WAEC · 2022 · Private, 2nd series · General Maths · Paper 2

1. Write each fact as an equation.
2. The third term: $T_3 = a + 2d = 23$.
3. The sum of the first seven terms: $S_7 = \frac72[2a + 6d] = 7(a + 3d) = 210$, so $a + 3d = 30$.

(a)

1. Take the first equation from the second: $(a + 3d) - (a + 2d) = 30 - 23$, so the common difference is $d = 7$.

(b)

1. Put $d = 7$ into $a + 2d = 23$: $a = 23 - 14 = 9$.

(c)

1. $S_{20} = \frac{20}{2}[2a + 19d]$
   $= 10 \times (18 + 133)$
   $= 10 \times 151$
   $= 1510$.

(a)

1. Let a packed lunch cost $$x$ and a packed breakfast $$y$.
2. The total cost: $2x + 3y = 40$ (1).
3. Two lunches cost $12 less than one breakfast: $2x = y - 12$ (2).
4. Substitute (2) into (1): $(y - 12) + 3y = 40$.
5. So $4y = 52$ and $y = 13$.
6. A packed breakfast costs $13.00.

(b)

1. The root $x = -\frac13$ gives $3x = -1$, so the factor $(3x + 1)$.
2. The root $x = 5$ gives the factor $(x - 5)$.
3. The equation is $(3x + 1)(x - 5) = 0$.
4. Expand: $3x^2 - 15x + x - 5 = 0$, so $3x^2 - 14x - 5 = 0$.

1. Draw the tower $AO$ (110 m) and the building $BC$ on the same ground, with the engineer at the top $O$ of the tower.
2. From $O$, draw a horizontal line across to the building.
3. The angles are measured up ($38^\circ$) and down ($22^\circ$) from it.

(a)

1. The foot of the building is 110 m below the horizontal, at an angle of depression of $22^\circ$: $\tan 22^\circ = \frac{110}{d}$, so $d = \frac{110}{0.4040} \approx 272.3$ m.

(b)

1. Above the horizontal, the top of the building rises $d\tan 38^\circ \approx 272.3 \times 0.7813$
   $\approx 212.7$ m.
2. So the building is $110 + 212.7 = 322.7$ m high.

(a)

1. $PQRS$ is a cyclic quadrilateral, so $\angle QPS = 180^\circ - 70^\circ$
   $= 110^\circ$.
2. $T$, $P$, $U$ are on a straight line, so the angles at $P$ add up to $180^\circ$: $\angle QPT = 180^\circ - 110^\circ - 30^\circ$
   $= 40^\circ$.

(b)

1. $|QR| = |RS|$, so triangle $QRS$ is isosceles and $\angle RSQ = \frac{180^\circ - 70^\circ}{2}$
   $= 55^\circ$.
2. By the alternate segment theorem, the angle between the tangent $PT$ and the chord $PQ$ equals the angle in the alternate segment: $\angle PSQ = \angle QPT = 40^\circ$.
3. So $\angle PSR = \angle PSQ + \angle QSR$
   $= 40^\circ + 55^\circ$
   $= 95^\circ$.

(a)

1. Total surface area of a closed cone $= \pi r^2 + \pi r l$.
2. With $r = 3.8$: $\frac{22}{7} \times 3.8^2 \approx 45.383$ and $\frac{22}{7} \times 3.8 \approx 11.943$, so $45.383 + 11.943l = 374$.
3. Then $11.943l \approx 328.617$ and $l \approx 27.516$ cm.
4. Height: $h = \sqrt{27.516^2 - 3.8^2}$
   $= \sqrt{757.13 - 14.44}$
   $\approx 27.252$ cm.
5. Volume $= \frac13 \times \frac{22}{7} \times 3.8^2 \times 27.252$
   $\approx 412.26\text{ cm}^3$.

(a)

1. Let the numbers be $a$ and $8 - a$ (so their sum is 8).
2. Their product is $-33$: $a(8 - a) = -33$.
3. So $8a - a^2 = -33$, which rearranges to $a^2 - 8a - 33 = 0$.
4. Factorise: $(a - 11)(a + 3) = 0$.
5. So $a = 11$ or $a = -3$.
6. The numbers are 11 and $-3$.

(b)

1. Let $Q$ overtake $P$ at $t$ hours after 10.30 am.
2. By then $P$ has travelled for $t$ hours, and $Q$ (which passed the point 30 minutes later) for $t - \frac12$ hours.
3. When $Q$ overtakes, they have gone the same distance from the point: $54t = 90\left(t - \frac12\right)$.
4. So $54t = 90t - 45$, $36t = 45$ and $t = 1\frac14$ hours $= 1$ hour 15 minutes.
5. $Q$ overtakes $P$ at 11.45 am.

(a)

1. Each year the value falls by $14\%$ of what it was at the start of that year, so it's multiplied by 0.86.
2. After 3 years: $65\,000 \times 0.86^3 = 65\,000 \times 0.636056$
   $\approx ₦41,343.64$.

(b)(i)

1. 240 oranges at 3 for GH¢ 1.00 cost $\frac{240}{3} =$ GH¢ 80.00.

(ii)

1. Sold at GH¢ 0.50 each: $240 \times 0.50 =$ GH¢ 120.00.

(iii)

1. Profit $= 120 - 80 = 40$, and $\frac{40}{80} \times 100\% = 50\%$.

(a)

1. Triangle $PRS$ is right-angled at $R$: $|SR| = \sqrt{24.12^2 - 20^2}$
   $= \sqrt{181.77}$
   $\approx 13.48$ cm, which is 13.5 cm to one decimal place.

(b)

1. $\angle RQS$ is an exterior angle of triangle $PQS$, so $\angle RQS = 34^\circ + 20^\circ$
   $= 54^\circ$.
2. In right-angled triangle $QRS$: $|QR| = \frac{13.482}{\tan 54^\circ}$
   $\approx 9.796$ cm.
3. So $|PQ| = 20 - 9.796 \approx 10.2$ cm.

(c)

1. Triangle $PQS$ has base $PQ$ and height $SR$ (perpendicular to the line $PR$): area $= \frac12 \times 10.204 \times 13.482$
   $\approx 68.8\text{ cm}^2$.

(a)

1. Draw three overlapping circles F, V and H in a rectangle of 50 students, with 15 outside all the circles.
2. Let $x$ play all three.
3. The 35 who play at least one game satisfy $18 + 20 + 21 - 9 - 6 - 11 + x = 35$, so $33 + x = 35$ and $x = 2$.
4. Two games only: F and V only $9 - 2 = 7$, F and H only $6 - 2 = 4$, V and H only $11 - 2 = 9$.
5. One game only: football $18 - 7 - 4 - 2 = 5$, volleyball $20 - 7 - 9 - 2 = 2$, handball $21 - 4 - 9 - 2 = 6$.
6. Fill these into the diagram.

(b)(i)

1. All three: 2.

(ii)

1. Only one game: $5 + 2 + 6 = 13$.

(iii)

1. Only football: 5.

(c)

1. Only two games: $7 + 4 + 9 = 20$ students, so $P = \frac{20}{50} = \frac25$.

1. In 2.5 hours, $M$ sails $35 \times 2.5 = 87.5$ km on $230^\circ$ and $N$ sails $30 \times 2.5 = 75$ km on $320^\circ$.
2. The bearings differ by $320^\circ - 230^\circ = 90^\circ$, so the triangle $AMN$ is right-angled at $A$.

(a)

1. $|MN| = \sqrt{75^2 + 87.5^2}$
   $= \sqrt{13\,281.25}$
   $\approx 115.2$ km.

(b)

1. At $N$: $\tan\angle ANM = \frac{87.5}{75}$, so $\angle ANM \approx 49.4^\circ$.
2. At $N$, the direction back to $A$ is $320^\circ - 180^\circ = 140^\circ$, and $M$ is $49.4^\circ$ further round clockwise.
3. Bearing of $M$ from $N$ $= 140^\circ + 49.4^\circ = 189.4^\circ$.

(a)

1. $\sum f = (n - 2) + (n - 1) + (n - 3) + (2n - 6) + (8 - n)$
   $= 4n - 4$ and $\sum fx = 2(n - 2) + 3(n - 1) + 4(n - 3) + 5(2n - 6) + 6(8 - n)$
   $= 13n - 1$.
2. The mean is 3.75: $\frac{13n - 1}{4n - 4} = 3.75$, so $13n - 1 = 15n - 15$, $2n = 14$ and $n = 7$.

(b)(i)

1. With $n = 7$ the frequencies are $5, 6, 4, 8, 1$, a total of 24.
2. Running totals: $5, 11, 15, 23, 24$.
3. $Q_1$ is at position $\frac{24}{4} = 6$: the 6th mark is 3.
4. $Q_3$ is at position $\frac{3 \times 24}{4} = 18$: the 18th mark is 5.
5. Interquartile range $= 5 - 3 = 2$.

(ii)

1. "At least 3 marks" is everyone except the 5 who scored 2: $24 - 5 = 19$.
2. The probability is $\frac{19}{24}$.

(a)(i)

1. Draw $YZ = 8.4$ cm, construct $60^\circ$ at $Y$ and mark $YX = 7.2$ cm on the arm.
2. Join $XZ$.

(ii)

1. Construct the perpendicular bisector of $XY$.
2. It cuts $XY$ at its midpoint $M$.

(iii)

1. With centre $M$ and radius $|YZ| = 8.4$ cm, and with centre $Z$ and radius $|YM| = 3.6$ cm, draw arcs meeting at $N$.
2. Join $MN$ and $NZ$: $MNZY$ is a parallelogram, with $MN \parallel YZ$.

(b)

1. Measure: (i) $|XZ| \approx 7.9$ cm.

(ii)

1. $\angle NZX \approx 68^\circ$.
2. Check: by the cosine rule $|XZ|^2 = 7.2^2 + 8.4^2 - 2(7.2)(8.4)\cos 60^\circ$
   $= 61.92$, so $|XZ| \approx 7.87$ cm.
3. $ZN \parallel XY$, so $\angle NZX = \angle ZXY$, and the sine rule gives $\sin\angle ZXY = \frac{8.4\sin 60^\circ}{7.87}$, about $68^\circ$.

(a)

1. $x$ years ago the man was $42 - x$ and the son $12 - x$.
2. Their product was 304: $(42 - x)(12 - x) = 304$, so $504 - 54x + x^2 = 304$ and $x^2 - 54x + 200 = 0$.
3. Factorise: $(x - 4)(x - 50) = 0$.
4. The son is only 12, so $x = 50$ is impossible: it was 4 years ago.
5. Check: $38 \times 8 = 304$.

(b)

1. $\log_4 256 = 4$ (since $4^4 = 256$), so $\log_5 2x = 1 - 4 = -3$.
2. In index form: $2x = 5^{-3} = \frac{1}{125}$, so $x = \frac{1}{250} = 0.004$.