JAMB Mathematics Past Questions & Answers - Page 278

1,386.

Evaluate \(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\)

A.

0.013

B.

0.014

C.

0.14

D.

0.13

Correct answer is B

\(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\)

= \(\frac{36}{420} \times \frac{54}{90} \times \frac{63}{240}\)

= \(\frac{6}{70} \times \frac{18}{30} \times \frac{21}{80}\)

= \(\frac{27}{2000}\)

= 0.0135

\(\approx\) = 0.014

1,387.

Evaluate \(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\)

A.

\(\frac{28}{39}\)

B.

\(\frac{13}{39}\)

C.

\(\frac{39}{28}\)

D.

\(\frac{84}{13}\)

Correct answer is A

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\)

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\)

= \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\)

= \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\)

= \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\)

\(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\)

= \(\frac{28}{39}\)

1,388.

The mean of twelve positive numbers is 3. When another number is added, the mean becomes 5. Find the thirteenth number

A.

29

B.

26

C.

25

D.

24

Correct answer is A

Let the sum of the 12 numbers be x and the 13th number be y.

\(\frac{x}{12} = 3 \implies x = 36\)

\(\frac{36 + y}{13} = 5 \implies 36 + y = 65\)

\(y = 65 - 36 = 29\)

1,389.

Solve for x if \(25^{x} + 3(5^{x}) = 4\)

A.

1 or -4

B.

0

C.

1

D.

-4 or 0

Correct answer is B

\(25^{x} + 3(5^{x}) = 4\)

Let \(5^{x}\) = y.

\((5^{2})^{x} + 3(5^{x}) - 4 = 0\)

\(y^{2} + 3y - 4 = 0\)

\(y^{2} - y + 4y - 4 = 0\)

\(y(y - 1) + 4(y - 1) = 0\)

\((y + 4)(y - 1) = 0\)

\(y = -4 ; y = 1\)

y = -4 is not possible.

y = 1 \(\implies\) x = 0.

1,390.

A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases?

A.

\(\frac{5}{3}\)

B.

\(\frac{1}{9}\)

C.

\(\frac{4}{9}\)

D.

\(\frac{1}{36}\)

Correct answer is C

\(\begin{array}{c|c} & W & W & W & W & B & B \\ \hline W & WW & WW & WW & WW & WB & WB \\ W & WW & WW & WW & WW & WB & WB\\W & WW & WW & WW & WW & WB & WB\\ W & WW & WW & WW & WW & WB & WB\\ B & BW & BW & BW & BW & BB & BB \\ B & BW & BW & BW & BW & BB & BB\end{array}\)

P(WW) = \(\frac{16}{36}\)

= \(\frac{4}{9}\)