Advanced Math 4 — every item

Regroup $6{,}004$ as $5{,}000 + 900 + 90 + 14$. Then $14 - 7 = 7$, $9 - 5 = 4$, $9 - 3 = 6$, $5 - 1 = 4$, giving $\mathbf{4{,}647}$. Check: $4{,}647 + 1{,}357 = 6{,}004$.

Remove the extra $1{,}800$ from the total: $9{,}400 - 1{,}800 = 7{,}600$. That is two equal shares, so Sunday $= 7{,}600 \div 2 = \mathbf{3{,}800}$. Saturday is $3{,}800 + 1{,}800 = 5{,}600$, and $5{,}600 + 3{,}800 = 9{,}400$.

Liam has $4{,}000 - 2{,}350 = 1{,}650$ stickers. The difference is $2{,}350 - 1{,}650 = \mathbf{700}$. ($1{,}650$ is Liam's amount, not the difference; $800$ is a regrouping slip; $5{,}650$ adds instead of comparing.)

Factor pairs of $24$: $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$. That is $1, 2, 3, 4, 6, 8, 12, 24$ — $\mathbf{8}$ factors.

Pairs: $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, $6 \times 6$. Factors: $1, 2, 3, 4, 6, 9, 12, 18, 36$. Sum: $1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = \mathbf{91}$.

$53$ has no factor among $2, 3, 5, 7$, so it is prime: $\mathbf{53}$. ($1$ has only one factor, so it is not prime; $51 = 3 \times 17$ and $91 = 7 \times 13$ are odd but composite.)

$308 \times 7 = 300 \times 7 + 8 \times 7 = 2{,}100 + 56 = \mathbf{2{,}156}$.

Capacity: $24 \times 36 = 600 + 120 + 120 + 24 = 864$. Books that still fit: $864 - 250 = \mathbf{614}$. Estimate check: $25 \times 36 = 900$, and $900 - 250 = 650$, close to $614$.

$34 \times 26 = 34 \times 6 + 34 \times 20 = 204 + 680 = \mathbf{884}$. ($272$ uses $68$ instead of $680$ — the partial product landed in the wrong place; $804$ drops the $4 \times 20$ box of the area model; $900$ is only the estimate.)

$2{,}350 \div 6 = 391$ remainder $4$, because $6 \times 391 = 2{,}346$ and $2{,}350 - 2{,}346 = 4$. The remainder is $\mathbf{4}$.

$27 \div 6 = 4$ R $3$. The $3$ leftover pancakes are split six ways: $\frac{3}{6} = \frac{1}{2}$ each. So each friend gets $4\frac{1}{2} = \mathbf{\frac{9}{2}}$ pancakes.

$236 \div 9 = 26$ R $2$. Twenty-six vans leave $2$ children behind, so one more van is needed: $\mathbf{27}$ vans. ($26$ ignores the leftover children; $26 \text{ R } 2$ is a division result, not a number of vans; $2$ is the remainder; $28$ adds one van too many.)

$4 \times 3 = 12$, so multiply the top by $3$ too: $\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$. The missing numerator is $\mathbf{9}$.

Order from least to greatest: $\frac{1}{2},\ \frac{5}{8},\ \frac{2}{3},\ \frac{3}{4},\ \frac{9}{8}$ (since $\frac{15}{24} < \frac{16}{24} < \frac{18}{24}$). The middle one is $\mathbf{\frac{2}{3}}$.

$\frac{3}{4} = \frac{6}{8} > \frac{5}{8}$, so $\mathbf{\frac{3}{4} > \frac{5}{8}}$ is true. ($\frac{3}{5} > \frac{3}{4}$ is the denominator trap — fifths are smaller than quarters; $\frac{5}{12} < \frac{1}{2}$; $\frac{2}{3} = \frac{16}{24} > \frac{15}{24} = \frac{5}{8}$; $\frac{7}{6}$ is greater than $1$.)

$\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \mathbf{\frac{11}{12}}$.

$4\frac{1}{4} - 1\frac{5}{6} = 4\frac{3}{12} - 1\frac{10}{12} = 3\frac{15}{12} - 1\frac{10}{12} = 2\frac{5}{12}$. Left: $\mathbf{\frac{29}{12}}$ m, that is $2\frac{5}{12}$ m. Check: $1\frac{10}{12} + 2\frac{5}{12} = 3\frac{15}{12} = 4\frac{3}{12}$.

$\frac{2}{5} + \frac{1}{4} = \frac{8}{20} + \frac{5}{20} = \mathbf{\frac{13}{20}}$. ($\frac{3}{9}$ adds tops and bottoms — it is even less than $\frac{2}{5}$ alone; $\frac{3}{20}$ subtracts; $\frac{11}{20}$ renames $\frac{1}{4}$ as $\frac{3}{20}$ by mistake.)

$6 \times \frac{2}{3} = \frac{6 \times 2}{3} = \frac{12}{3} = \mathbf{4}$.

Blue: $\frac{5}{6} \times 36 = 30$. With stars: $\frac{2}{5} \times 30 = \mathbf{12}$.

One quarter of $20$ is $5$, so three quarters is $3 \times 5 = \mathbf{15}$ green marbles. ($3$ counts three marbles instead of three parts; $5$ is only one part; $60$ multiplies by $3$ without dividing by $4$.)

Shorter than $5$ inches: $4\frac{1}{2}$ ($2$ pencils) and $4\frac{3}{4}$ ($4$ pencils). $2 + 4 = \mathbf{6}$ pencils.

$2 \times 4\frac{1}{2} = 9$; $4 \times 4\frac{3}{4} = 19$; $3 \times 5 = 15$; $1 \times 5\frac{1}{4} = 5\frac{1}{4}$; $2 \times 5\frac{1}{2} = 11$. Total: $9 + 19 + 15 + 5\frac{1}{4} + 11 = 59\frac{1}{4} = \mathbf{\frac{237}{4}}$ inches.

$6\frac{3}{8} + 4\frac{4}{8} + 11\frac{7}{8} = 21\frac{14}{8} = 22\frac{6}{8} = \mathbf{22\frac{3}{4}}$ inches. ($10$ is the number of crayons, not a length; $6\frac{3}{4}$ adds the three lengths once each; $21\frac{3}{4}$ drops the regrouped whole.)

$3$ km $= 3{,}000$ m, so $3$ km $450$ m $= 3{,}000 + 450 = \mathbf{3{,}450}$ m.

Jug $= 2{,}000$ mL. Poured: $6 \times 250 = 1{,}500$ mL. Left: $2{,}000 - 1{,}500 = \mathbf{500}$ mL.

13:50 $\to$ 15:50 is $2$ h; 15:50 $\to$ 16:00 is $10$ min; 16:00 $\to$ 16:25 is $25$ min. Total 2 h 35 min. ($3$ h $25$ min counts 13 to 16 as three full hours and ignores that the trip starts at 50 minutes past; $2$ h $25$ min forgets the $10$ minutes before 16:00; $2$ h $45$ min adds $10$ minutes twice.)

$P = 2(9 + 4) = 2 \times 13 = \mathbf{26}$ cm. (The area, $9 \times 4 = 36$ cm², is a different question.)

Whole rectangle: $12 \times 7 = 84$ m². Cut-out corner: $(12 - 5) \times (7 - 3) = 7 \times 4 = 28$ m². Patio: $84 - 28 = \mathbf{56}$ m². Check by splitting instead: a $12 \times 3$ strip along the bottom ($36$) plus a $5 \times 4$ block above it ($20$) gives $56$.

The new rectangle is $12$ cm by $8$ cm, so its area is $12 \times 8 = \mathbf{96}$ cm² — four times the original $24$. ($48$ doubles the old area, which is the trap; $40$ is the new perimeter, not an area; $24$ is the original area.)

$2 + 0.3 + 0.00 + 0.007 = \mathbf{2.307}$. Writing $2.37$ would put the $7$ in the hundredths place.

As hundredths: $0.60 < 0.68 < 0.70 < 0.75$. The longest is $\frac{3}{4}$ m and the second longest is $\mathbf{0.7}$ m.

$0.8 = 0.80$ and $80$ hundredths is more than $79$ hundredths, so $\mathbf{0.8 > 0.79}$. ($0.27 < 0.30$ and $0.36 < 0.40$ — comparing $27$ with $3$ ignores place value; $0.50 = 0.5$, the zero adds nothing; $0.07 < 0.10$.)

$6.25 + 3.90 = \mathbf{10.15}$. Estimate: $6 + 4 = 10$.

Spent: $13.45 + 2.80 = 16.25$. Left: $20.00 - 16.25 = \mathbf{3.75}$ dollars. Mental check: $16.25 + 3.75 = 20$.

$5.20 + 0.37 = \mathbf{5.57}$ L. ($0.89$ and $8.9$ come from lining up the last digits as if adding $52 + 37$; $5.47$ is a tenths slip.)

$24 \times 3 = 72$ tenths $= \mathbf{7.2}$. Estimate: $2 \times 3 = 6$, and $7.2$ is a bit more.

One piece: $5.4 \div 6 = 0.9$ m (since $54 \div 6 = 9$ tenths). Four pieces: $4 \times 0.9 = \mathbf{3.6}$ m.

$36 \times 5 = 180$ hundredths $= 1.80 = \mathbf{1.8}$ m. ($0.18$ places the point too far left, as if the answer had three decimal places; $18$ forgets the point entirely; $1.5$ multiplies $0.30$ instead of $0.36$.)

$115^\circ$ is more than $90^\circ$ and less than $180^\circ$, so it is obtuse. (Acute is under $90^\circ$; right is exactly $90^\circ$; straight is exactly $180^\circ$.)

$360 - (150 + 95) = 360 - 245 = \mathbf{115}$ degrees. (Using $180$ instead of $360$ would give a negative number, a sign that the wrong total was used.)

The first ray is on $0^\circ$ of the inner scale, so the inner reading is the measure: $\mathbf{40^\circ}$. It is acute, as the prompt says. ($140^\circ$ reads the wrong scale and would be obtuse; $50^\circ$ is $90 - 40$, a misread; $320^\circ$ is the reflex angle on the outside.)

Two lines join the midpoints of opposite sides and two are the diagonals: $\mathbf{4}$ lines of symmetry. (A rectangle that is not a square has only the first two; its diagonals do not work.)

The fold line cuts the $8$ cm sides into $4$ cm halves, so each piece is $6$ cm by $4$ cm. Perimeter $= 2(6 + 4) = \mathbf{20}$ cm. (Halving the original perimeter of $28$ gives $14$, which is wrong: the cut adds two new $6$ cm edges.)

A square has four right angles, so it meets the definition of a rectangle: every square is a rectangle. (Not every rectangle is a square — an $8$ by $6$ one is not; having equal sides is an extra property, not a reason to stop being a rectangle; every rectangle has four right angles, not two.)

Top $4$ + bottom $4$ + uprights $4$ = $\mathbf{12}$ edges. (A cuboid also has $6$ faces and $8$ vertices.)

From above you see the footprint: one square per stack. The stacks are $3, 2, 1, 1$ cubes tall, but that does not matter from above — there are $\mathbf{4}$ squares. (The front view would show $3 + 2 + 1 + 1 = 7$ squares.)

Front layer: $3 + 2 + 1 = 6$ cubes. Back layer, hidden behind it: another $6$. Total $6 + 6 = \mathbf{12}$. ($6$ counts only the visible front layer; $18$ treats every column as $3$ tall; $9$ is a miscount of one layer and a half.)