JAMB Mathematics Past Questions & Answers - Page 269

1,341.

Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\)

A.

\(\frac{p + q}{a(p - q)}\)

B.

\(\frac{p - q}{a(p + q)}\)

C.

\(\frac{p - q}{apq}\)

D.

\(\frac{pq}{a(p - q)}\)

Correct answer is B

\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication,

q(1 + ax) = p(1 - ax)

q + qax = p - pax

qax + pax = p - q

∴ x = \(\frac{p - q}{a(p + q)}\)

1,342.

If \(\sqrt{x^2 + 9}\) = x + 1, solve for x

A.

5

B.

4

C.

3

D.

2

E.

1

Correct answer is B

\(\sqrt{x^2 + 9}\) = x + 1

x2 + 9 = (x + 1)2 + 1

0 = x2 + 2x + 1 - x2 - 9

= 2x - 8 = 0

2(x - 4) = 0

x = 4

1,343.

If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x

A.

3

B.

4

C.

12

D.

17

Correct answer is A

Principal = N255.00, Interest = 27.00

year = x Rate: 4%

∴ 1 = \(\frac{PRT}{100}\)

27 = \(\frac{225 \times 4 \times T}{100}\)

2700 = 900T

T = 3 years

1,344.

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

1,345.

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