1983 Mathematics JAMB past Questions And Answers from 21 to 30

1983 mathematics 21 to 30

Click here to see 1983 Mathematics 11 to 20

21. Find X if (X ₄)² = 100100₂

A. 6 B. 12 C. 100 D. 210 E. 110

Solution:

First convert all to base 10.

X ₄ = X x 4° = X x 1 = X ₁₀

100100 ₂ = 1 x 2^⁵ + 1 x 2² = 32 + 4 = 36

X² = 36

x = √36 = 6

Convert it back to its base 4.

6 _ten = 12 _four

Click here to learn more on how to convert from one base to another

22. Simplify Log ₁₀ a^½ + ¼Log ₁₀ a – 1/12Log ₁₀ a⁷

A. 1 B. 1/6log ₁₀ a C. 0 D. 10 E. a

Solution:

= ½Log ₁₀ a + ¼Log ₁₀ a – 7(1/12)Log ₁₀ a

= ½Log ₁₀ a + ¼Log ₁₀ a – 7/12Log ₁₀ a

= (½ + ¼ – ⁷/12)Log ₁₀ a = (6 + 3 – 7)/12Log ₁₀ a

= ⅙Log ₁₀ a

23. If w varies inversely as V and u varies directly as w³, find the relationship between u and V given that u = 1, when V = 2

A. u = 8V³ B. u = 2√V C. V = 8/u² D. V = 8u² E. U = 8/v³

Solution:

Let “~” represents the proportionality sign

w ~ 1/V and u ~ w³

w = k/V………(1) and u = kw³………(2)

where k is constant.

w needs to be eliminated since we are looking for the relationship between u and V

Substitute w = k/V in equation 2

u = k(k/V)³ = k x k³/V³

Note that constant multiply by constant is equal to that same constant

u = k/V³

Now, when u = 1 and V = 2

1 = k/2³ = k/8

k = 8

The relationship is now u = 8/V³

24. Solve the simultaneous equations for x.
x² + y – 8 = 0……….i
y + 5x – 2 = 0………ii

A. –28, 7 B. 6,-28 C. 6,-1 D. –1, 7 E. 3, 2

y = 2 – 5x………iii

Substitute equation iii in equation i

x² + (2 – 5x) – 8 = 0

x² + 2 – 5x – 8 = 0

x² – 5x – 6 = 0

x² – 6x + x – 6 = 0

x(x – 6) + 1 (x – 6) = 0

either x + 1 = 0 or x – 6 = 0

x = – 1 or 6

25. Find the missing value in the following table.

A. -3 B. 3 C. –9 D. 13 E. 9

Solution:

y = x² – x + 3

When x = -2,

(-2)² – (-2) + 3 = 4 + 2 + 3 = 9

26. If O is the centre of the circle in the figure above. Find the value of x.

A. 50 B. 260 C. 100 D. 65 E. 130

Solution:

Angle O is two times 130 = 260

x° = 360 – 260 = 100°

x = 100°

27. Find the angle of the sectors representing each item in a pie chart of the following data. 6, 10, 14, 16, 26

A. 15°, 25°, 35°, 40°, 65° B. 60°, 100°, 140°, 160°, 260° C. 6°, 10°, 14°, 16°, 26° D. 30°, 50°, 70°, 80°, 130° E. None of the above.

Solution:

6 + 10 + 14 + 16 + 26 = 72

6/72 x 360 = 30°

10/72 x 360 = 50°

14/72 x 360 = 70°

16/72 x 360 = 80°

26/72 x 360 = 130°

The angle of the sectors are 30, 50, 70, 80 and 130 respectively

28. The scores of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65, 70. What is the sum of the median and modal scores?

A.125 B. 130 C. 140 D. 150 E. 137.5

Solution:

Arrange either in ascending order or descending order of magnitude.

55, 60, 60, 60, 60, 65, 65, 65, 65, 65, 65, 70, 70, 70, 70, 75

Median = (65 + 65) ÷ 2 = 65

Mode = 65

Sum = 65 + 65 = 130

29. The letters of the word MATRICULATION are cut and put into a box. One of the letter is drawn at random from the box. Find the probability of drawing a vowel.

A. 2/13 B. 5/13 C. 6/13 D. 8/13 E. 4/13

Solution:

Possible outcome = 13 = the total number of letters in the word MATRICULATION

Required outcome = number of vowel in the word MATRICULATION

“A” is a vowel, it appears two times

“I” is a vowel, it appears two times

“U” is a vowel

“O” is a vowel

So, the probability of drawing a vowel is 6/13

30. Correct each of the number 59.81789 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures.

A. 4.46 B. 4.48 C. 4.47 D. 4.49 E. 4.50

Solution:

59.81789 = 59.8 to 3 s.f

0.0746829 = 0.0747 to 3 s.f

59.8 x 0.0747 = 4.46706

= 4.47 to 3 s.f

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