JAMB Mathematics Past Questions & Answers - Page 272

1,356.

A man's initial salary is N540.00 a month and increases after each period of six months by N36.00 a month. Find his salary in the eighth month of the third year

A.

N828.00

B.

N756.00

C.

N720.00

D.

N684.00

Correct answer is C

Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 yrs and 6 months amount(increment) = N36 x 5 = N180 The man's new salary = N540 = N180 = N720.00

1,357.

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

A.

0, 8

B.

-1, \(\frac{5}{3}\)

C.

2, 3

D.

1, -1

Correct answer is B

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

\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)

\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)

\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)

\(k^{2} - 8k = 0\)

\(k(k - 8) = 0\)

\(\therefore \text{k = 0 or 8}\)

The terms of the sequence given k = 0: (1, -1, 1)

\(\implies \text{The common ratio r = -1}\)

The terms of the sequence given k = 8: (9, 15, 25)

\(\implies \text{The common ratio r = } \frac{5}{3}\)

The possible values of the common ratio are -1 and \(\frac{5}{3}\).

1,358.

A binary operation \(\ast\) is defined on a set of real numbers by x \(\ast\) y = xy for all real values of x and y. If x \(\ast\) 2 = x. Find the possible values of x

A.

0, 1

B.

1, 2

C.

2, 2

D.

0, 2

Correct answer is A

x \(\ast\) y = xy

x \(\ast\) 2 = x2

x \(\ast\) 2 = x

∴ x2 - x = 0

x(x - 1) = 0

x = 0 or 1

1,359.

If x is negative, what is the range of values of x within which \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\)

A.

3 < x < 4

B.

-4 < x < -3

C.

-2 < x < -1

D.

-3 < x < 0

Correct answer is B

\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\)

= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)

x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0

= x(x + 3)9x + 4) > 0

Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0

= x > -4 (solution only)

Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0

= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

1,360.

Solve the inequality y2 - 3y > 18

A.

-3 < y < 6

B.

y < -3 or y > 6

C.

y > -3 or y > 6

D.

y < 3 or y < 6

Correct answer is A

y2 - 3y > 18 = 3y - 18 > 0

y2 - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0

= (y + 3) (y - 6) > 0

Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0

= y > -3 y > 6

Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0

= y < -3, y < 6

Combining solution in case 1 and Case 2

= x < -3y < 6

= -3 < y < 6