JAMB Mathematics 1992

15 reviewed questions with answers and explanations.

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Question 6

Simplify 5√18 − 3√72 + 4√50.

  1. 17√4
  2. 4√17
  3. 17√2
  4. 12√4
Answer and explanation

C: 17√2

Since √18 = 3√2, √72 = 6√2 and √50 = 5√2, the expression is 15√2 − 18√2 + 20√2 = 17√2.

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Question 10

Make t the subject of the formula s = ut + ½at².

  1. (u ± √(u² − 2as))/a
  2. (−u ± √(u² − 2as))/a
  3. (u ± √(u² + 2as))/a
  4. (−u ± √(u² + 2as))/a
Answer and explanation

D: (−u ± √(u² + 2as))/a

Rearrange to at² + 2ut − 2s = 0. For a ≠ 0, the quadratic formula gives t = (−2u ± √(4u² + 8as))/(2a) = (−u ± √(u² + 2as))/a. Real solutions require u² + 2as ≥ 0; a physical time must also satisfy the conditions of the motion.

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Question 12

Solve the equation y − 11√y + 24 = 0.

  1. 8, 3
  2. 64, 9
  3. 6, 4
  4. 9, −8
Answer and explanation

B: 64, 9

Put z = √y, so z ≥ 0. Then z² − 11z + 24 = (z − 8)(z − 3) = 0. Thus √y is 8 or 3, giving y = 64 or 9. Both values satisfy the original equation.

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Question 15

Resolve 3/(x² + x − 2) into partial fractions.

  1. 1/(x−1) − 1/(x+2)
  2. 1/(x+2) − 1/(x−1)
  3. 1/(x+1) − 1/(x−2)
  4. 1/(x−2) + 1/(x+1)
Answer and explanation

A: 1/(x−1) − 1/(x+2)

Factor the denominator as (x−1)(x+2). Writing 3=A(x+2)+B(x−1), x=1 gives A=1 and x=−2 gives B=−1. Thus the partial fractions are 1/(x−1)−1/(x+2), with x≠1,−2.

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Question 19

What is the nth term of the sequence 2, 6, 12, 20, …?

  1. 4n − 2
  2. 2(3n − 1)
  3. n² + n
  4. n² + 3n + 2
Answer and explanation

C: n² + n

The terms are 1×2, 2×3, 3×4 and 4×5. The pattern is n(n + 1) = n² + n, starting at n = 1. It gives 2, 6, 12 and 20, matching all four terms.

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Question 21

For real numbers m and n, the binary operation * is defined by m*n = mn + m + n. Find its identity element.

  1. e = 1
  2. e = −1
  3. e = −2
  4. e = 0
Answer and explanation

D: e = 0

An identity e must satisfy m*e = m for every m. Thus me + m + e = m, so e(m + 1) = 0 for every m. Taking m = 0 gives e = 0. Substitution confirms m*0 = m and 0*m = m.

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Question 26

P and R are the endpoints of a diameter of a circle, and Q is a point on the semicircle. PQ and QR are chords. S lies on PR, with QS perpendicular to PR. Which expression gives the length QS?

  1. QS = PS × SR
  2. QS = √(PS × SR)
  3. QS = √2 × √(PS × SR)
  4. QS = (1/√2) × √(PS × SR)
Answer and explanation

B: QS = √(PS × SR)

The angle PQR is 90° because PR is a diameter. The altitude QS divides the right triangle into two similar triangles, PQS and QRS. Their corresponding sides give PS/QS = QS/SR, so QS² = PS × SR. Taking the positive square root gives QS = √(PS × SR).

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Question 27

Towns P and Q both lie on latitude 60°N. Their longitudes are 20°E and 25°W respectively. Calculate the distance between them along this latitude, taking Earth’s radius as 6400 km.

  1. 800π/9 km
  2. 800√(3π)/9 km
  3. 800π km
  4. 800√(3π) km
Answer and explanation

C: 800π km

The longitude difference is 20° + 25° = 45°. At latitude 60°N, the parallel has radius 6400 cos 60° = 3200 km. The required arc is 45/360 of its circumference: (45/360) × 2π × 3200 = 800π km.

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Question 30

Two perpendicular radii of a circle with centre O each have length 6 cm. A chord joins their endpoints. Find the area of the minor segment between this chord and the quarter-circle arc.

  1. 9π cm²
  2. 9(π − 2) cm²
  3. 18π cm²
  4. 36π cm²
Answer and explanation

B: 9(π − 2) cm²

The perpendicular radii form a 90° sector. Its area is (90/360) × π × 6² = 9π cm². The right triangle formed by the radii has area ½ × 6 × 6 = 18 cm². The segment area is therefore 9π − 18 = 9(π − 2) cm².

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Question 31

The locus of a point which is equidistant from two given fixed points is the

  1. perpendicular bisector of the straight line joining them
  2. parallel line to the straight line joining them
  3. transverse to the straight line joining them
  4. angle bisector of 90° which the straight line joining them makes with the horizontal
Answer and explanation

A: perpendicular bisector of the straight line joining them

Place the fixed points at (−a, 0) and (a, 0). Equal distances from (x, y) require (x + a)² + y² = (x − a)² + y². For two distinct fixed points, a ≠ 0, so x = 0. This is the perpendicular bisector of the segment joining the points.

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Question 33

Find the equation of the line through (5, 7) parallel to the line 7x + 5y = 12.

  1. 5x + 7y = 120
  2. 7x + 5y = 70
  3. x + y = 7
  4. 15x + 17y = 90
Answer and explanation

B: 7x + 5y = 70

A parallel line has the same coefficients of x and y, so write it as 7x + 5y = c. Substituting (5, 7) gives c = 7×5 + 5×7 = 70. Hence the required equation is 7x + 5y = 70.

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Question 43

A frequency distribution has values 2, 4, 6 and 8, with respective frequencies 4, y, 6 and 5. If its mean is 5.2, find y.

  1. 6.0
  2. 5.2
  3. 5.0
  4. 4.0
Answer and explanation

C: 5.0

The weighted total is 2×4 + 4y + 6×6 + 8×5 = 84 + 4y, and the total frequency is 15 + y. Thus (84 + 4y)/(15 + y) = 5.2. Multiplying out gives 84 + 4y = 78 + 5.2y, so 6 = 1.2y and y = 5.

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Question 44

The numbers of children in families are 0, 1, 2, 3, 4, 5 and 6, with respective frequencies 7, 11, 6, 7, 7, 5 and 3. Find the mode and median, respectively.

  1. 2,1
  2. 1,2
  3. 1,5
  4. 5,2
Answer and explanation

B: 1,2

The highest frequency is 11, corresponding to 1 child, so the mode is 1. There are 46 families. The median uses the 23rd and 24th observations. Cumulative frequencies are 7, 18 and 24 through 0, 1 and 2 children, so both middle observations are 2. The mode and median are therefore 1 and 2.

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Question 47

Two fair six-sided dice are thrown together. Determine the probability of obtaining a total score of 8.

  1. 1/12
  2. 5/36
  3. 1/8
  4. 7/36
Answer and explanation

B: 5/36

The dice have 6 × 6 = 36 equally likely ordered outcomes. A total of 8 occurs for (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2): five outcomes. The probability is 5/36.

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Question 48

Events P and Q have probabilities 3/4 and 1/6, respectively. If the probability that both occur is 1/12, what is the probability that at least one occurs?

  1. 1/96
  2. 1/8
  3. 5/6
  4. 11/12
Answer and explanation

C: 5/6

Add the probabilities and subtract the overlap counted twice: Pr(P ∪ Q) = Pr(P) + Pr(Q) − Pr(P ∩ Q) = 3/4 + 1/6 − 1/12 = 9/12 + 2/12 − 1/12 = 10/12 = 5/6.

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