Advanced Math 5 — every item

$(9 - 4) \times 6 + 2 = 5 \times 6 + 2 = 30 + 2 = \mathbf{32}$.

$4[30 - (7 + 5) \div 3] = 4[30 - 12 \div 3] = 4[30 - 4] = 4 \times 26 = \mathbf{104}$.

$4 \times 6 = 24$ beads used. $50 - 24 = \mathbf{26}$ beads left. (Subtracting first, $(50 - 4) \times 6 = 276$, is the classic order-of-operations error; $36$ and $30$ are subtraction slips.)

$72 = 8 \times 9 = (2 \times 2 \times 2) \times (3 \times 3) = 2^3 \times 3^2$. The exponent of $2$ is $\mathbf{3}$.

$\text{LCM} = 2^3 \cdot 3^2 \cdot 5 = 8 \times 9 \times 5 = \mathbf{360}$. Check: $\text{GCF}(24, 90) = 2 \cdot 3 = 6$, and $6 \times 360 = 2{,}160 = 24 \times 90$. ✓

$\text{LCM}(8, 12) = 2^3 \cdot 3 = \mathbf{24}$ seconds. ($4$ is the GCF; $96 = 8 \times 12$ is a common multiple but not the least — it counts the shared $2^2$ twice; $48 = 2^4 \cdot 3$ has one $2$ too many.)

$48 \times 25 = 12 \times (4 \times 25) = 12 \times 100 = \mathbf{1{,}200}$.

$148 \times 25 = 100 \times 25 + 48 \times 25 = 2{,}500 + 1{,}200 = 3{,}700$. Left over: $3{,}700 - 3{,}512 = \mathbf{188}$.

$312 \times 204 = 312 \times 200 + 312 \times 4 = 62{,}400 + 1{,}248 = \mathbf{63{,}648}$. ($7{,}488$ is $312 \times 24$ — the zero in $204$ was dropped, and the four-digit size should have been the warning; $62{,}400$ forgets the $\times 4$ part; $64{,}848$ adds $312 \times 8$ instead of $312 \times 4$.)

$34 \times 8 = 272$, remainder $13$; bring down the $6$: $136 = 34 \times 4$. So $2{,}856 \div 34 = \mathbf{84}$ exactly. Check: $34 \times 84 = 2{,}856$. ✓

$1{,}574 \div 22 = 71$ remainder $12$, since $22 \times 71 = 1{,}562$. Each piece is $71\frac{12}{22} = \mathbf{71\tfrac{6}{11}}$ cm.

$2{,}350 \div 36 = 65$ remainder $10$. Ten cookies do not fill a box, so there are $\mathbf{65}$ full boxes. ($66$ rounds up — right for “how many vans are needed,” wrong here; $65\frac{5}{18}$ is the exact quotient, not a number of boxes; $650$ is a tens-place estimating error.)

$3 \div 4 = \frac{3}{4}$. Each friend gets $\mathbf{\frac{3}{4}}$ of a pizza. (Cut every pizza into quarters; each friend takes one quarter from each of the $3$ pizzas.)

$4 \div 5 = \frac{4}{5} = \frac{8}{10} = \mathbf{0.8}$ m. Check: $5 \times 0.8 = 4$. ✓

$2 \div 5 = \frac{2}{5}$. Each friend gets $\mathbf{\frac{2}{5}}$ of a loaf. ($2\frac{1}{2}$ is $5 \div 2$, the division turned upside down — five people cannot each get more than the two loaves; $\frac{1}{5}$ shares only one loaf; $\frac{1}{2}$ and $10$ are not $2$ divided by $5$.)

$3\frac{3}{12} - 1\frac{8}{12} = 2\frac{15}{12} - 1\frac{8}{12} = \mathbf{1\tfrac{7}{12}}$. (As improper fractions: $\frac{39}{12} - \frac{20}{12} = \frac{19}{12}$.)

Bow: $\frac{1}{4} \times 60 = 15$ cm, leaving $45$ cm. Bookmark: $\frac{2}{3} \times 45 = 30$ cm. Left: $45 - 30 = \mathbf{15}$ cm. (In fractions of the whole: $\frac{3}{4}$ remains after the bow, the bookmark is $\frac{2}{3}$ of $\frac{3}{4} = \frac{1}{2}$, and $\frac{1}{4}$ of $60$ is left.)

Ali: $\frac{2}{5} \times 40 = 16$, leaving $24$. Bea: $\frac{1}{4} \times 24 = 6$. Jo keeps $24 - 6 = \mathbf{18}$. ($14$ takes $\frac{1}{4}$ of the original $40$ instead of the remainder; $24$ forgets Bea; $6$ is Bea's share.)

$\frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \mathbf{\frac{3}{10}}$. Cross-cancelling first: $\frac{3}{2} \times \frac{1}{5} = \frac{3}{10}$.

$4 \div \frac{1}{5} = 4 \times 5 = \mathbf{20}$ pieces. Check: $20 \times \frac{1}{5} = 4$. ✓ (Dividing by a unit fraction makes the answer bigger; $\frac{4}{5}$ would be multiplying.)

$\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \mathbf{\frac{8}{15}}$ m². ($\frac{3}{4} = \frac{6}{8}$ comes from adding the tops and adding the bottoms; $\frac{22}{15}$ is the sum $\frac{2}{3} + \frac{4}{5}$; $\frac{2}{5}$ is a cancelling slip.)

Total $= 55$ over five scores. Mean $= 55 \div 5 = \mathbf{11}$.

Total $= \frac{3}{2} + \frac{3}{2} + 3 = 6$ m over $3 + 2 + 3 = 8$ ribbons. Mean $= 6 \div 8 = \mathbf{\frac{3}{4}}$ m.

Total $= 17$ m; $10$ airplanes. Mean $= 17 \div 10 = \mathbf{1.7}$ m. ($2.5$ is $10$ airplanes $\div$ $4$ columns; $1.75$ averages the four column labels and ignores how many airplanes are in each; $4.25$ is $17 \div 4$; $17$ is the total.)

$4.80 + 3.75 = \mathbf{8.55}$.

$1.25 \div 0.4 = 12.5 \div 4 = \mathbf{3.125}$. Sense check: $0.4$ is less than $1$, so the answer is bigger than $1.25$; and $3 \times 0.4 = 1.2$, close to $1.25$. ✓

$0.3 \times 0.3 = 0.09$ m²: one place plus one place is two places. Sense check: $0.3 < 1$, so $0.3 \times 0.3$ must be less than $0.3$, which rules out $0.9$. Answer $\mathbf{0.09}$. ($0.6$ is $0.3 + 0.3$; $9$ and $0.009$ have the wrong number of decimal places.)

$180^\circ - 48^\circ = \mathbf{132}^\circ$.

$\angle P = 35^\circ$ (vertically opposite angles are equal). Then $\angle R = 180^\circ - 35^\circ - 60^\circ = \mathbf{85}^\circ$.

$\angle ACD = \angle A + \angle B = 50^\circ + 60^\circ = \mathbf{110}^\circ$. Check: $\angle ACB = 70^\circ$ and $180^\circ - 70^\circ = 110^\circ$. ✓ ($130^\circ$ is $180^\circ - 50^\circ$, subtracting only one interior angle; $70^\circ$ is the interior angle at $C$, not the exterior one; $100^\circ$ is an addition slip.)

$P = (4, 7)$. The $y$-coordinate is $\mathbf{7}$.

Sum $= 82 \times 4 = 328$. The known scores total $243$. Fourth score $= 328 - 243 = \mathbf{85}$.

Distance $= 3 \times 40 = 120$ km, so the point is $(\text{time}, \text{distance}) = \mathbf{(3, 120)}$. ($(120, 3)$ swaps the coordinates; $(3, 40)$ uses the speed instead of the distance; $(40, 120)$ uses no time at all.)

$1$ unit $= 4$ cups; white $= 4 \times 4 = \mathbf{16}$ cups.

Before: $30 \div 5 = 6$ per unit, so Ali $12$, Bea $18$. After: still $5$ units for $30$, so Ali has $1 \times 6 = 6$ and Bea $24$. Ali gave $12 - 6 = \mathbf{6}$ beads. Check: $6:24 = 1:4$. ✓

One part $= 40 \div 8 = 5$; red $= 3 \times 5 = \mathbf{15}$. ($24$ is $\frac{3}{5}$ of $40$ — the ratio read as a fraction of the whole; $25$ is the blue count; $5$ is one part.)

$180 \div 3 = \mathbf{60}$ km/h.

Combined rate $= \frac{1}{4} + \frac{1}{12} = \frac{1}{3}$ fence per hour, so the whole fence takes $\mathbf{3}$ hours. (Adding the times, $4 + 12 = 16$, cannot be right: together they must be faster than Ali alone.)

Rate $= \frac{3}{8}$ tank per hour. Time for $1$ tank $= 1 \div \frac{3}{8} = \frac{8}{3} = \mathbf{2\tfrac{2}{3}}$ hours. Or: each $\frac{1}{4}$ tank takes $\frac{2}{3}$ h, and four quarters take $\frac{8}{3}$ h. ($1\frac{1}{3}$ treats $\frac{3}{4}$ as the hourly rate; $1\frac{1}{2}$ is $\frac{3}{4} \times 2$; $2\frac{3}{4}$ just adds the two numbers.)

$10\%$ of $60$ is $6$ and $5\%$ is $3$, so $15\%$ is $6 + 3 = \mathbf{9}$. (Or $0.15 \times 60 = 9$.)

Sale price $= 60 - 15 = 45$. Tax $= 0.06 \times 45 = 2.70$. Final price $= 45 + 2.70 = \mathbf{47.70}$ dollars. (Tax on the original $60 would be $3.60 and give a wrong total of $48.60.)

$5\%$ of $80 = 4$, so the new price is $80 + 4 = 84$ dollars: $84. ($85 adds $5$ instead of $5\%$; $4 is only the increase; $88 is a $10\%$ increase.)