JAMB Mathematics Past Questions & Answers

991.

In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Z^o at the centre of the circle, what is Z?

View answer & explanation

Correct answer is E

Total students = 120

grade = 18

Z^o = $\frac{18}{120}$ x $\frac{360}{1}$

= 54^o

992.

Two points X and Y both on latitude 60^oS have longitude147^oE and 153^oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude(Take 2 $\pi$ R = 4 x 10⁴km, where R is the radius of the earth)

16667km

28850km

8333km

2233km

View answer & explanation

Correct answer is A

Length of an area = $\frac{\theta}{360}$ x 2 $\pi$ r

Longitude difference = 147 + 153 = 30N^oN

distance between xy = $\frac{\theta}{360}$ x 2 $\pi$ R cos60^o

= $\frac{300}{360}$ x 4 x 10⁴ x $\frac{1}{2}$

= 1.667 x 104km(1667 km)

993.

A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?

$\frac{2}{3}$

$\frac{2}{15}$

$\frac{1}{2}$

$\frac{1}{3}$

View answer & explanation

Correct answer is D

P(R¹) = $\frac{6}{10}$

= $\frac{2}{3}$

P(R¹ n R¹¹) = P(both red)

$\frac{3}{5}$ x $\frac{5}{9}$

= $\frac{1}{3}$

994.

PRSQ is a trapezium of area 14cm² in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm²

7.0

6.0

5.2

5.0

View answer & explanation

Correct answer is B

Area of trapezium = 14cm²

Area of trapezium = $\frac{1}{2}$ (a + b)h

14 = $\frac{1}{2}$ (4 + 3)h

14 = $\frac{7}{2}$ h

h = $\frac{14 \times 2}{7}$

= 4

Area of SQR = $\frac{1}{2}$ (3 x 4)

= $\frac{12}{2}$

= 6.0

995.

A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?

5kg

16kg

19kg

6kg

View answer & explanation

Correct answer is A

$\frac{1\sqrt{3}}{(\frac{1}{2})^2}$

= $\frac{4}{\sqrt{3}}$

= $\frac{\sqrt{3}}{\sqrt{3}}$

= $\frac{4\sqrt{3}}{\sqrt{3}}$

m = 64kg, V = $\frac{4\pi r^3}{3}$

= $\frac{4\pi(4)^3}{3}$

= $\frac{256\pi}{3}$ x 10^-6m³

density(P) = $\frac{\text{Mass}}{\text{Volume}}$

= $\frac{64}{\frac{256\pi}{3 \times 10^{-6}}}$

= $\frac{64 \times 3 \times 10^{-6}}{256}$

= $\frac{3}{4 \times 10^{-6}}$

m = PV = $\frac{3}{4 \pi \times 10^{-6}}$ x $\frac{4}{3}$ $\pi$ [3² - 2²] x 10^-6

$\frac{3}{4 \times 10^{-6}}$ x $\frac{4}{3}$ x 5 x 10^-6

= 5kg