Mathematics

(d) none

$\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 2 \times 4 = 4\ \text{cm}^2$ This is not among options (a)-(c).

(c) 4

$A \subset U$, so $U \cap A = A$. The multiples of 4 below 20 are $4, 8, 12, 16$, hence $n(U \cap A) = 4$.

(b) 3

If the product of four consecutive integers equals one of them and that one is non-zero, the other three would need a product of 1, which is impossible for consecutive integers. So the integers must include 0, and then the product is $0$, equal to that integer. The largest such set is $0, 1, 2, 3$, so the largest value is 3.

(b) $x^2 + 5x + 5$

Pair the factors: $(x+1)(x+4) = x^2+5x+4,\quad (x+2)(x+3) = x^2+5x+6$ Let $t = x^2+5x+5$. Then the expression is $(t-1)(t+1)+1 = t^2$, so the square root is $x^2+5x+5$.

(d) 11.4 yrs

$\text{Average} = \frac{30 \times 11 + 20 \times 12}{50} = \frac{330 + 240}{50} = \frac{570}{50} = 11.4\ \text{yrs}$

(c) 2400

Men are $100 - 35 = 65\%$ of the population: $0.65 \times P = 1560 \Rightarrow P = \frac{1560}{0.65} = 2400$

(b) 39

Let the tens digit be $t$; the units digit is $3t$. Then $t + 3t = 12 \Rightarrow t = 3$, units digit $9$, so the number is 39.

(a) $x + 1$

$x^4 + x = x(x^3+1) = x(x+1)(x^2-x+1),\qquad x^2-3x-4 = (x-4)(x+1)$ The common factor is $x+1$.

(a) $\frac{1}{8}$

A cube has 12 edges, so $12a = 6 \Rightarrow a = \tfrac12$ cm. Volume $= a^3 = \tfrac18$ cm³.

(c) 2

$y^2 - 4y + 4 = (y-2)^2 = 0 \Rightarrow y = 2$ (a repeated root).

(a) primary

Data collected first-hand by the investigator for a specific purpose are primary data. Data taken from existing sources (books, reports, records) are secondary data.

(d) 15

Let the angle be $x$. Complementary angles sum to $90^\circ$, so $x + 5x = 90^\circ \implies 6x = 90^\circ \implies x = 15^\circ.$ Check: complement $= 75^\circ = 5 \times 15^\circ$.

(d) 3

Total work $= 7 \times 20 = 140$ worker-days. Fewer days need more workers (inverse proportion): $\text{workers needed} = \frac{140}{14} = 10.$ Additional workers $= 10 - 7 = 3$.

(c) degree 8

The highest power of a variable in $20 + x^8 - y^5 + x^8 + x^2$ is $x^8$. Combining like terms gives $20 + 2x^8 - y^5 + x^2$, whose degree is $8$.

(d) 70

"$m\%$ of $m$ is 49" gives $\frac{m}{100}\times m = 49 \implies m^2 = 4900 \implies m = 70.$

(c) 2

The series is geometric with $a = 32$ and $r = \frac{16}{32} = \frac{1}{2}$. $t_5 = ar^{4} = 32 \times \left(\frac{1}{2}\right)^4 = \frac{32}{16} = 2.$ Check: $32, 16, 8, 4, 2$.

(d) none

Since $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt2}$, $\frac{x+1}{\sqrt2} = 24 \implies x = 24\sqrt2 - 1 \approx 32.94.$ This matches none of 40, 50, 65.

(d) none

With $r = 0.2$ cm and $h = 4$ cm:

• Curved surface area $= 2\pi rh = 2\pi(0.2)(4) = 1.6\pi \approx 5.03\ \text{cm}^2$.
• Total surface area $= 2\pi r(r+h) = 2\pi(0.2)(4.2) = 1.68\pi \approx 5.28\ \text{cm}^2$.

Neither equals 49, 33 or 16 $\text{cm}^2$, so the answer is none of these.

(d) none

$\text{Area} = 4 \times 4 = 16\ \text{cm}^2.$ None of the options (70, $110-x$, 45) equals 16.

(d) none

For a cube, $a^3 = 1664$, so $a = \sqrt[3]{1664} \approx 11.85$. Then $\text{Surface area} = 6a^2 \approx 6 \times 140.4 \approx 842.5.$ This is none of 16, 64, 96.