Year : 
1993
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 51 Questions

# Question Ans
21.

A binary operation \(\ast\) is defined on a set of real numbers by x \(\ast\) y = xy for all real values of x and y. If x \(\ast\) 2 = x. Find the possible values of x

A. 0, 1

B. 1, 2

C. 2, 2

D. 0, 2

Detailed Solution

x \(\ast\) y = xy

x \(\ast\) 2 = x2

x \(\ast\) 2 = x

∴ x2 - x = 0

x(x - 1) = 0

x = 0 or 1
22.

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

A. 0, 8

B. -1, \(\frac{5}{3}\)

C. 2, 3

D. 1, -1

Detailed Solution

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\)
\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)
\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)
\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)
\(k^{2} - 8k = 0\)
\(k(k - 8) = 0\)
\(\therefore \text{k = 0 or 8}\)
The terms of the sequence given k = 0: (1, -1, 1)
\(\implies \text{The common ratio r = -1}\)
The terms of the sequence given k = 8: (9, 15, 25)
\(\implies \text{The common ratio r = } \frac{5}{3}\)
The possible values of the common ratio are -1 and \(\frac{5}{3}\).
23.

A man's initial salary is N540.00 a month and increases after each period of six months by N36.00 a month. Find his salary in the eighth month of the third year

A. N828.00

B. N756.00

C. N720.00

D. N684.00

Detailed Solution

Initial salary = N540

increment = N36 (every 6 months)

Period of increment = 2 yrs and 6 months

amount(increment) = N36 x 5 = N180

The man's new salary = N540 = N180

= N720.00
24.

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

A. 12

B. 10

C. 9

D. 8

Detailed Solution

A rectangular polygon has each interior angle to be 150o

let the polygon has n-sides

therefore, Total interior angle 150 x n = 150n

hence 150n = (2n - 4)90

150n = 180n - 360

360 = (180 - 150)n

30n = 360

n = 12
25.

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of 22\(\frac{1}{2}\)o at the centre of the circle

A. 2\(\pi\)

B. \(\pi\)

C. \(\frac{2}{3}\)

D. \(\frac{\pi}{2}\)

Detailed Solution

Diameter = 8cm

∴ Radius = 4cm

Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)

∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4

= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)

= \(\frac{180}{360}\)

= \(\frac{\pi}{2}\)
26.

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

A. 5

B. 1

C. 3

D. 6

Detailed Solution

2x + 3 \(\neq\) 2x - 3 for any value of x

∴ for the \(\bigtriangleup\) to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6
27.

Find the radius of a sphere whose surface area is 154cm2 (\(\pi = \frac{22}{7}\))

A. 7.00cm

B. 3.50cm

C. 3.00cm

D. 1.75cm

Detailed Solution

Surface area = 154cm2 (area of sphere)

4\(\pi\)r2 = 154

r\(\sqrt{\frac{154}{4\pi}}\)

= 3.50cm
28.

Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60o

A. 4.0m2

B. 1m2

C. 4.7m2

D. 5.0m2

Detailed Solution

Area of sector

\(\frac{\theta}{360}\) x \(\pi\)r2, \(\theta\) = 60o, r = 3m

= \(\frac{60}{360}\) x \(\frac{12}{7}\) x 3 x 3

\(\frac{1}{6}\) x \(\frac{22}{7}\) x 9

= \(\frac{33}{7}\)

= 4.7m2
29.

The angle between latitudes 30oS and 13oN is

A. 17o

B. 33o

C. 43o

D. 53o

Detailed Solution

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.

∴ Total angle difference = (30 + 13) = 43o
30.

If sin \(\theta\) = cos \(\theta\), find \(\theta\) between 0o and 360o

A. 45o, 225o

B. 135o, 315o

C. 45o, 315o

D. 135o, 225o

Detailed Solution

sin \(\theta\) = cos \(\theta\) 0 \(\leq\) \(\theta\) \(\leq\) 360o

The acute angle where sin \(\theta\) = cos \(\theta\) = 45o

But at the third Quadrant Cos \(\theta\) = -ve; sin \(\theta\) = -ve.

at the 3rd quadrant, value with respect to Q is

(180 + Q) where Q = acute angle

(180 + 45) = 225°

The two solution are 45°, 225°
21.

A binary operation \(\ast\) is defined on a set of real numbers by x \(\ast\) y = xy for all real values of x and y. If x \(\ast\) 2 = x. Find the possible values of x

A. 0, 1

B. 1, 2

C. 2, 2

D. 0, 2

Detailed Solution

x \(\ast\) y = xy

x \(\ast\) 2 = x2

x \(\ast\) 2 = x

∴ x2 - x = 0

x(x - 1) = 0

x = 0 or 1
22.

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

A. 0, 8

B. -1, \(\frac{5}{3}\)

C. 2, 3

D. 1, -1

Detailed Solution

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\)
\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)
\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)
\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)
\(k^{2} - 8k = 0\)
\(k(k - 8) = 0\)
\(\therefore \text{k = 0 or 8}\)
The terms of the sequence given k = 0: (1, -1, 1)
\(\implies \text{The common ratio r = -1}\)
The terms of the sequence given k = 8: (9, 15, 25)
\(\implies \text{The common ratio r = } \frac{5}{3}\)
The possible values of the common ratio are -1 and \(\frac{5}{3}\).
23.

A man's initial salary is N540.00 a month and increases after each period of six months by N36.00 a month. Find his salary in the eighth month of the third year

A. N828.00

B. N756.00

C. N720.00

D. N684.00

Detailed Solution

Initial salary = N540

increment = N36 (every 6 months)

Period of increment = 2 yrs and 6 months

amount(increment) = N36 x 5 = N180

The man's new salary = N540 = N180

= N720.00
24.

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

A. 12

B. 10

C. 9

D. 8

Detailed Solution

A rectangular polygon has each interior angle to be 150o

let the polygon has n-sides

therefore, Total interior angle 150 x n = 150n

hence 150n = (2n - 4)90

150n = 180n - 360

360 = (180 - 150)n

30n = 360

n = 12
25.

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of 22\(\frac{1}{2}\)o at the centre of the circle

A. 2\(\pi\)

B. \(\pi\)

C. \(\frac{2}{3}\)

D. \(\frac{\pi}{2}\)

Detailed Solution

Diameter = 8cm

∴ Radius = 4cm

Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)

∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4

= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)

= \(\frac{180}{360}\)

= \(\frac{\pi}{2}\)
26.

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

A. 5

B. 1

C. 3

D. 6

Detailed Solution

2x + 3 \(\neq\) 2x - 3 for any value of x

∴ for the \(\bigtriangleup\) to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6
27.

Find the radius of a sphere whose surface area is 154cm2 (\(\pi = \frac{22}{7}\))

A. 7.00cm

B. 3.50cm

C. 3.00cm

D. 1.75cm

Detailed Solution

Surface area = 154cm2 (area of sphere)

4\(\pi\)r2 = 154

r\(\sqrt{\frac{154}{4\pi}}\)

= 3.50cm
28.

Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60o

A. 4.0m2

B. 1m2

C. 4.7m2

D. 5.0m2

Detailed Solution

Area of sector

\(\frac{\theta}{360}\) x \(\pi\)r2, \(\theta\) = 60o, r = 3m

= \(\frac{60}{360}\) x \(\frac{12}{7}\) x 3 x 3

\(\frac{1}{6}\) x \(\frac{22}{7}\) x 9

= \(\frac{33}{7}\)

= 4.7m2
29.

The angle between latitudes 30oS and 13oN is

A. 17o

B. 33o

C. 43o

D. 53o

Detailed Solution

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.

∴ Total angle difference = (30 + 13) = 43o
30.

If sin \(\theta\) = cos \(\theta\), find \(\theta\) between 0o and 360o

A. 45o, 225o

B. 135o, 315o

C. 45o, 315o

D. 135o, 225o

Detailed Solution

sin \(\theta\) = cos \(\theta\) 0 \(\leq\) \(\theta\) \(\leq\) 360o

The acute angle where sin \(\theta\) = cos \(\theta\) = 45o

But at the third Quadrant Cos \(\theta\) = -ve; sin \(\theta\) = -ve.

at the 3rd quadrant, value with respect to Q is

(180 + Q) where Q = acute angle

(180 + 45) = 225°

The two solution are 45°, 225°