JAMB Mathematics Past Questions & Answers - Page 353

1,761.

Simplify \(\frac{5+\sqrt{7}}{3+\sqrt{7}}\)

A.

17-√7

B.

4-√7

C.

15+√7

D.

7-√7

Correct answer is B

\(\frac{5+\sqrt{7}}{3+\sqrt{7}}=\frac{5+\sqrt{7}}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}}\\
=\frac{(5+\sqrt{7})(3-\sqrt{7})}{3^2 - \sqrt{7}^2}\\
=\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}\\
=\frac{8-2\sqrt{7}}{2}\)
Factorize then divide by 2
\(=\frac{2(4-\sqrt{7}}{2}\\
=4-\sqrt{7}\)

1,762.

If log\(_{10}\)2 = 0.3010 and log\(_{10}\)7 = 0.8451, evaluate log\(_{10}\)280

A.

3.4471

B.

2.4471

C.

1.4471

D.

1.4071

Correct answer is B

Log 2 = 0.3010, Log 7 = 0.8451
∴Log 280 = Log 28 x 10
= Log 7x4x10
=Log7 + Log4 + Log10
= Log7 + Log2\(^2\) + Log10
= Log7 + 2Log2 + Log10
= 0.8451 + 2(0.3010) + 1
= 0.8451 + 0.6020 + 1
= 2.4471

1,763.

A student spent 1/5 of his allowance on books, 1/2 of remainder on food and kept the rest for contingencies. What fraction was kept?

A.

7/15

B.

8/15

C.

2/5

D.

4/5

Correct answer is C

Let his allowance = y.

\(Books = \frac{y}{5}\)

\(Remainder: y - \frac{y}{5} = \frac{4y}{5}\)

\(Food: \frac{1}{2} \times \frac{4y}{5} = \frac{2y}{5}\)

\(Contingencies: \frac{4y}{5} - \frac{2y}{5} = \frac{2y}{5}\)

Therefore, he kept \(\frac{2}{5}\) of his allowance for contingencies.

1,764.

A man bought a second-hand photocopying machine for N34,000. He serviced it at a cost of N2,000 and then sold it at a profit of 15%. What was the selling price?

A.

N37,550

B.

N40,400

C.

N41,400

D.

N42,400

Correct answer is C

C.P = N34000 + N2000 = N36000
Gain = 100 + 15 = 115%
S.P = \(\frac{115}{100}\times \frac{N3600}{1}\)
= N41400

1,765.

Evaluate \(\frac{81.81+99.44}{20.09+36.16}\) correct to 3 significant figures.

A.

6.24

B.

3.22

C.

2.78

D.

2.13

Correct answer is B

\(\frac{81.81 + 99.44}{20.09 + 36.16}\\
=\frac{181.25}{56.25}\\
=\frac{18125}{5625}\\
\frac{29}{9}\\
=3\frac{2}{9}\\
≅ 3.22\)