WAEC · 2022 · Private · General Maths · Paper 2

(a)

1. Draw three overlapping circles F, V and H in a rectangle of 100.
2. Put 10 in the centre, 12 in F and V only, and 18, 6 and 4 in the "only" regions.
3. Football holds 48: $18 + 12 + 10 + (\text{F and H only}) = 48$, so F and H only $= 8$.
4. Volleyball holds 42: $6 + 12 + 10 + (\text{V and H only}) = 42$, so V and H only $= 14$.
5. Check with hockey: $4 + 8 + 14 + 10 = 36$, which matches.

(b)(i)

1. Two sports only: $12 + 8 + 14 = 34$.

(ii)

1. Inside the circles: $18 + 6 + 4 + 12 + 8 + 14 + 10 = 72$.
2. So $100 - 72 = 28$ liked none of the three.

(a)

1. Add $b$ to both sides: $f + b = \frac{am}{y^3}$.
2. Multiply both sides by $y^3$ and divide by $(f + b)$: $y^3 = \frac{am}{f + b}$.
3. Take the cube root: $y = \sqrt[3]{\frac{am}{f + b}}$.

(b)

1. Let $x$ books cost GH¢ 5.00 each.
2. The other $15 - x$ cost GH¢ 3.00 each.
3. The total cost is GH¢ 67.00: $5x + 3(15 - x) = 67$.
4. So $5x + 45 - 3x = 67$, $2x = 22$ and $x = 11$. 11 books cost GH¢ 5.00.
5. Check: $11 \times 5 + 4 \times 3 = 55 + 12 = 67$ ✓.

(a)

1. $3x + 5y - 2 = 0$ gives $5y = -3x + 2$, so $y = -\frac35x + \frac25$ and the gradient is $-\frac35$.
2. Gradient of $PQ = \frac{k - 2}{7 - (-3)}$
   $= \frac{k - 2}{10}$.
3. The gradients are equal: $\frac{k - 2}{10} = -\frac35$, so $k - 2 = -6$ and $k = -4$.

(b)

1. With $Q(7, -4)$ and $R(2, -5)$: $|QR| = \sqrt{(7 - 2)^2 + (-4 - (-5))^2}$
   $= \sqrt{25 + 1}$
   $= \sqrt{26}$
   $\approx 5.1$ units.

(a)

1. Find the height from the volume: $\frac{22}{7} \times 14^2 \times h = 9240$, so $616h = 9240$ and $h = 15$ cm.
2. Curved surface area $= 2\pi rh$
   $= 2 \times \frac{22}{7} \times 14 \times 15$
   $= 1320\text{ cm}^2$.

(b)

1. An exterior angle and its interior angle make $180^\circ$: the interior angles at $P$ and $R$ are $180^\circ - y$ and $180^\circ - x$.
2. The angles of the triangle add up to $180^\circ$: $(180^\circ - y) + (180^\circ - x) + n = 180^\circ$.
3. So $n = x + y - 180^\circ$
   $= 210^\circ - 180^\circ$
   $= 30^\circ$.

(a)(i)

1. Draw the slanted tree as the hypotenuse, 12.8 m, from its foot on the ground to the bird at the top, which is 10 m vertically above the ground.

(ii)

1. The angle of depression from the bird equals the angle of elevation of the bird from the foot of the tree (alternate angles).
2. The 10 m height is opposite that angle and the tree is the hypotenuse: $\sin\theta = \frac{10}{12.8}$
   $\approx 0.7813$, so $\theta \approx 51.4^\circ$
   $\approx 51^\circ$.

(b)

1. Total number of students: $11 + 12 + 10 + 17 + 16 + 15 + 14 + 13 + 12 = 120$. "At most 7 marks" means 7 marks or fewer, so leave out those who scored 8 or 9: $120 - (13 + 12) = 95$.
2. Probability $= \frac{95}{120} = \frac{19}{24}$.

(a)

1. List the integers: $P = \{-2, -1, 0, 1, \ldots\}$, $Q = \{2, 3, 4, 5\}$, $R = \{\ldots, 0, 1, 2\}$.

(i)

1. $Q \cup R = \{\ldots, 0, 1, 2, 3, 4, 5\}$, so $P \cap (Q \cup R) = \{-2, -1, 0, 1, 2, 3, 4, 5\}$.

(ii)

1. $P \cap Q = \{2, 3, 4, 5\}$ and $P \cap R = \{-2, -1, 0, 1, 2\}$, so $(P \cap Q) \cup (P \cap R) = \{-2, -1, 0, 1, 2, 3, 4, 5\}$, the same set.

(b)

1. The interest equals the amount borrowed: $\frac{P \times 15 \times T}{100} = P$, so $15T = 100$ and $T = 6\frac23 \approx 7$ years.

(a)

1. Call the other two sides $a$ and $b$.
2. The perimeter is 90 cm and the hypotenuse is 39 cm, so $a + b = 90 - 39 = 51$, and $b = 51 - a$.
3. By Pythagoras' theorem, $a^2 + b^2 = 39^2 = 1521$.
4. Put in $b = 51 - a$: $a^2 + (51 - a)^2 = 1521$, so $a^2 + 2601 - 102a + a^2 = 1521$.
5. Simplify: $2a^2 - 102a + 1080 = 0$.
6. Divide by 2: $a^2 - 51a + 540 = 0$.
7. Factorise: $(a - 15)(a - 36) = 0$, so $a = 15$ or $a = 36$.
8. Either way the two sides are 15 cm and 36 cm.
9. (Check: $15 + 36 + 39 = 90$ and $15^2 + 36^2 = 225 + 1296 = 1521 = 39^2$ ✓.)

(b)(i)

1. $P$: no Mathematics student is outside the clever students, so every Mathematics student is clever: draw the Mathematics circle inside the Clever circle.
2. $Q$: every Physics student studies Mathematics: draw the Physics circle inside the Mathematics circle.
3. So the universal set of students holds three circles, Physics inside Mathematics inside Clever.

(ii)

1. (I)** A clever student can be outside the Physics circle (for example, outside the Mathematics circle altogether).
2. So "every clever student is a Physics student" is not valid.
3. (II) The Physics circle is inside the Mathematics circle, which is inside the Clever circle, so every Physics student is clever: valid.
4. (III) A Mathematics student can be in the ring between the Physics and Mathematics circles, studying Mathematics but not Physics.
5. So "every Mathematics student studies Physics" is not valid.

(a)

1. Every student reads at least one subject, so $30 + 35$ counts the students who read both twice: $30 + 35 - x = 60$, and $x = 5$ read both.
2. Literature only $= 30 - 5 = 25$.
3. Government only $= 35 - 5 = 30$.
4. Only one subject: $25 + 30 = 55$ students, so $P = \frac{55}{60} = \frac{11}{12}$.

(b)(i)

1. When the sector is folded into a cone, its arc becomes the circumference of the base.
2. Arc length $= \frac{220}{360} \times 2\pi \times 63$ and base circumference $= 2\pi r$, so $r = \frac{220}{360} \times 63 = 38.5$ cm.

(ii)

1. The radius of the sheet, 63 cm, becomes the slant height of the cone.
2. Height: $h = \sqrt{63^2 - 38.5^2}$
   $= \sqrt{3969 - 1482.25}$
   $= \sqrt{2486.75}$
   $\approx 49.867$ cm.
3. Volume $= \frac13\pi r^2 h$
   $= \frac13 \times \frac{22}{7} \times 38.5^2 \times 49.867$
   $\approx 77\,435.6\text{ cm}^3$.

(a)

1. Substitute each $x$ into $y = \frac{6}{x - 1} + 1$.
2. $x = 2.5$: $\frac{6}{1.5} + 1 = 5$.
3. $x = 3$: $\frac62 + 1 = 4$.
4. $x = 5$: $\frac64 + 1 = 2.5$.
5. $x = 6$: $\frac65 + 1 = 2.2$.
6. The full row is $7, 5, 4, 3, 2.5, 2.2, 2$.

(b)

1. (i) Plot the points and join them with a smooth curve.

(ii)

1. $y = 8 - x$ is a straight line: for example $(2, 6)$ and $(7, 1)$.
2. Draw it with a ruler on the same axes.

(c)

1. Rearrange the first equation: $(x - 1)(y - 1) = 6$ gives $y - 1 = \frac{6}{x - 1}$, which is the curve in (b)(i).
2. And $x + y = 8$ is the line $y = 8 - x$.
3. So the solutions are the points where the curve and the line cross.
4. Reading from the graph: $x \approx 2.3$, $y \approx 5.7$ and $x \approx 5.7$, $y \approx 2.3$.
5. (By algebra, $x = 4 \pm \sqrt3$.)

(a)

1. The four tankers take $4 \times 13\,500 = 54\,000$ litres.
2. Since $1000$ litres $= 1\text{ m}^3$, that is $54\text{ m}^3$.
3. The tank's radius is 3.5 m, so a drop of $h$ m removes $\pi r^2h = \frac{22}{7} \times 3.5^2 \times h$
   $= 38.5h\text{ m}^3$.
4. So $38.5h = 54$ and $h \approx 1.40$ m.

(b)

1. Multiply both sides by $\frac35$: $\sin(x - 29^\circ) = \frac12 \times \frac35$
   $= \frac{3}{10}$.
2. So $x - 29^\circ = \sin^{-1} 0.3$
   $\approx 17.46^\circ$, and $x \approx 46.46^\circ \approx 46^\circ$.

1. Use an assumed mean of 5 years, with $d = x - 5$:

2. Age $x$	1	2	3	4	5	6	7	8	9	10	Total
   $f$	12	14	20	45	36	20	23	16	8	6	200
   $d$	$-4$	$-3$	$-2$	$-1$	0	1	2	3	4	5
   $fd$	$-48$	$-42$	$-40$	$-45$	0	20	46	48	32	30	1
   $fd^2$	192	126	80	45	0	20	92	144	128	150	977

(a)

1. Mean $= 5 + \frac{1}{200} = 5.005 \approx 5.0$ years.

(b)

1. Standard deviation $= \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}$
   $= \sqrt{\frac{977}{200} - \left(\frac{1}{200}\right)^2}$
   $= \sqrt{4.885}$
   $\approx 2.2$ years.

(a)

1. $WXYZ$ is a cyclic quadrilateral, so $\angle WZY = 180^\circ - 111^\circ$
   $= 69^\circ$.
2. The angle at the centre is twice the angle at the circumference on the same arc $WXY$: $\angle WOY = 2 \times 69^\circ$
   $= 138^\circ$.
3. The angles of quadrilateral $OWXY$ add up to $360^\circ$: $\angle OWX = 360^\circ - 138^\circ - 111^\circ - 43^\circ$
   $= 68^\circ$.

(b)

1. $|WZ| = |YZ|$ and $OW = OY$, so $OZ$ is a line of symmetry and cuts $\angle WZY$ in half: $\angle OZY = \frac12 \times 69^\circ$
   $= 34.5^\circ$.
2. $OY = OZ$ (radii), so triangle $OYZ$ is isosceles and $\angle OYZ = \angle OZY = 34.5^\circ$.

(a)

1. The first 87 multiples of 5 are $5, 10, 15, \ldots$, an A.P. with $a = 5$ and $d = 5$.
2. $S_n = \frac n2[2a + (n - 1)d]$, so $S_{87} = \frac{87}{2}[10 + 86 \times 5]$
   $= \frac{87}{2} \times 440$
   $= 87 \times 220$
   $= 19\,140$.

(b)

1. Multiply every term by $15p$, the LCM of the denominators: $15(p - 5) = 25 - 3p$.
2. Expand: $15p - 75 = 25 - 3p$.
3. So $18p = 100$ and $p = \frac{100}{18}$
   $= \frac{50}{9}$
   $= 5\frac59$.

(c)

1. Gradient $= \frac{6 - 4}{3 - (-1)}$
   $= \frac24$
   $= \frac12$.
2. Using the point $(-1, 4)$: $y - 4 = \frac12(x + 1)$.
3. Multiply by 2: $2y - 8 = x + 1$.
4. So $2y - x - 9 = 0$.