JAMB 1987 · UME · Stepwise

Question 1

Number bases

Convert $241_5$ to base 8.

Worked solution (try it first)

Change to base ten: $241_5 = 2 \times 25 + 4 \times 5 + 1 = 71$ .
Divide by 8: $71 = 8 \times 8 + 7$ , and $8 = 1 \times 8 + 0$ .
Read the digits from the top power down: $107_8$ , option B.

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Question 2

Number foundations & fractions

Find the least length of a rod which can be cut into exactly equal strips, each of either 40 cm or 48 cm in length.

Worked solution (try it first)

The length must divide exactly by 40 and by 48, so you want their LCM.
As primes: $40 = 2^3 \times 5$ and $48 = 2^4 \times 3$ .
The LCM is $2^4 \times 3 \times 5 = 240$ .
So the least length is 240 cm, option B.

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Question 3

A rectangular lawn has an area of 1815 square yards. If its length is 50 metres, find its width in metres, given that 1 metre equals 1.1 yards.

Worked solution (try it first)

Change the length to yards to match the area: $50 \times 1.1 = 55$ yards.
Width $=$ area $\div$ length $= 1815 \div 55 = 33$ yards.
Change back to metres: $33 \div 1.1 = 30$ m, option D.

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Question 4

Approximation & error

Reduce each number to two significant figures and then evaluate $\dfrac{0.02174 \times 1.2047}{0.023789}$ .

Worked solution (try it first)

Round each number to 2 significant figures: $0.02174 \to 0.022$ , $1.2047 \to 1.2$ and $0.023789 \to 0.024$ .
Top: $0.022 \times 1.2 = 0.0264$ .
Divide: $0.0264 \div 0.024 = 1.1$ , option C.

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A train moves from $P$ to $Q$ at an average speed of 90 km/h and immediately returns from $Q$ to $P$ along the same route at an average speed of 45 km/h. Find the average speed for the entire journey.

Worked solution (try it first)

Let each way be $d$ km.
Time is distance over speed, so the total time is $\dfrac{d}{90} + \dfrac{d}{45} = \dfrac{d + 2d}{90}$
$= \dfrac{d}{30}$ hours.
Average speed is total distance over total time: $2d \div \dfrac{d}{30} = 60$ .
So the average speed is 60.00 km/h, option B.

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Question 6

If the length of a square is increased by $20\%$ while its width is decreased by $20\%$ to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

Worked solution (try it first)

Take the square's side as 1.
The rectangle is $1.2$ by $0.8$ , so its area is $1.2 \times 0.8 = 0.96$ .
The square's area is 1, so the ratio is $0.96 : 1 = 96 : 100$ , which simplifies to $24 : 25$ , option D.

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Question 7

Commercial arithmetic

Two brothers invested a total of ₦5,000.00 in a farm project. The yield was sold for ₦15,000.00 at the end of the season. If the profit was shared in the ratio $2 : 3$ , what is the difference in the amount of profit received by the brothers?

Worked solution (try it first)

The profit is the sale minus the investment: $15\,000 - 5000 =$ ₦10,000.
The ratio $2 : 3$ has 5 parts, so one part is $10\,000 \div 5 =$ ₦2,000.
The shares are ₦4,000 and ₦6,000.
The difference is $3 - 2 = 1$ part, ₦2,000, option A.

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Question 8

Statistics: data & averages

Peter's weekly wages are ₦20.00 for the first 20 weeks and ₦36.00 for the next 24 weeks. Find his average weekly wage for the remaining 8 weeks of the year if his average weekly wage for the whole year is ₦30.00.

Worked solution (try it first)

A year has 52 weeks, so the year's total at an average of ₦30 is $52 \times 30 = 1560$ naira.
The first 44 weeks earn $20 \times 20 + 24 \times 36 = 400 + 864$ , which is ₦1264.
That leaves $1560 - 1264 = 296$ naira for the last 8 weeks.
So the average there is $296 \div 8 = 37$ , which is ₦37.00, option A.

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Question 9

Commercial arithmetic

A man invests a sum of money at $4\%$ per annum simple interest. After 3 years, the amount is ₦7,000.00. Find the sum invested.

Worked solution (try it first)

Simple interest for 3 years at $4\%$ is $3 \times 4\% = 12\%$ of the principal $P$ .
The amount is principal plus interest, so $1.12P = 7000$ .
Divide both sides by 1.12: $P = 6250$ .
The sum invested was ₦6,250.00, option B.

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Question 10

Commercial arithmetic

By selling 20 oranges for ₦1.35 a trader makes a profit of $8\%$ . What is his percentage gain or loss if he sells the same 20 oranges for ₦1.10?

Worked solution (try it first)

₦1.35 is $108\%$ of the cost price, so the cost is $1.35 \div 1.08 =$ ₦1.25.
Selling for ₦1.10 is a loss of $1.25 - 1.10 =$ ₦0.15.
As a percentage of the cost: $\dfrac{0.15}{1.25} \times 100\% = 12\%$ loss, option C.

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Question 13

Instead of writing $\frac{35}{6}$ as a decimal correct to 3 significant figures, a student wrote it correct to 3 decimal places. Find his error in standard form.

Worked solution (try it first)

Divide: $\frac{35}{6} = 5.8333\ldots$
To 3 significant figures this is 5.83.
To 3 decimal places it is 5.833.
The error is $5.833 - 5.83 = 0.003$ .
In standard form that is $3.0 \times 10^{-3}$ , option B.

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Question 14

Logarithms

Simplify without using tables $\dfrac{\log_2 6 - \log_2 3}{\log_2 8 - 2\log_2 \frac12}$ .

Worked solution (try it first)

Top: subtracting logs divides, so $\log_2 6 - \log_2 3 = \log_2 2 = 1$ .
Bottom: $\log_2 8 = 3$ and $\log_2 \frac12 = -1$ , so the bottom is $3 - 2(-1)$ , which is $3 + 2 = 5$ .
So the value is $\frac15$ , option A.

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Question 15

Surds

Simplify without using tables $\dfrac{2\sqrt{14} \times 3\sqrt{21}}{7\sqrt{24} \times 2\sqrt{98}}$ .

Worked solution (try it first)

Multiply out the top: $2\sqrt{14} \times 3\sqrt{21} = 6\sqrt{294}$ .
Since $294 = 49 \times 6$ , this is $42\sqrt6$ .
Simplify the bottom surds: $7\sqrt{24} = 14\sqrt6$ and $2\sqrt{98} = 14\sqrt2$ .
Their product is $196\sqrt{12}$ , which is $392\sqrt3$ .
Divide: $\dfrac{42\sqrt6}{392\sqrt3} = \dfrac{42}{392}\sqrt2$ , because $\frac{\sqrt6}{\sqrt3} = \sqrt2$ .
Cancel by 14: $\frac{42}{392} = \frac{3}{28}$ .
So the value is $\dfrac{3\sqrt2}{28}$ , option D.

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Question 16

Expressions, formulae & change of subject

If $p = \dfrac{2}{3} \cdot \dfrac{1 - r^2}{n^2}$ , find $n$ when $r = \frac{1}{\sqrt3}$ and $p = 1$ .

Worked solution (try it first)

With $r = \frac{1}{\sqrt3}$ , $r^2 = \frac13$ , so $1 - r^2 = \frac23$ .
Substitute: $1 = \frac23 \times \frac{2/3}{n^2}$
$= \frac{4}{9n^2}$ .
Multiply by $9n^2$ : $9n^2 = 4$ , so $n^2 = \frac49$ .
Take the square root: $n = \frac23$ , option D.

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Question 18

The formula $Q = 15 + 0.5n$ gives the cost $Q$ (in naira) of feeding $n$ people for a week. Find, in kobo, the extra cost of feeding one additional person.

Worked solution (try it first)

In $Q = 15 + 0.5n$ , each extra person increases $n$ by 1, so $Q$ goes up by 0.5.
The cost is in naira: ₦0.50, and ₦1 is 100 kobo, so this is 50k, option D.

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Question 19

Variation

If $P$ varies inversely as $V$ and $V$ varies directly as $R^2$ , find the relationship between $P$ and $R$ , given that $R = 7$ when $P = 2$ .

Worked solution (try it first)

$P = \dfrac{c}{V}$ and $V = mR^2$ , so $P = \dfrac{c}{mR^2}$ : $P$ varies inversely as $R^2$ .
So $P = \dfrac{k}{R^2}$ , which means $PR^2 = k$ .
Put in $R = 7$ , $P = 2$ : $k = 2 \times 49 = 98$ .
So $PR^2 = 98$ , option B.

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Question 20

Expressions, formulae & change of subject

Make $y$ the subject of the formula $Z = x^2 + \dfrac{1}{y^3}$ .

Worked solution (try it first)

Subtract $x^2$ from both sides: $\frac{1}{y^3} = Z - x^2$ .
Turn both sides upside down: $y^3 = \frac{1}{Z - x^2}$ .
Take the cube root: $y = \dfrac{1}{(Z - x^2)^{1/3}}$ , option C.

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Question 21

Quadratics & their graphs

Find the values of $m$ which make the quadratic function $x^2 + 2(m + 1)x + m + 3$ a perfect square.

Worked solution (try it first)

A perfect square has equal roots, so $b^2 = 4ac$ : $[2(m + 1)]^2 = 4(m + 3)$ .
Divide both sides by 4: $(m + 1)^2 = m + 3$ , so $m^2 + 2m + 1 = m + 3$ .
Rearrange: $m^2 + m - 2 = 0$ , which factorises as $(m + 2)(m - 1) = 0$ .
So $m = 1$ or $m = -2$ , option C.

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Question 23

Find two values of $y$ which satisfy the simultaneous equations $x + y = 5$ and $x^2 - 2y^2 = 1$ .

Worked solution (try it first)

Make $x$ the subject of the linear equation: $x = 5 - y$ .
Substitute: $(5 - y)^2 - 2y^2 = 1$ .
Expanding, $25 - 10y + y^2 - 2y^2 = 1$ .
Collect terms and multiply by $-1$ : $y^2 + 10y - 24 = 0$ .
Factorise: $(y + 12)(y - 2) = 0$ , so $y = -12$ or $y = 2$ , option C.

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Question 24

Angles, triangles & polygons

An $(n - 2)^2$ -sided figure has $n$ diagonals (according to this rule). Find the number $n$ of diagonals for a 25-sided figure.

Worked solution (try it first)

By the rule, a 25-sided figure has $(n - 2)^2 = 25$ .
Take the square root: $n - 2 = 5$ (the number is positive).
Add 2: $n = 7$ , option A.

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Question 25

Polynomials & algebraic division

A cubic function $f(x)$ is specified by the graph shown. The values of the independent variable for which the function vanishes are

Worked solution (try it first)

A function vanishes where its value is 0, that is, where the curve meets the $x$ -axis.
The graph crosses the $x$ -axis at $x = -1$ , $x = 0$ and $x = 1$ .
So the values are $-1, 0, 1$ , option A.

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Question 26

Inequalities

Solve the inequality $x - 1 > 4(x + 2)$ .

Worked solution (try it first)

Expand the bracket: $x - 1 > 4x + 8$ .
Subtract $4x$ and add 1 to both sides: $-3x > 9$ .
Divide by $-3$ and reverse the sign, because you are dividing by a negative: $x < -3$ , option B.

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Question 27

Quadratics & their graphs

Simplify $\dfrac{x^2 - y^2}{2x^2 + xy - y^2}$ .

Worked solution (try it first)

The top is a difference of two squares: $x^2 - y^2 = (x - y)(x + y)$ .
Factorise the bottom: $2x^2 + xy - y^2 = 2x^2 + 2xy - xy - y^2$ , which is $(2x - y)(x + y)$ .
Cancel the common factor $x + y$ .
This leaves $\dfrac{x - y}{2x - y}$ , option C.

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Question 28

Quadratics & their graphs

The minimum value of $y$ in the equation $y = x^2 - 6x + 8$ is

Worked solution (try it first)

Complete the square: half of $-6$ is $-3$ , so $x^2 - 6x = (x - 3)^2 - 9$ .
So $y = (x - 3)^2 - 9 + 8 = (x - 3)^2 - 1$ .
A square is never negative, so the least value of $y$ is $-1$ (when $x = 3$ ), option D.

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Question 29

Sequences & series (AP, GP)

Find the sum of the first 21 terms of the progression $-10, -8, -6, \dots$

Worked solution (try it first)

This is an A.P. with first term $a = -10$ and common difference $d = -8 - (-10) = 2$ .
Use $S_n = \frac n2\big(2a + (n - 1)d\big)$ with $n = 21$ : the bracket is $2(-10) + 20 \times 2$ , which is $-20 + 40 = 20$ .
So $S_{21} = \frac{21}{2} \times 20 = 210$ , option D.

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Question 30

Sequences & series (AP, GP)

Find the eleventh term of the progression $4, 8, 16, \dots$

Worked solution (try it first)

This is a G.P. with $a = 4$ and $r = 8 \div 4 = 2$ .
The $n$ th term is $ar^{n - 1}$ , so the eleventh term is $4 \times 2^{10}$ .
Write 4 as $2^2$ and add the powers: $2^2 \times 2^{10} = 2^{12}$ , option B.

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Question 32

Angles, triangles & polygons

In the diagram, $QR \parallel TS$ and $QR : TS = 2 : 3$ . Find the ratio of the area of triangle $PQR$ to the area of the trapezium $QRST$ .

Worked solution (try it first)

$QR \parallel TS$ , so triangles $PQR$ and $PTS$ are similar with lengths in the ratio $2 : 3$ .
Areas of similar figures are in the ratio of the squares of the lengths: $2^2 : 3^2 = 4 : 9$ .
The trapezium is the big triangle minus the small one: $9 - 4 = 5$ parts.
So the ratio is $4 : 5$ , option B.

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Question 33

Angles, triangles & polygons

Three angles of a nonagon are equal and the sum of the six other angles is $1110^\circ$ . Calculate the size of one of the equal angles.

Worked solution (try it first)

The interior angles of a nonagon (9 sides) add up to $(9 - 2) \times 180^\circ = 1260^\circ$ .
The three equal angles share what is left: $1260^\circ - 1110^\circ = 150^\circ$ .
So each one is $150^\circ \div 3 = 50^\circ$ , option D.

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Question 34

In the figure, $\angle XYZ = \angle YTZ = 90^\circ$ , $XT = 9$ cm and $TZ = 16$ cm. Find $YZ$ .

Worked solution (try it first)

The whole hypotenuse is $XZ = 9 + 16 = 25$ cm.
Triangles $YTZ$ and $XYZ$ are similar (they share the angle at $Z$ and each has a right angle), so $\dfrac{YZ}{XZ} = \dfrac{TZ}{YZ}$ , which gives $YZ^2 = TZ \times XZ$ .
So $YZ^2 = 16 \times 25 = 400$ .
So $YZ = 20$ cm, option B.

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Question 35

Circle geometry

Two chords $QR$ and $NP$ of a circle intersect inside the circle at $X$ . If $\angle RQP = 37^\circ$ , $\angle RQN = 49^\circ$ and $\angle QPN = 35^\circ$ , find $\angle PRQ$ .

Worked solution (try it first)

Angles in the same segment are equal.
$\angle RPN$ and $\angle RQN$ both stand on arc $RN$ , so $\angle RPN = 49^\circ$ .
So $\angle QPR$ is $\angle QPN + \angle NPR$ , which is $35^\circ + 49^\circ = 84^\circ$ .
The angles of triangle $PQR$ add up to $180^\circ$ : $\angle PRQ = 180^\circ - 37^\circ - 84^\circ$
$= 59^\circ$ , option D.

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Question 36 ✱✱

Angles, triangles & polygons

In the figure, two parallel lines are cut by a transversal, and the two interior angles on the same side of the transversal are bisected. The bisectors meet at the angle marked $x$ . Find the value of $x$ .

Worked solution (try it first)

Co-interior angles between parallel lines add up to $180^\circ$ .
The lower angle is $2a$ and the upper is $2b$ , so $2a + 2b = 180^\circ$ .
Halve it: $a + b = 90^\circ$ .
The two bisectors and the transversal form a triangle with angles $a$ , $b$ and $x$ , so $x = 180^\circ - (a + b) = 90^\circ$ , option C.

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Question 37

In the figure, $PQRS$ is a rectangle $2h$ cm wide and $3h$ cm high. The shaded U-shaped region is made of 2 cm strips along the two vertical sides and along the top. If the shaded area is $72\text{ cm}^2$ , find $h$ .

Worked solution (try it first)

The two side strips run the full height: each is $2 \times 3h$ , so together they are $12h\text{ cm}^2$ .
The top strip fits between them, so its length is $2h - 4$ : its area is $2(2h - 4) = 4h - 8$ .
Shaded area: $12h + 4h - 8 = 72$ , so $16h = 80$ and $h = 5$ cm, option D.

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Question 38 ✱✱

Trigonometric ratios

The sine, cosine and tangent of $210^\circ$ are respectively

Worked solution (try it first)

$210^\circ = 180^\circ + 30^\circ$ , so it lies in the third quadrant, with reference angle $30^\circ$ .
In the third quadrant sine and cosine are negative and tangent is positive.
So $\sin210^\circ = -\sin30^\circ = -\frac12$ and $\cos210^\circ = -\cos30^\circ$
$= -\frac{\sqrt3}{2}$ .
And $\tan210^\circ = \tan30^\circ$
$= \frac{1}{\sqrt3}$ , which is $\frac{\sqrt3}{3}$ .
That is option A.

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Question 39

Trigonometric ratios

If $\tan\theta = \dfrac{m^2 - n^2}{2mn}$ , find $\sec\theta$ .

Worked solution (try it first)

Draw a right-angled triangle with the opposite side $m^2 - n^2$ and the adjacent side $2mn$ .
Pythagoras: the hypotenuse squared is $(m^2 - n^2)^2 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4$
$= (m^2 + n^2)^2$ , so the hypotenuse is $m^2 + n^2$ .
$\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}$
$= \dfrac{m^2 + n^2}{2mn}$ , option B.

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Question 40 ✱✱

Elevation, depression & bearings

From two points $X$ and $Y$ , 8 m apart and in line with a pole, the angles of elevation of the top of the pole are $30^\circ$ and $60^\circ$ respectively. Find the height of the pole, assuming that $X$ , $Y$ and the foot of the pole are on the same horizontal plane.

Worked solution (try it first)

Let the height be $h$ .
$Y$ has the larger angle, so it is nearer: its distance to the foot is $\frac{h}{\tan60^\circ} = \frac{h}{\sqrt3}$ .
From $X$ the distance is $\frac{h}{\tan30^\circ} = \sqrt3\,h$ .
The two distances differ by 8 m: $\sqrt3\,h - \frac{h}{\sqrt3} = 8$ , so $\frac{2h}{\sqrt3} = 8$ .
So $h = 4\sqrt3$ m, option C.

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Question 41

A room is 12 m long, 9 m wide and 8 m high. Find the cosine of the angle which a diagonal of the room makes with the floor.

Worked solution (try it first)

The diagonal of the floor: $\sqrt{12^2 + 9^2} = \sqrt{225} = 15$ m.
The room diagonal rises 8 m above that floor diagonal, so it is $\sqrt{15^2 + 8^2} = \sqrt{289} = 17$ m.
The angle with the floor has the floor diagonal as its adjacent side: $\cos\theta = \frac{15}{17}$ , option A.

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Question 42

Longitude & latitude

What is the circumference of the circle of latitude $\theta^\circ$ on the earth, whose radius is $R$ ?

Worked solution (try it first)

The circle of latitude $\theta$ has its centre on the earth's axis, and its radius is $r = R\cos\theta$ (from the right-angled triangle with hypotenuse $R$ ).
Its circumference is $2\pi r = 2\pi R\cos\theta$ , option B.

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Question 43 ✱✱

Solid mensuration

The base of a pyramid is a square of side 8 cm. If its vertex is directly above the centre, find the height, given that each slant edge is $4\sqrt3$ cm.

Worked solution (try it first)

The diagonal of the square base is $8\sqrt2$ cm, so each corner is $4\sqrt2$ cm from the centre.
The height, that distance and a slant edge form a right-angled triangle, so $h^2 = (4\sqrt3)^2 - (4\sqrt2)^2$ , which is $48 - 32 = 16$ .
So $h = 4$ cm, option C.

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Question 44

Construction & loci

The figure is an example of the construction of a

Worked solution (try it first)

The arc is centred at $P$ , a point off the line, and it cuts the line at $Q$ and $R$ , so $PQ = PR$ .
Equal arcs from $Q$ and $R$ meet at $X$ , so $X$ is also equidistant from $Q$ and $R$ .
The line $PX$ is then perpendicular to $QR$ .
So it is the perpendicular from a given point to a given line, option B.

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Question 45

What is the locus of the mid-points of all chords of length 6 cm within a circle of radius 5 cm and centre $O$ ?

Worked solution (try it first)

The line from the centre to the mid-point of a chord is perpendicular to the chord.
Half of the 6 cm chord is 3 cm.
By Pythagoras, with the radius 5 cm as hypotenuse, the distance from $O$ to the mid-point is $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm.
Every such mid-point is 4 cm from $O$ , so the locus is a circle of radius 4 cm and centre $O$ , option A.

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Question 46

Statistics: data & averages

Taking the period of daylight on a certain day to be from 5.30 a.m. to 7.00 p.m., calculate the angles (in a pie chart) for the periods of daylight and of darkness on that day.

Worked solution (try it first)

From 5.30 a.m. to noon is $6\frac12$ hours, and from noon to 7.00 p.m. is 7 hours, so daylight lasts $13\frac12$ hours.
Daylight angle: $\frac{13.5}{24} \times 360^\circ = 202.5^\circ$
$= 202^\circ30'$ .
Darkness takes the rest: $360^\circ - 202^\circ30' = 157^\circ30'$ , option C.

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Question 47

Statistics: data & averages

The goals scored by 40 football teams are recorded below. What is the total number of goals scored by all the teams?

Number of goals	0	1	2	3	4	5	6
Frequency	4	3	15	16	1	0	1

Worked solution (try it first)

Each row gives goals × number of teams: $0(4) + 1(3) + 2(15) + 3(16) + 4(1) + 5(0) + 6(1)$ .
That is $0 + 3 + 30 + 48 + 4 + 0 + 6 = 91$ goals, option C.

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Question 48

Statistics: data & averages

The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If $P$ is the mean and $Q$ the median, then $P + Q$ is

Worked solution (try it first)

The marks add up to 60, so the mean is $P = \frac{60}{8} = 7.5$ .
In order: 2, 3, 5, 7, 8, 9, 12, 14.
The median is halfway between the 4th and 5th: $Q = \frac{7 + 8}{2} = 7.5$ .
So $P + Q = 7.5 + 7.5 = 15$ , option D.

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Question 49

Dispersion & cumulative frequency

The scores of a group of students in a music test are shown. If $CF(x)$ is the number of students with scores less than or equal to $x$ , find $CF(6)$ .

Score	1	2	3	4	5	6	7	8	9
No. of students	3	6	10	8	6	5	2	4	12

Worked solution (try it first)

$CF(6)$ counts every student who scored 6 or less, so add the numbers for scores 1 to 6.
$3 + 6 + 10 + 8 + 6 + 5 = 38$ , so $CF(6) = 38$ , option B.

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Question 50

Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium.

Worked solution (try it first)

A parallelogram has both pairs of opposite sides parallel.
The square, rectangle and rhombus all have this.
The kite has no parallel sides and the trapezium only one pair.
So 3 of the 5 figures work, and the probability is $\frac35$ , option A.

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