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JAMB Mathematics Past Questions & Answers - Page 5

21.

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

A.

600

B.

450

C.

900

D.

300

Correct answer is A

PR2=PQ2+RQ22(PQ)(RQ)cosQ

cosQ=PQ2+RQ2PR22(PQ)(RQ)

cosQ=82+52722×8×5

cosQ=64+254980

cosQ=4080=0.5

Q=cos1(0.5)=600

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

⇒ -4x = -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%