JAMB 2010 · UTME · Stepwise

Question 2

Number bases

Find $r$ , if $6r7_8 = 511_9$ .

Worked solution (try it first)

Change the right side to base ten: $511_9 = 5 \times 81 + 9 + 1 = 415$ .
The place values in base eight are 64, 8 and 1, so $6r7_8 = 384 + 8r + 7 = 391 + 8r$ .
Set $391 + 8r = 415$ : $8r = 24$ , so $r = 3$ , option C.

Question 4

Approximation & error

A student measures a piece of rope and found that it was $1.26\text{ m}$ long. If the actual length of the rope was $1.25\text{ m}$ , what was the percentage error in the measurement?

Worked solution (try it first)

The error is $1.26 - 1.25 = 0.01$ m.
Divide by the actual length and multiply by 100: $\frac{0.01}{1.25} \times 100\% = \frac{1}{1.25}\%$ .
$1 \div 1.25 = 0.8$ , so the percentage error is 0.80%, option D.

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Question 5

Commercial arithmetic

At what rate will the interest on ₦400 increase to ₦24 in 3 years reckoning in simple interest?

Worked solution (try it first)

Use $I = \dfrac{PRT}{100}$ with $I = 24$ , $P = 400$ and $T = 3$ : $24 = \dfrac{400 \times R \times 3}{100}$ .
The right side simplifies to $12R$ , so $12R = 24$ .
Divide both sides by 12: $R = 2$ .
The rate is $2\%$ , option B.

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Question 6

Number foundations & fractions

If $p : q = \frac23 : \frac56$ and $q : r = \frac34 : \frac12$ , find $p : q : r$ .

Worked solution (try it first)

Clear the fractions.
Multiply $\frac23 : \frac56$ by 6: $p : q = 4 : 5$ .
Multiply $\frac34 : \frac12$ by 4: $q : r = 3 : 2$ .
Make the two $q$ values the same.
The LCM of 5 and 3 is 15, so $p : q = 12 : 15$ and $q : r = 15 : 10$ .
So $p : q : r = 12 : 15 : 10$ , option C.

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Question 7

Indices & standard form

Evaluate $\left(\dfrac{81}{16}\right)^{-\frac14} \times 2^{-1}$ .

Worked solution (try it first)

A negative index turns the fraction upside down: $\left(\frac{81}{16}\right)^{-\frac14} = \left(\frac{16}{81}\right)^{\frac14}$ .
Take fourth roots: $\sqrt[4]{16} = 2$ and $\sqrt[4]{81} = 3$ , so this is $\frac23$ .
$2^{-1} = \frac12$ , so the value is $\frac23 \times \frac12 = \frac13$ , option A.

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Question 8

Logarithms

Given that $\log 2 = 0.3010$ and $\log 7 = 0.8451$ , evaluate $\log 112$ .

Worked solution (try it first)

Write 112 using 2 and 7: $112 = 16 \times 7 = 2^4 \times 7$ .
So $\log 112 = 4\log 2 + \log 7$ .
That is $4(0.3010) + 0.8451 = 1.2040 + 0.8451$
$= 2.0491$ , option B.

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Question 9

Surds

Rationalise $\dfrac{2\sqrt3 + \sqrt5}{\sqrt5 - \sqrt3}$ .

Worked solution (try it first)

Multiply the top and bottom by the conjugate of the bottom, $\sqrt5 + \sqrt3$ .
Bottom: $(\sqrt5 - \sqrt3)(\sqrt5 + \sqrt3) = 5 - 3$ , which is 2.
Top: $(2\sqrt3 + \sqrt5)(\sqrt5 + \sqrt3) = 2\sqrt{15} + 6 + 5 + \sqrt{15}$ , which is $3\sqrt{15} + 11$ .
So the value is $\dfrac{3\sqrt{15} + 11}{2}$ , option A.

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Question 10

Indices & standard form

Express the product of 0.21 and 0.34 in standard form.

Worked solution (try it first)

Multiply without the decimal points: $21 \times 34 = 714$ .
There are $2 + 2 = 4$ decimal places altogether, so the product is 0.0714.
Move the point two places right to get 7.14, so the product is $7.14 \times 10^{-2}$ , option C.

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Question 12

Sets & Venn diagrams

In a survey of 50 newspaper readers, 40 read Champion and 30 read Guardian. How many read both papers?

Worked solution (try it first)

Take each reader to read at least one of the papers, so $n(C \cup G) = 50$ .
Both papers: $40 + 30 - 50 = 20$ , option D.

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Question 13

Expressions, formulae & change of subject

Make $Q$ the subject of the formula $P = \dfrac{M}{5}(X + Q) + 1$ .

Worked solution (try it first)

Subtract 1 from both sides: $P - 1 = \frac{M}{5}(X + Q)$ .
Multiply by 5 and expand: $5P - 5 = MX + MQ$ .
Subtract $MX$ and divide by $M$ : $Q = \dfrac{5P - MX - 5}{M}$ , option B.

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Question 14

If $9x^2 + 6xy + 4y^2$ is a factor of $27x^3 - 8y^3$ , find the other factor.

Worked solution (try it first)

Both terms are cubes: $27x^3 = (3x)^3$ and $8y^3 = (2y)^3$ .
A difference of cubes factorises as $A^3 - B^3 = (A - B)(A^2 + AB + B^2)$ .
With $A = 3x$ and $B = 2y$ : $(3x - 2y)(9x^2 + 6xy + 4y^2)$ .
The other factor is $3x - 2y$ , option A.

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Question 15

Factorize completely $\dfrac{x^3 + 3x^2 - 10x}{2x^2 - 8}$ .

Worked solution (try it first)

Take out the common factor $x$ on top and factorise the quadratic: $x^3 + 3x^2 - 10x = x(x + 5)(x - 2)$ .
Take out 2 on the bottom and use the difference of two squares: $2x^2 - 8 = 2(x - 2)(x + 2)$ .
Cancel the common bracket $x - 2$ : $\dfrac{x(x + 5)}{2(x + 2)}$ , option B.

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Question 16

Solve for $x$ and $y$ if $x - y = 2$ and $x^2 - y^2 = 8$ .

Worked solution (try it first)

Factorise the difference of two squares: $x^2 - y^2 = (x - y)(x + y)$ .
So $2(x + y) = 8$ , which gives $x + y = 4$ .
Add this to $x - y = 2$ : $2x = 6$ , so $x = 3$ .
Then $y = 1$ .
So $(x, y) = (3, 1)$ , option B.

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Question 17

Variation

If $y$ varies directly as the square root of $x$ and $y = 3$ when $x = 16$ , calculate $y$ when $x = 64$ .

Worked solution (try it first)

$y = k\sqrt x$ .
Put in $y = 3$ , $x = 16$ : $3 = 4k$ , so $k = \frac34$ .
When $x = 64$ : $y = \frac34 \times 8 = 6$ , option C.

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Question 18

Variation

If $x$ is inversely proportional to $y$ and $x = 2\frac12$ when $y = 2$ , find $x$ if $y = 4$ .

Worked solution (try it first)

Inverse proportion means $xy$ is constant: $k = 2\frac12 \times 2 = 5$ .
When $y = 4$ : $x = \frac54 = 1\frac14$ , option D.

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Question 19

Inequalities

For what range of values of $x$ is $\frac12x + \frac14 > \frac13x + \frac12$ ?

Worked solution (try it first)

Subtract $\frac13x$ and $\frac14$ from both sides: $\frac12x - \frac13x > \frac12 - \frac14$ .
Simplify each side: $\frac16x > \frac14$ .
Multiply both sides by 6: $x > \frac64$ , which is $\frac32$ .
So option B.

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Question 20

Inequalities

Solve the inequalities $-6 \le 4 - 2x < 5 - x$ .

Worked solution (try it first)

Split it into two inequalities.
First, $-6 \le 4 - 2x$ : add $2x$ and 6 to both sides to get $2x \le 10$ , so $x \le 5$ .
Second, $4 - 2x < 5 - x$ : add $2x$ and subtract 5 from both sides to get $-1 < x$ .
Both must hold: $-1 < x \le 5$ , option B.

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Question 21

Sequences & series (AP, GP)

Find the sum to infinity of the series $0.5 + 0.05 + 0.005 + 0.0005 + \ldots$

Worked solution (try it first)

Each term is a tenth of the one before, so this is a G.P. with $a = 0.5$ and $r = 0.05 \div 0.5 = 0.1$ .
Since $r$ is between $-1$ and 1, $S_\infty = \dfrac{a}{1 - r}$
$= \dfrac{0.5}{0.9}$ .
Multiply top and bottom by 10: $\frac59$ , option A.

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Question 22

Sequences & series (AP, GP)

The 3rd term of an arithmetic progression is $-9$ and the 7th term is $-29$ . Find the 10th term of the progression.

Worked solution (try it first)

From the 3rd term to the 7th term is 4 steps of $d$ : $4d = -29 - (-9) = -20$ , so $d = -5$ .
Go back two steps from the 3rd term: $a = -9 - 2(-5) = 1$ .
The 10th term is $a + 9d = 1 + 9(-5) = -44$ , option C.

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Question 23

Binary operations

If $x * y = x + y^2$ , find the value of $(2 * 3) * 5$ .

Worked solution (try it first)

Work out the bracket first, squaring the second number: $2 * 3 = 2 + 3^2 = 11$ .
Then $11 * 5 = 11 + 5^2 = 11 + 25 = 36$ , option A.

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Question 24

Linear & simultaneous equations

If $p$ and $q$ are two non-zero numbers and $18(p + q) = (18 + p)q$ , which of the following must be true?

Worked solution (try it first)

Expand both sides: $18p + 18q = 18q + pq$ .
Take $18q$ from both sides: $18p = pq$ .
Since $p \neq 0$ , divide both sides by $p$ : $q = 18$ , option A.

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Question 25

Matrices & determinants

If $\begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15$ , find the value of $x$ .

Worked solution (try it first)

A $2 \times 2$ determinant is $ad - bc$ : $7x - 3 \times 2 = 7x - 6$ .
So $7x - 6 = 15$ .
Add 6 to both sides: $7x = 21$ .
Divide by 7: $x = 3$ , option A.

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Question 27

Matrices & determinants

If $P = \begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}$ , what is $P^{-1}$ ?

Worked solution (try it first)

The determinant is $|P| = 2 \times 1 - (-3) \times 1 = 5$ .
Swap the two diagonal entries (2 and 1) and change the signs of the other two ( $-3$ becomes 3, and 1 becomes $-1$ ): $\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}$ .
Divide every entry by 5: $P^{-1} = \begin{pmatrix} \frac15 & \frac35 \\ -\frac15 & \frac25 \end{pmatrix}$ , option B.

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Question 28

Circle geometry

In the diagram, $O$ is the centre of the circle, $ROS$ is a straight line produced to $U$ , $UT$ is a tangent to the circle at $T$ and $\angle TRS = 25^\circ$ . Find $x$ .

Worked solution (try it first)

$RS$ is a diameter, so $\angle RTS = 90^\circ$ (angle in a semicircle).
The angles of triangle $RTS$ add up to $180^\circ$ : $\angle TSR = 180^\circ - 90^\circ - 25^\circ$
$= 65^\circ$ .
$x$ is between the tangent and the chord $TR$ , so it equals the angle in the alternate segment: $x = \angle TSR = 65^\circ$ , option A.

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Question 29

Angles, triangles & polygons

The interior angles of a quadrilateral are $(x + 15)^\circ$ , $(2x - 45)^\circ$ , $(x - 30)^\circ$ and $(x + 10)^\circ$ . Find the value of the least interior angle.

Worked solution (try it first)

The angles of a quadrilateral add up to $360^\circ$ : $(x + 15) + (2x - 45) + (x - 30) + (x + 10) = 360$ .
Collect terms: $5x - 50 = 360$ , so $5x = 410$ and $x = 82$ .
The angles are $97^\circ$ , $119^\circ$ , $52^\circ$ and $92^\circ$ .
The least is $52^\circ$ , option B.

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Question 30

Circle geometry

From the cyclic quadrilateral $TUVW$ , where $\angle TUV = (3x + 20)^\circ$ and $\angle TWV = 88^\circ$ , find the value of $x$ .

Worked solution (try it first)

$\angle TUV$ and $\angle TWV$ are opposite angles of the cyclic quadrilateral, so they add up to $180^\circ$ : $3x + 20 + 88 = 180$ .
Simplify: $3x + 108 = 180$ , so $3x = 72$ .
Divide both sides by 3: $x = 24$ , option C.

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Question 31

If the two smaller sides of a right-angled triangle are $4\text{ cm}$ and $5\text{ cm}$ , find its area.

Worked solution (try it first)

The two smaller sides of a right-angled triangle are the ones that meet at the right angle, so one is the base and the other the height.
Area $= \frac12 \times \text{base} \times \text{height}$
$= \frac12 \times 4 \times 5$
$= 10\text{ cm}^2$ , option A.

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Question 32

Plane mensuration

An arc subtends an angle of $50^\circ$ at the centre of a circle of radius $6\text{ cm}$ . Calculate the area of the sector formed. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

Area of a sector $= \frac{\theta}{360} \times \pi r^2$
$= \frac{50}{360} \times \frac{22}{7} \times 36$ .
Simplify $\frac{50 \times 36}{360} = 5$ , so the area is $5 \times \frac{22}{7} = \frac{110}{7}\text{ cm}^2$ , option B.

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Question 33

Solid mensuration

A cylindrical pipe $50\text{ m}$ long with radius $7\text{ m}$ has one end open. What is the total surface area of the pipe?

Worked solution (try it first)

One end open means one end is closed: the curved surface plus one circle.
Curved surface: $2\pi rh = 2\pi \times 7 \times 50$
$= 700\pi$ .
One end: $\pi r^2 = 49\pi$ .
Total: $700\pi + 49\pi = 749\pi\text{ m}^2$ , option D.

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Question 34

What is the locus of a point that is equidistant from the points $P(1, 3)$ and $Q(3, 5)$ ?

Worked solution (try it first)

Points equidistant from $P$ and $Q$ lie on the perpendicular bisector of $PQ$ .
The mid-point of $PQ$ is $\left(\frac{1 + 3}{2}, \frac{3 + 5}{2}\right) = (2, 4)$ .
The gradient of $PQ$ is $\frac{5 - 3}{3 - 1} = 1$ .
Perpendicular lines have gradients that multiply to $-1$ , so the bisector has gradient $-1$ .
Through $(2, 4)$ with gradient $-1$ : $y - 4 = -(x - 2)$ , so $y = -x + 6$ , option A.

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Question 35

Coordinate geometry

Find the distance between the points $\left(\frac12, \frac12\right)$ and $\left(-\frac12, -\frac12\right)$ .

Worked solution (try it first)

The changes are $\frac12 - \left(-\frac12\right) = 1$ in $x$ and 1 in $y$ .
By Pythagoras the distance is $\sqrt{1^2 + 1^2} = \sqrt2$ , option A.

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Question 36

Coordinate geometry

Find the gradient of the line passing through the points $P(1, 1)$ and $Q(2, 5)$ .

Worked solution (try it first)

Gradient is the change in $y$ over the change in $x$ : $\dfrac{5 - 1}{2 - 1}$ .
That is $\frac41 = 4$ , option A.

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Question 37

Coordinate geometry

Find the equation of a line parallel to $y = -4x + 2$ passing through $(2, 3)$ .

Worked solution (try it first)

Parallel lines have the same gradient, so the new line has gradient $-4$ .
Through $(2, 3)$ : $y - 3 = -4(x - 2)$ , so $y - 3 = -4x + 8$ and $y = -4x + 11$ .
Move everything to one side: $y + 4x - 11 = 0$ , option D.

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Question 38

Trigonometric ratios

If $\cot\theta = \frac{8}{15}$ , where $\theta$ is acute, find $\sin\theta$ .

Worked solution (try it first)

$\cot\theta = \frac{\text{adjacent}}{\text{opposite}}$
$= \frac{8}{15}$ , so draw a right-angled triangle with adjacent side 8 and opposite side 15.
Pythagoras: the hypotenuse is $\sqrt{64 + 225} = \sqrt{289} = 17$ .
So $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$
$= \frac{15}{17}$ , option B.

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Question 39

If the area of $\triangle PQR$ is $12\sqrt3\text{ cm}^2$ , $|PQ| = 8\text{ cm}$ , $|PR| = q$ and $\angle QPR = 60^\circ$ , find the value of $q$ .

Worked solution (try it first)

Use area $= \frac12 ab\sin C$ with the sides $PQ = 8$ and $PR = q$ and the angle $60^\circ$ between them.
Area $= \frac12 \times 8 \times q \times \frac{\sqrt3}{2}$
$= 2\sqrt3\,q$ .
Set $2\sqrt3\,q = 12\sqrt3$ and divide both sides by $2\sqrt3$ : $q = 6$ cm, option A.

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Question 40

Calculus (JAMB bridge)

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

Worked solution (try it first)

Chain rule: bring down the power 3 and reduce it by one, giving $3(2x + 1)^2$ .
Multiply by the derivative of the inside, $2x + 1$ , which is 2: $\frac{dy}{dx} = 6(2x + 1)^2$ , option D.

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Question 41

Calculus (JAMB bridge)

If $y = x\sin x$ , find $\dfrac{dy}{dx}$ .

Worked solution (try it first)

Product rule with $u = x$ and $v = \sin x$ : $u' = 1$ and $v' = \cos x$ .
$\frac{dy}{dx} = u'v + uv'$
$= \sin x + x\cos x$ , option B.

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Question 42

At what value of $x$ does the function $y = -3 - 2x + x^2$ attain a minimum value?

Worked solution (try it first)

At a turning point $\frac{dy}{dx} = 0$ : $-2 + 2x = 0$ , so $x = 1$ .
$\frac{d^2y}{dx^2} = 2 > 0$ , so it is a minimum.
So $x = 1$ , option A.

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Question 43

Calculus (JAMB bridge)

Evaluate $\displaystyle\int_0^2 (x^3 + x^2)\,dx$ .

Worked solution (try it first)

Integrate: $\left[\frac{x^4}{4} + \frac{x^3}{3}\right]_0^2$ .
At $x = 2$ : $\frac{16}{4} + \frac83 = 4 + \frac83$
$= 6\frac23$ .
At $x = 0$ it is 0.
So the integral is $6\frac23$ , option B.

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Question 44

Calculus (JAMB bridge)

Find $\displaystyle\int (\sin x + 2)\,dx$ .

Worked solution (try it first)

$\sin x$ integrates to $-\cos x$ , and the constant 2 integrates to $2x$ .
So $\int(\sin x + 2)\,dx = -\cos x + 2x + k$ , option C.

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Question 45

Statistics: data & averages

Marks	2	3	4	5	6	7	8
Students	3	1	5	2	4	2	3

From the table, if the pass mark is 5, how many students failed the test?

Worked solution (try it first)

With a pass mark of 5, a student fails with a mark below 5: marks 2, 3 and 4.
Add those students: $3 + 1 + 5 = 9$ , option D.

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Question 48

Dispersion & cumulative frequency

Find the standard deviation of 2, 3, 5 and 6.

Worked solution (try it first)

The mean is $\frac{2 + 3 + 5 + 6}{4} = 4$ .
The squared deviations are 4, 1, 1, 4, which add up to 10.
The variance is $\frac{10}{4} = \frac52$ , so the standard deviation is $\sqrt{\frac52}$ , option A.

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Question 49

Counting

In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

Worked solution (try it first)

Two women from five: $^5C_2 = 10$ ways.
Three men from six: $^6C_3 = 20$ ways.
Each choice of women goes with each choice of men, so multiply: $10 \times 20 = 200$ , option B.

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Question 50

Probability

If three unbiased coins are tossed, find the probability that they are all heads.

Worked solution (try it first)

Each coin shows a head with probability $\frac12$ , and the coins are independent.
Multiply for the three coins: $\frac12 \times \frac12 \times \frac12 = \frac18$ , option A.

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