Secondary 4 Additional Mathematics Numbers Ratio Proportion Quiz

Secondary 4 Additional Mathematics Quiz - Numbers Ratio Proportion

ANSWER KEY

Total Marks: 40

---

Section A: Direct Proportion and Inverse Proportion

Question 1 [2 marks]

(a) Since $y \propto x^2$, we write $y = kx^2$ for some constant $k$.

Substituting $y = 72$ and $x = 6$: $72 = k \times 6^2 = 36k$ $k = 2$

Therefore, $y = 2x^2$ [1 mark]

(b) When $x = 10$: $y = 2 \times 10^2 = 2 \times 100 = \boxed{200}$ [1 mark]

Common mistake: Forgetting to square the $x$ value, or solving $72 = 6k$ to get $k = 12$.

---

Question 2 [2 marks]

(a) Since $p \propto \frac{1}{\sqrt{q}}$, we write $p = \frac{k}{\sqrt{q}}$.

Substituting $p = 4$ and $q = 9$: $4 = \frac{k}{\sqrt{9}} = \frac{k}{3}$ $k = 12$

Therefore, $p = \frac{12}{\sqrt{q}}$ [1 mark]

(b) When $p = 12$: $12 = \frac{12}{\sqrt{q}}$ $\sqrt{q} = 1$ $q = \boxed{1}$ [1 mark]

---

Question 3 [2 marks]

Since $P \propto \frac{1}{V}$, we write $P = \frac{k}{V}$.

When $V = 2$, $P = 300$: $300 = \frac{k}{2}$ $k = 600$

So $P = \frac{600}{V}$

When $V = 5$: $P = \frac{600}{5} = \boxed{120 \text{ Pa}}$ [2 marks]

Alternative: Using $P_1V_1 = P_2V_2$ (constant product for inverse proportion): $300 \times 2 = P_2 \times 5$ $P_2 = \frac{600}{5} = 120 \text{ Pa}$

---

Question 4 [2 marks]

Since $A \propto r^2$, we write $A = kr^2$.

When $r = 3$, $A = 36\pi$: $36\pi = k \times 9$ $k = 4\pi$

So $A = 4\pi r^2$ (this is the standard formula $4\pi r^2$)

When $r = 5$: $A = 4\pi \times 25 = \boxed{100\pi \text{ cm}^2}$ [2 marks]

---

Question 5 [2 marks]

(a) Since $T \propto \frac{1}{s}$, we have $T = \frac{k}{s}$.

When $s = 60$, $T = 2$: $2 = \frac{k}{60}$ $k = 120$

So $T = \frac{120}{s}$

When $s = 50$: $T = \frac{120}{50} = 2.4 \text{ hours} = \boxed{2 \text{ hours } 24 \text{ minutes}}$ [1 mark]

(b) For $T = 1.5$: $1.5 = \frac{120}{s}$ $s = \frac{120}{1.5} = \boxed{80 \text{ km/h}}$ [1 mark]

---

Question 6 [2 marks]

$y \propto \frac{(x-2)}{\sqrt{z}}$, so $y = \frac{k(x-2)}{\sqrt{z}}$

When $x = 5$, $z = 9$, $y = 8$: $8 = \frac{k(5-2)}{\sqrt{9}} = \frac{3k}{3} = k$

So $k = 8$ and $y = \frac{8(x-2)}{\sqrt{z}}$

When $x = 8$, $z = 4$: $y = \frac{8(8-2)}{\sqrt{4}} = \frac{8 \times 6}{2} = \frac{48}{2} = \boxed{24}$ [2 marks]

Common mistake: Forgetting to subtract 2 from $x$ before multiplying.

---

Question 7 [2 marks]

$R \propto \frac{1}{d^2}$, so $R = \frac{k}{d^2}$

When $d = 2$, $R = 10$: $10 = \frac{k}{4}$ $k = 40$

So $R = \frac{40}{d^2}$

When $d = 4$: $R = \frac{40}{16} = \frac{5}{2} = \boxed{2.5 \text{ ohms}}$ [2 marks]

Teaching note: When diameter doubles, resistance becomes $\frac{1}{4}$ (inverse square law). This is important in electrical engineering for choosing wire thickness.

---

Section B: Ratio and Its Applications

Question 8 [2 marks]

$a : b = 3 : 5$ and $b : c = 5 : 8$

Since $b$ is 5 in both ratios, we can combine directly:

$a : b : c = \boxed{3 : 5 : 8}$ [2 marks]

---

Question 9 [2 marks]

(a) $x : y = 2 : 3 = 8 : 12$ (multiplying by 4) $y : z = 4 : 5 = 12 : 15$ (multiplying by 3)

So $x : y : z = \boxed{8 : 12 : 15}$ [1 mark]

(b) If $z = 15$ parts $= 30$, then 1 part $= 2$

$x = 8$ parts $= 8 \times 2 = \boxed{16}$ [1 mark]

---

Question 10 [2 marks]

Let original boys $= 7k$ and original girls $= 5k$.

After adding 6 to each: $\frac{7k + 6}{5k + 6} = \frac{5}{4}$

Cross-multiplying: $4(7k + 6) = 5(5k + 6)$ $28k + 24 = 25k + 30$ $3k = 6$ $k = 2$

Original total $= 7k + 5k = 12k = 12 \times 2 = \boxed{24}$ [2 marks]

Common mistake: Adding 6 only to boys, or using wrong ratio after addition.

---

Question 11 [2 marks]

(a) Scale: 1 cm represents 50,000 cm = 0.5 km

Actual distance $= 8 \times 0.5 = \boxed{4 \text{ km}}$ [1 mark]

(b) Actual dimensions: $3.2 \times 0.5 = 1.6$ km and $2.5 \times 0.5 = 1.25$ km

Actual area $= 1.6 \times 1.25 = \boxed{2 \text{ km}^2}$ [1 mark]

Alternative using area scale factor: Area scale $= (1 : 50\,000)^2 = 1 : 2.5 \times 10^9$

Map area $= 3.2 \times 2.5 = 8$ cm$^2$

Actual area $= 8 \times 2.5 \times 10^9$ cm$^2$ $= 2 \times 10^{10}$ cm$^2$ $= 2$ km$^2$

---

Question 12 [2 marks]

Let present ages be $4k$, $5k$, $6k$.

In 8 years: $\frac{4k + 8}{5k + 8} = \frac{6}{7}$

Cross-multiplying: $7(4k + 8) = 6(5k + 8)$ $28k + 56 = 30k + 48$ $8 = 2k$ $k = 4$

Charles's present age $= 6k = 6 \times 4 = \boxed{24}$ [2 marks]

---

Question 13 [2 marks]

(a) Total parts $= 5 + 3 + 2 = 10$

Mass of zinc $= \frac{3}{10} \times 80 = \boxed{24 \text{ kg}}$ [1 mark]

(b) Original masses: Copper $= 40$ kg, Zinc $= 24$ kg, Tin $= 16$ kg

After adding 10 kg tin: Tin becomes 26 kg

New ratio: $40 : 24 : 26 = \boxed{20 : 12 : 13}$ [1 mark]

---

Question 14 [2 marks]

(a) Difference between Aaron's and Brenda's shares: $3x - 2x = x = 24$

So $x = 24$ [1 mark]

(b) Colin's share = 5 = \5$? No wait — let me re-read: the ratio is$2x : 3x : 5$, where the third term is constant 5, not$5x$.

This means Colin's share is $5$ in ratio units, but these are not monetary values directly.

Actually, re-interpreting: The ratio terms are $2x$, $3x$, and $5$ (where 5 is a constant, not multiplied by $x$).

Total "parts" interpretation is tricky here. Let's use the difference: $3x - 2x = x = 24$

So the ratio values are $48 : 72 : 5$ — but this gives Colin only 5, which seems inconsistent with $x = 24$ being monetary.

Re-reading: The difference is $24, and this equals $(3x - 2x) = x$ parts. So 1 part (in ratio terms where $x$ represents a scaling) corresponds to $24... but the third term is fixed at 5.

Actually: if total is divided in ratio $2x : 3x : 5$, then the actual shares are proportional to these. Let total be $T$.

Aaron gets $\frac{2x}{2x + 3x + 5} \times T$, etc.

Difference condition: Aaron and Brenda differ by $x$ parts, and this equals $24 in value. But we need to be careful what "parts" mean.

Let each "unit" of ratio be worth $k$ dollars. Then:

• Aaron: $2xk$
• Brenda: $3xk$
• Colin: $5k$

Brenda - Aaron $= 3xk - 2xk = xk = 24$

So $xk = 24$. We need another relation to find $x$ and $k$ separately... but we only have one equation.

Given the problem as stated with $x$ as the variable to find: if ratio is $2x : 3x : 5$, then difference is $x$ (ratio units). If this equals 24 (in same units), then $x = 24$ ratio-units.

But then Colin's share $= 5$ ratio-units $= \frac{5}{x} \times 24$... this gets convoluted.

Clarification: The standard interpretation is that $x$ is a common multiplier for the first two terms, with third fixed. The "value of $x$" refers to the variable itself, and since difference is $x$ parts = 24, then $x = 24$.

Colin's share: The ratio is $2(24) : 3(24) : 5 = 48 : 72 : 5$.

Total parts $= 125$. Colin's fraction $= \frac{5}{125} = \frac{1}{25}$.

But we need total value... Actually, let's use: if $x$ parts corresponds to 24 in value for the difference, and Colin has 5 (fixed ratio units), then:

Value per ratio unit $= \frac{24}{x} = \frac{24}{24} = 1$ (in some currency unit).

Actually simpler: Since $xk = 24$ and we found $x = 24$, then $k = 1$.

So Colin's share = 5k = 5 \times 1 = \5$? This seems small.

Let me re-approach: The ratio $2x : 3x : 5$ means the shares are $2x \cdot k$, $3x \cdot k$, $5k$ for some constant $k$.

Difference: $3xk - 2xk = xk = 24$.

We have two unknowns ($x$ and $k$) in one equation. The problem asks for "value of $x$" — this suggests $x$ can be determined uniquely, implying $k = 1$ or the ratio terms are actual currency amounts.

If ratio terms are actual amounts: $2x$, $3x$, $5$ in dollars. Then $3x - 2x = x = 24$, so $x = 24$.

Colin's share = \boxed{\5} or if we check: shares are \48, $72, $5 — but these don't have a sensible total.

Given context of such problems, likely: ratio is $2x : 3x : 5$ where the whole ratio uses $x$ as scale, and "5" means $5x$ was intended, or there's a typo in my generation.

Standard form would be $2 : 3 : 5$ with $x$ as common multiplier, giving $2x : 3x : 5x$.

Assuming standard interpretation $2x : 3x : 5x$:

Difference: $3x - 2x = x = 24$

(a) $x = \boxed{24}$ [1 mark]

(b) Colin's share = 5x = 5 \times 24 = \boxed{\120}$ [1 mark]

Note to student: The original ratio notation $2x : 3x : 5$ is ambiguous. Standard problems use $2x : 3x : 5x$. If the third term truly has no $x$, the problem has insufficient constraints for a unique numerical answer.

---

Section C: Proportionality in Real-World Contexts and Combined Variation

Question 15 [2 marks]

(a) $E \propto v^2$, so $E = kv^2$.

When $v = 10$, $E = 250$: $250 = k \times 100$ $k = 2.5$

So $E = 2.5v^2$

When $v = 16$: $E = 2.5 \times 256 = \boxed{640 \text{ J}}$ [1 mark]

(b) If $v$ increases by 20%, new $v = 1.2v$, so new $E = k(1.2v)^2 = 1.44kv^2 = 1.44E$

Percentage increase $= (1.44 - 1) \times 100\% = \boxed{44\%}$ [1 mark]

Teaching note: This shows kinetic energy grows faster than velocity. A 20% speed increase yields 44% energy increase, explaining why high-speed crashes are disproportionately dangerous.

---

Question 16 [2 marks]

$T \propto \sqrt{\ell}$, so $T = k\sqrt{\ell}$

When $\ell = 100$, $T = 2$: $2 = k\sqrt{100} = 10k$ $k = 0.2$

So $T = 0.2\sqrt{\ell}$

When $T = 3$: $3 = 0.2\sqrt{\ell}$ $\sqrt{\ell} = 15$ $\ell = \boxed{225 \text{ cm}}$ [2 marks]

---

Question 17 [2 marks]

$F \propto \frac{m_1 m_2}{d^2}$, so $F = \frac{Gm_1 m_2}{d^2}$

Given: $G = 6.67 \times 10^{-11}$ when $m_1 = m_2 = d = 1$.

For $m_1 = 2$, $m_2 = 3$, $d = 2$: $F = 6.67 \times 10^{-11} \times \frac{2 \times 3}{2^2} = 6.67 \times 10^{-11} \times \frac{6}{4}$ $= 6.67 \times 10^{-11} \times 1.5 = \boxed{1.00 \times 10^{-10} \text{ N}} \text{ (to 3 s.f.)}$ [2 marks]

---

Question 18 [2 marks]

From the expected graph (straight line through origin with points (4,16) and (9,36)):

(a) Straight line through origin $\Rightarrow$ $y \propto x^2$ (direct proportion between $y$ and $x^2$)

So $y$ is directly proportional to $x^2$, or $y \propto x^2$ [1 mark]

(b) Gradient $= \frac{36 - 16}{9 - 4} = \frac{20}{5} = 4$

Equation: $y = 4x^2$ [1 mark]

Visual check: The line passes through $(4, 16)$: $4 \times 4 = 16$ ✓ and $(9, 36)$: $4 \times 9 = 36$ ✓

---

Question 19 [2 marks]

Taking logs: $\log_{10} y = \log_{10} A + n\log_{10} x$

From expected graph with points $(0, 1)$ and $(2, 5)$:

(a) Gradient $= n = \frac{5 - 1}{2 - 0} = \frac{4}{2} = \boxed{2}$ [½ mark]

When $\log_{10} x = 0$, $\log_{10} y = 1$, so $\log_{10} A = 1$, thus $A = \boxed{10}$ [½ mark]

(b) Equation: $y = 10x^2$

When $x = 5$: $y = 10 \times 25 = \boxed{250}$ [1 mark]

---

Question 20 [2 marks]

(a) $C = a + bn$ where $a$ is fixed cost, $b$ is cost per book.

When $n = 500$, $C = 12000$: $12000 = a + 500b$

When $n = 800$, $C = 18000$: $18000 = a + 800b$

Subtracting: $6000 = 300b$ $b = 20$

Substituting: $a = 12000 - 500 \times 20 = 12000 - 10000 = 2000$

$C = 2000 + 20n$ [1 mark]

(b) When $n = 1000$: $C = 2000 + 20 \times 1000 = 2000 + 20000 = \boxed{\$22,000}$ [½ mark]

(c) For very large $n$: The model assumes fixed costs remain constant and unit cost doesn't change with scale. In reality, bulk discounts may reduce unit cost, or additional fixed costs (larger premises, more staff) may be needed. The linear model is only valid for a limited range. [½ mark]

---

END OF ANSWER KEY