JAMB Mathematics Past Questions & Answers - Page 176

876.

Two cars X and Y start at the same point and travel towards a point P which is 150km away. If the average speed of Y is 60km per hour and x arrives at P 25 minutes earlier than Y. What is the average speed of X?

A.

51\(\frac{3}{7}\)km per hour

B.

72km per hour

C.

37\(\frac{1}{2}\) km per hour

D.

66km per hour

E.

75km per hour

Correct answer is B

Distance travelled by both x and y = 150 km.

Speed of y = 60km per hour, Time = \(\frac{dist.}{Speed}\)

Time spent by y = \(\frac{150}{60}\)

= 2\(\frac{3}{6}\)hrs

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

if x arrived at p 25 minutes earlier than y, then time spent by x = 2 hour 5 mins

= 125 mins

x average speed = \(\frac{150}{125}\) x \(\frac{60}{1}\)

= \(\frac{9000}{125}\)

= 72 km\hr

877.

Given that 10x = 0.2 and log102 = 0.3010, find x

A.

-1.3010

B.

-0.6990

C.

0.6990

D.

1.3010

E.

0.02

Correct answer is B

Given that log102 = 0.3010, If 10x = 0.2

log1010x = log10 0.2 = log10

xlog10 10 = log10 2 - log1010 since log10 10 = 1

x = 0.3010 - 1

x = -0.6990

878.

If \(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\), find y when x = 4.

A.

8

B.

\(\sqrt{65}\)

C.

4\(\sqrt{2}\)

D.

3

E.

1

Correct answer is E

\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\)

When x = 4,

\(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\)

\(\log_{2} y = 3 - \log_{2} 2^{3}\)

\(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\)

\(\log_{2} y = 0 \implies y = 2^{0} = 1\)

879.

In \(\bigtriangleup\)XYZ, XY = 3cm, XZ = 5cm and YZ = 7cm. If the bisector of XYZ meets XZ at W, what is the length of XW?

A.

1.5cm

B.

2.5cm

C.

3cm

D.

4cm

E.

None of the above

Correct answer is A

Let XW = a, a :7

\(\frac{a}{5 - a}\) = \(\frac{3}{7}\)

7a = 15 - 3a

10a = 15

a = \(\frac{15}{10}\)

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

= 1.5cm

880.

Solve \(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)

A.

x = -1 or 3

B.

x = 1 or 3

C.

x = 1 or -5

D.

x = -1 or 5

E.

x = -1 or -3

Correct answer is C

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

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

\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)

= x2 + 4x + 3 = 8

x2 + 4x - 5 = 0

= (x - 1)(x + 5) = 0

x = 1 or -5