1983 mathematics 11 to 20

Click here to see 1983 Mathematics 1 to 10

## 11. Make T the subject of the equation

Solution:

## 12. In a class of 60 pupils, the statistical distribution of the number of pupils offering Biology, History, French, Geography and Additional Mathematics is as shown in the pie chart above. How many pupils offer Additional Mathematics?

A. 15 B. 10 C. 18 D. 12 E. 28

Solution:

Additional Mathematics = (2x – 24)°

Biology = (3x – 18)

History = (2x + 12)°

French = (x +12)°

Geography = x°

(2x – 24)° + (3x – 18) + (2x + 12)° + (x + 12)° + x = 360°

2x + 3x + 2x + x + x – 24 – 18 + 12 + 12 = 360

9x – 18 = 360

9x = 360 + 18

9x = 378

x = Geography = 378/9 = 42°

Additional Mathematics = (2x – 24)

= [2(42) – 24] = 84 – 24 = 60°

Number of pupils that offer Additional Mathematics = 60/360 X 60

= 3600/360 = 10 pupils

## 13. The value of (0.03)³ – (0.02)³ is

A. 0.019 B. 0.0019 C. 0.00019 D. 0.000019 E. 0.000035

Solution:

If we use the method of difference of two cubes

a³ – b³ = (a – b)(a² + ab + b²)

Where a = 0.03

b = 0.02

(a – b)(a² + ab + b²) = (0.03 – 0.02)(0.03² + 0.03 x 0.02 + 0.02²)

= 0.01(0.0009 + 0.0006 + 0.0004)

= 0.01(0.0019)

= 0.000019

## 14. y varies partly as the square of x and y partly as the inverse of the square root of x. write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.

Solution:

Let ~ represent the proportionality sign

y ~ x² and y ~ 1/√x

y = ax² and y = k/√x

where a and k are constant

y = ax² + k/√x

when y = 2 and x = 1

2 = a(1)² + k/√1

2 = a + k……………………..i

when y = 6 and x = 4

6 = a(4)² + k/√4

6 = 16a + k/2

12 = 32a + k………………..ii

Substitute a = 2 – k in equa. (ii)

12 = 32(2 – k) + k

12 = 64 – 32k + k

-31k = -52

k = 52/31

Substitute K = 52/31 in equa. (i)

a + 52/31 = 2

a = 2 – (52/31)

a = (62 – 52)/31 = 10/31

Therefore the expression will now be:

## 15. Simplify (x – 7) /(x² – 9) ( x² – 3x)/( x² – 49)

A. x/(x-3)(x+7) B. (x+3)(x+7)/x C. x/(x-3)(x – 7) D. x/(x+3)(x+7) E. x/(x+4)(x+7)

Solution:

(x – 7) /(x² – 9) X ( x² – 3x)/( x² – 49)

= (x – 7)/(x – 3)(x + 3) X x(x – 3)/(x – 7)(x + 7)

= 1/(x + 3) X x/(x + 7)

= x/(x + 3)(x + 7)

## 16. The lengths of the sides of a right-angled triangle at (3x + 1)cm, (3x – 1)cm and x cm. Find the value of x.

A. 2 B. 6 C. 18 D. 12 E. 0

Solution:

Using Pythagoras theorem

(3x + 1)² = (3x – 1)² + x²

(3x + 1)(3x + 1) = (3x – 1)(3x – 1) + x²

9x² + 3x + 3x + 1 = 9x² – 3x – 3x + 1 + x²

9x² – 9x² – x² + 6x + 6x + 1 – 1 = 0

-x² + 12x = 0

x² – 12x = 0

x² = 12x

x = 12

## 17. The scores of a set of a final year students in the first semester examination in a paper are 41,29,55,21,47,70,70,40,43,56,73,23,50,50. Find the median of the scores.

A. 47 B. 48½ C. 50 D. 48 E. 49

Solution:

Arrange the scores in ascending order of magnitude

21, 23, 29, 40, 41, 43, (47, 50), 50, 55, 56, 70, 70, 73.

Median = (47 + 50)/2

= 97/2 = 48½

## 18. Which of the following equations represents the above

graph?

A. y = 1 + 2x + 3x² B. y = 1 – 2x + 3x² C. y = 1 + 2x – 3x² D. y = 1 – 2x – 3x²

E. y = 3x² + 2x – 1

Solution:

From the graph above, you can see that the curve cut X axis at -1 and 1/3

x = -1 and x = 1/3

Either (x + 1) = 0 or (x – 1/3) = 0

(x + 1)(3x – 1) = 0

3x² – x + 3x – 1 = 0

3x² + 2x – 1 = 0

## 19. The above figure FGHK is a rhombus. What is the value of the angle x? A.90° B.30° C.150° D.120° E.60°

Solution:

Angle FKG = angle FKH/2

30 = FKH/2

FKH = 60°

angle FKH + angle KHG = 180°

60 + KHG = 180

x = KHG = 180 – 60 = 120°

## 20. PQRS is a desk of dimensions 2m x 0.8m which is inclined at 30° to the horizontal. Find the inclination of the diagonal PR to the horizontal. A.23° 35’ B.30° C.15° 36’ D.10° E.10° 42’

Solution:

PS = QR = 0.8m

Using Pythagoras theorem, PR² = PQ² + QR²

PR² = 2² + 0.8² = 4 + 0.64 = 4.64

PR = √4.64 = 2.15m

Sin30 = R/0.8

R = 0.8 x sin30 = 0.4m

Now we have to calculate the required angle.

Sin@ = 0.4/PR = 0.4/2.15 = 0.1860

@ = Sin-¹0.1860 = 10.719°

Knowing the minutes of 0.71 = 0.72 x 60 = 43.2

so the angle of inclination= 10° 42′

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