JAMB Mathematics Past Questions & Answers - Page 275

1,371.

Solve without using tables log5(62.5) - log5(\(\frac{1}{2}\))

A.

3

B.

4

C.

5

D.

8

Correct answer is A

log5(62.5) - log5(\(\frac{1}{2}\))

= log5\(\frac{(62.5)}{\frac{1}{2}}\) - log5(2 x 62.5)

= log5(125)

= log553 - 3log55

= 3

1,372.

If 2log3 y + log3 x2 = 4, then y is

A.

4 - log3

B.

\(\frac{4}{log_3 x}\)

C.

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

D.

\(\pm\) \(\frac{9}{x}\)

Correct answer is D

2log3y + log3x2 = 4

log3y2 + log3x2 = 4

∴ log3 (x2y2) = log381(correct all to base 4)

x2y2 = 81

∴ xy = \(\pm\)9

∴ y = \(\pm\)\(\frac{9}{x}\)

1,373.

simplify \(\frac{1}{√3 - 2}\) - \(\frac{1}{√3 + 2}\)

A.

3

B.

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

C.

7

D.

-4

Correct answer is D

\(\frac{1}{√3 - 2}\) - \(\frac{1}{√3 + 2}\)

L.C.M = (3- 2) (3 + 2)

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


\(\frac{√3 + 2 - √3 + 2}{3 - 2√3 + 2√3 - 4}\) = \(\frac{4}{3 - 2}\)

= \(\frac{4}{-1}\)

= -4

1,374.

Solve for y in the equation 10^1 x 5(2y - 2) x 4(y - 1) = 1

A.

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

B.

\(\frac{5}{4}\)

C.

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

D.

5

Correct answer is C

10y x 5(2y - 2) x 4(y - 1) = 1

but 10y - (5 x 2)y = 5y x 2y

= (Law of indices)

5y x 2y x 5(2y - 2) x 4(y - 1) = 1

but 4(y - 1) = 22(y - 1)

= 2y - 2 (Law of indices)

5y x 5(2y -2) x 2(- 2) = 1

5(3y -2) x 2y x 2(2y -2) = 1

= 5(3y -2) x 2(3y -2) = 1

But ao = 1

10(3y -2) = 10o

3y - 2 = 0

∴ y = \(\frac{2}{3}\)

1,375.

If 9\(^{(x - \frac{1}{2})} = 3^{x2}\)

Find x

A.

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

B.

1

C.

2

D.

3

Correct answer is B

9(x - \(\frac{1}{2}\)) 3x2 = 32(x - \(\frac{1}{2}\)) = 3x2
∴ 2(x - \(\frac{1}{2}\)) = x2

2x - 1 = x2

hence x2 - 2x + 1 = 0

(x - 1)(x - 1) = 0

x = 1