NECO · 2024 · SSCE · General Maths · Paper 2

1. Number the equations: $4a + 12b = 16$ (1) and $2a + 3b = 7$ (2).
2. Divide (1) by 4 to make it simpler: $a + 3b = 4$ (3).
3. Both (2) and (3) contain $3b$, so take (3) from (2): $a = 3$.
4. Put $a = 3$ into (3): $3 + 3b = 4$, so $3b = 1$ and $b = \frac13$.
5. Now $K^2 = a + b = 3 + \frac13 = \frac{10}{3}$.
6. The positive value is $K = \sqrt{\frac{10}{3}} \approx 1.826$.

(a)(i)

1. In a G.P., $T_n = ar^{n - 1}$, so $T_4 = ar^3 = 8$ and $T_7 = ar^6 = \frac{64}{27}$.
2. Divide: $r^3 = \frac{64/27}{8} = \frac{8}{27}$, so $r = \frac23$.

(ii)

1. $a = \frac{8}{r^3}$
   $= 8 \times \frac{27}{8}$
   $= 27$.

(b)

1. The place values in base five are 125, 25, 5 and 1: $4231_{\text{five}} = 4 \times 125 + 2 \times 25 + 3 \times 5 + 1$
   $= 500 + 50 + 15 + 1$
   $= 566$.

(a)(i)

1. Gradient of $PQ = \frac{3 - 8}{4 - (-3)}$
   $= \frac{-5}{7}$
   $= -\frac57$.
2. Using the point $Q(4, 3)$: $y - 3 = -\frac57(x - 4)$.
3. Multiply by 7: $7y - 21 = -5x + 20$, so $5x + 7y = 41$, that is $y = \frac{41 - 5x}{7}$.
4. (Check with $P$: $5(-3) + 7(8) = 41$ ✓.)

(ii)

1. Gradient of $QR = \frac{2 - 3}{1 - 4}$
   $= \frac{-1}{-3}$
   $= \frac13$.
2. Using $R(1, 2)$: $y - 2 = \frac13(x - 1)$, so $3y - 6 = x - 1$ and $x - 3y + 5 = 0$, that is $y = \frac{x + 5}{3}$.

(b)

1. The interior angles of a hexagon add up to $(6 - 2) \times 180^\circ = 720^\circ$.
2. So $(x + 5) + (2x + 4) + (x + 10) + (2x + 2) + (3x + 2) + (x + 15) = 720$.
3. Collect terms: $10x + 38 = 720$, so $10x = 682$ and $x = 68.2$.

1. First the mean: $2 + 6 + 9 + 8 + 7 + 9 + 8 + 7 = 56$, and there are 8 numbers, so $\bar x = \frac{56}{8} = 7$.

(i)

1. The distances from 7, ignoring signs, are $|x - 7| = 5, 1, 2, 1, 0, 2, 1, 0$, which add up to 12.
2. Mean deviation $= \frac{12}{8} = 1.50$.

(ii)

1. Square each deviation: $(x - 7)^2 = 25, 1, 4, 1, 0, 4, 1, 0$, which add up to 36.
2. Variance $= \frac{36}{8} = 4.5$, so the standard deviation is $\sqrt{4.5} \approx 2.12$.

(a)

1. Use the product rule: if $y = uv$, then $\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$.
2. Here $u = 3x^3 + 2x^2 + 1$, so $\frac{du}{dx} = 9x^2 + 4x$, and $v = 3x^2 + 4$, so $\frac{dv}{dx} = 6x$.
3. Then $\frac{dy}{dx} = (9x^2 + 4x)(3x^2 + 4) + (3x^3 + 2x^2 + 1)(6x)$.
4. Expand: $(27x^4 + 12x^3 + 36x^2 + 16x) + (18x^4 + 12x^3 + 6x)$.
5. Collect like terms: $\frac{dy}{dx} = 45x^4 + 24x^3 + 36x^2 + 22x$.

(b)

1. The sine of an angle equals the cosine of its complement, so when $\sin A = \cos B$ (both acute), $A + B = 90^\circ$.
2. So $(x + 30) + (2x + 33) = 90$.
3. Then $3x + 63 = 90$, $3x = 27$ and $x = 9$.

(a)

1. Set out the working with logarithms from four-figure tables:

2. No.	Log
   $\sqrt[4]{4000}$	$3.6021 \div 4 = 0.9005$
   $75.9$	$1.8802$
   numerator	$2.7807$
   $5.61$	$0.7490$
   $4.39$	$0.6425$
   denominator	$1.3915$
   result	$2.7807 - 1.3915 = 1.3892$

3. The antilog of $.3892$ is 2450, and the characteristic 1 means $\times 10$: the answer is about $24.50$.

(b)

1. First change to base ten: $4553_{\text{six}} = 4 \times 216 + 5 \times 36 + 5 \times 6 + 3$
   $= 864 + 180 + 30 + 3$
   $= 1077$.
2. Then divide by 3 repeatedly, keeping the remainders: $1077 \div 3 = 359$ r 0.
3. $359 \div 3 = 119$ r 2.
4. $119 \div 3 = 39$ r 2.
5. $39 \div 3 = 13$ r 0.
6. $13 \div 3 = 4$ r 1.
7. $4 \div 3 = 1$ r 1.
8. $1 \div 3 = 0$ r 1.
9. Read the remainders from the bottom up: $1110220_{\text{three}}$.

(a)

1. Move the known fractions to one side: $\frac{3}{4(x - 4)} = \frac{19}{12} - \frac56$.
2. With denominator 12, $\frac56 = \frac{10}{12}$, so the right side is $\frac{9}{12} = \frac34$.
3. Now $\frac{3}{4(x - 4)} = \frac34$.
4. The tops are both 3, so the bottoms must be equal: $4(x - 4) = 4$.
5. So $x - 4 = 1$ and $x = 5$.

(b)

1. "Partly constant and partly varies as the distance" means $C = a + bd$, where $C$ is the cost in naira and $d$ the distance in km.
2. From the two journeys: $a + 220b = 1025$ and $a + 300b = 1345$.
3. Take the first from the second: $80b = 320$, so $b = 4$.
4. Then $a = 1025 - 220 \times 4 = 145$.
5. So $C = 145 + 4d$.
6. For 982 km: $C = 145 + 4 \times 982$
   $= 145 + 3928$
   $= 4073$.
7. The cost is ₦4,073.00.

(i)

1. Total surface area of a solid cone $= \pi r l + \pi r^2$
   $= \pi r(l + r)$
   $= \frac{22}{7} \times 10 \times 36$
   $\approx 1131.4\text{ cm}^2$.

(ii)

1. The height comes from Pythagoras: $h = \sqrt{26^2 - 10^2} = \sqrt{576} = 24$ cm.
2. Volume $= \frac13 \times \frac{22}{7} \times 10^2 \times 24$
   $\approx 2514.29\text{ cm}^3$.

(iii)

1. The cone was made from a sector of radius 26 cm (the slant height), whose arc became the base circumference: $\frac{\theta}{360} \times 2\pi \times 26 = 2\pi \times 10$.
2. So $\theta = \frac{10}{26} \times 360^\circ$
   $\approx 138^\circ$.

(a)

1. The $n$th term of an A.P. is $a + (n - 1)d$.
2. So $T_9 = a + 8d = 61$ and $T_{21} = a + 20d = 145$.
3. Take the first equation from the second: $12d = 84$, so $d = 7$.
4. Then $a = 61 - 8 \times 7 = 61 - 56 = 5$.
5. $S_n = \frac{n}{2}[2a + (n - 1)d]$, so $S_{48} = \frac{48}{2}[2 \times 5 + 47 \times 7]$
   $= 24 \times (10 + 329)$
   $= 24 \times 339$
   $= 8136$.

(b)

1. Integrate term by term: $\frac32x^2$ becomes $\frac32 \times \frac{x^3}{3} = \frac{x^3}{2}$, and $5$ becomes $5x$.
2. So $\displaystyle\int_0^2 \left(\tfrac32x^2 + 5\right)dx = \left[\frac{x^3}{2} + 5x\right]_0^2$.
3. Put in the limits: $\left(\frac{8}{2} + 10\right) - (0 + 0) = 4 + 10$
   $= 14$.

(a)

1. Put each $x$ into $y = x^2 + x - 12$.
2. For $x = -3$: $9 - 3 - 12 = -6$.
3. For $x = -2$: $4 - 2 - 12 = -10$.
4. For $x = -1$: $1 - 1 - 12 = -12$.
5. For $x = 1$: $1 + 1 - 12 = -10$.
6. For $x = 2$: $4 + 2 - 12 = -6$.
7. The missing values are $-6, -10, -12, -10, -6$.

(b)

1. Plot the nine points $(-4, 0), (-3, -6), \ldots, (4, 8)$ and join them with a smooth U-shaped curve.
2. For the line $y = 2x + 1$, three points are enough: $(-4, -7)$, $(0, 1)$ and $(4, 9)$.

(c)

1. Where the line meets the curve, $x^2 + x - 12 = 2x + 1$.
2. Taking $2x + 1$ from both sides gives $x^2 - x - 13 = 0$, so the roots are the $x$-values of the two crossing points.
3. Read them from the graph: $x \approx -3.1$ and $x \approx 4.1$.
4. (The exact values are $\frac{1 \pm \sqrt{53}}{2}$, about $-3.14$ and $4.14$.
5. The second is just past $x = 4$, so extend the curve and line slightly to read it.)

(d)

1. The lowest point of the curve is halfway between the roots $-4$ and $3$ of $x^2 + x - 12 = 0$, at $x = -0.5$.
2. There $y = 0.25 - 0.5 - 12 = -12.25$, so the minimum value of $y$ is $-12.25$.

(a)

1. Draw $AB = 6.0$ cm and construct $60^\circ$ at $A$.
2. With centre $B$ and radius 7.5 cm, cut the arm at $D$ (so $|AD| \approx 8.4$ cm).
3. Through $B$ construct a line parallel to $AD$.
4. With centre $D$ and radius 7.0 cm, cut it at $C$, on the far side from $A$ so that $ABCD$ is a trapezium.
5. Join $DC$.

(b)

1. Construct the perpendicular bisectors of $AB$ and $BC$.
2. They meet at the centre $O$.
3. With centre $O$ and radius $OA$, draw the circle through $A$, $B$ and $C$.

(c)

1. Measure the radius: about 8.1 cm.
2. (Check by calculation: with $A$ at the origin and $B$ at $(6, 0)$, $C \approx (11.05, 8.75)$, and the circumradius is about 8.14 cm.)
3. Circumference $= 2\pi r$
   $\approx 2 \times \frac{22}{7} \times 8.14$
   $\approx 51$ cm.

(a)(i)

1. Use the class marks (mid-points) $5.5, 15.5, 25.5, 35.5, 45.5, 55.5$.
2. Then $\sum f = 150$ and $\sum fx = 26(5.5) + 36(15.5) + 38(25.5) + 30(35.5) + 15(45.5) + 5(55.5)$
   $= 143 + 558 + 969 + 1065 + 682.5 + 277.5$
   $= 3695$.
3. Mean $= \frac{3695}{150} \approx 24.6$.

(ii)

1. The median is the $\frac{150}{2} = 75$th score.
2. The running totals of the frequencies are $26, 62, 100, \ldots$, so the 75th score is in the class 21–30.
3. Its lower boundary is $20.5$, 62 scores come before it, it holds 38 and its width is 10: median $= 20.5 + \frac{75 - 62}{38} \times 10$
   $= 20.5 + 3.42$
   $\approx 23.9$.

(iii)

1. The modal class is 21–30 (frequency 38).
2. It is $38 - 36 = 2$ more than the class before and $38 - 30 = 8$ more than the class after: mode $= 20.5 + \frac{2}{2 + 8} \times 10 = 22.5$.

(b)

1. Admission goes to the highest scores.
2. The top two classes, 51–60 and 41–50, hold exactly $5 + 15 = 20$ candidates.
3. So the 20 admitted are those who scored 41 or more: the cut-off mark is 41 (the class boundary 40.5).