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Objective paper · 39 questions · partial

JAMB 1995 · UME

Topics include Number bases, Approximation & error, Commercial arithmetic, Indices & standard form, Logarithms, Surds.

Our copy of this paper is missing questions 12, 13, 17, 24, 29, 32, 43, 44, 47, 48, 49.

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Answer every question in order, timed if you like (suggested 25 min). You're marked when you hand in, then you see where to focus and the working for each question.

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Or read it here: every question below has a worked solution.

Question 1

Calculate 33105−144253310_5 - 1442_5.

Worked solution (try it first)
  1. Change to base ten: 33105=375+75+5=4553310_5 = 375 + 75 + 5 = 455 and 14425=125+100+20+2=2471442_5 = 125 + 100 + 20 + 2 = 247.
  2. Subtract: 455−247=208455 - 247 = 208.
  3. Change back: 208=1×125+3×25+1×5+3208 = 1 \times 125 + 3 \times 25 + 1 \times 5 + 3, so the answer is 131351313_5, option A.

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Question 2

Write 3.1415926 correct to 5 decimal places.

Worked solution (try it first)
  1. Keep five decimals: 3.14159.
  2. The sixth decimal is 2.
  3. 2 is less than 5, so round down: 3.14159, option B.

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Question 3

The length of a notebook, 15 cm, was measured as 16.8 cm. Calculate the percentage error to 2 significant figures.

Worked solution (try it first)
  1. The error is 16.8−15=1.816.8 - 15 = 1.8 cm.
  2. Divide by the true length and multiply by 100: 1.815×100%=0.12×100%\frac{1.8}{15} \times 100\% = 0.12 \times 100\%.
  3. So the percentage error is 12%, option A.

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Question 4

A worker's present salary is ₦24,000 per annum. His annual increment is 10%10\% of his basic salary. What would be his annual salary at the beginning of the third year?

Worked solution (try it first)
  1. Each increment is 10%10\% of the basic salary: 0.1×24 000=0.1 \times 24\,000 = ₦2,400.
  2. At the beginning of the third year he has had two increments: 2×2400=2 \times 2400 = ₦4,800.
  3. So his salary is 24 000+4800=24\,000 + 4800 = ₦28,800, option A.

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Question 5

Express the product of 0.0014 and 0.011 in standard form.

Worked solution (try it first)
  1. Write each number in standard form: 0.0014=1.4×10−30.0014 = 1.4 \times 10^{-3} and 0.011=1.1×10−20.011 = 1.1 \times 10^{-2}.
  2. Multiply the numbers, 1.4×1.1=1.541.4 \times 1.1 = 1.54, and add the powers, 10−3×10−2=10−510^{-3} \times 10^{-2} = 10^{-5}.
  3. So the product is 1.54×10−51.54 \times 10^{-5}, option D.

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Question 6

Evaluate 8134−27133×23\dfrac{81^{\frac34} - 27^{\frac13}}{3 \times 2^3}.

Worked solution (try it first)
  1. 813481^{\frac34}: take the fourth root, then cube.
  2. 814=3\sqrt[4]{81} = 3 and 33=273^3 = 27.
  3. 271327^{\frac13} is the cube root of 27, which is 3.
  4. So the top is 27−3=2427 - 3 = 24.
  5. The bottom is 3×8=243 \times 8 = 24, so the value is 2424=1\frac{24}{24} = 1, option B.

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Question 7

Find the value of 1632+log⁡100.0001+log⁡23216^{\frac32} + \log_{10} 0.0001 + \log_2 32.

Worked solution (try it first)
  1. 163216^{\frac32} is the square root of 16, cubed: 43=644^3 = 64.
  2. 0.0001=10−40.0001 = 10^{-4}, so log⁡100.0001=−4\log_{10} 0.0001 = -4.
  3. And 32=2532 = 2^5, so log⁡232=5\log_2 32 = 5.
  4. So the value is 64−4+5=6564 - 4 + 5 = 65, option D.

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Question 8

Simplify 12−312+3\dfrac{\sqrt{12} - \sqrt3}{\sqrt{12} + \sqrt3}.

Worked solution (try it first)
  1. Take out the square factor: 12=4×3\sqrt{12} = \sqrt4 \times \sqrt3, which is 232\sqrt3.
  2. Top: 23−3=32\sqrt3 - \sqrt3 = \sqrt3.
  3. Bottom: 23+3=332\sqrt3 + \sqrt3 = 3\sqrt3.
  4. So the value is 333=13\dfrac{\sqrt3}{3\sqrt3} = \frac13, option A.

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Question 9

Four members of a school's first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?

Worked solution (try it first)
  1. Cricket has 11 players and rugby has 14.
  2. Adding gives 25, but the 4 boys in both teams are counted twice.
  3. Take them away once: 11+14−4=2111 + 14 - 4 = 21, option B.

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Question 10

If S={x:x2=9,x>4}S = \{x : x^2 = 9, x > 4\}, then SS is equal to

Worked solution (try it first)
  1. x2=9x^2 = 9 gives x=3x = 3 or x=−3x = -3.
  2. Neither is greater than 4, so no number satisfies both conditions and SS is the empty set, ∅\varnothing, option C.

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Question 11

If x−1x - 1 and x+1x + 1 are both factors of x3+px2+qx+6x^3 + px^2 + qx + 6, evaluate pp and qq.

Worked solution (try it first)
  1. By the factor theorem, the expression is 0 at x=1x = 1: 1+p+q+6=01 + p + q + 6 = 0, so p+q=−7p + q = -7.
  2. It is 0 at x=−1x = -1: −1+p−q+6=0-1 + p - q + 6 = 0, so p−q=−5p - q = -5.
  3. Add the equations: 2p=−122p = -12, so p=−6p = -6 and q=−1q = -1, option A.

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Question 14

The graph of f(x)=x2−5x+6f(x) = x^2 - 5x + 6 crosses the xx-axis at the points

Worked solution (try it first)
  1. The graph crosses the xx-axis where f(x)=0f(x) = 0: x2−5x+6=0x^2 - 5x + 6 = 0.
  2. Factorise: (x−2)(x−3)=0(x - 2)(x - 3) = 0, so x=2x = 2 or x=3x = 3.
  3. So the points are (2,0)(2, 0) and (3,0)(3, 0), option D.

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Question 15

Factorize completely the expression abx2+6y−3ax−2byxabx^2 + 6y - 3ax - 2byx.

Worked solution (try it first)
  1. Group the terms with axax and those with yy: (abx2−3ax)−(2bxy−6y)(abx^2 - 3ax) - (2bxy - 6y).
  2. Take out the common factors: ax(bx−3)−2y(bx−3)ax(bx - 3) - 2y(bx - 3).
  3. Take out the common bracket: (ax−2y)(bx−3)(ax - 2y)(bx - 3), option A.

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Question 16

Solve the inequality (x−3)(x−4)≤0(x - 3)(x - 4) \le 0.

Worked solution (try it first)
  1. The roots are x=3x = 3 and x=4x = 4, where the product is zero.
  2. The product is negative between the roots, because one bracket is positive and the other negative there.
  3. "Less than or equal to 0" includes the roots, so 3≤x≤43 \le x \le 4, option A.

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Question 18

Simplify x2−1x3+2x2−x−2\dfrac{x^2 - 1}{x^3 + 2x^2 - x - 2}.

Worked solution (try it first)
  1. Group the bottom in pairs: x3+2x2−x−2=x2(x+2)−(x+2)x^3 + 2x^2 - x - 2 = x^2(x + 2) - (x + 2), which is (x+2)(x2−1)(x + 2)(x^2 - 1).
  2. So the fraction is x2−1(x+2)(x2−1)\dfrac{x^2 - 1}{(x + 2)(x^2 - 1)}.
  3. Cancel x2−1x^2 - 1.
  4. This leaves 1x+2\dfrac{1}{x + 2}, option A.

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Question 19

Express 5x−12(x−2)(x−3)\dfrac{5x - 12}{(x - 2)(x - 3)} in partial fractions.

Worked solution (try it first)
  1. Write 5x−12(x−2)(x−3)=Ax−2+Bx−3\frac{5x - 12}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}, so 5x−12=A(x−3)+B(x−2)5x - 12 = A(x - 3) + B(x - 2).
  2. Put x=2x = 2: −2=−A-2 = -A, so A=2A = 2.
  3. Put x=3x = 3: 3=B3 = B.
  4. So the partial fractions are 2x−2+3x−3\frac{2}{x - 2} + \frac{3}{x - 3}, option B.

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Question 20✱✱

Use the graph of the curve y=f(x)y = f(x) to solve the inequality f(x)>0f(x) > 0.

−1120xy
Worked solution (try it first)
  1. f(x)>0f(x) > 0 where the curve is above the xx-axis.
  2. The curve crosses the axis at −1-1, 11 and 22.
  3. It is above the axis between −1-1 and 11, and to the right of 22.
  4. The inequality is strict and f(x)=0f(x) = 0 at the crossings, so leave them out: −1<x<1-1 < x < 1 or x>2x > 2, option A.

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Question 21

Which of the following binary operations is commutative in the set of integers?

Worked solution (try it first)
  1. An operation is commutative when a∗b=b∗aa * b = b * a, that is, when swapping aa and bb leaves the formula the same.
  2. In a+b−aba + b - ab, swapping gives b+a−bab + a - ba, which is the same, because addition and multiplication are commutative.
  3. The others change: for example, with option A, 1∗2=51 * 2 = 5 but 2∗1=42 * 1 = 4.
  4. So the answer is option B.

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Question 22

If a∗b=+aba * b = +\sqrt{ab}, evaluate 2∗(12∗27)2 * (12 * 27).

Worked solution (try it first)
  1. Work out the bracket first: 12∗27=12×2712 * 27 = \sqrt{12 \times 27}
    =324= \sqrt{324}
    =18= 18.
  2. Then 2∗18=2×182 * 18 = \sqrt{2 \times 18}
    =36= \sqrt{36}
    =6= 6, option C.

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Question 23

Find the sum to infinity of the sequence 1,910,(910)2,(910)3,…1, \frac9{10}, \left(\frac9{10}\right)^2, \left(\frac9{10}\right)^3, \dots

Worked solution (try it first)
  1. This is a G.P. with a=1a = 1 and r=910r = \frac{9}{10}.
  2. Since rr is between −1-1 and 1, S∞=a1−rS_\infty = \dfrac{a}{1 - r}, and 1−910=1101 - \frac{9}{10} = \frac{1}{10}.
  3. So S∞=1÷110=10S_\infty = 1 \div \frac{1}{10} = 10, option D.

Also set as JAMB 2015 · UTME · Q26

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Question 25

If X=(1203)X = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} and Y=(2143)Y = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}, find XYXY.

Worked solution (try it first)
  1. For XYXY, multiply each row of XX by each column of YY.
  2. Row 1 of XX is (1,2)(1, 2): with column 1 it gives 2+8=102 + 8 = 10, and with column 2, 1+6=71 + 6 = 7.
  3. Row 2 of XX is (0,3)(0, 3): with column 1 it gives 0+12=120 + 12 = 12, and with column 2, 0+9=90 + 9 = 9.
  4. So XY=(107129)XY = \begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}, option A.

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Question 26

Determine the value of xx in the figure.

81°53°x
Worked solution (try it first)
  1. Angles in triangle ABCABC add up to 180∘180^\circ: ∠ABC=180∘−81∘−53∘\angle ABC = 180^\circ - 81^\circ - 53^\circ
    =46∘= 46^\circ.
  2. ABCDABCD is a cyclic quadrilateral, and BB and DD are opposite vertices.
  3. Opposite angles of a cyclic quadrilateral add up to 180∘180^\circ: x=180∘−46∘=134∘x = 180^\circ - 46^\circ = 134^\circ, option A.

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Question 27

PTPT is a tangent to the circle TYZXTYZX, YT=YXYT = YX and ∠PTX=50∘\angle PTX = 50^\circ. Calculate ∠TZY\angle TZY.

50°?TXYZP
Worked solution (try it first)
  1. The angle between tangent TPTP and chord TXTX equals the angle in the alternate segment, so ∠TYX=∠PTX=50∘\angle TYX = \angle PTX = 50^\circ.
  2. YT=YXYT = YX, so triangle TYXTYX is isosceles: ∠YXT=180∘−50∘2\angle YXT = \dfrac{180^\circ - 50^\circ}{2}
    =65∘= 65^\circ.
  3. Angles in the same segment are equal.
  4. ∠TZY\angle TZY and ∠TXY\angle TXY both stand on arc TYTY, so ∠TZY=65∘\angle TZY = 65^\circ, option B.

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Question 28

In a triangle XYZXYZ, ∠YXZ=44∘\angle YXZ = 44^\circ and ∠XYZ=112∘\angle XYZ = 112^\circ. Calculate the acute angle between the internal bisectors of ∠XYZ\angle XYZ and ∠XZY\angle XZY.

Worked solution (try it first)
  1. The angles of triangle XYZXYZ add up to 180∘180^\circ: ∠XZY=180∘−44∘−112∘\angle XZY = 180^\circ - 44^\circ - 112^\circ
    =24∘= 24^\circ.
  2. The bisectors make 112∘÷2=56∘112^\circ \div 2 = 56^\circ and 24∘÷2=12∘24^\circ \div 2 = 12^\circ with YZYZ.
  3. In the triangle they form with YZYZ, the angle where they meet is 180∘−56∘−12∘=112∘180^\circ - 56^\circ - 12^\circ = 112^\circ.
  4. The acute angle between the lines is 180∘−112∘=68∘180^\circ - 112^\circ = 68^\circ, option C.

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Question 30

Two perpendicular lines PQPQ and QRQR intersect at (1,−1)(1, -1). If the equation of PQPQ is x−2y+4=0x - 2y + 4 = 0, find the equation of QRQR.

Worked solution (try it first)
  1. Rearrange PQPQ: 2y=x+42y = x + 4, so y=12x+2y = \frac12x + 2 and its gradient is 12\frac12.
  2. Perpendicular gradients multiply to −1-1, so QRQR has gradient −2-2.
  3. QRQR passes through (1,−1)(1, -1): y−(−1)=−2(x−1)y - (-1) = -2(x - 1), so y+1=−2x+2y + 1 = -2x + 2.
  4. Collect terms on one side: 2x+y−1=02x + y - 1 = 0, option D.

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Question 31

PP is on the locus of points equidistant from two given points XX and YY. UVUV is a straight line through YY parallel to the locus. If ∠PYU=40∘\angle PYU = 40^\circ, find ∠XPY\angle XPY.

Worked solution (try it first)
  1. The locus of points equidistant from XX and YY is the perpendicular bisector of XYXY.
  2. UVUV is parallel to it, so UVUV is perpendicular to XYXY.
  3. So the angle between YUYU and YXYX is 90∘90^\circ, and ∠PYX=90∘−40∘\angle PYX = 90^\circ - 40^\circ
    =50∘= 50^\circ.
  4. PP is on the locus, so PX=PYPX = PY and triangle PXYPXY is isosceles: ∠PXY=∠PYX=50∘\angle PXY = \angle PYX = 50^\circ.
  5. Angles in a triangle add up to 180∘180^\circ: ∠XPY=180∘−100∘\angle XPY = 180^\circ - 100^\circ
    =80∘= 80^\circ, option B.

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Question 33

The base diameter of a cylinder is 14 cm and its height is 12 cm. Calculate the total surface area if the cylinder has both a base and a top. [π=227]\left[\pi = \frac{22}{7}\right]

Worked solution (try it first)
  1. Radius =14÷2=7= 14 \div 2 = 7 cm.
  2. With a base and a top, the total surface area is 2πr(r+h)2\pi r(r + h).
  3. 2×227×7×(7+12)=44×192 \times \frac{22}{7} \times 7 \times (7 + 12) = 44 \times 19
    =836 cm2= 836\text{ cm}^2, option A.

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Question 34

In the diagram, PR=10PR = 10 cm and ∠QPR=30∘\angle QPR = 30^\circ. Find PQPQ if the area of triangle PQRPQR is 35 cm235\text{ cm}^2.

10 cm30°PQR
Worked solution (try it first)
  1. Use area =12absin⁡C= \frac12ab\sin C with the 30∘30^\circ angle between PQPQ and PRPR: 12×PQ×10×sin⁡30∘=35\frac12 \times PQ \times 10 \times \sin30^\circ = 35.
  2. sin⁡30∘=12\sin30^\circ = \frac12, so the left side is 2.5×PQ2.5 \times PQ.
  3. So PQ=35÷2.5=14PQ = 35 \div 2.5 = 14 cm, option C.

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Question 35✱✱

A schoolboy lying on the ground 30 m away from the foot of a water tower observes that the angle of elevation of the top of the tower is 60∘60^\circ. Calculate the height of the water tower.

Worked solution (try it first)
  1. The height is opposite the 60∘60^\circ angle and the 30 m is adjacent, so tan⁡60∘=height30\tan60^\circ = \frac{\text{height}}{30}.
  2. Multiply both sides by 30: height =30tan⁡60∘= 30\tan60^\circ.
  3. Since tan⁡60∘=3\tan60^\circ = \sqrt3, the height is 30330\sqrt3 m, option B.

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Question 36

QRSQRS is a triangle with QS=12QS = 12 m, ∠RQS=30∘\angle RQS = 30^\circ and ∠QRS=45∘\angle QRS = 45^\circ. Calculate the length of RSRS.

Worked solution (try it first)
  1. Sine rule: RSRS faces the 30∘30^\circ at QQ, and QS=12QS = 12 faces the 45∘45^\circ at RR.
  2. So RSsin⁡30∘=12sin⁡45∘\dfrac{RS}{\sin30^\circ} = \dfrac{12}{\sin45^\circ}.
  3. So RS=12×1222RS = \dfrac{12 \times \frac12}{\frac{\sqrt2}{2}}
    =122= \dfrac{12}{\sqrt2}.
  4. Rationalise: 1222=62\dfrac{12\sqrt2}{2} = 6\sqrt2 m, option C.

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Question 37

Which of the following is a sketch of y=3sin⁡xy = 3\sin x?

A−π−π/2π/2π3−3B−π−π/2π/2π3−3C−π−π/2π/2π3−3D−π−π/2π/2π3−3
Worked solution (try it first)
  1. At x=0x = 0, y=3sin⁡0=0y = 3\sin 0 = 0, so the curve passes through the origin.
  2. That rules out B and D.
  3. At x=π2x = \frac{\pi}{2}, y=3sin⁡π2=3y = 3\sin\frac{\pi}{2} = 3, a maximum.
  4. At x=−π2x = -\frac{\pi}{2}, y=−3y = -3, a minimum.
  5. Only sketch C rises through the origin to 3 at π2\frac{\pi}{2}, so the answer is option C.

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Question 38

The derivative of cosec x\text{cosec}\,x is

Worked solution (try it first)
  1. Write cosec x=(sin⁡x)−1\text{cosec}\,x = (\sin x)^{-1} and use the chain rule: the derivative is −(sin⁡x)−2cos⁡x=−cos⁡xsin⁡2x-(\sin x)^{-2}\cos x = -\dfrac{\cos x}{\sin^2 x}.
  2. Split it as −cos⁡xsin⁡x×1sin⁡x=−cot⁡x cosec x-\dfrac{\cos x}{\sin x} \times \dfrac{1}{\sin x} = -\cot x\,\text{cosec}\,x, option B.

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Question 39

For what value of xx is the tangent to the curve y=x2−4x+3y = x^2 - 4x + 3 parallel to the xx-axis?

Worked solution (try it first)
  1. A tangent parallel to the xx-axis has gradient 0.
  2. dydx=2x−4=0\frac{dy}{dx} = 2x - 4 = 0, so x=2x = 2, option B.

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Question 40

Two variables xx and yy are such that dydx=4x−3\frac{dy}{dx} = 4x - 3 and y=5y = 5 when x=2x = 2. Find yy in terms of xx.

Worked solution (try it first)
  1. Integrate: y=2x2−3x+cy = 2x^2 - 3x + c.
  2. Put in x=2x = 2, y=5y = 5: 5=8−6+c5 = 8 - 6 + c, so c=3c = 3.
  3. So y=2x2−3x+3y = 2x^2 - 3x + 3, option B.

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Question 41

Find the area bounded by the curve y=3x2−2x+1y = 3x^2 - 2x + 1, the ordinates x=1x = 1 and x=3x = 3, and the xx-axis.

Worked solution (try it first)
  1. The area is ∫13(3x2−2x+1) dx=[x3−x2+x]13\int_1^3 (3x^2 - 2x + 1)\,dx = \left[x^3 - x^2 + x\right]_1^3.
  2. At x=3x = 3: 27−9+3=2127 - 9 + 3 = 21.
  3. At x=1x = 1: 1−1+1=11 - 1 + 1 = 1.
  4. Subtract: 21−1=2021 - 1 = 20, option D.

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Question 42

The frequency distribution below shows the ages of students in a secondary school. In a pie chart of the data, the angle for the 15-year-olds is

Age in years 13 14 15 16 17
No. of students 3 10 30 42 15
Worked solution (try it first)
  1. Total students: 3+10+30+42+15=1003 + 10 + 30 + 42 + 15 = 100.
  2. The 15-year-olds' share: 30100×360∘=108∘\frac{30}{100} \times 360^\circ = 108^\circ, option D.

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Question 45

Find the standard deviation of the data in the table below.

Class 1–3 4–6 7–9
Frequency 5 8 5
Worked solution (try it first)
  1. Use the class mid-points 2, 5 and 8.
  2. The total is 5×2+8×5+5×8=905 \times 2 + 8 \times 5 + 5 \times 8 = 90 over 18 values, so the mean is 9018=5\frac{90}{18} = 5.
  3. The deviations are −3-3, 0 and 3, so ∑fd2=5×9+0+5×9\sum f d^2 = 5 \times 9 + 0 + 5 \times 9
    =90= 90.
  4. The variance is 9018=5\frac{90}{18} = 5, so the standard deviation is 5\sqrt5, option D.

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Question 46

The variance of the scores 1, 2, 3, 4, 5 is

Worked solution (try it first)
  1. The mean of 1, 2, 3, 4, 5 is 3.
  2. The squared deviations are 4, 1, 0, 1, 4, which add up to 10.
  3. The variance is 105=2.0\frac{10}{5} = 2.0, option C.

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Question 50

A die has four of its faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive times, the top face will be white in both cases?

Worked solution (try it first)
  1. On one throw, 4 of the 6 faces are white, so P(white)=46=23P(\text{white}) = \frac46 = \frac23.
  2. The two throws are independent, so multiply: 23×23=49\frac23 \times \frac23 = \frac49, option C.

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