Advanced Math 7 — every item

$49 < 50 < 64$ gives $7 < \sqrt{50} < 8$. The smaller whole number is $\mathbf{7}$ (and $\sqrt{50} \approx 7.07$, just above it).

$4{,}560{,}000 = 4.56 \times 10^6$, so $a = \mathbf{4.56}$. ($45.6 \times 10^5$ is equal but is not scientific notation, because $45.6 \ge 10$.)

$\sqrt{49} = 7$, $0.\overline{3} = \frac{1}{3}$, and $-\frac{5}{8}$ are all fractions of integers, so the last list is all rational: $\mathbf{\sqrt{49},\ 0.\overline{3},\ -\tfrac{5}{8}}$. (Each other list contains an irrational: $\sqrt{50}$; $\pi$ — note $\frac{22}{7}$ and $3.14$ are rational approximations, not $\pi$; $\sqrt{2}$ and $\sqrt{8}$.)

$\sqrt{\frac{16}{25}} = \frac{4}{5}$, a fraction of integers, so it is rational — the third statement is true. ($\frac{22}{7}$ is only an approximation of $\pi$; $\pi$ is irrational. $0.\overline{3} = \frac{1}{3}$ is rational: every repeating decimal is a fraction.)

$3 < \sqrt{10} < 4$ and $9 < \sqrt{90} < 10$, so $n$ can be $4, 5, 6, 7, 8, 9$: $\mathbf{6}$ integers. ($80 = 90 - 10$ counts the wrong thing; $7$ wrongly includes $3$ or $10$, which lie outside the range.)

The $x$-term is $-3x$, so the coefficient is $\mathbf{-3}$. ($5$ is the coefficient of $x^2$; $7$ is the constant term.)

Twice $n$ is $2n$; seven less than that is $\mathbf{2n - 7}$. ($7 - 2n$ reads the words in order and gets the subtraction backwards.)

(iii): $2(x + 3) = 2x + 6$ is true for every $x$ (expand the left side), so it is an identity. (i) is true only at $x = 3$ — an equation to solve. (ii) is a formula. (iv) is an expression. Answer: (iii).

$\mathbf{\tfrac{1}{2}x - 4}$ has no equals sign, so it is an expression. $x = 5$ is an equation (its solution is $5$), and so are $2x + 3 = 9$ and $y = 3x$.

$(-2)^3 - 2(-2)^2 + 3(-2) - 4 = -8 - 8 - 6 - 4 = \mathbf{-26}$. (Reading $-2x^2$ as $(-2x)^2 = 16$ gives $-8 + 16 - 6 - 4 = -2$, the classic sign error.)

$\frac{2}{4} + \frac{3}{4} - \frac{8}{4} = -\frac{3}{4}$, so the expression is $-\frac{3}{4}x$ and the coefficient is $\mathbf{-\tfrac{3}{4}}$.

$(2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15 = \mathbf{2x^2 + 7x - 15}$. Check at $x = 1$: $(-1)(6) = -6$ and $2 + 7 - 15 = -6$. ✓

$xy + 2x + 3y + 6 = x(y + 2) + 3(y + 2) = \mathbf{(x + 3)(y + 2)}$. Check by expanding: $xy + 2x + 3y + 6$. ✓

$-2(x - 5) = -2x + 10$: $\mathbf{-2x + 10}$. ($-2x - 10$ forgets that $-2 \times -5$ is positive; $2x - 10$ flips the wrong sign; $-2x - 5$ never multiplies the $5$.)

$(2x + 3)(x - 4) = 2x^2 - 5x - 12$, so $a + b + c = 2 - 5 - 12 = \mathbf{-15}$. Shortcut: at $x = 1$ the product is $(5)(-3) = -15$. ($-10$ comes from $2x^2 - 12$, the “first times first, last times last” error.)

$5x - 7 = 3x + 9 \Rightarrow 2x = 16 \Rightarrow x = \mathbf{8}$. Check: $40 - 7 = 33$ and $24 + 9 = 33$. ✓

$6 \times \frac{x + 5}{3} = 2(x + 5)$ and $6 \times \frac{x - 1}{2} = 3(x - 1)$, so $2x + 10 = 3x - 3$ and $x = \mathbf{13}$. Check: $\frac{18}{3} = 6 = \frac{12}{2}$. ✓

$3b + 8 = 2b + 16 \Rightarrow b = \mathbf{8}$. Ben is $8$ and Maya is $24$; in $8$ years they are $16$ and $32$, and $32 = 2 \times 16$. ✓

$12\left(\frac{x}{3} + 2\right) = 4x + 24$ and $12\left(\frac{x}{4} + 5\right) = 3x + 60$, so $x = \mathbf{36}$. Check: $12 + 2 = 14$ and $9 + 5 = 14$. ✓ ($3$ comes from $4x + 2 = 3x + 5$, forgetting to multiply the constants; $84$ adds $60 + 24$ instead of subtracting.)

$2(3w + 3) = 48 \Rightarrow 3w + 3 = 24 \Rightarrow w = 7$. Length $= 2(7) + 3 = 17$. Area $= 7 \times 17 = \mathbf{119}$ cm². ($238 = 14 \times 17$ uses $2w$ as the width; $144 = 12^2$ treats the rectangle as a square with perimeter $48$.)

$2$ h $15$ min $= 2.25$ h. Distance $= 80 \times 2.25 = \mathbf{180}$ km. (Using $2.15$ hours is the common slip: minutes are not decimals.)

Total distance $= 120$ km. Total time $= 2 + 1 = 3$ h. Average speed $= \frac{120}{3} = \mathbf{40}$ km/h — not $45$, the mean of the two speeds, because twice as long was spent at the slow speed.

Area $= \frac{1}{2}(10)(20) + 30 \times 20 = 100 + 600 = \mathbf{700}$ m. (Using $20 \times 40 = 800$ ignores that the tram was still speeding up for the first $10$ s.)

With $120$ km each way: $240$ km in $3 + 2 = 5$ h gives $\mathbf{48}$ km/h. ($50$ is the trap: averaging the speeds is only right when the times are equal. More time is spent at $40$, so the answer must be below $50$.)

Gap at 9:30: $6$ km. Closing speed: $8$ km/h. Time: $\frac{6}{8}$ h $= 45$ min, so the cyclist catches up at 10:15 a.m. Check: walker $2.25 \times 4 = 9$ km; cyclist $0.75 \times 12 = 9$ km. ✓ (10:00 divides the $6$ km gap by $12$ instead of by the closing speed $8$.)

$I = Prt = 500 \times 0.04 \times 3 = \mathbf{60}$ dollars. (The account then holds $500 + 60 = 560$ dollars.)

$0.8P = 72 \Rightarrow P = 72 \div 0.8 = \mathbf{90}$. Check: $20\%$ of $90$ is $18$, and $90 - 18 = 72$. ✓

Interest needed $= 300$. Per year: $1{,}200 \times 0.05 = 60$. Years $= 300 \div 60 = \mathbf{5}$. Or with $A = P(1 + rt)$: $1{,}500 = 1{,}200(1 + 0.05t) \Rightarrow 1.25 = 1 + 0.05t \Rightarrow t = 5$.

One quarter of the original is $20$, so the original is $4 \times 20 = \mathbf{80}$ dollars. Check: $25\%$ of $80$ is $20$; $80 - 20 = 60$. ✓ ($75 adds $25\%$ of $60 — the wrong base. $45 discounts the sale price again.)

$800 + 48t = 1{,}000 + 40t \Rightarrow 8t = 200 \Rightarrow t = \mathbf{25}$ years. Check: A holds $800 + 1{,}200 = 2{,}000$ and B holds $1{,}000 + 1{,}000 = 2{,}000$. ✓ ($5$ divides the $200 gap by B's $40$ per year instead of by the $8$ per year difference.)

$180 - 37 = \mathbf{143}$ degrees. ($53$ is the complement, not the supplement.)

Exterior angle $= \frac{360}{9} = 40^\circ$, so interior $= 180 - 40 = \mathbf{140}$ degrees. Same by the sum: $\frac{7 \times 180}{9} = \frac{1260}{9} = 140$. ✓

$\angle ABC = 112 - 47 = \mathbf{65}$ degrees. Check by the long route: $\angle ACB = 180 - 112 = 68$, and $47 + 65 + 68 = 180$. ✓

Co-interior angles are supplementary: $\angle DQP = 180 - 64 = \mathbf{116}$ degrees. ($64^\circ$ would be right for alternate or corresponding angles, not co-interior; $26^\circ$ subtracts from $90^\circ$; $296^\circ$ subtracts from $360^\circ$.)

$5x + 100 = 540 \Rightarrow x = 88$. Largest $= 88 + 40 = \mathbf{128}$ degrees. ($92$ uses $360^\circ$ as the interior sum — the exterior-angle trap; $108$ is a regular pentagon's angle; $118$ stops at $x + 30$.)

$T_{10} = 4 + (10 - 1) \times 7 = 4 + 63 = \mathbf{67}$. (Adding $d$ ten times, $4 + 70 = 74$, gives the $11$th term.)

$T_n = 7 + 5(n - 1) = 7 + 5n - 5 = \mathbf{5n + 2}$. Check: $n = 1 \to 7$, $n = 2 \to 12$, $n = 3 \to 17$. ✓ ($n + 5$ is not a formula for $T_n$; it only restates the step.)

Figure $8$ is the $9$th triangular number: $\frac{9 \times 10}{2} = \mathbf{45}$. By differences: $15 + 6 = 21$, $+7 = 28$, $+8 = 36$, $+9 = 45$. ✓

$d = 4$ gives $T_n = 4n + b$, and $T_1 = 3$ forces $b = -1$. So $\mathbf{T_n = 4n - 1}$: $3, 7, 11, 15$. ✓ ($n + 4$ gives $5, 6, 7$ — the “add the difference” trap; $4n + 3$ uses $a + dn$ and gives $7, 11, 15$, one term early.)

$3n + 4 < 100 \Rightarrow n < 32$, so $n = 1, 2, \ldots, 31$: $\mathbf{31}$ terms. Check: $T_{31} = 97 < 100$ and $T_{32} = 100$, which is not less than $100$. ($32$ counts the term equal to $100$.)

$m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = \mathbf{3}$.

$c = 3$ and $m = \frac{1 - 3}{4 - 0} = -\frac{1}{2}$, so $\mathbf{y = -\tfrac{1}{2}x + 3}$. Check $(4, 1)$: $-\frac{1}{2}(4) + 3 = 1$. ✓

$h = -4t + 60$: the tank starts at $60$ cm and falls $4$ cm per minute — the first statement. (It is empty when $60 - 4t = 0$, at $t = 15$ minutes, not $4$.)

$m = \frac{8 - 2}{4 - 1} = \frac{6}{3} = \mathbf{2}$. ($\frac{1}{2}$ is $\frac{\Delta x}{\Delta y}$, upside down; $-2$ subtracts in opposite orders; $6$ is just the rise.)

$m = \frac{2 - 11}{1 - (-2)} = \frac{-9}{3} = -3$. From $(1, 2)$ to $(4, k)$ the run is $3$, so the rise is $-9$: $k = 2 - 9 = \mathbf{-7}$. Check: $\frac{-7 - 2}{4 - 1} = -3$. ✓

$3x < 15 \Rightarrow x < 5$. The largest integer less than $5$ is $\mathbf{4}$ ($5$ gives $20 < 20$, which is false).

$5 - 2x > 11 \Rightarrow -2x > 6 \Rightarrow x < -3$. The largest integer less than $-3$ is $\mathbf{-4}$. Check: $5 - 2(-4) = 13 > 11$ ✓, while $x = -3$ gives $11 > 11$, false.

$-1$ is included and $4$ is not: $\mathbf{-1 \le x < 4}$.

$-2x > 6 \Rightarrow x < -3$: $\mathbf{x < -3}$. Check $x = -10$: $20 + 7 = 27 > 13$ ✓. ($x > -3$ forgets to reverse; test $x = 0$: $7 > 13$ is false.)

$-6 \le -2x < 8 \Rightarrow 3 \ge x > -4$, so $x \in \{-3, -2, -1, 0, 1, 2, 3\}$: $\mathbf{7}$ integers. Check the ends: $x = 3$ gives $7 - 6 = 1$ ✓; $x = -4$ gives $15$, which is not $< 15$.

$C = \pi d = 14\pi$, so $k = \mathbf{14}$. ($28$ treats $14$ as the radius.)

$A = \frac{120}{360} \times \pi \times 81 = 27\pi$, so $k = \mathbf{27}$. (The arc length would be $\frac{1}{3} \times 18\pi = 6\pi$ — a different question.)

Quarter circle $= \frac{1}{4}\pi(8^2) = 16\pi \approx 50.24$. Remaining $= 64 - 50.24 = \mathbf{13.76}$ cm².

$A = \frac{1}{6} \times 36\pi = \mathbf{6\pi}$ cm². ($2\pi$ is $\frac{1}{6} \times 2\pi r$ — the arc length, not the area; $\frac{3}{2}\pi$ uses $r = 3$, half the radius; $36\pi$ is the whole circle.)

$r = 6$, so the corners have area $144 - 36\pi$ and $a + b = 144 + 36 = \mathbf{180}$. ($288$ uses $r = 12$, the diameter; $156$ uses $12\pi$ for the circle; $144$ forgets the circle.)