NECO · 2023 · SSCE · General Maths · Paper 2

(i)

1. $P$ varies directly as $Q^2$ (on top) and inversely as $Z^3$ (underneath): $P = \dfrac{kQ^2}{Z^3}$.
2. Put in $P = 5$, $Q = 3$, $Z = 1$: $5 = \dfrac{9k}{1}$, so $k = \frac59$.
3. The relationship is $P = \dfrac{5Q^2}{9Z^3}$.

(ii)

1. Put in $P = 3$, $Q = 5$: $3 = \dfrac{5 \times 25}{9Z^3}$
   $= \dfrac{125}{9Z^3}$.
2. So $27Z^3 = 125$, $Z^3 = \dfrac{125}{27}$ and $Z = \sqrt[3]{\dfrac{125}{27}} = \dfrac53$.

(a)(i)

1. The interior angles of an $n$-sided polygon add up to $(n - 2) \times 180^\circ$.
2. So $(n - 2) \times 180 = 1440$, $n - 2 = 8$ and $n = 10$.

(ii)

1. The exterior angles of any polygon add up to $360^\circ$, so each is $\frac{360^\circ}{10} = 36^\circ$.

(b)

1. $I = \frac{PRT}{100}$: $29\,880 = \frac{P \times 3 \times 15}{100}$
   $= 0.45P$, so $P = \frac{29\,880}{0.45} = ₦66,400$.

(a)(i)

1. The angle at the centre is twice the angle at the circumference on the same arc: $\theta = 2 \times 30^\circ = 60^\circ$.

(ii)

1. $OA = OB$ (radii), so triangle $AOB$ is isosceles.
2. With a $60^\circ$ angle between the equal sides, the other two angles are also $60^\circ$.
3. So it is equilateral, and the radius equals the chord: $r = 12$ cm.

(b)(i)

1. The shaded region is the minor segment: sector minus triangle.
2. Sector $= \frac{60}{360} \times \frac{22}{7} \times 12^2$
   $\approx 75.43\text{ cm}^2$.
3. Triangle $= \frac12 \times 12 \times 12 \times \sin 60^\circ$
   $\approx 62.35\text{ cm}^2$.
4. Segment $\approx 75.43 - 62.35 = 13.08$, which is $13.1\text{ cm}^2$ to three significant figures.

(ii)

1. Triangle $AOB$ is equilateral.

(a)

1. Write every number as a power of 3: $27 = 3^3$, $\frac13 = 3^{-1}$ and $9 = 3^2$.
2. Then $27^{2x - 1} = 3^{6x - 3}$, $\left(\frac13\right)^{-(3x + 2)} = 3^{3x + 2}$ and $9^{x + 3} = 3^{2x + 6}$.
3. Multiplying powers of 3 adds the indices: $3^{(6x - 3) + (3x + 2)} = 3^{9x - 1}$.
4. So $3^{9x - 1} = 3^{2x + 6}$, and the indices are equal: $9x - 1 = 2x + 6$, $7x = 7$, $x = 1$.

(b)

1. Chain rule: bring down the power, reduce it by one, then multiply by the derivative of the inside.
2. $\frac{d}{dx}(2x^2 + 5)^4 = 4(2x^2 + 5)^3 \times 4x$
   $= 16x(2x^2 + 5)^3$.

(a)

1. There are $10 + 8 + 4 + 6 + 12 = 40$ teachers, and the whole circle is $360^\circ$, so each teacher gets $\frac{360^\circ}{40} = 9^\circ$.
2. The sector angles are English $10 \times 9^\circ = 90^\circ$, Mathematics $72^\circ$, Physics $36^\circ$, French $54^\circ$ and Biology $108^\circ$.
3. Check: $90 + 72 + 36 + 54 + 108 = 360$.
4. Draw the sectors with a protractor and label each with its subject and angle.

(b)

1. The two teachers are chosen one after the other from 40, so the second is chosen from the 39 left.
2. Both from Physics: $\frac{4}{40} \times \frac{3}{39} = \frac{12}{1560}$.
3. Both from Biology: $\frac{12}{40} \times \frac{11}{39} = \frac{132}{1560}$.
4. These can't both happen, so add: $\frac{12 + 132}{1560} = \frac{144}{1560}$
   $= \frac{6}{65}$.

(a)

1. $231_{\text{four}} = 2 \times 16 + 3 \times 4 + 1$
   $= 45$.
2. In base $k$, $135_k = k^2 + 3k + 5$.
3. So $k^2 + 3k + 5 = 45$, which gives $k^2 + 3k - 40 = 0$ and $(k + 8)(k - 5) = 0$.
4. A base must be positive, so $k = 5$.

(b)(i)

1. The sector is $\frac{120}{360} = \frac13$ of the circle.
2. Arc $= \frac13 \times 2 \times \frac{22}{7} \times 21$
   $= 44$ cm.
3. Perimeter $= 44 + 21 + 21 = 86$ cm.

(ii)

1. Area $= \frac13 \times \frac{22}{7} \times 21^2$
   $= 462\text{ cm}^2$.

(c)

1. Rationalise each fraction.
2. $\frac{3}{\sqrt5 + \sqrt2} \times \frac{\sqrt5 - \sqrt2}{\sqrt5 - \sqrt2} = \frac{3(\sqrt5 - \sqrt2)}{3}$
   $= \sqrt5 - \sqrt2$, and $\frac{1}{\sqrt5 - \sqrt2} = \frac{\sqrt5 + \sqrt2}{3}$.
3. So the expression is $\sqrt5 - \sqrt2 - \frac{\sqrt5 + \sqrt2}{3} = \frac{3\sqrt5 - 3\sqrt2 - \sqrt5 - \sqrt2}{3}$
   $= \frac{2\sqrt5 - 4\sqrt2}{3}$.

(a)

1. Let the man's age now be $m$ years and his daughter's age be $d$ years.
2. Their ages add up to 60, so $m + d = 60$.
3. Six years ago they were $m - 6$ and $d - 6$, and the man was three times as old: $m - 6 = 3(d - 6)$.
4. Expanding, $m - 6 = 3d - 18$, so $m = 3d - 12$.
5. Substitute into the first equation: $3d - 12 + d = 60$, so $4d = 72$ and $d = 18$.
6. Then $m = 60 - 18 = 42$.
7. The man is 42 and his daughter is 18.
8. Check: six years ago they were 36 and 12, and $36 = 3 \times 12$ ✓.

(b)

1. A root $x = -\frac34$ gives $4x = -3$, so the factor $(4x + 3)$.
2. A root $x = \frac56$ gives $6x = 5$, so the factor $(6x - 5)$.
3. The equation is $(4x + 3)(6x - 5) = 0$.
4. Expanding: $24x^2 - 20x + 18x - 15 = 0$, so $24x^2 - 2x - 15 = 0$.

(c)

1. Inside the bracket: $1 - \frac{144}{169} = \frac{25}{169}$.
2. The power $\frac12$ is a square root: $\left(\frac{25}{169}\right)^{\frac12} = \frac{5}{13}$.
3. The power $-1$ turns a fraction upside down: $\left(\frac{2}{13}\right)^{-1} = \frac{13}{2}$.
4. So the value is $4 \times \frac{5}{13} \times \frac{13}{2} = 10$.

(a)

1. Move the 5 inside as a power: $5\log 2 = \log 2^5 = \log 32$.
2. Then combine: $\log 32 + \log 40 - \log 12.8 = \log\frac{32 \times 40}{12.8}$
   $= \log\frac{1280}{12.8}$
   $= \log 100$
   $= 2$.

(b)

1. The gradient is $\frac{dy}{dx}$, so integrate it: $y = \frac{6x^3}{3} + \frac{8x^2}{2} - 3x + c$
   $= 2x^3 + 4x^2 - 3x + c$.
2. The curve passes through $(-2, 5)$: $2(-8) + 4(4) - 3(-2) + c = 5$, so $-16 + 16 + 6 + c = 5$ and $c = -1$.
3. The curve is $y = 2x^3 + 4x^2 - 3x - 1$.

(c)

1. $\frac{dy}{dx} = 6x + 4$ (the constant $-5$ gives 0).

(a)(i)

1. Multiply each matrix by its number, then subtract entry by entry: $3A - 2B = \begin{pmatrix} 9 - 8 & 6 - 4 & -3 + 6 \\ 3 - 6 & 0 + 2 & 3 - 2 \\ 6 - 0 & -6 + 4 & 0 - 4 \end{pmatrix}$
   $= \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \\ 6 & -2 & -4 \end{pmatrix}$.

(ii)

1. Expand along the first row: $1\big(2(-4) - 1(-2)\big) - 2\big((-3)(-4) - 1(6)\big) + 3\big((-3)(-2) - 2(6)\big) = 1(-6) - 2(6) + 3(-6)$
   $= -36$.

(b)(i)

1. Compound interest at $12\%$ for 3 years: $120\,000 \times 1.12^3 = 120\,000 \times 1.404928$
   $= ₦168,591.36$.

(ii)

1. Profit $= 350\,000 - 168\,591.36 = ₦181,408.64$.

1. The bucket is a frustum: a cone with its tip cut off.
2. Its volume is the big cone minus the small cone that was removed.
3. Let the full cone have height $H$.
4. The radii are 20 cm (top) and 12 cm (bottom), and the small cone's height is $H - 32$.
5. By similar triangles, $\frac{H}{20} = \frac{H - 32}{12}$, so $12H = 20H - 640$ and $H = 80$ cm.
6. The small cone is $80 - 32 = 48$ cm high.
7. Volume of water $= \frac13\pi(20^2 \times 80) - \frac13\pi(12^2 \times 48)$
   $= \frac13 \times \frac{22}{7} \times (32\,000 - 6912)$
   $\approx 26\,282.7\text{ cm}^3$.
8. (The frustum formula $\frac{\pi h}{3}(R^2 + Rr + r^2)$ gives the same.)
9. The drum's base is $\frac{22}{7} \times 28^2 = 2464\text{ cm}^2$.
10. The water keeps its volume, so its depth is $\frac{26\,282.7}{2464} \approx 10.67$ cm, which is 11 cm to the nearest whole number.

(a)

1. Draw $AC = 7$ cm.
2. At $A$ construct $120^\circ$ (two $60^\circ$ angles side by side), and with the compasses set to 5 cm cut the arm at $B$.
3. Join $BC$.

(b)(i)

1. $l_1$, the points equidistant from $A$ and $C$, is the perpendicular bisector of $AC$: equal arcs from $A$ and $C$, and the line through their crossings.

(ii)

1. $l_2$, the points 4.5 cm from $C$, is the circle with centre $C$ and radius 4.5 cm.

(c)

1. $N_1$ and $N_2$ are the two points where the circle cuts the bisector.

(d)

1. Measure: (i) $|N_1N_2| \approx 5.7$ cm.

(ii)

1. $|BC| \approx 10.4$ cm.
2. Check by calculation: the bisector is 3.5 cm from $C$, so $|N_1N_2| = 2\sqrt{4.5^2 - 3.5^2}$
   $= 2\sqrt8$
   $\approx 5.7$ cm.
3. And by the cosine rule $|BC|^2 = 5^2 + 7^2 - 2(5)(7)\cos 120^\circ$
   $= 109$, so $|BC| \approx 10.4$ cm.

(a)

1. Turn the mean into a total.
2. There are 12 scores and the mean is 18, so they add up to $12 \times 18 = 216$.
3. The known scores add up to $12 + 15 + 25 + 12 + 30 + 15 + 25 + 12 + 16 + 12 = 174$, and the unknown ones to $x + 2x = 3x$.
4. So $174 + 3x = 216$, $3x = 42$ and $x = 14$.
5. The question asks for $2x$: $2x = 28$.

(b)

1. The scores are now $12, 14, 15, 28, 25, 12, 30, 15, 25, 12, 16, 12$.
2. Their distances from the mean 18, ignoring signs, are $6, 4, 3, 10, 7, 6, 12, 3, 7, 6, 2, 6$, which add up to 72.
3. Mean deviation $= \frac{\sum |x - \bar x|}{n}$
   $= \frac{72}{12}$
   $= 6$.

(c)

1. Square each distance: $36, 16, 9, 100, 49, 36, 144, 9, 49, 36, 4, 36$, which add up to 524.
2. Variance $= \frac{524}{12} \approx 43.67$, so the standard deviation is $\sqrt{43.67} \approx 6.61$.