Stepwise
Start practisingPractise

Objective paper · 37 questions · partial

JAMB 1988 · UME

Topics include Number foundations & fractions, Approximation & error, Commercial arithmetic, Logarithms, Indices & standard form, Quadratics & their graphs.

Our copy of this paper is missing questions 3, 8, 10, 13, 15, 17, 18, 19, 20, 26, 36, 41, 43.

Sit this paper

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.

Sit it as an examPractise it, marked as you go

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

Question 1

Simplify 112÷(2÷14 of 32)1\frac12 \div \left(2 \div \frac14 \text{ of } 32\right).

Worked solution (try it first)
  1. Work inside the bracket, "of" first: 14\frac14 of 32 is 8.
  2. Then the division in the bracket: 2÷8=142 \div 8 = \frac14.
  3. Finally, 112÷14=32×41\frac12 \div \frac14 = \frac32 \times 4
    =6= 6, option C.

Report a problem with this question

Question 2

If xx is the sum of the prime numbers between 1 and 6, and yy is the H.C.F. of 6, 9 and 15, find the product of xx and yy.

Worked solution (try it first)
  1. The prime numbers between 1 and 6 are 2, 3 and 5, so x=2+3+5=10x = 2 + 3 + 5 = 10.
  2. The largest number that divides 6, 9 and 15 is 3, so y=3y = 3.
  3. So xy=10×3=30xy = 10 \times 3 = 30, option B.

Report a problem with this question

Question 4

Find, correct to one decimal place, 0.24633÷0.03060.24633 \div 0.0306.

Worked solution (try it first)
  1. Multiply top and bottom by 10 000 to clear the decimals: 0.24633÷0.0306=2463.3÷3060.24633 \div 0.0306 = 2463.3 \div 306.
  2. Divide: 2463.3÷306=8.052463.3 \div 306 = 8.05.
  3. To 1 decimal place, the second decimal is 5, so round up: 8.1, option D.

Report a problem with this question

Question 5

Two sisters, Taiwo and Kehinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:911 : 9. Later Kehinde sells 23\frac23 of her share to Taiwo for ₦720.00. Find the value of the store.

Worked solution (try it first)
  1. The ratio 11:911 : 9 has 20 parts, so Kehinde owns 920\frac{9}{20} of the store.
  2. She sells 23\frac23 of that: 23×920=310\frac23 \times \frac{9}{20} = \frac{3}{10} of the store, for ₦720.
  3. So one tenth is 720÷3=720 \div 3 = ₦240, and the store is worth 10×240=10 \times 240 = ₦2,400, option B.

Report a problem with this question

Question 6

A basket contains green, black and blue balls in the ratio 5:2:15 : 2 : 1. If there are 10 blue balls, find the new ratio when 10 green and 10 black balls are removed.

Worked solution (try it first)
  1. Blue is 1 part and there are 10 blue balls, so one part is 10 balls: 50 green, 20 black and 10 blue.
  2. Remove 10 green and 10 black: 40 green, 10 black and 10 blue are left.
  3. Divide by 10: 40:10:10=4:1:140 : 10 : 10 = 4 : 1 : 1, option D.

Report a problem with this question

Question 7

A taxpayer is allowed 18\frac18 of his income tax-free and pays 20%20\% on the remainder. If he pays ₦490.00 tax, what is his income?

Worked solution (try it first)
  1. 18\frac18 of his income II is tax-free, so he is taxed on the other 78\frac78 of it.
  2. The tax is 20%20\% of that: 0.2×78I=4900.2 \times \frac78 I = 490, which is 740I=490\frac{7}{40}I = 490.
  3. Multiply both sides by 407\frac{40}{7}: I=2800I = 2800.
  4. His income is ₦2,800.00, option C.

Report a problem with this question

Question 9

If log⁡102=0.3010\log_{10} 2 = 0.3010 and log⁡103=0.4771\log_{10} 3 = 0.4771, evaluate log⁡104.5\log_{10} 4.5 without using logarithm tables.

Worked solution (try it first)
  1. Write 4.5 using 2 and 3: 4.5=92=3224.5 = \frac92 = \frac{3^2}{2}.
  2. So log⁡4.5=2log⁡3−log⁡2\log 4.5 = 2\log 3 - \log 2.
  3. That is 2(0.4771)−0.3010=0.9542−0.30102(0.4771) - 0.3010 = 0.9542 - 0.3010
    =0.6532= 0.6532, option C.

Report a problem with this question

Question 11

The thickness of an 800-page book is 18 mm. Calculate the thickness of one leaf of the book, giving your answer in metres and in standard form.

Worked solution (try it first)
  1. Each leaf has a page on each side, so 800 pages make 800÷2=400800 \div 2 = 400 leaves.
  2. One leaf is 18÷400=0.04518 \div 400 = 0.045 mm thick.
  3. Divide by 1000 to change mm to metres: 0.0000450.000045 m, which is 4.50×10−54.50 \times 10^{-5} m, option D.

Report a problem with this question

Question 12

Simplify x+2x+1−x−2x+2\dfrac{x + 2}{x + 1} - \dfrac{x - 2}{x + 2}.

Worked solution (try it first)
  1. Use the common denominator (x+1)(x+2)(x + 1)(x + 2).
  2. The top becomes (x+2)2−(x−2)(x+1)(x + 2)^2 - (x - 2)(x + 1).
  3. Expand: (x+2)2=x2+4x+4(x + 2)^2 = x^2 + 4x + 4 and (x−2)(x+1)=x2−x−2(x - 2)(x + 1) = x^2 - x - 2.
  4. Subtract the whole second product: x2+4x+4−x2+x+2=5x+6x^2 + 4x + 4 - x^2 + x + 2 = 5x + 6.
  5. So the answer is 5x+6(x+1)(x+2)\dfrac{5x + 6}{(x + 1)(x + 2)}, option C.

Report a problem with this question

Question 14

If xx varies inversely as the cube root of yy, and x=1x = 1 when y=8y = 8, find yy when x=3x = 3.

Worked solution (try it first)
  1. Inverse variation means xy3x\sqrt[3]{y} is constant.
  2. With x=1x = 1, y=8y = 8: k=1×2=2k = 1 \times 2 = 2.
  3. When x=3x = 3: 3y3=23\sqrt[3]{y} = 2, so y3=23\sqrt[3]{y} = \frac23.
  4. Cube both sides: y=(23)3=827y = \left(\frac23\right)^3 = \frac{8}{27}, option C.

Report a problem with this question

Question 16

If g(y)=y−311+11y2−9g(y) = \frac{y - 3}{11} + \frac{11}{y^2 - 9}, what is g(y+3)g(y + 3)?

Worked solution (try it first)
  1. Replace every yy by y+3y + 3: g(y+3)=(y+3)−311+11(y+3)2−9g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^2 - 9}.
  2. The first top is yy.
  3. Expand the bottom: (y+3)2−9=y2+6y+9−9(y + 3)^2 - 9 = y^2 + 6y + 9 - 9, which is y(y+6)y(y + 6).
  4. So g(y+3)=y11+11y(y+6)g(y + 3) = \frac{y}{11} + \frac{11}{y(y + 6)}, option A.

Report a problem with this question

Question 21

Given that 3x−5y−3=03x - 5y - 3 = 0 and 2y−6x+5=02y - 6x + 5 = 0, the value of (x,y)(x, y) is

Worked solution (try it first)
  1. Write both in standard form: 3x−5y=33x - 5y = 3 and −6x+2y=−5-6x + 2y = -5.
  2. Double the first so the xx terms cancel: 6x−10y=66x - 10y = 6.
  3. Add the second: −8y=1-8y = 1, so y=−18y = -\frac18.
  4. Put this into 3x−5y=33x - 5y = 3: 3x+58=33x + \frac58 = 3, so 3x=1983x = \frac{19}{8} and x=1924x = \frac{19}{24}.
  5. So (x,y)=(1924,−18)(x, y) = \left(\frac{19}{24}, -\frac18\right), option D.

Report a problem with this question

Question 22✱✱

The solution of the quadratic equation px2+qx+b=0px^2 + qx + b = 0 is

Worked solution (try it first)
  1. Match px2+qx+b=0px^2 + qx + b = 0 with ax2+bx+c=0ax^2 + bx + c = 0: the x2x^2 coefficient is pp, the xx coefficient is qq and the constant is bb.
  2. The formula is x=−(x coefficient)±(x coefficient)2−4(x2 coefficient)(constant)2(x2 coefficient)x = \dfrac{-(x\text{ coefficient}) \pm \sqrt{(x\text{ coefficient})^2 - 4(x^2\text{ coefficient})(\text{constant})}}{2(x^2\text{ coefficient})}.
  3. Put in qq, pp and bb: x=−q±q2−4bp2px = \dfrac{-q \pm \sqrt{q^2 - 4bp}}{2p}, option C.

Report a problem with this question

Question 23

Simplify 1x2+5x+6+1x2+3x+2\dfrac{1}{x^2 + 5x + 6} + \dfrac{1}{x^2 + 3x + 2}.

Worked solution (try it first)
  1. Factorise the bottoms: x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x + 2)(x + 3) and x2+3x+2=(x+1)(x+2)x^2 + 3x + 2 = (x + 1)(x + 2).
  2. The common denominator is (x+1)(x+2)(x+3)(x + 1)(x + 2)(x + 3).
  3. Each top is multiplied by the factor its bottom lacks, so the top is (x+1)+(x+3)=2x+4(x + 1) + (x + 3) = 2x + 4.
  4. Factorise the top as 2(x+2)2(x + 2) and cancel x+2x + 2.
  5. This leaves 2(x+1)(x+3)\dfrac{2}{(x + 1)(x + 3)}, option C.

Report a problem with this question

Question 24

Evaluate 4a2−49b22a2+5ab−7b2\dfrac{4a^2 - 49b^2}{2a^2 + 5ab - 7b^2}.

Worked solution (try it first)
  1. The top is a difference of two squares: 4a2−49b2=(2a−7b)(2a+7b)4a^2 - 49b^2 = (2a - 7b)(2a + 7b).
  2. Factorise the bottom: 2a2+5ab−7b2=2a2−2ab+7ab−7b22a^2 + 5ab - 7b^2 = 2a^2 - 2ab + 7ab - 7b^2, which is (2a+7b)(a−b)(2a + 7b)(a - b).
  3. Cancel the common factor 2a+7b2a + 7b.
  4. This leaves 2a−7ba−b\dfrac{2a - 7b}{a - b}, option D.

Report a problem with this question

Question 25

The figure shows the graph of y=x2+3x+4y = \frac{x^2 + 3}{x + 4} and the line y=1y = 1. Use it to find the solution of the equation x2−x−1=0x^2 - x - 1 = 0.

−3−2−1121234xy0y = (x2 + 3)/(x + 4)y = 1
Worked solution (try it first)
  1. Where the curve meets the line y=1y = 1: x2+3x+4=1\dfrac{x^2 + 3}{x + 4} = 1.
  2. Multiply by x+4x + 4: x2+3=x+4x^2 + 3 = x + 4, which rearranges to x2−x−1=0x^2 - x - 1 = 0.
  3. So the roots are the xx-values where the line y=1y = 1 cuts the curve: about x=−0.6x = -0.6 and x=1.6x = 1.6, option A.
  4. Check with the formula: x=1±52x = \frac{1 \pm \sqrt5}{2}, which is 1.6181.618 or −0.618-0.618.

Report a problem with this question

Question 27

The solutions of x2−2x−1=0x^2 - 2x - 1 = 0 are the points of intersection of two graphs. If one of the graphs is y=2+x−x2y = 2 + x - x^2, find the second graph.

Worked solution (try it first)
  1. At the intersections, 2+x−x22 + x - x^2 equals the second graph's yy.
  2. That equation must be the same as x2−2x−1=0x^2 - 2x - 1 = 0.
  3. Rewrite the given equation as 0=−x2+2x+10 = -x^2 + 2x + 1, and subtract it from the curve: (2+x−x2)−(−x2+2x+1)=1−x(2 + x - x^2) - (-x^2 + 2x + 1) = 1 - x.
  4. So the second graph is y=1−xy = 1 - x.
  5. Check: 2+x−x2=1−x2 + x - x^2 = 1 - x rearranges to x2−2x−1=0x^2 - 2x - 1 = 0, option A.

Report a problem with this question

Question 28

If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is −6-6, find the common difference.

Worked solution (try it first)
  1. The 8th and 9th terms are a+7da + 7d and a+8da + 8d, so their sum gives 2a+15d=722a + 15d = 72.
  2. The 4th term gives a+3d=−6a + 3d = -6.
  3. Double it: 2a+6d=−122a + 6d = -12.
  4. Subtract the second equation from the first: 9d=72−(−12)=849d = 72 - (-12) = 84.
  5. So d=849=913d = \frac{84}{9} = 9\frac13, option D.

Report a problem with this question

Question 29

If 7 and 189 are the first and fourth terms of a geometric progression respectively, find the sum of the first three terms.

Worked solution (try it first)
  1. The 4th term of a G.P. is ar3ar^3, so 7r3=1897r^3 = 189 and r3=27r^3 = 27.
  2. Take the cube root: r=3r = 3.
  3. The first three terms are 7, 21 and 63, and their sum is 91, option B.

Report a problem with this question

Question 30

In the figure, PP, QQ, RR and SS lie on a circle and the chords QRQR and RSRS are equal. SRSR is produced to TT with ∠QRT=120∘\angle QRT = 120^\circ, and the exterior angle at SS (between SPSP and RSRS produced) is 100∘100^\circ. Calculate the angle x=∠PRSx = \angle PRS.

120°100°xPQRST
Worked solution (try it first)
  1. Angles on a straight line: ∠QRS=180∘−120∘\angle QRS = 180^\circ - 120^\circ
    =60∘= 60^\circ.
  2. With QR=RSQR = RS, triangle QRSQRS is isosceles with a 60∘60^\circ apex, so it is equilateral and ∠SQR=60∘\angle SQR = 60^\circ.
  3. Angles in the same segment: ∠SPR\angle SPR and ∠SQR\angle SQR both stand on arc SRSR, so ∠SPR=60∘\angle SPR = 60^\circ.
  4. Angles on a straight line at SS: ∠PSR=180∘−100∘\angle PSR = 180^\circ - 100^\circ
    =80∘= 80^\circ.
  5. The angles of triangle PSRPSR add up to 180∘180^\circ: x=180∘−80∘−60∘x = 180^\circ - 80^\circ - 60^\circ
    =40∘= 40^\circ, option D.

Report a problem with this question

Question 31

In the figure, PQPQ is parallel to STST, ∠QRS=40∘\angle QRS = 40^\circ, the angle at QQ between QPQP and QRQR is 3x3x and ∠TSR=x\angle TSR = x. Find the value of xx.

3x40°xPQRST
Worked solution (try it first)
  1. Draw a line through RR parallel to PQPQ.
  2. Angles on a straight line at QQ: QRQR makes 180∘−3x180^\circ - 3x with the rightward direction there, so RQRQ makes 180∘−3x180^\circ - 3x with the parallel line at RR (alternate angles).
  3. STST is parallel too, so RSRS makes xx with the parallel line at RR (alternate angles with ∠TSR\angle TSR).
  4. Both arms lie on the same side of that line, so ∠QRS\angle QRS is the difference: x−(180∘−3x)=40∘x - (180^\circ - 3x) = 40^\circ, which gives 4x=220∘4x = 220^\circ.
  5. So x=55x = 55, option A.

Report a problem with this question

Question 32

For which of the following exterior angles is a regular polygon possible? (i) 35∘35^\circ (ii) 18∘18^\circ (iii) 115∘115^\circ

Worked solution (try it first)
  1. The exterior angles of a regular polygon are equal and add up to 360∘360^\circ, so the number of sides is 360÷(exterior angle)360 \div (\text{exterior angle}), which must be a whole number.
  2. 360÷35≈10.3360 \div 35 \approx 10.3 and 360÷115≈3.1360 \div 115 \approx 3.1: neither is whole.
  3. 360÷18=20360 \div 18 = 20, a polygon with 20 sides.
  4. So only (ii) works, option B.

Report a problem with this question

Question 33

In the figure, PS=7PS = 7 cm and RY=9RY = 9 cm. If the area of parallelogram PQRSPQRS is 56 cm256\text{ cm}^2, find the area of trapezium PQTSPQTS.

7 cm9 cmPQRSYT
Worked solution (try it first)
  1. The area of parallelogram PQRSPQRS is base × height, so the height is 56÷7=856 \div 7 = 8 cm.
  2. QR=PS=7QR = PS = 7 cm, and PYTSPYTS is also a parallelogram, so YT=PS=7YT = PS = 7 cm.
  3. So QT=7+9+7=23QT = 7 + 9 + 7 = 23 cm.
  4. Trapezium PQTSPQTS has parallel sides 7 and 23 cm: area =12(7+23)×8= \frac12(7 + 23) \times 8
    =120 cm2= 120\text{ cm}^2, option C.

Report a problem with this question

Question 34

A quadrant of a circle of radius 6 cm is cut away from each corner of a rectangle 25 cm long and 18 cm wide. Find the perimeter of the remaining figure.

Worked solution (try it first)
  1. Each corner cut takes 6 cm off both sides that meet there.
  2. The straight edges left are 2(25−12)+2(18−12)=26+12=382(25 - 12) + 2(18 - 12) = 26 + 12 = 38 cm.
  3. Each corner now has a quarter-circle arc of radius 6.
  4. Four quarters make a full circle: 2π×6=12π2\pi \times 6 = 12\pi cm.
  5. Perimeter =(38+12π)= (38 + 12\pi) cm, option B.

Report a problem with this question

Question 35

In the figure, ∠STQ=∠SRP\angle STQ = \angle SRP, PT=TQ=6PT = TQ = 6 cm and QS=5QS = 5 cm. Find SRSR.

665PQRST
Worked solution (try it first)
  1. Triangles QTSQTS and QRPQRP share ∠Q\angle Q and have ∠QTS=∠QRP\angle QTS = \angle QRP, so they are similar, with TT matching RR and SS matching PP.
  2. So QTQR=QSQP\dfrac{QT}{QR} = \dfrac{QS}{QP}.
  3. Here QT=6QT = 6, QS=5QS = 5 and QP=6+6=12QP = 6 + 6 = 12, so 6QR=512\dfrac{6}{QR} = \dfrac{5}{12}.
  4. Cross-multiply: 5×QR=725 \times QR = 72, so QR=725QR = \frac{72}{5}.
  5. Then SR=QR−QSSR = QR - QS
    =725−255= \frac{72}{5} - \frac{25}{5}
    =475= \frac{47}{5}, option A.

Report a problem with this question

Question 37

In the figure, PS=RS=QSPS = RS = QS and ∠QSR=50∘\angle QSR = 50^\circ. Find ∠QPR\angle QPR.

50°?SQRP
Worked solution (try it first)
  1. SP=SQ=SRSP = SQ = SR, so PP, QQ and RR lie on a circle with centre SS.
  2. ∠QSR\angle QSR is the angle at the centre on arc QRQR, and ∠QPR\angle QPR is the angle at the circumference on the same arc.
  3. The angle at the circumference is half the angle at the centre: ∠QPR=12×50∘\angle QPR = \frac12 \times 50^\circ
    =25∘= 25^\circ, option A.

Report a problem with this question

Question 38

In the figure, XRXR and YQYQ are tangents to the circle YZXPYZXP. If ∠ZXR=45∘\angle ZXR = 45^\circ and ∠YZX=55∘\angle YZX = 55^\circ, find ∠ZYQ\angle ZYQ.

45°55°?XYZPRQ
Worked solution (try it first)
  1. Tangent–chord at XX: the angle between tangent XRXR and chord XZXZ equals the angle in the alternate segment, so ∠ZYX=45∘\angle ZYX = 45^\circ.
  2. The angles of triangle XYZXYZ add up to 180∘180^\circ: ∠YXZ=180∘−55∘−45∘\angle YXZ = 180^\circ - 55^\circ - 45^\circ
    =80∘= 80^\circ.
  3. Tangent–chord at YY: the tangent makes 80∘80^\circ with chord YZYZ on the side away from QQ.
  4. So ∠ZYQ=180∘−80∘\angle ZYQ = 180^\circ - 80^\circ
    =100∘= 100^\circ, option C.

Report a problem with this question

Question 39

From a point 14314\sqrt3 m away from a tree, a man finds that the angle of elevation of the top of the tree is 30∘30^\circ. If he measures this angle from a point 2 m above the ground, how high is the tree?

Worked solution (try it first)
  1. The angle is measured at his eye, so first find the height of the top above eye level: 143×tan⁡30∘14\sqrt3 \times \tan30^\circ.
  2. Since tan⁡30∘=13\tan30^\circ = \frac{1}{\sqrt3}, that is 14 m.
  3. Add the 2 m from the ground to his eye: the tree is 14+2=1614 + 2 = 16 m, option D.

Report a problem with this question

Question 40

Alero starts a 3 km walk from PP on a bearing of 023∘023^\circ. She then walks 4 km on a bearing of 113∘113^\circ to QQ. What is the bearing of QQ from PP?

Worked solution (try it first)
  1. The bearings 023∘023^\circ and 113∘113^\circ differ by 90∘90^\circ, so the two legs meet at a right angle.
  2. The angle at PP between the first leg and PQPQ has tan⁡=43\tan = \frac{4}{3}, so it is 53∘8′53^\circ8'.
  3. QQ is clockwise from the first leg, so add: the bearing is 023∘+53∘8′=076∘8′023^\circ + 53^\circ8' = 076^\circ8', option C.

Report a problem with this question

Question 42

In triangle PQRPQR, PQ=1PQ = 1 cm, QR=2QR = 2 cm and ∠PQR=120∘\angle PQR = 120^\circ. Find the longest side of the triangle.

Worked solution (try it first)
  1. The longest side faces the largest angle, the 120∘120^\circ at QQ.
  2. So the longest side is PRPR.
  3. Cosine rule: PR2=12+22−2(1)(2)cos⁡120∘PR^2 = 1^2 + 2^2 - 2(1)(2)\cos120^\circ.
  4. cos⁡120∘=−12\cos120^\circ = -\frac12, so PR2=5+2=7PR^2 = 5 + 2 = 7 and PR=7PR = \sqrt7 cm, option D.

Report a problem with this question

Question 44

A metal pipe 10 cm long has an external diameter of 12 cm and a thickness of 1 cm. Find the volume of the metal used in making the pipe.

Worked solution (try it first)
  1. External radius: 12÷2=612 \div 2 = 6 cm.
  2. The wall is 1 cm thick, so the internal radius is 6−1=56 - 1 = 5 cm.
  3. The metal is the ring between the circles times the length: π(62−52)×10\pi(6^2 - 5^2) \times 10.
  4. That is π×11×10=110π cm3\pi \times 11 \times 10 = 110\pi\text{ cm}^3, option B.

Report a problem with this question

Question 45

A solid consists of a hemisphere surmounted by a right circular cone, both of radius 3.0 cm; the cone has height 6.0 cm. Find the volume of the solid.

Worked solution (try it first)
  1. Hemisphere: half of 43πr3\frac43\pi r^3, which is 23π×27=18π cm3\frac23\pi \times 27 = 18\pi\text{ cm}^3.
  2. Cone: 13πr2h=13π×9×6\frac13\pi r^2h = \frac13\pi \times 9 \times 6
    =18π cm3= 18\pi\text{ cm}^3.
  3. Total: 18π+18π=36π cm318\pi + 18\pi = 36\pi\text{ cm}^3, option B.

Report a problem with this question

Question 46

PQRPQR is a triangle in which PQ=10PQ = 10 cm and ∠QPR=60∘\angle QPR = 60^\circ. SS is a point equidistant from PP and QQ, and also equidistant from PQPQ and PRPR. If UU is the foot of the perpendicular from SS to PRPR, find SUSU to one decimal place.

Worked solution (try it first)
  1. SS is equidistant from PP and QQ, so it lies on the perpendicular bisector of PQPQ.
  2. Its foot on PQPQ is 5 cm from PP.
  3. SS is equidistant from PQPQ and PRPR, so it lies on the bisector of ∠QPR\angle QPR.
  4. That line makes 60∘÷2=30∘60^\circ \div 2 = 30^\circ with PQPQ.
  5. In the right-angled triangle at PP, the distance from SS to PQPQ is 5tan⁡30∘≈2.8875\tan30^\circ \approx 2.887 cm.
  6. On the angle bisector, the distances to PQPQ and PRPR are equal, so SU≈2.9SU \approx 2.9 cm, option B.

Report a problem with this question

Question 47

In a class of 150 students, the sector of a pie chart representing the students offering Physics has angle 12∘12^\circ. How many students are offering Physics?

Worked solution (try it first)
  1. The Physics sector is 12360\frac{12}{360} of the circle, so it is 12360\frac{12}{360} of the class.
  2. 12360×150=15030\frac{12}{360} \times 150 = \frac{150}{30}
    =5= 5 students, option D.

Report a problem with this question

Question 48

If xx and yy are the mean and the median respectively of 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find xy\frac xy correct to one decimal place.

Worked solution (try it first)
  1. The numbers add up to 156, so the mean is x=15610=15.6x = \frac{156}{10} = 15.6.
  2. They are already in order.
  3. With 10 numbers the median is halfway between the 5th and 6th: y=15+162=15.5y = \frac{15 + 16}{2} = 15.5.
  4. So xy=15.615.5\frac xy = \frac{15.6}{15.5}
    ≈1.006\approx 1.006, which is 1.0 to one decimal place, option D.

Report a problem with this question

Question 49

In the distribution below, the mode and the median respectively are

Score (xx) 0 1 2 3 4 5 6
Frequency 7 11 6 7 7 5 3
Worked solution (try it first)
  1. The mode is the score with the highest frequency: score 1, which occurs 11 times.
  2. There are 7+11+6+7+7+5+3=467 + 11 + 6 + 7 + 7 + 5 + 3 = 46 scores, so the median is halfway between the 23rd and 24th.
  3. Running totals: 7 (score 0), 18 (score 1), 24 (score 2).
  4. The 19th to 24th scores are all 2, so the median is 2.
  5. So the mode and median are 1 and 2, option B.

Report a problem with this question

Question 50

If two dice are thrown together, what is the probability of obtaining a score of at least 10?

Worked solution (try it first)
  1. Two dice give 36 equally likely outcomes.
  2. A score of at least 10 means 10, 11 or 12: (4,6),(5,5),(6,4)(4, 6), (5, 5), (6, 4), then (5,6),(6,5)(5, 6), (6, 5), then (6,6)(6, 6).
  3. That is 6 outcomes.
  4. So the probability is 636=16\frac{6}{36} = \frac16, option A.

Report a problem with this question