JAMB Mathematics Past Questions & Answers - Page 362

1,806.
1,807.

If 125\(_x\) = 20\(_10\) find x

A.

2

B.

3

C.

4

D.

5

Correct answer is B

125\(_x\) = 20\(_10\)
1 x X\(^2\) + 2 x X\(^1\) + 5x \(^0\) = 20
X2 + 2X + 5 = 20
X\(^2\) + 2X - 15 = 0
(X + 5)( X - 3) = 0
X + 5 implies X = -5
X - 3 implies X = 3
But X cannot be negative
X = 3

1,808.

Add 11012,101112 and 1112

A.

1110112

B.

1101102

C.

1010112

D.

1010102

Correct answer is C

11012 + 101112 + 1112 = 101011

1,809.

The area of a square is 144 sqcm. Find the length of the diagonal.

A.

13 cm

B.

12√2 cm

C.

12 cm

D.

11√3 cm

Correct answer is B

No explanation has been provided for this answer.

1,810.

Find the value of \(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}\)

A.

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

B.

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

C.

1

D.

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

Correct answer is D

\(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}= \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\\
=\left(\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}\right)\div \left(\frac{\sqrt{3}\sqrt{3}+1}{\sqrt{3}}\right)\\
=\frac{(3-1)}{(3+1)}\\
=\frac{2}{4}=\frac{1}{2}\)