JAMB Mathematics Past Questions & Answers - Page 180

896.

Find the value of x satisfying \(\frac{x}{2}\) - \(\frac{1}{3}\) < \(\frac{2x}{5}\) + \(\frac{1}{6}\)

A.

x < 5

B.

x < 7\(\frac{1}{2}\)

C.

x > 5

D.

x > 7\(\frac{1}{2}\)

Correct answer is A

\(\frac{x}{2} - \frac{1}{3} < \frac{2x}{5} + \frac{1}{6}\)

\(\frac{x}{2} - \frac{2x}{5} < \frac{1}{6} + \frac{1}{3}\)

\(\frac{x}{10} < \frac{1}{2}\)

\(2x < 10 \implies x < 5\)

897.

Find a two-digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits

A.

24

B.

42

C.

74

D.

47

E.

72

Correct answer is D

Let the tens digits of the number be x and the unit digit be y

3x = 2y - 2

3x - 2y = -2.......(i)

If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20

(20x + 2y) - (10y + x) = 20

19x - 8y = 20.....(ii)

Multiply eqn.(i) by 8 and eqn.(ii) by 2

24x - 16y = -16......(iii)

38x - 16y = 40........(iv)

eqn(iv) - eqn(iii)

14x = 56

x = 4

Sub. for x = 4 in eqn(i)

3(4) - 2y = -2

14 = 2y

y = 7

So the original number is 10(4) + 7

i.e. 10x + y

= 47

898.

The ratio of the price of a loaf of bread to the price of a packet of sugar in 1975 was r : t. In 1980, the price of a loaf went up by 25% and that of a packet of sugar went up by 10%. Their new ratio is now

A.

40r:50t

B.

44r : 50t

C.

50r : 44t

D.

44r:55t

E.

55r:44t

Correct answer is C

Ratio of bread to sugar = r:t

25% increase in bread = \(\frac{125r}{100}\)

10% increase in sugar = \(\frac{100t}{100}\)

New ratio = \(\frac{125r}{100}\):\(\frac{110t}{100}\)

= 25r:22t

= 50r:44t

899.

If sine x equals cosine x, what is x in radians?

A.

\(\frac{\pi}{2}\)

B.

\(\frac{\pi}{3}\)

C.

\(\frac{\pi}{4}\)

D.

\(\frac{\pi}{6}\)

E.

\(\frac{\pi}{12}\)

Correct answer is C

\(\sin x = \cos x\)

\(\implies x = 45°\)

In radians, \(x = \frac{\pi}{4}\).

900.

Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = \(\frac{1}{3}\).

A.

81

B.

27

C.

\(\frac{1}{9}\)

D.

\(\frac{1}{27}\)

E.

\(\frac{1}{81}\)

Correct answer is A

\(x \propto \frac{1}{y}\)

\(x = \frac{k}{y}\)

\(y \propto t^{2}\)

\(y = ct^{2}\) 

k and c are constants.

\(x = \frac{k}{ct^{2}}\)

Let \(\frac{k}{c} = d\) (a constant)

\(x = \frac{d}{t^{2}}\)

\(1 = \frac{d}{3^{2}} \implies d = 9\)

\(\therefore x = \frac{9}{t^{2}}\)

\(x = 9 \div (\frac{1}{3})^{2} \)

= \( 9 \div \frac{1}{9} = 9 \times 9 = 81\)