WAEC · 2019 · Private · General Maths · Paper 2

(a)

1. Multiply every term by 14, the LCM of 2 and 7: $7(1 + 4x) - 2(5 + 2x) < 14(x - 2)$.
2. Expand: $7 + 28x - 10 - 4x < 14x - 28$.
3. So $24x - 3 < 14x - 28$.
4. Collect terms: $10x < -25$.
5. Divide by 10: $x < -2\frac12$.

(b)

1. $x : y = 3 : 5$ means $x = 3k$ and $y = 5k$ for some number $k$.
2. Substitute: $\frac{2(3k)^2 - (5k)^2}{(5k)^2 - (3k)^2} = \frac{18k^2 - 25k^2}{25k^2 - 9k^2}$
   $= \frac{-7k^2}{16k^2}$
   $= -\frac{7}{16}$.

1. The $n$th term of an A.P. is $a + (n - 1)d$.
2. So $T_2 = a + d = x - 1$, $T_4 = a + 3d = x + 1$ and $T_6 = a + 5d = 7$.

(a)

1. Take the second term from the fourth: $(a + 3d) - (a + d) = (x + 1) - (x - 1)$.
2. Both $a$ and $x$ cancel: $2d = 2$, so the common difference is $d = 1$.

(b)

1. Put $d = 1$ into $a + 5d = 7$: $a + 5 = 7$, so the first term is $a = 2$.

(c)

1. Put $a = 2$ and $d = 1$ into $a + d = x - 1$: $3 = x - 1$, so $x = 4$.
2. (Check: $T_4 = a + 3d = 5$ and $x + 1 = 5$ ✓.)

(a)

1. Each log is a small whole number: $\log_2 8 = 3$, $\log_2 16 = 4$, $\log_2 2 = 1$ and $\log_4 16 = 2$.
2. So the expression is $\frac{3 + 4 - 4 \times 1}{2} = \frac32$
   $= 1\frac12$.

(b)

1. Change both to base ten: $1342_{\text{five}} = 1 \times 125 + 3 \times 25 + 4 \times 5 + 2$
   $= 222$ and $241_{\text{five}} = 2 \times 25 + 4 \times 5 + 1$
   $= 71$.
2. So $x = 222 - 71 = 151$.

(a)

1. Add the logs to multiply: $\log_{10}(ab) = 1.3010 + 1.4771 = 2.7781$.
2. So $ab = 10^{2.7781} \approx 600$.
3. (In fact $\log 20 = 1.3010$ and $\log 30 = 1.4771$, so $ab = 20 \times 30 = 600$.)

(b)(i)

1. $ABE$ is a straight line, so $\angle ABC = 180^\circ - 62^\circ$
   $= 118^\circ$.
2. The angle at the centre is twice the angle at the circumference on the same arc, so the reflex angle $AOC = 2 \times 118^\circ = 236^\circ$, and the interior angle $AOC = 360^\circ - 236^\circ = 124^\circ$.

(ii)

1. In triangle $ABC$: $\angle BAC = 180^\circ - 39^\circ - 118^\circ$
   $= 23^\circ$.

(a)

1. "Partly constant and partly varies as the number of tourists": $C = a + bn$, where $C$ is the cost in dollars and $n$ the number of tourists.
2. 20 tourists: $a + 20b = 740$.
3. The number increased by 10, so 30 tourists: $a + 30b = 960$.
4. Take the first equation from the second: $10b = 220$, so $b = 22$.
5. Then $a = 740 - 20 \times 22 = 740 - 440 = 300$.
6. So $C = 300 + 22n$.
7. For 15 tourists: $C = 300 + 22 \times 15 = 300 + 330 = 630$.
8. The dinner costs $630.00.

(a)

1. He sold the car for $90\%$ of $5,600: $0.9 \times 5600 = 5040$, so he received $5,040.
2. After spending $1,310, he invested $5040 - 1310 = 3730$, that is $3,730.
3. Simple interest for 3 years at $6\%$: $\frac{3730 \times 6 \times 3}{100} = 671.40$.
4. The interest earned is $671.40.

(b)

1. Write everything as powers of the same base.
2. $4^{-7} = (2^2)^{-7} = 2^{-14}$, so the first equation is $2^x \times 2^{-14} = 2^1$.
3. Add the powers: $2^{x - 14} = 2^1$, so $x - 14 = 1$ and $x = 15$.
4. Similarly $9^{2y} = (3^2)^{2y} = 3^{4y}$, so $3^{-x} \times 3^{4y} = 3^1$ gives $-x + 4y = 1$.
5. With $x = 15$: $4y = 16$, so $y = 4$.

(a)

1. $OM$ and $ON$ are radii, so the perimeter is $2r + 22.42 = 55.6$.
2. So $2r = 33.18$ and $r = 16.59$ cm.
3. The perpendicular from $O$ bisects the chord and the angle: $\sin\frac{\angle MON}{2} = \frac{11.21}{16.59}$
   $\approx 0.6757$.
4. So $\frac{\angle MON}{2} \approx 42.5^\circ$ and $\angle MON \approx 85^\circ$.

(b)(i)

1. Draw north at $T$, with $P$ on $060^\circ$ and $Q$ on $130^\circ$, the same distance from $T$.

(ii)

1. $\angle PTQ = 130^\circ - 60^\circ$
   $= 70^\circ$.
2. $|TP| = |TQ|$, so the triangle is isosceles and $\angle TPQ = \frac{180^\circ - 70^\circ}{2}$
   $= 55^\circ$.
3. At $P$, the direction back to $T$ is $060^\circ + 180^\circ = 240^\circ$, and $Q$ is $55^\circ$ further round anticlockwise.
4. Bearing of $Q$ from $P$ $= 240^\circ - 55^\circ = 185^\circ$.

(a)

1. Draw three overlapping circles, M, P and C, in a rectangle for the 40 students, who each study at least one of the subjects.
2. Put 3 in the centre.
3. The "two subjects" numbers include the centre, so the "exactly two" regions are: M and P only $5 - 3 = 2$, M and C only $8 - 3 = 5$, P and C only $5 - 3 = 2$.
4. Mathematics only $= 23 - 2 - 5 - 3 = 13$.
5. Let Chemistry only be $x$.
6. Then Physics only is $2x$.
7. All the regions add up to 40: $13 + 2 + 5 + 2 + 3 + x + 2x = 40$, so $25 + 3x = 40$ and $x = 5$.

(i)

1. Physics only $= 2x = 10$.

(ii)

1. Only one subject $= 13 + 10 + 5 = 28$.

(b)

1. Exactly two subjects $= 2 + 5 + 2 = 9$, so $P = \frac{9}{40}$.

(a)

1. The same amount of rice is eaten either way, so the number of "boy-days" is the same: $m$ boys for 10 days equals $(m + 4)$ boys for 8 days.
2. So $10m = 8(m + 4)$.
3. Then $10m = 8m + 32$, $2m = 32$ and $m = 16$.

(b)

1. Let the number be $x$.
2. $\frac56$ of it is 10 more than $\frac13$ of it: $\frac56x = \frac13x + 10$.
3. Multiply by 6: $5x = 2x + 60$.
4. So $3x = 60$ and $x = 20$.

(c)

1. Gradient $= \frac{-\frac12 - \frac12}{-1 - 2}$
   $= \frac{-1}{-3}$
   $= \frac13$.
2. Using the point $\left(2, \frac12\right)$: $y - \frac12 = \frac13(x - 2)$.
3. Multiply by 6: $6y - 3 = 2x - 4$.
4. So $2x - 6y - 1 = 0$.

(a)

1. Draw the wall vertical and the ground horizontal, with the 11 m ladder making $75^\circ$ with the ground.
2. Then draw the new position, reaching 0.2 m higher up the wall, with its foot nearer the wall.

(b)(i)

1. At first the ladder reaches $11\sin 75^\circ \approx 10.625$ m up the wall.
2. Pushed 0.2 m up, it reaches $10.825$ m.
3. The ladder is still 11 m long: $\sin\theta = \frac{10.825}{11}$
   $\approx 0.9841$, so $\theta \approx 79.8^\circ$, which is $80^\circ$ to the nearest whole number.

(ii)

1. The foot was $11\cos 75^\circ \approx 2.847$ m from the wall and is now $11\cos 79.8^\circ \approx 1.95$ m from it.
2. It moved $2.847 - 1.95 \approx 0.9$ m, which is 1 m to the nearest whole number.

(a)

1. Put each $x$ into $y = 4x^2 - 8x - 21$.
2. For $x = -1.5$: $9 + 12 - 21 = 0$.
3. For $x = -0.5$: $1 + 4 - 21 = -16$.
4. For $x = 1.0$: $4 - 8 - 21 = -25$.
5. For $x = 1.5$: $9 - 12 - 21 = -24$.
6. For $x = 2.5$: $25 - 20 - 21 = -16$.
7. For $x = 4.0$: $64 - 32 - 21 = 11$.
8. The row is $11, 0, -9, -16, -21, -24, -25, -24, -21, -16, -9, 0, 11$.

(b)(i)

1. With 2 cm to 1 unit across and 2 cm to 5 units up, plot the thirteen points and join them with a smooth U-shaped curve.
2. It is symmetrical about $x = 1$, where $y = -25$.

(ii)

1. (α)** Take 21 from both sides of $4x^2 - 8x = 3$: $4x^2 - 8x - 21 = -18$.
2. Draw the line $y = -18$.
3. It cuts the curve at $x \approx -0.3$ and $x \approx 2.3$.
4. The solution set is $\{-0.3, 2.3\}$.

(β)

1. Take $x$ from both sides of $4x^2 - 7x - 21 = 0$: it becomes $4x^2 - 8x - 21 = -x$.
2. So draw the line $y = -x$ through $(-2, 2)$, $(0, 0)$ and $(4, -4)$.
3. It cuts the curve at $x \approx -1.6$ and $x \approx 3.3$.
4. The solution set is $\{-1.6, 3.3\}$.

(a)

1. The curved surface of a cone is $\pi r l$.
2. The first cone has $\pi \times 5 \times 12 = 60\pi$.
3. For the second, $\pi \times 6 \times l = 60\pi$, so its slant height is $l = 10$ cm.
4. Height of the second cone: $h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ cm.

(b)(i)

1. Multiply rows of $\mathbf B$ by columns of $\mathbf A$: $\mathbf{BA} = \begin{pmatrix} 3(2) + (-2)(-1) & 3(5) + (-2)(-3) \\ 4(2) + 1(-1) & 4(5) + 1(-3) \end{pmatrix}$
   $= \begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix}$.

(ii)

1. $|\mathbf{BA}| = 8 \times 17 - 21 \times 7$
   $= 136 - 147$
   $= -11$.