JAMB Mathematics Past Questions & Answers - Page 38

186.

What is the product of \(\frac{27}{5}\), \((3)^{-3}\) and (\(\frac{1}{5})^{-1}\)?

A.

5

B.

3

C.

1

D.

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

Correct answer is C

\(\frac{27}{5} \times 3^{-3} \times \frac{(1)^{-1}}{5}\)

= \(\frac{27}{5} \times \frac{1}{3^3} \times \frac{1}{\frac{1}{5}}\)

\(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{5}{1}\) = 1

187.

The angle of a sector of a circle radius 10.5 cm is 48\(^o\). Calculate the perimeter of the sector.

A.

25.4cm

B.

25.4cm

C.

25.6cm

D.

29.8cm

Correct answer is D

The lenght of Arc AB = \(\frac{Q}{360}\) 2\(\pi\)r

= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5 = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)

= \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm 

Perimeter = 8.8 + 2r = 8.8 + 2r 

= 8.8 + 2(10.5)

= 8.8 + 21

= 29.8cm

188.

In preparing rice cutlets, a cook used 75g of rice, 40g of margarine, 105g of meat and 20g of bread crumbs. Find the angle of the sector which represent meat in a pie chart? 

A.

30\(^o\)

B.

60\(^o\)

C.

112.5\(^o\)

D.

157.5\(^o\)

Correct answer is D

Rice = 75g, Margarine = 40g, Meat = 105g

Bread = 20g

Total = 240

Angle of sector represented by meat 

= \(\frac{105}{240} \times \frac{360^o}{1}\) 

= 157.5 

189.

Simplify \(\frac{324 - 4x^2}{2x + 18}\) 

A.

2(x - 9)

B.

2(9 + x)

C.

81 - x\(^2\)

D.

-2(x - 9)

Correct answer is D

234 - 4x\(^2\) = 18\(^2\) - (2x)\(^2\) = (18 - 2x)(18 + 2x)

2x + 18 = 2x + 18 = (2x + 18)

18 - 2x = 2(a - x) or -2(x - a) 

 

190.

If log\(_{10}\)2 = 0.3010 and log\(_{10}\)3 = 0.4771, eventually without using the logarithm tables, log\(_{10}\)4.5

A.

0.3010

B.

0.4771

C.

0.6532

D.

0.9542

Correct answer is C

log\(_{10}\)2 = 0.3010 and log\(_{10}\)3 = 04771

log\(_{10} 4.5 = log_{10}\) (\(\frac{3 \times 3}{2}\))

log\(_{10}\) 3 + log\(_{10}\) 3 - log\(_{10}\)2 = 0.4471 + 0.771 - 0.3010

= 0.6532