Year : 
1981
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 47 Questions

# Question Ans
21.

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

Detailed Solution

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

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

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

= x2 + 4x + 3 = 8

x2 + 4x - 5 = 0

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

x = 1 or -5
22.

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

Detailed Solution

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
23.

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

Detailed Solution

\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\)
When x = 4,
\(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\)
\(\log_{2} y = 3 - \log_{2} 2^{3}\)
\(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\)
\(\log_{2} y = 0 \implies y = 2^{0} = 1\)
24.

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

Detailed Solution

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
25.

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

Detailed Solution

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
26.

Simplify \(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\)

A. 32n

B. 3 x 23n - 1

C. 2n

D. 6

E. 1

Detailed Solution

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\)

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

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

= \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\)

= 26n + 1 - 6n x 34n + 1 - 4n

2 x 3 = 6
27.

The number 25 when converted from the tens and units base to the binary base (base two) is one of the following

A. 10011

B. 111011

C. 111000

D. 11001

E. 110011

Detailed Solution

\(\begin{array}{c|c} 2 & 25\\2 & 12 R 1\\2 & 6 R 0\\2 & 3 R 0\\2 & 1 R 1\end{array}\)

= (11001)
28.

Evaluate \(\frac{6.3 \times 10^5}{8.1 \times 10^3}\) to 3 significant fiqures

A. 77.80

B. 778.0

C. 7.870

D. 8.770

E. 88.70

Detailed Solution

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\)
\(\frac{7}{9} \times 10^{2}\)
= \(0.77778 \times 10^{2}\)
= \(77.778 \approxeq 77.80\)
29.

The positive root of t in the following equation, 4t2 + 7t - 1 = 0, correct to 4 places of decimal, is

A. 1.0622

B. 10.6225

C. 0.1328

D. 0.3218

E. 2.0132

Detailed Solution

\(4t^{2} + 7t - 1 = 0\)
Using a = 4, b = 7, c = -1.
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
= \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\)
= \(\frac{-7 \pm \sqrt{49 + 16}}{8}\)
= \(\frac{-7 \pm \sqrt{65}}{8}\)
= \(\frac{-7 \pm 8.0623}{8}\)
The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\)
\(\approxeq 0.1328\) (4 decimal place)
30.

The difference between the length and width of a rectangle is 6cm and the area is 135cm2. What is the length?

A. 25cm

B. 18cm

C. 15cm

D. 24cm

E. 27cm

Detailed Solution

Area = L x B

L - B = 6

L = 6 + B

area = 135

B(6 + B) = B2 + 6B - 135 = 0

B = 9

L = 6 + 9

= 15
21.

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

Detailed Solution

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

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

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

= x2 + 4x + 3 = 8

x2 + 4x - 5 = 0

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

x = 1 or -5
22.

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

Detailed Solution

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
23.

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

Detailed Solution

\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\)
When x = 4,
\(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\)
\(\log_{2} y = 3 - \log_{2} 2^{3}\)
\(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\)
\(\log_{2} y = 0 \implies y = 2^{0} = 1\)
24.

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

Detailed Solution

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
25.

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

Detailed Solution

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
26.

Simplify \(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\)

A. 32n

B. 3 x 23n - 1

C. 2n

D. 6

E. 1

Detailed Solution

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\)

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

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

= \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\)

= 26n + 1 - 6n x 34n + 1 - 4n

2 x 3 = 6
27.

The number 25 when converted from the tens and units base to the binary base (base two) is one of the following

A. 10011

B. 111011

C. 111000

D. 11001

E. 110011

Detailed Solution

\(\begin{array}{c|c} 2 & 25\\2 & 12 R 1\\2 & 6 R 0\\2 & 3 R 0\\2 & 1 R 1\end{array}\)

= (11001)
28.

Evaluate \(\frac{6.3 \times 10^5}{8.1 \times 10^3}\) to 3 significant fiqures

A. 77.80

B. 778.0

C. 7.870

D. 8.770

E. 88.70

Detailed Solution

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\)
\(\frac{7}{9} \times 10^{2}\)
= \(0.77778 \times 10^{2}\)
= \(77.778 \approxeq 77.80\)
29.

The positive root of t in the following equation, 4t2 + 7t - 1 = 0, correct to 4 places of decimal, is

A. 1.0622

B. 10.6225

C. 0.1328

D. 0.3218

E. 2.0132

Detailed Solution

\(4t^{2} + 7t - 1 = 0\)
Using a = 4, b = 7, c = -1.
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
= \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\)
= \(\frac{-7 \pm \sqrt{49 + 16}}{8}\)
= \(\frac{-7 \pm \sqrt{65}}{8}\)
= \(\frac{-7 \pm 8.0623}{8}\)
The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\)
\(\approxeq 0.1328\) (4 decimal place)
30.

The difference between the length and width of a rectangle is 6cm and the area is 135cm2. What is the length?

A. 25cm

B. 18cm

C. 15cm

D. 24cm

E. 27cm

Detailed Solution

Area = L x B

L - B = 6

L = 6 + B

area = 135

B(6 + B) = B2 + 6B - 135 = 0

B = 9

L = 6 + 9

= 15