# 1983 Mathematics JAMB past Questions And Answers from 21 to 30

1983 mathematics 21 to 30

## 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 + 1 x 2² = 32 + 4 = 36

X² = 36

x = √36 = 6

Convert it back to its base 4.

6 ten = 12 four

## 22. Simplify Log 10 a½ + 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……….iy + 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