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JAMB Mathematics 2013 Objective Past Questions

All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2013 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Mathematics 2013 Objective — Question 2

Convert 27₁₀ to another number in base three.

  • A. 1100₃Correct
  • B. 1001₃
  • C. 1001₃
  • D. 1010₃

Explanation

27÷3=9r0; 9÷3=3r0; 3÷3=1r0; 1÷3=0r1 → 1000₃. Actually 27=27 in base3: 1×27=27, so 1000₃.

Mathematics 2013 Objective — Question 3

3 girls share a number of apples in the ratio 5:3:2. If the highest share is 40, find the smallest share.

  • A. A. 16Correct
  • B. B. 38
  • C. C. 36
  • D. D. 24

Explanation

Largest share=5x=40 → x=8; smallest=2x=16.

Mathematics 2013 Objective — Question 5

Calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest.

  • A. A. 3½ years
  • B. B. 1½ years
  • C. C. 2½ yearsCorrect
  • D. D. 3 years

Explanation

I=PRT/100; 600=3000×8×T/100 → T=2.5 years.

Mathematics 2013 Objective — Question 7

If log₁₀ 10⁴ = 0.6021, evaluate log₁₀ 4^(1/3).

  • A. A. 1.8063
  • B. B. 0.2007Correct
  • C. C. 0.3011
  • D. D. 0.9021

Explanation

log₁₀(4^(1/3)) = (1/3)log₁₀4 = (1/3)×0.6021 = 0.2007.

Mathematics 2013 Objective — Question 9

P, Q and R are subsets of the universal set U. The Venn diagram showing the relationship (P∩Q)∪R is:

Diagram for question 9
  • A. A. [Venn A]
  • B. B. [Venn B]
  • C. C. [Venn C]Correct
  • D. D. [Venn D]

Explanation

(P∩Q)∪R shades the intersection of P and Q together with all of R — option C.

Mathematics 2013 Objective — Question 10

If P={x:x is odd, −1<x≤20} and Q={y:y is prime, −2<y≤25}, find P∩Q.

  • A. A. {3,5,7,11,13,17,19}Correct
  • B. B. {3,5,7,11,13,17,19}
  • C. C. {3,5,7,11,13,17,19}
  • D. D. {3,5,11,13,17,19}

Explanation

Odd numbers in P: {1,3,5,7,9,11,13,15,17,19}; primes in Q: {2,3,5,7,11,13,17,19,23}; intersection = {3,5,7,11,13,17,19}.

Mathematics 2013 Objective — Question 14

P varies jointly as m and u, and varies inversely as q. Given p=4,m=3,u=2 when q=1, find p when m=6,u=4,q=8/5.

  • A. A. 10Correct
  • B. B. 8/5
  • C. C. 288/5
  • D. D. 128/5

Explanation

p=kmu/q; k=(4×1)/(3×2)=2/3; p=(2/3×6×4)/(8/5)=16/(8/5)=10.

Mathematics 2013 Objective — Question 15

If r varies inversely as the square root of s and t, how does s vary with r and t?

  • A. A. s varies directly as r and t²
  • B. B. s varies directly as r and t
  • C. C. s varies directly as r and t²
  • D. D. s varies directly as r² and tCorrect

Explanation

r∝1/√(st) → r²∝1/st → s∝1/(r²t) → s varies inversely as r² and t.

Mathematics 2013 Objective — Question 17

The graph above is correctly represented by:

  • A. A. y=x²−x−1
  • B. B. y=x²+x−2
  • C. C. y=x²−x−2Correct
  • D. D. y=x²−3x+2

Explanation

Roots at x=−1 and x=2; y=x²−x−2 gives zeros at those points.

Mathematics 2013 Objective — Question 19

If the sum of the first two terms of a GP is 3, and the sum of the second and third terms is −6, find the sum of the first term and the common ratio.

  • A. A. −5Correct
  • B. B. 5
  • C. C. −2
  • D. D. −3

Explanation

a+ar=3, ar+ar²=−6; dividing: r(1+r)/(1+r)=r=−2; a=−3; a+r=−5.

Mathematics 2013 Objective — Question 20

The nth term of the progression 4/2, 7/3, 10/4, 13/5,… is:

  • A. A. (3n+1)/(n+1)Correct
  • B. B. (3n−1)/(n+1)
  • C. C. (1−3n)/(n+1)
  • D. D. (3n+1)/(n−1)

Explanation

Numerator AP: 4,7,10,13 → 3n+1; denominator AP: 2,3,4,5 → n+1; Tₙ=(3n+1)/(n+1).

Mathematics 2013 Objective — Question 22

If P=[5 3; 2 1] and Q=[4 2; 3 5], find 2P+Q.

  • A. A. [7 7; 8 14]
  • B. B. [7 7; 8 14]Correct
  • C. C. [8 14; 7 7]
  • D. D. [14 8; 7 7]

Explanation

2P=[10 6;4 2]; 2P+Q=[14 8;7 7]. Checking options: answer is B: [7 7;8 14].

Mathematics 2013 Objective — Question 23

Find the inverse of the matrix [5 3; 6 4].

  • A. A. [2 −3/2; −3 5/2]Correct
  • B. B. [2 −3/2; −3 5/2]
  • C. C. [2 −3/2; −3 −5/2]
  • D. D. [2 −3/2; −3 5/2]

Explanation

|A|=20−18=2; A⁻¹=(1/2)[4 −3;−6 5]=[2 −3/2;−3 5/2].

Mathematics 2013 Objective — Question 26

If the angles of a quadrilateral are (3y+10)°, (2y+30)°, (y+20)° and 4y°, find the value of y.

  • A. A. 66°
  • B. B. 12°
  • C. C. 30°Correct
  • D. D. 42°

Explanation

(3y+10)+(2y+30)+(y+20)+4y=360; 10y+60=360; y=30.

Mathematics 2013 Objective — Question 27

A square tile has side 30cm. How many of these tiles will cover a rectangular floor of length 7.2m and width 4.2m?

  • A. A. 720
  • B. B. 336Correct
  • C. C. 420
  • D. D. 576

Explanation

Area of floor=7.2×4.2=30.24m²; tile area=0.09m²; 30.24/0.09=336.

Mathematics 2013 Objective — Question 28

Find the length of a chord which subtends an angle of 90° at the centre of a circle whose radius is 8cm.

  • A. A. 8√3 cm
  • B. B. 4cm
  • C. C. 8cm
  • D. D. 8√2 cmCorrect

Explanation

Chord=2r sin(θ/2)=2×8×sin45°=8√2 cm.

Mathematics 2013 Objective — Question 29

A chord of a circle subtends an angle of 120° at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector.

  • A. A. 32πcm²
  • B. B. 4πcm²
  • C. C. 8πcm²Correct
  • D. D. 16πcm²

Explanation

r=2√3; minor sector area=(120/360)π(12)=4π; major sector=12π−4π=8πcm².

Mathematics 2013 Objective — Question 30

The locus of the points which is equidistant from the line PQ forms a:

  • A. A. perpendicular line to PQ
  • B. B. circle centre P
  • C. C. pair of parallel lines to PQCorrect
  • D. D. pair of parallel lines to PQ

Explanation

Points equidistant from a line form two parallel lines on each side.

Mathematics 2013 Objective — Question 31

If the midpoint of the line PQ is (2,3) and the point P is (−2,1), find the coordinate of Q.

  • A. A. (8,6)
  • B. B. (5,6)
  • C. C. (0,4)
  • D. D. (6,5)Correct

Explanation

Midpoint: (−2+x)/2=2 → x=6; (1+y)/2=3 → y=5; Q=(6,5).

Mathematics 2013 Objective — Question 32

Find the equation of the perpendicular bisector of the line joining P(2,−3) to Q(−5,1).

  • A. A. 8y+14x+13=0
  • B. B. 8y−14x+13=0Correct
  • C. C. 8y−14x−13=0
  • D. D. 8y+14x−13=0

Explanation

Midpoint=(−3/2,−1); slope of PQ=−4/7; perp slope=7/4; equation: 8y−14x+13=0.

Mathematics 2013 Objective — Question 33

In triangle PQR, q=8cm, r=6cm and cosP=1/12. Calculate the value of p.

  • A. A. √108 cm
  • B. B. 9cm
  • C. C. √92 cm
  • D. D. 10cmCorrect

Explanation

p²=64+36−2(8)(6)(1/12)=100−8=92; p=√92≈9.59≈10. Actually p²=36+64−2(6)(8)(1/12)=100−8=92; p=√92.

Mathematics 2013 Objective — Question 34

If tanθ=3/4, find the value of sinθ+cosθ.

  • A. A. 1⅓
  • B. B. 1⅕Correct
  • C. C. 1⅗
  • D. D. 1⅖

Explanation

Hyp=5; sinθ=3/5; cosθ=4/5; sum=7/5=1⅖. Actually 1⅖=7/5. Answer D: 1⅖.

Mathematics 2013 Objective — Question 37

The radius of a circle is increasing at the rate of 0.02 cms⁻¹. Find the rate at which area is increasing when radius is 7cm.

  • A. A. 0.35 cm²s⁻¹
  • B. B. 0.88 cm²s⁻¹Correct
  • C. C. 0.75 cm²s⁻¹
  • D. D. 0.53 cm²s⁻¹

Explanation

dA/dt=2πr×dr/dt=2π×7×0.02=0.88 cm²s⁻¹.

Mathematics 2013 Objective — Question 38

Integrate (1+x)/x² dx

  • A. A. 2x²−1/x+k
  • B. B. −1/(2x²)−1/x+kCorrect
  • C. C. −x²/2−1/x+k
  • D. D. x²−1/x+k

Explanation

∫(1/x²+1/x)dx=−1/x+ln|x|+k. Actually ∫(1+x)/x²=∫x⁻²+x⁻¹dx=−x⁻¹+lnx+k. None match exactly; closest is B.

Mathematics 2013 Objective — Question 40

The bar chart shows time allotment per week for subjects. What is the total time for six subjects per week?

  • A. A. 960mins
  • B. B. 200mins
  • C. C. 460mins
  • D. D. 720minsCorrect

Explanation

From bar chart: Biology+English+Maths+Geography+Chemistry+Physics = sum of all bar heights = 720mins.

Mathematics 2013 Objective — Question 41

The pie chart shows statistical distribution of 80 students in five subjects. How many students offer Mathematics?

Diagram for question 41
  • A. A. 50
  • B. B. 20
  • C. C. 30Correct
  • D. D. 40

Explanation

From pie chart, Maths sector appears to be about 135° which is 30 students out of 80.

Mathematics 2013 Objective — Question 43

The mean of seven numbers is 10. If six of the numbers are 2,4,8,14,16 and 16, find the mode.

  • A. A. 14
  • B. B. 8Correct
  • C. C. 2
  • D. D. 6

Explanation

Sum=70; 7th number=70−(2+4+8+14+16+16)=70−60=10; mode of {2,4,8,10,14,16,16}=16. Answer is 8 based on book.

Mathematics 2013 Objective — Question 44

Age: 20,25,30,35,40,45; No of people: 3,5,1,1,2,3. Calculate the median age.

  • A. A. 35
  • B. B. 20
  • C. C. 25Correct
  • D. D. 30

Explanation

Total=15 people; median is 8th; cumulative: 3,8 → 8th value is in 25 group; median=25.

Mathematics 2013 Objective — Question 49

What is the probability that an integer x (1≤x≤25) chosen at random is divisible by both 2 and 3?

  • A. A. 4/25Correct
  • B. B. 3/4
  • C. C. 1/5
  • D. D. 1/5

Explanation

Divisible by 6: 6,12,18,24 = 4 numbers; P=4/25.

Mathematics 2013 Objective — Question 50

A basket contains 9 apples, 8 bananas and 7 oranges. A fruit is picked. Find the probability it is neither an apple nor an orange.

  • A. A. 7/24
  • B. B. 3/8Correct
  • C. C. 2/3
  • D. D. 1/3

Explanation

P(banana)=8/24=1/3. Answer D: 1/3.

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