JAMB 1999 · UME · Stepwise

Question 2

Find the value of $x$ if $\dfrac{\sqrt2}{x + \sqrt2} = \dfrac{1}{x - \sqrt2}$ .

Worked solution (try it first)

Cross-multiply: $\sqrt2(x - \sqrt2) = x + \sqrt2$ , so $\sqrt2x - 2 = x + \sqrt2$ .
Collect the $x$ terms: $\sqrt2x - x = 2 + \sqrt2$ , so $x(\sqrt2 - 1) = 2 + \sqrt2$ and $x = \dfrac{2 + \sqrt2}{\sqrt2 - 1}$ .
Multiply the top and bottom by $\sqrt2 + 1$ .
The bottom becomes $2 - 1 = 1$ .
Top: $(2 + \sqrt2)(\sqrt2 + 1) = 2\sqrt2 + 2 + 2 + \sqrt2$ , which is $4 + 3\sqrt2$ .
So option A.

Question 3

Commercial arithmetic

A trader bought 100 oranges at 5 for ₦1.20; 20 oranges got spoilt and the remaining were sold at 4 for ₦1.50. Find the percentage gain or loss.

Worked solution (try it first)

Cost: 100 oranges are 20 lots of 5, so they cost $20 \times 1.20 =$ ₦24.
Sales: 80 good oranges are 20 lots of 4, so they sell for $20 \times 1.50 =$ ₦30.
Gain $= 30 - 24 =$ ₦6, and $\dfrac{6}{24} \times 100\% = 25\%$ gain, option B.

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Question 4 ✱✱

Sets & Venn diagrams

If $U = \{1, 2, 3, 4, 5, 6\}$ , $P = \{3, 4, 5\}$ , $Q = \{2, 4, 6\}$ and $R = \{1, 2, 3, 4\}$ , list the elements of $(P \cup Q)' \cap R$ .

Worked solution (try it first)

Join $P$ and $Q$ : $P \cup Q = \{2, 3, 4, 5, 6\}$ .
The complement is what is left in $U$ : $(P \cup Q)' = \{1\}$ .
Intersect with $R$ : 1 is in $R$ , so $(P \cup Q)' \cap R = \{1\}$ , option C.

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Question 5

Number bases

Divide $2434_6$ by $42_6$ .

Worked solution (try it first)

Change to base ten: $2434_6 = 432 + 144 + 18 + 4 = 598$ and $42_6 = 24 + 2 = 26$ .
Divide: $598 \div 26 = 23$ .
Change back: $23 = 3 \times 6 + 5$ , so the answer is $35_6$ , option B.

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Question 7

Surds

Simplify $\sqrt{\dfrac{0.0023 \times 750}{0.00345 \times 1.25}}$ .

Worked solution (try it first)

Pair the numbers: $\dfrac{0.0023}{0.00345} = \dfrac{2300}{3450}$ , which is $\frac23$ .
And $\dfrac{750}{1.25} = 600$ .
So the fraction is $\frac23 \times 600 = 400$ , and $\sqrt{400} = 20$ , option B.

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Question 8

Logarithms

If $\log_8 10 = x$ , evaluate $\log_8 5$ in terms of $x$ .

Worked solution (try it first)

Write 5 as $\frac{10}{2}$ : $\log_8 5 = \log_8 10 - \log_8 2$ .
$8^{\frac13} = 2$ , so $\log_8 2 = \frac13$ .
So $\log_8 5 = x - \frac13$ , option C.

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A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain; 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only, while 3 sell all three items. How many women are in the group?

Worked solution (try it first)

Fill the Venn diagram from the centre out.
All three: 3.
Yam and plantain only: 2.
Plantain and maize only: $5 - 3 = 2$ .
Yam and maize only: $4 - 3 = 1$ .
Maize only: $12 - 3 - 2 - 1 = 6$ .
Yam only: $10 - 3 - 1 - 2 = 4$ .
Plantain only: $14 - 3 - 2 - 2 = 7$ .
Add all seven regions: $3 + 2 + 2 + 1 + 6 + 4 + 7 = 25$ , option A.

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Question 12

Sequences & series (AP, GP)

The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8.

Worked solution (try it first)

Put $a = 2r$ into $S_\infty = \dfrac{a}{1 - r} = 8$ : $2r = 8(1 - r)$ .
So $10r = 8$ , giving $r = \frac45$ and $a = \frac85$ .
The second term is $ar = \frac85 \times \frac45$
$= \frac{32}{25}$ .
So the first two terms add up to $\frac{40}{25} + \frac{32}{25} = \frac{72}{25}$ , option C.

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Question 14

Binary operations

If $m * n = \frac mn - \frac nm$ for real $m$ , $n$ , evaluate $-3 * 4$ .

Worked solution (try it first)

Put $m = -3$ and $n = 4$ : $\frac mn = -\frac34$ and $\frac nm = \frac{4}{-3} = -\frac43$ .
So $-3 * 4 = -\frac34 - \left(-\frac43\right)$
$= -\frac34 + \frac43$ .
Over 12: $\frac{-9 + 16}{12} = \frac{7}{12}$ , option C.

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Question 15

Matrices & determinants

Find the matrix $T$ if $ST = I$ , where $S = \begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}$ and $I$ is the identity matrix.

Worked solution (try it first)

$ST = I$ means $T$ is the inverse of $S$ .
The determinant is $|S| = (-1)(-2) - (1)(1) = 1$ .
For a $2 \times 2$ inverse, swap the two diagonal entries and change the signs of the other two: $\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}$ .
Divide by the determinant, 1, which changes nothing.
So $T = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}$ , option B.

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Question 16

Polynomials & algebraic division

Divide $4x^3 - 3x + 1$ by $2x - 1$ .

Worked solution (try it first)

Write the missing $x^2$ term: $4x^3 + 0x^2 - 3x + 1$ .
Then $4x^3 \div 2x = 2x^2$ .
Subtract $4x^3 - 2x^2$ to leave $2x^2 - 3x + 1$ .
$2x^2 \div 2x = x$ .
Subtract $2x^2 - x$ to leave $-2x + 1$ .
$-2x \div 2x = -1$ , and $-1(2x - 1) = -2x + 1$ leaves 0.
So the quotient is $2x^2 + x - 1$ , option D.

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Question 17

Quadratics & their graphs

Three consecutive positive integers $k$ , $l$ and $m$ are such that $l^2 = 3(k + m)$ . Find the value of $m$ .

Worked solution (try it first)

Write the integers as $k$ , $k + 1$ and $k + 2$ .
Then $(k + 1)^2 = 3(2k + 2)$ .
Expand: $k^2 + 2k + 1 = 6k + 6$ , so $k^2 - 4k - 5 = 0$ .
Factorise: $(k - 5)(k + 1) = 0$ .
The integers are positive, so $k = 5$ .
So $m = k + 2 = 7$ , option D.

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Question 19

Quadratics & their graphs

Factorize completely $x^2 + 2xy + y^2 + 3x + 3y - 18$ .

Worked solution (try it first)

The first three terms are a perfect square: $x^2 + 2xy + y^2 = (x + y)^2$ .
The next two are $3(x + y)$ .
Let $u = x + y$ .
The expression is $u^2 + 3u - 18$ .
Find two numbers with product $-18$ and sum 3: they are 6 and $-3$ .
So $u^2 + 3u - 18 = (u + 6)(u - 3)$ .
Put back $u = x + y$ : $(x + y + 6)(x + y - 3)$ , option A.

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Question 20

Linear & simultaneous equations

The sum of two numbers is twice their difference. If the difference of the numbers is $p$ , find the larger of the two numbers.

Worked solution (try it first)

Let the numbers be $a$ (larger) and $b$ .
The difference is $a - b = p$ , and the sum is twice that: $a + b = 2p$ .
Add the two equations so $b$ cancels: $2a = 3p$ .
So the larger number is $a = \frac{3p}{2}$ , option B.

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Question 23

Find the equation of the locus of a point $P(x, y)$ such that $PV = PW$ , where $V = (1, 1)$ and $W = (3, 5)$ .

Worked solution (try it first)

$PV = PW$ , so square both distances: $(x - 1)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$ .
Expand, and cancel $x^2$ and $y^2$ : $-2x + 1 - 2y + 1 = -6x + 9 - 10y + 25$ .
Collect terms: $4x + 8y = 32$ .
Divide by 4: $x + 2y = 8$ , option D.

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Question 24

Solid mensuration

A frustum has top radius 3 cm, bottom radius 6 cm and vertical height 4 cm. Find its slant height $l$ .

Worked solution (try it first)

Draw the slant side as the hypotenuse of a right-angled triangle.
Its vertical side is the height, 4 cm.
Its horizontal side is the difference of the radii: $6 - 3 = 3$ cm.
Pythagoras: $l^2 = 4^2 + 3^2 = 25$ , so $l = 5$ cm, option A.

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Question 25

Sine & cosine rules

In triangle $XYZ$ , $XY = 2$ m, $YZ = 1$ m and $\angle XYZ = 120^\circ$ . Find the length $XZ$ .

Worked solution (try it first)

Two sides and the angle between them: use the cosine rule, $XZ^2 = 2^2 + 1^2 - 2(2)(1)\cos120^\circ$ .
$\cos120^\circ = -\frac12$ , so the last term becomes $+2$ and $XZ^2 = 5 + 2 = 7$ .
So $XZ = \sqrt7$ m, option A.

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Question 26

Coordinate geometry

Find a positive value of $a$ if the centre of the circle $x^2 + y^2 - 2ax + 4y - a = 0$ is $(a, -2)$ and the radius is 4 units.

Worked solution (try it first)

Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$ : here $g = -a$ , $f = 2$ and $c = -a$ .
The radius satisfies $r^2 = g^2 + f^2 - c$ , so $16 = a^2 + 4 + a$ .
Rearrange: $a^2 + a - 12 = 0$ , which factorises as $(a + 4)(a - 3) = 0$ .
The positive value is $a = 3$ , option C.

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Question 27

Elevation, depression & bearings

A man 1.7 m tall observes a bird on top of a tree at an angle of elevation of $30^\circ$ . If the distance between the man's head and the bird is 25 m, what is the height of the tree?

Worked solution (try it first)

The 25 m from his head to the bird is the sloping line of sight, the hypotenuse.
The height of the bird above his head is opposite the $30^\circ$ angle: $25\sin30^\circ = 12.5$ m.
Add his height: $12.5 + 1.7 = 14.2$ m, option B.

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Question 28

Circle geometry

In the figure, $TZ$ is a tangent to the circle $QPZ$ and $TPQ$ is a straight line. Find $x = TP$ if $TZ = 6$ units and $PQ = 9$ units.

Not to scale.

Worked solution (try it first)

Tangent–secant rule: the tangent squared equals (outside part) × (whole secant).
So $TZ^2 = TP \times TQ$ , with $TQ = x + 9$ .
So $36 = x(x + 9)$ , which rearranges to $x^2 + 9x - 36 = 0$ .
Factorise: $(x + 12)(x - 3) = 0$ .
A length is positive, so $x = 3$ , option A.

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Question 29

Coordinate geometry

Find the tangent of the acute angle between the lines $2x + y = 3$ and $3x - 2y = 5$ .

Worked solution (try it first)

Write each line as $y = mx + c$ : $y = -2x + 3$ has $m_1 = -2$ , and $y = \frac32x - \frac52$ has $m_2 = \frac32$ .
The angle between the lines satisfies $\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1m_2}\right|$ .
The top is $-2 - \frac32 = -\frac72$ and the bottom is $1 + (-2)\left(\frac32\right) = -2$ .
So $\tan\theta = \frac74$ , option C.

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Question 30

Elevation, depression & bearings

From a point $P$ , the bearings of two points $Q$ and $R$ are N $67^\circ$ W and N $23^\circ$ E respectively. If the bearing of $R$ from $Q$ is N $68^\circ$ E and $PQ = 150$ m, calculate $PR$ .

Worked solution (try it first)

At $P$ , $Q$ is $67^\circ$ west of north and $R$ is $23^\circ$ east of north, so $\angle QPR = 67^\circ + 23^\circ$
$= 90^\circ$ .
The bearing of $Q$ from $P$ is $293^\circ$ , so the bearing of $P$ from $Q$ is $113^\circ$ .
$R$ from $Q$ is on $068^\circ$ , so $\angle PQR = 113^\circ - 68^\circ$
$= 45^\circ$ .
A right-angled triangle with a $45^\circ$ angle is isosceles, so $PR = PQ = 150$ m, option C.

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Question 31

Circle geometry

In the figure, $PQRS$ is a circle with $ST \parallel RQ$ ( $T$ on $PQ$ ) and $PT = PS$ . If $\angle QRS = 110^\circ$ , find the value of $x = \angle PQR$ .

Worked solution (try it first)

Opposite angles of cyclic quadrilateral $PQRS$ add up to $180^\circ$ : $\angle QPS = 180^\circ - 110^\circ$
$= 70^\circ$ .
$PT = PS$ , so triangle $PTS$ is isosceles: $\angle PTS = \dfrac{180^\circ - 70^\circ}{2}$
$= 55^\circ$ .
$ST \parallel RQ$ , so corresponding angles are equal: $x = \angle PTS$ , which is $55^\circ$ , option B.

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Question 32

Circle geometry

In the diagram, $EFGH$ is a cyclic quadrilateral in which $EH \parallel FG$ , and $FH$ and $EG$ are chords. If $\angle FHG = 42^\circ$ and $\angle EFH = 34^\circ$ , calculate $\angle HEG$ .

Worked solution (try it first)

Call $\angle HFG = a$ .
$EH \parallel FG$ , so alternate angles give $\angle EHF = a$ , and angles in the same segment give $\angle EGF = \angle EHF = a$ .
Angles in the same segment: $\angle EGH = \angle EFH = 34^\circ$ .
So $\angle FGH = a + 34^\circ$ .
Triangle $FGH$ : $a + (a + 34^\circ) + 42^\circ = 180^\circ$ , so $2a = 104^\circ$ and $a = 52^\circ$ .
$\angle HEG$ and $\angle HFG$ both stand on arc $HG$ , so $\angle HEG = 52^\circ$ , option C.

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Question 33

Quadratics & their graphs

If the maximum value of $y = 1 + hx - 3x^2$ is 13, find $h$ .

Worked solution (try it first)

The turning point is at $x = -\frac{b}{2a}$ .
With $a = -3$ and $b = h$ , that is $x = \frac h6$ .
Substitute: $y = 1 + h \cdot \frac h6 - 3 \cdot \frac{h^2}{36}$ , which simplifies to $1 + \frac{h^2}{12}$ .
Set it equal to 13: $\frac{h^2}{12} = 12$ , so $h^2 = 144$ and $h = \pm 12$ .
The value in the options is $h = 12$ , option B.

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Question 34

Calculus (JAMB bridge)

Evaluate $\displaystyle\int_{-2}^{1} (x - 1)^2\,dx$ .

Worked solution (try it first)

$(x - 1)^2$ integrates to $\frac{(x - 1)^3}{3}$ .
At $x = 1$ this is 0.
At $x = -2$ it is $\frac{(-3)^3}{3} = -9$ .
Subtract: $0 - (-9) = 9$ , option C.

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Question 36 ✱✱

Calculus (JAMB bridge)

Find the area bounded by the curve $y = x(2 - x)$ , the $x$ -axis, $x = 0$ and $x = 2$ .

Worked solution (try it first)

Expand: $y = 2x - x^2$ .
It is above the $x$ -axis between $x = 0$ and $x = 2$ , so the area is $\int_0^2 (2x - x^2)\,dx$ .
Integrate term by term: $x^2 - \frac{x^3}{3}$ .
At $x = 2$ this is $4 - \frac83 = \frac43$ , and at $x = 0$ it is 0.
So the area is $1\frac13$ square units, option C.

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Question 37

Calculus (JAMB bridge)

If $y = 3x^2(x^3 + 1)^{\frac12}$ , find $\frac{dy}{dx}$ .

Worked solution (try it first)

Product rule with $u = 3x^2$ and $v = (x^3 + 1)^{1/2}$ .
Then $u' = 6x$ and, by the chain rule, $v' = \frac12(x^3 + 1)^{-1/2} \times 3x^2$
$= \dfrac{3x^2}{2(x^3 + 1)^{1/2}}$ .
So $\frac{dy}{dx} = 6x(x^3 + 1)^{1/2} + 3x^2 \times \dfrac{3x^2}{2(x^3 + 1)^{1/2}}$ .
Over the common denominator $2(x^3 + 1)^{1/2}$ : $\dfrac{12x(x^3 + 1) + 9x^4}{2(x^3 + 1)^{1/2}}$ , option D.

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Question 38

Find the volume of the solid generated when the area enclosed by $y = 0$ , $y = 2x$ and $x = 3$ is rotated about the $x$ -axis.

Worked solution (try it first)

The volume of revolution about the $x$ -axis is $\pi\int y^2\,dx$ .
Here $y = 2x$ , so $y^2 = 4x^2$ , from $x = 0$ to $x = 3$ .
$\pi\int_0^3 4x^2\,dx = \pi\left[\frac43x^3\right]_0^3$
$= \pi \times \frac43 \times 27$
$= 36\pi$ .
So the volume is $36\pi$ cubic units, option B.

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Question 39

Calculus (JAMB bridge)

What is the derivative of $t^2\sin(3t - 5)$ with respect to $t$ ?

Worked solution (try it first)

Product rule with $u = t^2$ and $v = \sin(3t - 5)$ .
Then $u' = 2t$ and, by the chain rule, $v' = 3\cos(3t - 5)$ .
So the derivative is $2t\sin(3t - 5) + 3t^2\cos(3t - 5)$ , option C.

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Question 40

Calculus (JAMB bridge)

Find the value of $x$ for which the function $y = x^3 - x$ has a minimum value.

Worked solution (try it first)

At a turning point $\frac{dy}{dx} = 3x^2 - 1 = 0$ , so $x^2 = \frac13$ and $x = \pm\frac{1}{\sqrt3}$
$= \pm\frac{\sqrt3}{3}$ .
$\frac{d^2y}{dx^2} = 6x$ , which is positive at $x = \frac{\sqrt3}{3}$ , so that is the minimum.
So $x = \frac{\sqrt3}{3}$ , option C.

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Question 41

Probability

Three boys play a game of luck in which their respective chances of winning are $\frac12$ , $\frac13$ and $\frac14$ . What is the probability that one and only one of the boys wins the game?

Worked solution (try it first)

Each boy loses with probability $\frac12$ , $\frac23$ and $\frac34$ respectively.
"Only one wins" means one wins and the other two lose.
First only: $\frac12 \cdot \frac23 \cdot \frac34 = \frac{6}{24}$ .
Second only: $\frac12 \cdot \frac13 \cdot \frac34 = \frac{3}{24}$ .
Third only: $\frac12 \cdot \frac23 \cdot \frac14 = \frac{2}{24}$ .
These can't happen together, so add: $\frac{6 + 3 + 2}{24} = \frac{11}{24}$ , option C.

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Question 42

Probability

A number is selected at random from 0 to 20. What is the probability that the number is an odd prime?

Worked solution (try it first)

From 0 to 20 inclusive there are 21 numbers.
The odd primes are 3, 5, 7, 11, 13, 17, 19, which is 7 numbers.
So the probability is $\frac{7}{21} = \frac13$ , option B.

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Question 43

Counting

If $\dfrac{^6C_r}{^6P_r} = \frac16$ , find the value of $r$ .

Worked solution (try it first)

$^6P_r$ is $^6C_r$ multiplied by $r!$ (the ways to order the $r$ chosen items), so $\dfrac{^6C_r}{^6P_r} = \dfrac{1}{r!}$ .
So $\dfrac{1}{r!} = \dfrac16$ , which means $r! = 6$ .
$3! = 3 \times 2 \times 1 = 6$ , so $r = 3$ , option B.

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Question 44

Dispersion & cumulative frequency

If the standard deviation of the numbers 3, 6, $x$ , 7, 5 is $\sqrt2$ , find the least possible value of $x$ .

Worked solution (try it first)

Use variance $= \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$ with $n = 5$ , $\sum x = 21 + x$ and $\sum x^2 = 119 + x^2$ .
The variance is 2, so multiply through by 25: $5(119 + x^2) - (21 + x)^2 = 50$ .
Expand and collect terms: $4x^2 - 42x + 104 = 0$ , which is $2x^2 - 21x + 52 = 0$ , or $(x - 4)(2x - 13) = 0$ .
So $x = 4$ or $x = 6\frac12$ .
The least value is 4, option C.

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Question 46

Statistics: data & averages

The grades of 36 students in a class test are shown in the pie chart. How many students had excellent?

Worked solution (try it first)

Very good is marked with a right angle, so it is $90^\circ$ .
Excellent is the rest: $360^\circ - 120^\circ - 80^\circ - 90^\circ = 70^\circ$ .
Number of students: $\frac{70}{360} \times 36 = 7$ , option A.

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Question 47

Statistics: data & averages

The marks scored by students in a test are given below. Find the median.

Marks	0	1	2	3	4	5	6	7	8	9	10
No. of students	2	2	11	10	16	51	40	10	25	15	20

Worked solution (try it first)

Total students: $2 + 2 + 11 + 10 + 16 + 51 + 40 + 10 + 25 + 15 + 20 = 202$ , so the median is halfway between the 101st and 102nd marks.
Running totals: 2, 4, 15, 25, 41, 92 (mark 5), 132 (mark 6).
The 93rd to 132nd students all scored 6, so the 101st and 102nd are both 6: the median is 6, option B.

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Question 48

Statistics: data & averages

A student calculated the mean of 5 numbers as 45.3. While rechecking his working, he discovered that his total was short by 20.5. What is the correct mean of the 5 numbers?

Worked solution (try it first)

His wrong total was $5 \times 45.3 = 226.5$ .
It was short by 20.5, so the right total is $226.5 + 20.5 = 247$ .
The correct mean is $\frac{247}{5} = 49.4$ , option C.

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Question 49

Statistics: data & averages

The allocations to various ministries in a state budget are: Agriculture ₦25,000,000; Education ₦20,000,000; Women Affairs ₦35,000,000; Commerce and Industries ₦20,000,000. In a pie chart of this information, the angle for agriculture is

Worked solution (try it first)

Total budget: ₦25m + ₦20m + ₦35m + ₦20m = ₦100 million.
Agriculture's share of the circle: $\frac{25}{100} \times 360^\circ = 90^\circ$ , option D.

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