JAMB Mathematics Past Questions & Answers - Page 392

1,956.

A bag contains 5 blacks balls and 3 red balls. Two balls are picked at random without replacement. What is the probability that a black and red balls are picked?

15/28

13/28

5/14

3/14

View answer & explanation

Correct answer is A

Black balls = 5
Red balls = 3/8
P(B) = 5/8, P(R) = 3/8
P( a Black and a Red) = BR + RB
= (5/8) * (3/7) * (3/8) * (5/7)
= (15/56) + (15/56)
= 30/56
= 15/28

1,957.

Find the number of committees of three that can be formed consisting of two men and one woman from four men and three women

View answer & explanation

Correct answer is C

\(^{4}C_2 \times ^{3}C_1 = \left(\frac{4!}{(4-2)!2!}\right)\times\left(\frac{3!}{(3-1)!1!}\right)\\=\frac{4!}{2!2!}\times \frac{3!}{2!}\\=\frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{3 \times 2!}{2!}\\=18\)

1,958.

If \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), find the value of n.

View answer & explanation

Correct answer is C

\(^{n}P_3 - 6(^{n}C_{4})=0\\\frac{n!}{(n-3)!}-6\left( \frac{n!}{(n-4)!4!}\right)=0\\\frac{n!}{(n-3)!}=6\left(\frac{n!}{(n-4)!4!}\right)\\n!((n-4)!4!)=6n!(n-3)!\\((n-4)!4!)=6(n-3)!\\\frac{(n-4)!}{(n-3)!}=\frac{6}{4!}\\\frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4 \times 3\times 2\times 1}\\\frac{1}{(n-3)}=]\frac{1}{4}\\n-3=4\\n=4+3\\n=7\)

1,959.

On a pie chart, there are four sectors of which three angles are 45°, 90° and 135°. If the smallest sector represents N28.00, how much is the largest sector?

N96.00

N84.00

N48.00

N42.00

View answer & explanation

Correct answer is B

Let the 4th angle be = x
∴ x + 45 + 90 + 135 = 360
x + 270 = 360
x = 360 - 270
x = 90
∴ smallest angle = 45o
45o = N28.00
1o = 28.00/45
135o = (28.00/45) * (135/1)
= N84.00