JAMB Mathematics Past Questions & Answers - Page 218

1,086.

A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2 : 3 : 5 respectively. If Bisi got 5, how many were share out?

A.

15

B.

25

C.

30

D.

50

Correct answer is B

Let x r3epresent total number of pencils shared

B : S : T = 2 + 3 + 5 = 10

2 : 3 : 5

= \(\frac{2}{10}\) x y

= 5

2y =5

2y = 50

∴ y = \(\frac{50}{2}\)

= 25

1,087.

Udoh deposited N150.00 in the bank. At the end of 5 years the simple interest on the principal was N55.00. At what rate per annum was the interest paid?

A.

11%

B.

7\(\frac{1}{3}\)%

C.

5

D.

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

Correct answer is B

S.I = \(\frac{PTR}{100}\)

R = \(\frac{100}{PT}\)

\(\frac{100 \times 50}{150 \times 6}\) = \(\frac{22}{3}\)

= 7\(\frac{1}{3}\)%

1,088.

Simplify \(\frac{0.0324 \times 0.00064}{0.48 \times 0.012}\)

A.

3.6 x 102

B.

3.6 x 10-2

C.

3.6 x 10-3

D.

3.6 x 10-4

Correct answer is C

\(\frac{0.0324 \times 0.00064}{0.48 \times 0.012}\) = \(\frac{324 \times 10^{-4} \times 10^{-5}}{48 x 10^{-2} x 12 x 10^{-3}}\)

\(\frac{324 \times 64 \times 1 0^{-9}}{48 \times 12 \times 10^{-5}}\) = 36 x 10-4

= 3.6 x 10-3

1,089.

If P = 18, Q = 21, R = -6 and S = -4, Calculate \(\frac{(P- Q)^3 + S^2}{R^3}\) 

A.

\(\frac{-11}{216}\)

B.

\(\frac{11}{216}\)

C.

\(\frac{-43}{116}\)

D.

\(\frac{43}{116}\)

Correct answer is B

\(\frac{(P- Q)^3 + S^2}{R^3}\) = \(\frac{(18 - 21)^3 + (-4)^2}{(-6)^3}\)

= \(\frac{-27 + 16}{R^3}\)

= \(\frac{-11}{-216}\)

= \(\frac{11}{216}\)

1,090.

Three boys shared some oranges. The first received \(\frac{1}{3}\) of the oranges, the second received \(\frac{2}{3}\) of the remainder. If the third boy received the remaining 12 oranges, how many oranges did they share?

A.

60

B.

54

C.

48

D.

42

Correct answer is B

let x represent the total number of oranges shared, let the three boys be A, B and C respectively. A received \(\frac{1}{3}\) of x, Remainder = \(\frac{2}{3}\) of x . B received \(\frac{2}{3}\) of remainder (i.e.) \(\frac{2}{3}\) of x

∴ C received \(\frac{2}{3}\) of remainder (\(\frac{2}{3}\) of x) = 12

\(\frac{1}{3}\) x \(\frac{2x}{3}\) = 12

2x = 108

x = 54