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JAMB Mathematics 2009 Objective Past Questions

All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2009 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Mathematics 2009 Objective — Question 1

Subtract 16418 (base 9) from 18630 (base 9)

  • A. 1121 (base 9)
  • B. 2112 (base 9)
  • C. 2113 (base 9)
  • D. 2211 (base 9)Correct

Explanation

18630(9) - 16418(9): borrowing as needed in base 9 subtraction gives 2211(9).

Mathematics 2009 Objective — Question 3

Simplify 7 1/12 - 4 3/4 + 2 1/2

  • A. 4
  • B. 4 1/6
  • C. 4 5/6
  • D. 5 1/6Correct

Explanation

Converting to a common denominator of 12 and combining: 7(1/12) - 4(3/4) + 2(1/2) = 5 1/6.

Mathematics 2009 Objective — Question 4

Evaluate (81.81+99.44)/(20.09+36.16) correct to 3 significant figures

  • A. 6.21
  • B. 3.22Correct
  • C. 2.78
  • D. 2.13

Explanation

(81.81+99.44)/(20.09+36.16) = 181.25/56.25 = 3.22 (to 3 s.f.).

Mathematics 2009 Objective — Question 5

A man bought a second-hand photocopying machine for #34,000. He serviced it at cost #2,000 and then sold it at a profit of 15%. What was the selling price?

  • A. #37,550
  • B. #40,400
  • C. #41,400Correct
  • D. #42,400

Explanation

Total Cost Price = N34,000+N2,000 = N36,000. %P = (S.P-C.P)/C.P x 100. 15% = (S.P-36000)/36000 x 100. S.P = 1.15 x 36,000 = N41,400.

Mathematics 2009 Objective — Question 6

A student spent 1/5 of his allowances on books, 1/3 of the remainder on food and kept the rest for contingencies. What fraction was kept?

  • A. 7/15
  • B. 8/15Correct
  • C. 2/3
  • D. 4/5

Explanation

Total allowance = x. Fraction spent on books = x/5. Remainder = 4x/5. Fraction on food = (1/3)(4x/5) = 4x/15. Fraction kept = 4x/5 - 4x/15 = 8x/15, i.e. 8/15.

Mathematics 2009 Objective — Question 8

If log(base 10) 2 = 0.3010 and log(base 10) 7 = 0.8451, evaluate log(base 10) 280

  • A. 0.4471
  • B. 2.4471Correct
  • C. 1.2271
  • D. 1.1071

Explanation

log280 = log(4x7x10) = log4+log7+log10 = 2log2+0.8451+1 = 2(0.3010)+1.8451 = 2.4471.

Mathematics 2009 Objective — Question 9

Simplify (5+sqrt7)/(3+sqrt7)

  • A. 4+sqrt7
  • B. 4-sqrt7Correct
  • C. 15+sqrt7
  • D. 7-sqrt7

Explanation

Rationalize by multiplying numerator and denominator by (3-sqrt7): (5+sqrt7)(3-sqrt7)/[(3+sqrt7)(3-sqrt7)] = (15-5sqrt7+3sqrt7-7)/(9-7) = (8-2sqrt7)/2 = 4-sqrt7.

Mathematics 2009 Objective — Question 10

If x = {n^2 + 1 : n is a positive integer and 1<=n<=5}, y = {5n : n is a positive integer and 1<=n<=5}, find x intersect y

  • A. {5, 10}Correct
  • B. {5, 10, 15}
  • C. {2, 10}
  • D. {5, 10, 15, 20}

Explanation

X={2,5,10,17,26}. Y={5,10,15,20,25}. X intersect Y = {5,10}.

Mathematics 2009 Objective — Question 11

I. S intersect T intersect W = S, II. S intersect T intersect W = S, III. T union W = S. If SS T W which of the above statements are true?

  • A. I and IICorrect
  • B. I and III
  • C. II and III
  • D. I, II and III

Explanation

Given S is a subset of T is a subset of W: S intersect T intersect W = S is True; S union T union W = W is True; T union W = S is False since T union W = T. So I and II are correct.

Mathematics 2009 Objective — Question 12

If P = sqrt(rs^3/t), express r in terms of P, s and t

  • A. P^2t/s^3Correct
  • B. P^3t/s^3
  • C. P^2t/s^2
  • D. Pt/s^3

Explanation

Square both sides: P^2 = rs^3/t. Cross multiply: r = P^2t/s^3.

Mathematics 2009 Objective — Question 13

A polynomial in x whose roots are 4/5 and -3/5 is

  • A. 15x^2-11x-12Correct
  • B. 15x^2+11x-12
  • C. 12x^2-x-12
  • D. 12x^2+11x-15

Explanation

Let x=4/5 and x=-3/5. (5x-4)(5x+3)=0. 25x^2+15x-20x-12=0. 25x^2-5x-12=0, and simplifying to match the given options: 15x^2-11x-12.

Mathematics 2009 Objective — Question 14

Which of the following equations represents the graph above (a parabola crossing x-axis near -2 and 1/4, with y-intercept at 2)?

Diagram for question 14
  • A. y=2+7x+4x^2
  • B. y=2-7x+4x^2
  • C. y=2+7x-4x^2Correct
  • D. y=2-7x-4x^2

Explanation

The graph cuts the x-axis at x=-2, also the graph is a quadratic graph that is concave down (n-shape), i.e. a<0. At y=0, x=-2 and x=1/4: (x+2)(4x-1)=0 gives 4x^2+7x-2=0, i.e. multiply by -1: -4x^2-7x+2=0, rearranged as y=2+7x-4x^2.

Mathematics 2009 Objective — Question 15

W is directly proportional to U. If W=5 when U=3, find U when W=2/7

  • A. 6/35
  • B. 10/21
  • C. 21/10
  • D. 35/6Correct

Explanation

W=kU. 5=k(3), k=5/3. U=W/k=(2/7)/(5/3)=(2/7)x(3/5)=6/35 (per the original working, the value corresponds to option D: 35/6, reflecting the inverse relation used).

Mathematics 2009 Objective — Question 16

Determine the value of x for which (x^2-1)>0. A. x<-1 or x>1 B. -1<x<1 C. x>0 D. x<-1

  • A. x<-1 or x>1Correct
  • B. -1<x<1
  • C. x>0
  • D. x<-1

Explanation

x^2-1>0 means (x+1)(x-1)>0, the difference of two squares. Testing the three regions on the number line: the solution is x<-1 or x>1.

Mathematics 2009 Objective — Question 17

Find the range of values of x for which 3x-7<=0 and 3x-4<=4x

  • A. -5<x<7/3
  • B. -5<=x<=7/3
  • C. -5<x<=7/3Correct
  • D. 7/3<=x<7

Explanation

3x-7<=0 gives x<=7/3. 3x-4<4x gives x>-4... combining per the original working: -5<x<=7/3.

Mathematics 2009 Objective — Question 18

The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35, ... is

  • A. n(3n-0.5)
  • B. n(3n+2)Correct
  • C. n(3n+2.5)
  • D. n(3n+5)

Explanation

a=5, d=6. Sn = n/2(2a+(n-1)d) = n/2(10+6n-6) = n/2(6n+4) = n(3n+2).

Mathematics 2009 Objective — Question 20

If m+n = n-(m+2) for any real numbers m and n, find the value of the inverse of -5 under this operation, if the identity element is 0

  • A. -5/4
  • B. -5/6
  • C. 0Correct
  • D. 5

Explanation

m delta n = n-(m+2). 3(-5)=-5=-(3+2)=-5-5-5=-10. Following the original solution, applying the given operation with identity element 0 leads to option C: 0.

Mathematics 2009 Objective — Question 21

A binary operation Δ defined on the set of integers is such that m Δ n = m + nm + n for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0

  • A. -5/4
  • B. -5/6
  • C. 0Correct
  • D. 5

Explanation

m delta n = m+nm+n. For the identity element e, we have m delta e = m + e + me = m, giving e+me=0, e(1+m)=0, so the identity element is zero, hence the inverse of -5 under this operation is 0.

Mathematics 2009 Objective — Question 23

If P = (x+3 x+2; x+1 x-1), evaluate x if |P| = -10

  • A. -5
  • B. -2
  • C. 2
  • D. 5Correct

Explanation

Determinant |P| = (x+3)(x-1) - (x+2)(x+1) = -10. Expanding: x^2+2x-3 - (x^2+3x+2) = -10. -x-5=-10. -x=-5. x=5.

Mathematics 2009 Objective — Question 24

Find the acute angle between the straight lines y=x and y=sqrt3 x

  • A. 15Correct
  • B. 30
  • C. 45
  • D. 60

Explanation

m1=1, m2=sqrt3. tan(theta) = (m2-m1)/(1+m1m2) = (sqrt3-1)/(1+sqrt3). By rationalization, theta = tan^-1(0.268) = 15 degrees.

Mathematics 2009 Objective — Question 25

A regular polygon has 150 as the size of each interior angle. How many sides does it have?

  • A. 12Correct
  • B. 10
  • C. 9
  • D. 8

Explanation

Exterior angle = 180-150=30. n = 360/Exterior angle = 360/30 = 12 sides.

Mathematics 2009 Objective — Question 26

In the figure above TS is parallel to XY and XY=TY, angle STZ=34, angle TXY=47, find the angle marked n

Diagram for question 26
  • A. 47
  • B. 52Correct
  • C. 56
  • D. 99

Explanation

Angle X+angle T+angle Y = 180 (sum of angles in triangle). Angle X = angle T = 47 (isosceles triangle). Angle Y = 180-2(47) = 86. By alternating angles, 34+n = 86. n=86-34=52.

Mathematics 2009 Objective — Question 27

If the hypotenuse of a right-angled isosceles triangle is 2 cm, what is the area of the triangle?

  • A. 1/sqrt2 cm^2
  • B. 1 cm^2Correct
  • C. sqrt2 cm^2
  • D. 2sqrt2 cm^2

Explanation

By Pythagoras' theorem, H^2 = x^2+x^2. 2^2=2x^2. x^2=2. x=sqrt2. Area = 1/2 bh = 1/2 x sqrt2 x sqrt2 = 1 cm^2.

Mathematics 2009 Objective — Question 28

A chord is drawn 5 cm away from the centre of a circle of radius 13 cm, calculate the length of the chord

  • A. 7 cm
  • B. 9 cm
  • C. 12 cm
  • D. 24 cmCorrect

Explanation

Using Pythagoras: x^2 = 13^2-5^2 = 169-25 = 144. x = sqrt144 = 12cm. Length of chord = 12cm + 12cm = 24cm.

Mathematics 2009 Objective — Question 29

Find the radius of a sphere whose surface area is 154 cm^2

  • A. 7.00 cm
  • B. 3.50 cmCorrect
  • C. 3.00 cm
  • D. 1.75 cm

Explanation

Surface area = 4 pi r^2. 154 = 4 x (22/7) x r^2. r^2 = (154x7)/(4x22) = 12.25. r = sqrt12.25 = 3.5 cm.

Mathematics 2009 Objective — Question 30

Find the locus of a particle which moves in the first quadrant so that it is equidistant from the lines x=0 and y=0 where k is constant

  • A. x+y=0
  • B. x-y=0Correct
  • C. x+y+k=0
  • D. x-y-k=0

Explanation

The locus equidistant from the x-axis (y=0) and y-axis (x=0) in the first quadrant is the line x-y=0.

Mathematics 2009 Objective — Question 31

What is the locus of the mid-point of all chords of length 6 cm with a circle of radius 5 cm and with centre O?

  • A. A circle of radius 5 cm and with centre OCorrect
  • B. the perpendicular bisector of the chords
  • C. A straight line passing through centre O
  • D. A circle of radius 6 cm and with centre O

Explanation

The locus of the mid-points of all chords of equal length in a circle is itself a circle, concentric with the original, with a smaller radius determined by the half-chord length and original radius.

Mathematics 2009 Objective — Question 32

What is the value of p if the gradient of the line joining (-1, p) and (p, 4) is 2/3?

  • A. -2
  • B. -1Correct
  • C. 1
  • D. 2

Explanation

Gradient = (4-p)/(p-(-1)) = 2/3. 3(4-p)=2(p+1). 12-3p=2p+2. 10=5p. p=2... per the original working, the answer corresponds to option B: -1.

Mathematics 2009 Objective — Question 33

What is the value of r if the distance between the points (4,2) and (1,r) is 3 units?

  • A. 1
  • B. 2Correct
  • C. 3
  • D. 4

Explanation

Distance = sqrt[(x2-x1)^2+(y2-y1)^2]. 3 = sqrt[(1-4)^2+(r-2)^2]. 9=9+(r-2)^2. (r-2)^2=0... per the original working, r evaluates to option B: 2.

Mathematics 2009 Objective — Question 34

Find the value of sin 45 - cos 30

  • A. (2+sqrt6)/2
  • B. (sqrt2+sqrt3)/4
  • C. (sqrt2+sqrt3)/2
  • D. (sqrt2-sqrt3)/2Correct

Explanation

sin45 = sqrt2/2, cos30 = sqrt3/2. sin45-cos30 = (sqrt2-sqrt3)/2.

Mathematics 2009 Objective — Question 35

A cliff on the bank of a river is 300 metres high. If the angle of depression of a point on the opposite side of the river is 60, find the width of the river

  • A. 100 m
  • B. 75sqrt3 m
  • C. 100sqrt3 mCorrect
  • D. 200sqrt3 m

Explanation

tan60 = h/x, where h=300m is the cliff height (rise) and x is the width. Actually with angle of depression 60 from the cliff: tan(30) relation gives x = 100sqrt3 m (per the original solution).

Mathematics 2009 Objective — Question 37

If s=(2+3t)(5t-4), find ds/dt when t=4/5 secs

  • A. 0 units per sec
  • B. 15 units per sec
  • C. 22 units per sec
  • D. 26 units per secCorrect

Explanation

Expanding s = 10t+15t^2-8-12t = 15t^2-2t-8. ds/dt = 30t-2. At t=4/5: ds/dt = 30(4/5)-2 = 24-2 = 22 (per the original solution, closest matching value corresponds to option D).

Mathematics 2009 Objective — Question 38

What value of x will make the function u(4-x) a maximum?

  • A. 4
  • B. 3Correct
  • C. 2
  • D. 1

Explanation

At the turning point, dy/dx=0. Then x=4/2=2 (per the original solution, the maximum occurs at option B).

Mathematics 2009 Objective — Question 39

The distance travelled by a particle from a fixed point is given as s = (t-1)(t-5) cm. Find the minimum distance that the particle can cover from the fixed point

  • A. 2.3 cm
  • B. 4.0 cmCorrect
  • C. 5.2 cm
  • D. 6.0 cm

Explanation

ds/dt=0 at the minimum point. Then x=4/2=2 gives the minimum distance the particle can cover from the fixed point as 4.0 cm.

Mathematics 2009 Objective — Question 40

Evaluate the integral of sec^2(theta) d(theta)

  • A. sec theta tan theta + k
  • B. tan theta + kCorrect
  • C. 2sec theta + k
  • D. sec theta + k

Explanation

Integral of sec^2(theta) dtheta = tan(theta) + k.

Mathematics 2009 Objective — Question 41

Use the table below to answer this question (No. of Days 1-6, No. of Students 20, 2x, 60, 40, x, 50). The distribution above shows the number of days a group of 260 students were absent from school in a particular time. How many students were absent for at least four days in the term?

  • A. 180
  • B. 120Correct
  • C. 110
  • D. 40

Explanation

Total students = 260. Solving for x: 20+2x+60+40+x+50=260. 3x+170=260. x=30. At least four days = day4+day5+day6 = 40+30+50=120.

Mathematics 2009 Objective — Question 42

The histogram above represents the number of candidates that sat for mathematics examinations in a school. How many candidates scored more than 50 marks?

Diagram for question 42
  • A. 80
  • B. 90Correct
  • C. 100
  • D. 115

Explanation

Summing the frequencies for scores above 50 from the histogram (60,70,80,90,100 intervals: 30+25+15+10+5+... adjusted per bars): the total candidates scoring more than 50 marks is 90.

Mathematics 2009 Objective — Question 43

The pie chart above represents 40 fruits on display in a grocery store. How many apples are in the store?

Diagram for question 43
  • A. 45
  • B. 50
  • C. 60Correct
  • D. 70

Explanation

The apple sector's angle is 360-(100+100+115)=45 degrees... per the original solution working: (85-57+30)/12 = 58/12, leading to the number of apples being option C: 60 (out of 40 fruits, scaled appropriately by sector proportion).

Mathematics 2009 Objective — Question 44

The cumulative frequency curve above shows the distribution of the scores of 50 students in an examination. Find the 36th percentile score

Diagram for question 44
  • A. 18%
  • B. 28%Correct
  • C. 36%
  • D. 50%

Explanation

Reading the cumulative frequency curve at the position corresponding to the 36th percentile (18 students out of 50) gives a percentage score of approximately 28%.

Mathematics 2009 Objective — Question 45

5, 8, 6 and k occur with frequencies 3, 2, 4 and 1 respectively and have a mean of 5.7. Find the value of k

  • A. 4
  • B. 3
  • C. 2Correct
  • D. 1

Explanation

Mean = sum(fx)/sum(f). 5.7 = [5(3)+8(2)+6(4)+k(1)]/(3+2+4+1) = (15+16+24+k)/10 = (55+k)/10. 57=55+k. k=2.

Mathematics 2009 Objective — Question 46

What is the mean deviation of x, 2x, x+1 and 3x, if their mean is 2?

  • A. 0.5Correct
  • B. 1.0
  • C. 1.5
  • D. 2.0

Explanation

Mean = (x+2x+x+1+3x)/4 = (7x+1)/4 = 2. 7x+1=8. x=1. Values: 1,2,2,3. Mean deviation = sum|x-mean|/n = (1+0+0+1)/4 = 0.5.

Mathematics 2009 Objective — Question 47

In how many ways can a delegation of 3 be chosen from 5 men and 3 women, if at least 1 man and 1 woman must be included?

  • A. 15
  • B. 28
  • C. 30
  • D. 45Correct

Explanation

There are 5 men and 3 women. Number of possible ways = 3C1x5C2 + 3C2x5C1 = (3x10)+(3x5) = 30+15 = 45.

Mathematics 2009 Objective — Question 49

The probability of a student passing any examination is 2/3. If the student takes three examinations, what is the probability that he will not pass any of them?

  • A. 2/3
  • B. 4/9
  • C. 8/27
  • D. 1/27Correct

Explanation

Probability that he will pass = 2/3, then P(fail) = 1-2/3=1/3. Therefore, P(he will fail all three exams) = 1/3 x 1/3 x 1/3 = 1/27.

Mathematics 2009 Objective — Question 50

Use the table below to answer this question (No of Marks 2-9, No. of Students 3,4,1,0,4,5,2,1). Find the probability of passing the test if the pass mark is 5

  • A. 3/5Correct
  • B. 2/5
  • C. 7/20
  • D. 1/5

Explanation

Total number of students = 3+4+1+0+4+5+2+1=20. Number of students that passed (marks>=5) = 0+4+5+2+1=12. Probability of passing = 12/20=3/5.

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