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Objective paper · 45 questions · partial

JAMB 2000 · UME

Topics include Sets & Venn diagrams, Commercial arithmetic, Surds, Approximation & error, Indices & standard form, Number bases.

Our copy of this paper is missing questions 7, 15, 33, 35, 38.

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Answer every question in order, timed if you like (suggested 30 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

Let P={1,2,u,v,w,x}P = \{1, 2, u, v, w, x\}, Q={2,3,u,v,w,5,6,y}Q = \{2, 3, u, v, w, 5, 6, y\} and R={2,3,4,v,x,y}R = \{2, 3, 4, v, x, y\}. Determine (P−Q)∩R(P - Q) \cap R.

Worked solution (try it first)
  1. P−QP - Q is the elements of PP that are not in QQ.
  2. Of 1,2,u,v,w,x1, 2, u, v, w, x, only 1 and xx are not in QQ, so P−Q={1,x}P - Q = \{1, x\}.
  3. Of these, only xx is in RR, so (P−Q)∩R={x}(P - Q) \cap R = \{x\}, option C.

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Question 2

If the population of a town was 240,000 in January 1998 and it increased by 2%2\% each year, what would be the population of the town in January 2000?

Worked solution (try it first)
  1. Each year the population is multiplied by 1.021.02, and January 1998 to January 2000 is 2 years.
  2. So the population is 240 000×1.022=240 000×1.0404240\,000 \times 1.02^2 = 240\,000 \times 1.0404.
  3. That is 249,696, option B.

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Question 3

If 23−23+22=m+n6\dfrac{2\sqrt3 - \sqrt2}{\sqrt3 + 2\sqrt2} = m + n\sqrt6, find the values of mm and nn respectively.

Worked solution (try it first)
  1. Multiply the top and bottom by the conjugate of the bottom, 3−22\sqrt3 - 2\sqrt2.
  2. Bottom: (3+22)(3−22)=3−8(\sqrt3 + 2\sqrt2)(\sqrt3 - 2\sqrt2) = 3 - 8, which is −5-5.
  3. Top: (23−2)(3−22)=6−46−6+4(2\sqrt3 - \sqrt2)(\sqrt3 - 2\sqrt2) = 6 - 4\sqrt6 - \sqrt6 + 4, which is 10−5610 - 5\sqrt6.
  4. Divide by −5-5: −2+6-2 + \sqrt6.
  5. So m=−2m = -2 and n=1n = 1, option B.

Also set as JAMB 2017 · UTME (set 2) · Q33

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Question 4

In a youth club with 94 members, 60 like modern music and 50 like traditional music. The number of members who like both is three times the number who like neither. How many members like only one type of music?

Worked solution (try it first)
  1. Let nn members like neither, so 3n3n like both.
  2. Members who like at least one: 60+50−3n=94−n60 + 50 - 3n = 94 - n.
  3. So 110−3n=94−n110 - 3n = 94 - n, which gives 2n=162n = 16 and n=8n = 8.
  4. So 24 like both.
  5. Modern only: 60−24=3660 - 24 = 36.
  6. Traditional only: 50−24=2650 - 24 = 26.
  7. Only one type: 36+26=6236 + 26 = 62, option C.

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Question 5

Evaluate (2.813×10−3)×1.0635.637×10−2\dfrac{(2.813 \times 10^{-3}) \times 1.063}{5.637 \times 10^{-2}}, reducing each number to two significant figures and leaving your answer in two significant figures.

Worked solution (try it first)
  1. Round each number to 2 significant figures: 2.8×10−32.8 \times 10^{-3}, 1.11.1 and 5.6×10−25.6 \times 10^{-2}.
  2. Top: 2.8×1.1=3.082.8 \times 1.1 = 3.08, so the top is 3.08×10−33.08 \times 10^{-3}.
  3. Divide: 3.08÷5.6=0.553.08 \div 5.6 = 0.55, and 10−3÷10−2=10−110^{-3} \div 10^{-2} = 10^{-1}.
  4. So the value is 0.55×10−1=0.0550.55 \times 10^{-1} = 0.055, option B.

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Question 6

A man wishes to keep some money in a savings deposit at 25%25\% compound interest so that after 3 years he can buy a car for ₦150,000. How much does he need to deposit now?

Worked solution (try it first)
  1. At 25%25\% compound interest the money is multiplied by 1.251.25 each year, so after 3 years a deposit PP becomes P×1.253P \times 1.25^3.
  2. Set this equal to the car price: 1.953125P=150 0001.953125P = 150\,000.
  3. Divide: P=150 000÷1.953125=76 800P = 150\,000 \div 1.953125 = 76\,800.
  4. He needs to deposit ₦76,800.00, option D.

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Question 8

Audu bought an article for ₦50,000 and sold it to Femi at a loss of x%x\%. Femi later sold the article to Oche at a profit of 40%40\%. If Femi made a profit of ₦10,000, find the value of xx.

Worked solution (try it first)
  1. Femi's 40%40\% profit is of what he paid, CC: 0.4C=10 0000.4C = 10\,000, so C=C = ₦25,000.
  2. So Audu sold for ₦25,000 an article that cost him ₦50,000, a loss of ₦25,000.
  3. As a percentage of his cost: 25 00050 000×100%=50%\dfrac{25\,000}{50\,000} \times 100\% = 50\%, so x=50x = 50, option B.

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Question 9

Simplify 3(2n+1)−4(2n−1)2n+1−2n\dfrac{3(2^{n + 1}) - 4(2^{n - 1})}{2^{n + 1} - 2^n}.

Worked solution (try it first)
  1. Write each power in terms of 2n2^n: 2n+1=2×2n2^{n + 1} = 2 \times 2^n and 2n−1=12×2n2^{n - 1} = \frac12 \times 2^n.
  2. Top: 3×2×2n−4×12×2n=6×2n−2×2n3 \times 2 \times 2^n - 4 \times \frac12 \times 2^n = 6 \times 2^n - 2 \times 2^n
    =4×2n= 4 \times 2^n.
  3. Bottom: 2×2n−2n=2n2 \times 2^n - 2^n = 2^n.
  4. So the fraction is 4×2n2n=4\frac{4 \times 2^n}{2^n} = 4, option C.

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Question 10

If P3446−23P26=2PP26P344_6 - 23P2_6 = 2PP2_6, find the value of the digit PP.

Worked solution (try it first)
  1. Units: 4−2=24 - 2 = 2, which matches.
  2. Sixes: 4−P4 - P must end in PP.
  3. Without a borrow, 4−P=P4 - P = P gives P=2P = 2.
  4. But then the 36s column is 3−3=03 - 3 = 0, not 2, so that fails.
  5. With a borrow of 6: 10−P=P10 - P = P (in base ten), so P=5P = 5.
  6. Check the 36s: 2+6−3=52 + 6 - 3 = 5, and the 216s: 4−2=24 - 2 = 2.
  7. Both match.
  8. So P=5P = 5, option D.

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Question 11

Evaluate 5−3log⁡52×22log⁡235^{-3\log_5 2} \times 2^{2\log_2 3}.

Worked solution (try it first)
  1. Move the −3-3 inside: −3log⁡52=log⁡52−3-3\log_5 2 = \log_5 2^{-3}
    =log⁡518= \log_5 \frac18.
  2. Since alog⁡aN=Na^{\log_a N} = N, the first factor is 18\frac18.
  3. In the same way 2log⁡23=log⁡292\log_2 3 = \log_2 9, so the second factor is 2log⁡29=92^{\log_2 9} = 9.
  4. So the value is 18×9=98=118\frac18 \times 9 = \frac98 = 1\frac18, option B.

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Question 12

A binary operation ∗* is defined by a∗b=aba * b = a^b. If a∗2=2−aa * 2 = 2 - a, find the possible values of aa.

Worked solution (try it first)
  1. a∗2=a2a * 2 = a^2, so the equation is a2=2−aa^2 = 2 - a.
  2. Bring everything to one side: a2+a−2=0a^2 + a - 2 = 0, which factorises as (a−1)(a+2)=0(a - 1)(a + 2) = 0.
  3. So a=1a = 1 or a=−2a = -2, option D.

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Question 13

The 3rd term of an A.P. is 4x−2y4x - 2y and the 9th term is 10x−8y10x - 8y. Find the common difference.

Worked solution (try it first)
  1. From the 3rd term to the 9th term is 6 steps of dd: (a+8d)−(a+2d)=6d(a + 8d) - (a + 2d) = 6d.
  2. Subtract the terms: (10x−8y)−(4x−2y)=6x−6y(10x - 8y) - (4x - 2y) = 6x - 6y.
  3. So 6d=6x−6y6d = 6x - 6y and d=x−yd = x - y, option C.

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Question 14

Find the inverse of pp under the binary operation p∗q=p+q−pqp * q = p + q - pq, where pp and qq are real numbers and zero is the identity.

Worked solution (try it first)
  1. The inverse qq of pp combines with pp to give the identity 0: p+q−pq=0p + q - pq = 0.
  2. Collect the qq terms: q−pq=−pq - pq = -p, so q(1−p)=−pq(1 - p) = -p.
  3. Divide by 1−p1 - p: q=−p1−p=pp−1q = \dfrac{-p}{1 - p} = \dfrac{p}{p - 1}, option C.

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Question 16

Evaluate (12−14+18−116+… )−1\left(\frac12 - \frac14 + \frac18 - \frac1{16} + \dots\right) - 1.

Worked solution (try it first)
  1. The bracket is a G.P. with a=12a = \frac12 and r=−14÷12=−12r = -\frac14 \div \frac12 = -\frac12.
  2. Its sum to infinity is a1−r=1/23/2\dfrac{a}{1 - r} = \dfrac{1/2}{3/2}
    =13= \frac13.
  3. So the expression is 13−1=−23\frac13 - 1 = -\frac23, option C.

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Question 17✱✱

Which of the graphs represents the solution of the simultaneous inequalities 2x−2≤y2x - 2 \le y and 2y−2≤x2y - 2 \le x?

−2−222A.−2−222B.−2−222C.−2−222D.
Worked solution (try it first)
  1. 2x−2≤y2x - 2 \le y is the region on or above the line y=2x−2y = 2x - 2, through (1,0)(1, 0) and (0,−2)(0, -2).
  2. Rearrange 2y−2≤x2y - 2 \le x: add 2 and divide by 2 to get y≤12x+1y \le \frac12x + 1, the region on or below the line through (−2,0)(-2, 0) and (0,1)(0, 1).
  3. The lines meet at (2,2)(2, 2).
  4. Test the origin: −2≤0-2 \le 0 and −2≤0-2 \le 0 are both true, so the region contains (0,0)(0, 0).
  5. Graph B shades the wedge between these two lines that contains the origin, so the answer is option B.

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Question 18

Find the values of tt for which the determinant of the matrix (t−400−1t+1134t−2)\begin{pmatrix} t - 4 & 0 & 0 \\ -1 & t + 1 & 1 \\ 3 & 4 & t - 2 \end{pmatrix} is zero.

Worked solution (try it first)
  1. The first row is (t−4,0,0)(t - 4, 0, 0), so expand along it: the determinant is (t−4)[(t+1)(t−2)−1×4](t - 4)\left[(t + 1)(t - 2) - 1 \times 4\right].
  2. Inside the bracket, t2−t−2−4=t2−t−6t^2 - t - 2 - 4 = t^2 - t - 6, which factorises as (t−3)(t+2)(t - 3)(t + 2).
  3. So the determinant is (t−4)(t−3)(t+2)(t - 4)(t - 3)(t + 2), which is zero when t=4t = 4, 3 or −2-2, option D.

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Question 19

If (x−1)(x - 1), (x+1)(x + 1) and (x−2)(x - 2) are factors of the polynomial ax3+bx2+cx−1ax^3 + bx^2 + cx - 1, find aa, bb, cc respectively.

Worked solution (try it first)
  1. By the factor theorem, f(1)=0f(1) = 0: a+b+c−1=0a + b + c - 1 = 0.
  2. And f(−1)=0f(-1) = 0: −a+b−c−1=0-a + b - c - 1 = 0.
  3. Add these: 2b−2=02b - 2 = 0, so b=1b = 1 and then a+c=0a + c = 0, so c=−ac = -a.
  4. f(2)=0f(2) = 0: 8a+4b+2c−1=08a + 4b + 2c - 1 = 0.
  5. Put in b=1b = 1 and c=−ac = -a: 6a+3=06a + 3 = 0, so a=−12a = -\frac12.
  6. Then c=12c = \frac12, so a,b,ca, b, c are −12,1,12-\frac12, 1, \frac12, option A.

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Question 20

A trader realizes 10x−x210x - x^2 naira profit from the sale of xx bags of corn. How many bags will give him the maximum profit?

Worked solution (try it first)
  1. The profit is greatest where its gradient is zero: ddx(10x−x2)=10−2x\frac{d}{dx}(10x - x^2) = 10 - 2x.
  2. 10−2x=010 - 2x = 0, so x=5x = 5.
  3. The second derivative is −2<0-2 < 0, so this is a maximum: 5 bags, option B.

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Question 21

Solve the inequality 2−x>x22 - x > x^2.

Worked solution (try it first)
  1. Bring everything to one side: 0>x2+x−20 > x^2 + x - 2, that is x2+x−2<0x^2 + x - 2 < 0.
  2. Factorise: (x+2)(x−1)<0(x + 2)(x - 1) < 0, so the roots are x=−2x = -2 and x=1x = 1.
  3. "Less than 0" means between the roots: −2<x<1-2 < x < 1, option D.

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Question 22

If α\alpha and β\beta are the roots of the equation 3x2+5x−2=03x^2 + 5x - 2 = 0, find the value of 1α+1β\frac1\alpha + \frac1\beta.

Worked solution (try it first)
  1. For ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −ba-\frac ba and the product is ca\frac ca.
  2. So α+β=−53\alpha + \beta = -\frac53 and αβ=−23\alpha\beta = -\frac23.
  3. Add the fractions: 1α+1β=α+βαβ\dfrac1\alpha + \dfrac1\beta = \dfrac{\alpha + \beta}{\alpha\beta}.
  4. So the value is −53÷(−23)=52-\frac53 \div \left(-\frac23\right) = \frac52, option D.

Also set as JAMB 2017 · UTME (set 2) · Q34

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Question 23

Find the minimum value of the function f(θ)=23−cos⁡θf(\theta) = \dfrac{2}{3 - \cos\theta} for 0≤θ≤2π0 \le \theta \le 2\pi.

Worked solution (try it first)
  1. The top is fixed at 2, so the fraction is smallest when the bottom, 3−cos⁡θ3 - \cos\theta, is largest.
  2. cos⁡θ\cos\theta is at least −1-1, so the bottom is at most 3−(−1)=43 - (-1) = 4, at θ=π\theta = \pi.
  3. So the minimum is 24=12\frac24 = \frac12, option A.

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Question 24

A frustum of a pyramid with a square base has its upper and lower sections as squares of sides 2 m and 5 m respectively, and the distance between them is 6 m. Find the height of the pyramid from which the frustum was obtained.

Worked solution (try it first)
  1. Let the full pyramid have height HH.
  2. The small pyramid cut off the top has height H−6H - 6 and base side 2 m.
  3. Similar pyramids: side is proportional to height, so 25=H−6H\dfrac{2}{5} = \dfrac{H - 6}{H}.
  4. Cross-multiply: 2H=5H−302H = 5H - 30, so 3H=303H = 30 and H=10H = 10 m, option D.

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Question 25

PP is a point on one side of the straight line UVUV and PP moves parallel to UVUV. If the straight line STST is the locus of PP and ∠VUS=50∘\angle VUS = 50^\circ, find ∠UST\angle UST.

Worked solution (try it first)
  1. PP moves parallel to UVUV, so its locus STST is parallel to UVUV, and USUS is a transversal.
  2. ∠VUS\angle VUS and ∠UST\angle UST lie between the parallels on the same side of USUS, so they are co-interior angles and add up to 180∘180^\circ.
  3. So ∠UST=180∘−50∘\angle UST = 180^\circ - 50^\circ
    =130∘= 130^\circ, option B.

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Question 26

A ship sails a distance of 50 km in the direction S50∘50^\circE and then 50 km in the direction N40∘40^\circE. Find the bearing of the ship from its original position.

Worked solution (try it first)
  1. As three-figure bearings, S50∘50^\circE is 130∘130^\circ and N40∘40^\circE is 040∘040^\circ.
  2. They differ by 90∘90^\circ, so the two legs meet at a right angle.
  3. The legs are equal, so the triangle is isosceles and the line back to the start makes 45∘45^\circ with the first leg.
  4. Turning 45∘45^\circ from 130∘130^\circ towards the second leg gives 085∘085^\circ, which is N85∘85^\circE, option D.

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Question 27

An equilateral triangle of side 3\sqrt3 cm is inscribed in a circle. Find the radius of the circle.

Worked solution (try it first)
  1. The height of an equilateral triangle of side 3\sqrt3 is 3sin⁡60∘=3×32\sqrt3 \sin60^\circ = \sqrt3 \times \frac{\sqrt3}{2}
    =32= \frac32 cm.
  2. The centre of the circle is where the medians meet, two-thirds of the way down each height from the vertex.
  3. So the radius is 23×32=1\frac23 \times \frac32 = 1 cm, option C.

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Question 28

3y=4x−13y = 4x - 1 and Ky=x+3Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find KK.

Worked solution (try it first)
  1. 3y=4x−13y = 4x - 1 gives y=43x−13y = \frac43x - \frac13, so its gradient is 43\frac43.
  2. Ky=x+3Ky = x + 3 gives y=1Kx+3Ky = \frac1Kx + \frac3K, so its gradient is 1K\frac1K.
  3. Perpendicular gradients multiply to −1-1: 43×1K=−1\frac43 \times \frac1K = -1, so 1K=−34\frac1K = -\frac34.
  4. Turn it upside down: K=−43K = -\frac43, option A.

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Question 29

In the diagram, ∠RPS=50∘\angle RPS = 50^\circ, ∠RPQ=30∘\angle RPQ = 30^\circ and PQ=QRPQ = QR. Find the value of ∠PRS\angle PRS.

50°30°?PQRS
Worked solution (try it first)
  1. PQ=QRPQ = QR, so triangle PQRPQR is isosceles and ∠QRP=∠QPR=30∘\angle QRP = \angle QPR = 30^\circ.
  2. So ∠PQR=180∘−60∘\angle PQR = 180^\circ - 60^\circ
    =120∘= 120^\circ.
  3. Opposite angles of cyclic quadrilateral PQRSPQRS add up to 180∘180^\circ: ∠PSR=180∘−120∘\angle PSR = 180^\circ - 120^\circ
    =60∘= 60^\circ.
  4. Triangle PRSPRS: ∠PRS=180∘−50∘−60∘\angle PRS = 180^\circ - 50^\circ - 60^\circ
    =70∘= 70^\circ, option B.

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Question 30

In the diagram, EFGHEFGH is a circle with centre OO. FHFH is a diameter and GEGE is a chord which meets FHFH at right angles at NN. If NH=8NH = 8 cm and EG=24EG = 24 cm, calculate FHFH.

OFHNEG
Worked solution (try it first)
  1. A diameter perpendicular to a chord bisects it, so EN=NG=12EN = NG = 12 cm.
  2. Intersecting chords: FN×NH=EN×NGFN \times NH = EN \times NG, so 8×FN=12×12=1448 \times FN = 12 \times 12 = 144 and FN=18FN = 18 cm.
  3. So FHFH is FN+NH=18+8=26FN + NH = 18 + 8 = 26 cm, option C.

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Question 31

If PP and QQ are fixed points and XX is a point which moves so that XP=XQXP = XQ, the locus of XX is

Worked solution (try it first)
  1. XP=XQXP = XQ means XX is always the same distance from PP as from QQ.
  2. Such points lie on the line through the mid-point of PQPQ at right angles to it.
  3. So the locus is the perpendicular bisector of PQPQ, option D.

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Question 32

In a regular polygon, each interior angle is double its corresponding exterior angle. Find the number of sides of the polygon.

Worked solution (try it first)
  1. Let the exterior angle be ee.
  2. The interior angle is 2e2e, and the two add up to 180∘180^\circ: 3e=180∘3e = 180^\circ, so e=60∘e = 60^\circ.
  3. The exterior angles add up to 360∘360^\circ, so the number of sides is 360÷60=6360 \div 60 = 6, option B.

Also set as JAMB 2017 · UTME (set 2) · Q39

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Question 34

The diagram shows the graph of y=x2y = x^2. Find the area of the shaded region, under the curve between x=0x = 0 and x=4x = 4.

4xyy = 16
The vertical scale is one fifth of the horizontal scale.
Worked solution (try it first)
  1. The area under the curve is ∫04x2 dx=[x33]04\int_0^4 x^2\,dx = \left[\frac{x^3}{3}\right]_0^4.
  2. That is 643−0=643\frac{64}{3} - 0 = \frac{64}{3} square units, option C.

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Question 36

If y=2xcos⁡2x−sin⁡2xy = 2x\cos2x - \sin2x, find dydx\frac{dy}{dx} when x=π4x = \frac\pi4.

Worked solution (try it first)
  1. Product rule on 2xcos⁡2x2x\cos2x: 2cos⁡2x+2x(−2sin⁡2x)=2cos⁡2x−4xsin⁡2x2\cos2x + 2x(-2\sin2x) = 2\cos2x - 4x\sin2x.
  2. Chain rule on sin⁡2x\sin2x: 2cos⁡2x2\cos2x.
  3. Subtract it: dydx=−4xsin⁡2x\frac{dy}{dx} = -4x\sin2x.
  4. At x=π4x = \frac\pi4, sin⁡2x=sin⁡π2=1\sin2x = \sin\frac\pi2 = 1, so dydx=−4×π4\frac{dy}{dx} = -4 \times \frac\pi4
    =−π= -\pi, option B.

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Question 37

A bowl is designed by revolving completely the area enclosed by y=x2−1y = x^2 - 1, y=0y = 0, y=3y = 3 and x≥0x \ge 0 around the yy-axis. What is the volume of this bowl?

Worked solution (try it first)
  1. About the yy-axis the volume is π∫x2 dy\pi\int x^2\,dy.
  2. From y=x2−1y = x^2 - 1, x2=y+1x^2 = y + 1.
  3. The limits are y=0y = 0 to y=3y = 3: V=π∫03(y+1) dyV = \pi\int_0^3 (y + 1)\,dy
    =π[y22+y]03= \pi\left[\frac{y^2}{2} + y\right]_0^3.
  4. That is π(92+3)=15π2\pi\left(\frac92 + 3\right) = \frac{15\pi}{2} cubic units, option B.

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Question 39

A function f(x)f(x) passes through the origin and its first derivative is 3x+23x + 2. What is f(x)f(x)?

Worked solution (try it first)
  1. Integrate the derivative: f(x)=32x2+2x+cf(x) = \frac32x^2 + 2x + c.
  2. The curve passes through the origin, so f(0)=0f(0) = 0 and c=0c = 0.
  3. So y=32x2+2xy = \frac32x^2 + 2x, option A.

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Question 40

The expression ax2+bx+cax^2 + bx + c equals 5 at x=1x = 1. If its derivative is 2x+12x + 1, what are the values of aa, bb, cc respectively?

Worked solution (try it first)
  1. The derivative of ax2+bx+cax^2 + bx + c is 2ax+b2ax + b.
  2. Match it with 2x+12x + 1: 2a=22a = 2 and b=1b = 1, so a=1a = 1 and b=1b = 1.
  3. At x=1x = 1 the expression is a+b+c=5a + b + c = 5, so 1+1+c=51 + 1 + c = 5 and c=3c = 3.
  4. So aa, bb, cc are 1, 1, 3, option D.

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Question 41

XX and YY are two events. The probability of XX or YY is 0.7 and the probability of XX is 0.4. If XX and YY are independent, find the probability of YY.

Worked solution (try it first)
  1. Let P(Y)=pP(Y) = p.
  2. For independent events, P(X∩Y)=P(X)×P(Y)=0.4pP(X \cap Y) = P(X) \times P(Y) = 0.4p.
  3. Use P(X∪Y)=P(X)+P(Y)−P(X∩Y)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y): 0.7=0.4+p−0.4p0.7 = 0.4 + p - 0.4p.
  4. So 0.6p=0.30.6p = 0.3 and p=0.5p = 0.5, option B.

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Question 42

If the mean of the numbers 00, x+2x + 2, 3x+63x + 6 and 4x+84x + 8 is 4, find their mean deviation.

Worked solution (try it first)
  1. The four numbers add up to 8x+168x + 16, and their mean is 4, so 8x+16=168x + 16 = 16 and x=0x = 0.
  2. The numbers are 0, 2, 6 and 8.
  3. Their distances from 4 are 4, 2, 2 and 4, which add up to 12.
  4. The mean deviation is 124=3\frac{12}{4} = 3, option C.

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Question 43

In how many ways can the word MATHEMATICS be arranged?

Worked solution (try it first)
  1. MATHEMATICS has 11 letters, so start with 11!11!.
  2. Three letters appear twice each: M, A and T.
  3. Swapping two equal letters gives the same word, so divide by 2!2! for each of them.
  4. So there are 11!2! 2! 2!\dfrac{11!}{2!\,2!\,2!} arrangements, option C.

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Question 44

A die is rolled 240 times with the results below. If a pie chart is constructed to represent the data, the angle corresponding to 4 is

No. 1 2 3 4 5 6
Frequency 30 43 54 40 41 32
Worked solution (try it first)
  1. The die was rolled 240 times, so each roll gets 360∘240=1.5∘\frac{360^\circ}{240} = 1.5^\circ.
  2. The number 4 came up 40 times: 40×1.5∘=60∘40 \times 1.5^\circ = 60^\circ, option D.

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Question 45

If U={x:x is an integer and 1≤x≤20}U = \{x : x \text{ is an integer and } 1 \le x \le 20\}, E1={x:x is a multiple of 3}E_1 = \{x : x \text{ is a multiple of 3}\} and E2={x:x is a multiple of 4}E_2 = \{x : x \text{ is a multiple of 4}\}, and an integer is picked at random from UU, find the probability that it is not in E2E_2.

Worked solution (try it first)
  1. E2E_2, the multiples of 4 up to 20, is {4,8,12,16,20}\{4, 8, 12, 16, 20\}, which has 5 members.
  2. So P(in E2)=520P(\text{in } E_2) = \frac{5}{20}
    =14= \frac14.
  3. Not in E2E_2 is the complement: 1−14=341 - \frac14 = \frac34, option A.

Also set as JAMB 2017 · UTME (set 2) · Q3

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Question 46

The cumulative frequency curve represents the ages of 100 students in a school. Which age group do 70%70\% of the students belong to?

15.516.517.518.519.520.520406080100Age (years)Cumulative frequency
Worked solution (try it first)
  1. The number of students in an age group is the cumulative frequency at its top end minus the cumulative frequency at its bottom end.
  2. From the curve, the cumulative frequency is 30 at 17.5 years and 100 at 20.5 years, so 100−30=70100 - 30 = 70 students are aged 17.5–20.5.
  3. That is 70%70\% of 100, option D.
  4. The other groups hold 60, 90 and 82 students.

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Question 47

The variance of xx, 2x2x, 3x3x, 4x4x and 5x5x is

Worked solution (try it first)
  1. The five numbers add up to 15x15x, so the mean is 15x5=3x\frac{15x}{5} = 3x.
  2. The deviations are −2x-2x, −x-x, 0, xx, 2x2x.
  3. Their squares add up to 4x2+x2+0+x2+4x2=10x24x^2 + x^2 + 0 + x^2 + 4x^2 = 10x^2.
  4. The variance is 10x25=2x2\frac{10x^2}{5} = 2x^2, option B.

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Question 48

Find the sum of the range and the mode of the numbers 10, 5, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 10, 9, 7, 10, 6, 5.

Worked solution (try it first)
  1. The range is highest minus lowest: 10−4=610 - 4 = 6.
  2. Count the values: 10 appears seven times, more than any other, so the mode is 10.
  3. So the sum is 6+10=166 + 10 = 16, option A.

Also set as JAMB 2017 · UTME (set 2) · Q14

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Question 49

In how many ways can a delegation of 3 be chosen from among 5 men and 3 women, if at least one man and at least one woman must be included?

Worked solution (try it first)
  1. Two men and one woman: 5C2×3C1=10×3=30^5C_2 \times {^3C_1} = 10 \times 3 = 30 ways.
  2. One man and two women: 5C1×3C2=5×3=15^5C_1 \times {^3C_2} = 5 \times 3 = 15 ways.
  3. These are the only mixes with at least one of each, so add them: 30+15=4530 + 15 = 45, option D.

Also set as JAMB 2016 · UTME · Q18

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Question 50

The table shows the frequency distribution of the ages (in years) of pupils in a secondary school. What percentage of the pupils are over 15 years but less than 21 years?

Interval (years) 10–12 13–15 16–18 19–20 21–23
No. of pupils 6 14 15 10 5
Worked solution (try it first)
  1. Over 15 but under 21 means the classes 16–18 and 19–20: 15+10=2515 + 10 = 25 pupils.
  2. Total pupils: 6+14+15+10+5=506 + 14 + 15 + 10 + 5 = 50.
  3. As a percentage: 2550×100=50%\frac{25}{50} \times 100 = 50\%, option C.

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