JAMB Mathematics Past Questions & Answers - Page 326

1,626.

Evaluate \(\frac{21}{9}\) to 3 significant figures

A.

2.30

B.

2.31

C.

2.32

D.

2.33

Correct answer is D

\(\frac{21}{9} = \frac{7}{3}\)

= 2.33 (to 3 sig. figs)

1,627.

Simply \(\frac{2\frac{2}{3} \times 1\frac{1}{2}}{4\frac{4}{5}}\)

A.

\(1\frac{1}{4}\)

B.

\(1\frac{1}{6}\)

C.

\(\frac{5}{6}\)

D.

\(\frac{4}{5}\)

Correct answer is C

\(\frac{2\frac{2}{3} \times 1\frac{1}{2}}{4\frac{4}{5}}\)

\(\frac{8}{3} \times \frac{3}{2} \div \frac{24}{5}\)

\(\frac{8}{3} \times \frac{3}{2} \times \frac{5}{24}\)

\(\frac{5}{6}\)

1,628.

Convert 112\(_6\) to a number in base three

A.

2211

B.

2121

C.

1212

D.

1122

Correct answer is D

First, convert to base 10

112\(_6\) = (1 x 6\(^2\)) + (1 x 6\(^1\)) + (2 x 6\(^0\))

= 36 + 6 + 2

= 44\(_{10}\)

\(\begin{array}{c|c} 3 & 44 \\ \hline 3 & 14 & r 2 \\ \hline 3 & 4 & r 2 \\ \hline 3 & 1 & r 1 \\ \hline 3 & 0 & r 1 \end{array}\)

Ans = 1122 = D

1,629.

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

A.

\(\frac{3}{8}\)

B.

\(\frac{1}{3}\)

C.

\(\frac{7}{24}\)

D.

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

Correct answer is B

n(apples) = 9
n(bananas) = 8
n(oranges) = 7

n(\(\varepsilon\)) = 24

Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)

1,630.

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

A.

\(\frac{1}{25}\)

B.

\(\frac{1}{5}\)

C.

\(\frac{4}{25}\)

D.

\(\frac{3}{4}\)

Correct answer is C

\((1 \leq x \leq 25)\) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}

Number N of x divisible by both 2 and 3 is 4.

n(\(\varepsilon\)) = 25

= \(\frac{N}{n(\varepsilon)}\)

= \(\frac{4}{25}\)