expands to $12x + 108$.

$x = 10 - 7 = \mathbf{3}$. Check: $3 + 7 = 10$ ✓. (Adding $7$ gives $17$, which does not check.)

$x = 1.2 + 3.5 = \mathbf{4.7}$. Check: $4.7 - 3.5 = 1.2$ ✓.

$x = \frac{18}{4} = \frac{9}{2} = \mathbf{4.5}$. Check: $4 \times 4.5 = 18$ ✓.

$x \ge 2$ includes $2$ (closed circle) and everything to its right (arrow right). Answer: closed circle at $2$, arrow right. An open circle would be for the strict inequality $x > 2$.

“More than $48$” is $\mathbf{h > 48}$ — strict, so $48$ is excluded. ($h \ge 48$ would be “at least $48$.”)

Negative $x$, positive $y$ is the upper left: Quadrant II. (Quadrant III is lower left, where both are negative.)

$\lvert 5 - (-1) \rvert = \lvert 6 \rvert = \mathbf{6}$. The common mistake is $5 - 1 = 4$, ignoring that $-1$ is on the other side of the axis.

Width $7$, height $3$: $P = 2(7 + 3) = \mathbf{20}$.

$x = 2$ gives $y = 3$, so $\mathbf{(2, 3)}$ is on the line. Checks: $(0,1)$ gives $-1 \ne 1$; $(-1,-2)$ gives $-3 \ne -2$.

$A = \frac{1}{2} \cdot 6 \cdot 4 = \mathbf{12}$ square units.

$A = \frac{1}{2} \times 10 \times 6 = \mathbf{30}$ cm².

$A = \frac{1}{2}(8 + 12)(5) = 10 \times 5 = \mathbf{50}$ cm².

$A = 9 \times 4 = \mathbf{36}$ cm². ($45$ uses the slanted side as the height; $18$ halves as if it were a triangle; $26$ is a perimeter-style sum.)

$\frac{1}{2} \cdot 8 \cdot h = 24 \Rightarrow 4h = 24 \Rightarrow h = \mathbf{6}$ cm.

Parallelogram $= 12 \times 7 = 84$. Triangle $= \frac{1}{2} \cdot 6 \cdot 4 = 12$. Shaded $= 84 - 12 = \mathbf{72}$ cm².

$S = 2(12 + 15 + 20) = 2 \times 47 = \mathbf{94}$ cm². (Using $6 \times 3 \times 4 = 72$ treats it as a cube.)

$\frac{5}{2} \times \frac{4}{3} \times 3 = \frac{5}{2} \times 4 = \mathbf{10}$ cm³.

One face: $150 \div 6 = 25$ cm², so the edge is $\sqrt{25} = \mathbf{5}$ cm.

Faces: $6$, $12$, $18$ (each appearing twice). $\mathbf{9}$ is not a face area — it would need a $3 \times 3$ face, and there is no pair of equal edges.

Base area $= \frac{5}{2} \times 2 = 5$. Height $= \frac{15}{2} \div 5 = \frac{15}{10} = \mathbf{1\tfrac{1}{2}}$ ft. Check: $5 \times 1.5 = 7.5$ ✓.

Hours of sleep for sixth graders varies from student to student, so you collect data and describe the distribution. The other three have one fixed answer.

Seven values; the $4$th is $\mathbf{12}$, with $3, 7, 8$ below and $15, 20, 21$ above.

$Q_1 = 7.5$, $Q_3 = 20.5$, $IQR = 20.5 - 7.5 = \mathbf{13}$. (Adding the quartiles gives $28$, a common error.)

Mean $= 52 \div 5 = 10.4$ — higher than four of the five scores, so it is not typical. The median, $3$, sits where the data cluster; the outlier $40$ does not move it. The range measures spread, not a typical value.

Mean $= 6$. Deviations: $\lvert 4-6 \rvert = 2$, $\lvert 8-6 \rvert = 2$, $\lvert 5-6 \rvert = 1$, $\lvert 7-6 \rvert = 1$, $\lvert 6-6 \rvert = 0$. $MAD = \frac{2+2+1+1+0}{5} = \frac{6}{5} = \mathbf{1.2}$.