JAMB Mathematics Past Questions & Answers - Page 184

916.

Simplify 3 - 2 \(\div\) \(\frac{4}{5}\) + \(\frac{1}{2}\)

A.

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

B.

-1

C.

1\(\frac{3}{10}\)

D.

1

E.

1\(\frac{9}{10}\)

Correct answer is D

3 - 2 \(\div\) (\(\frac{4}{5}\)) + \(\frac{1}{2}\)

3 - (2 x \(\frac{5}{4}\)) + \(\frac{1}{2}\) = 3 - \(\frac{10}{4}\) + \(\frac{1}{2}\)

= 3 - \(\frac{5}{2}\) + \(\frac{1}{2}\)

= \(\frac{6 - 5 + 1}{2}\)

= \(\frac{2}{2}\)

= 1

917.

Rationalize the expression \(\frac{1}{\sqrt{2} + \sqrt{5}}\)

A.

\(\frac{1}{3}\)(\(\sqrt{5} - \sqrt{2}\)

B.

\(\frac{\sqrt{2}}{3}\) + \(\frac{\sqrt{5}}{5}\)

C.

\(\sqrt{2} - \sqrt{5}\)

D.

5(\(\sqrt{2} - \sqrt{5}\)

E.

\(\frac{1}{3}\)(\(\sqrt{2} - \sqrt{5}\)

Correct answer is A

\(\frac{1}{\sqrt{2} + \sqrt{5}}\)

\(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\)

= \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\)

= \(\frac{\sqrt{2} - \sqrt{5}}{-3}\)

= \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\)

918.

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

A.

120km/hr

B.

60km/hr

C.

670km/hr

D.

40km/hr

Correct answer is B

Speed = \(\frac{distance}{time}\)

let x represent the speed, d represent distance

x = \(\frac{d}{4}\)

d = 4x

2x = \(\frac{600 - d}{3}\)

6x = 600 - d

6x = 600 - 4x

10x = 600

x = \(\frac{600}{10}\)

= 60km/hr

919.

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

A.

25.12

B.

25.30

C.

25.26

D.

25.50

E.

25.75

Correct answer is C

\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\)

mean = \(\frac{126.3}{5}\)

= 25.26

920.

In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30°. Find angles P and R

A.

P = 60o and R = 90o

B.

P = 30o and R = 120o

C.

P = 90o and R = 60o

D.

P = 60o and R 60o

E.

P = 45o and R = 105o

Correct answer is A

By using cosine formula, p2 = Q2 + R2 - 2QR cos p

Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\)

= \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\)

= \(\frac{3 + 12 - 9}{12}\)

= \(\frac{6}{12}\)

= \(\frac{1}{2}\)

= 0.5

Cos P = 0.5

p = cos-1 0.5 = 60°

= < P = 60°

If < P = 60° and < Q = 30

< R = 180° - 90°

angle P = 60° and angle R is 90°