JAMB Mathematics Past Questions & Answers - Page 5

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21.

Find the value of the angle marked x in the diagram above

A.

60\(^0\)

B.

45\(^0\)

C.

90\(^0\)

D.

30\(^0\)

Correct answer is A

\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\)

\(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\)

\(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\)

\(\implies cos Q = \frac{64 + 25 - 49}{80}\)

\(\implies cos Q = \frac{40}{80} = 0.5\)

\(\implies Q = cos^{-1} (0.5) = 60^0\)

\(\therefore x = 60^0\)

22.

The second term of a geometric series is \(^{-2}/_3\) and its sum to infinity is \(^3/_2\). Find its common ratio.

A.

\(^{-1}/_3\)

B.

2

C.

\(^{4}/_3\)

D.

\(^{2}/_9\)

Correct answer is A

\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)

\(T_n = ar^n - 1\)

∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)

\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = \(\frac{3 - 3r}{2}\)

Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)

= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)

= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)

= 3(3r - 3r\(^2\)) = -4

= 9r - 9r\(^2\) = -4

= 9r\(^2\) - 9r - 4 = 0

= 9r\(^2\) - 12r + 3r - 4 = 0

= 3r(3r - 4) + 1(3r - 4) = 0

= (3r - 4)(3r + 1) = 0

∴ r = \(\frac{4}{3} or - \frac{1}{3}\)

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)

23.

A rectangle has one side that is 6 cm shorter than the other. The area of the rectangle will increase by 68 cm\(^2\) if we add 2 cm to each side of the rectangle. Find the length of the shorter side

A.

15 cm

B.

19 cm

C.

13 cm

D.

21 cm

Correct answer is C

Let the length of the longer side = \(x\) cm

∴ The length of the shorter side = (\(x\) - 6) cm

If we increase each side's length by 2 cm, it becomes

(\(x\) + 2) cm and (\(x\) - 4) cm respectively

Area of a rectangle = L x B

\(A_1 = x(x - 6) = x^2 - 6x\)

\(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\)

\(A_1 + 68 = A_2\) (Given)

⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\)

⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68

⇒ -4\(x\) = -76

⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm

∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm

24.

Evaluate \(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} \times \frac{1}{4}\)

A.

- \(\frac{3}{40}\)

B.

\(\frac{3}{40}\)

C.

\(\frac{7}{40}\)

D.

-\(\frac{263}{40}\)

Correct answer is C

\(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} x \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{3}{4} ÷ \frac{5}{12}) \times \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\)

⇒ \(\frac{5}{8} - \frac{9}{20}\)

∴ \(\frac{7}{40}\)

25.

Bello buys an old bicycle for ₦9,200.00 and spends ₦1,500.00 on its repairs. If he sells the bicycle for ₦13,400.00, his gain percent is

A.

25.23%

B.

31.34%

C.

88.81%

D.

42.54%

Correct answer is A

Total cost price = ₦9,200.00 + ₦1,500.00 = ₦10,700.00

Selling price = ₦13,400.00

Gain = ₦13,400.00 - ₦10,700.00 = ₦2,700.00

∴ % gain = \(\frac{₦2,700.00}{₦10,700.00}\times\) 100% = 25.23%