1983 mathematics 41 to 50
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41. In the figure below find PRQ
A. 66½° B. 62½° C. 125° D. 105° E.65°
Solution:
POQ = 360 – 235 = 125°
PRQ x 2 = POQ = 125
PRQ = 125/2 = 62½°
42. Simplify 3√(27a⁹/8)
A. 9a²/2 B. 9a³/2 C. 2/3a² D. 2/3a² E. 3a³/2
Solution:
3√(27a⁹/8) = 3√[3³(a³)³/2³]
= 3a³/2
43. The farm yields of four crops on a piece of land in Ondo are represented on the pie chart above. What is the angle of the sector occupied by Okro in the chart?
A. 91½° B. 19⅓° C. 33⅓° D. 11° E. 91°
Solution:
Beans = 25.6kg, Okro = 14.5kg, Rice = 45.4kg, Yams = 184.5kg
Total kg of the crops = 25.6 + 14.5 + 45.4 + 184.5 = 270kg
Okro occupied 14.5kg
The angle of the sector = 14.5/270 x 360 = 5220/270
= 19⅓°
44. In the figure above, PQR is a straight line. Find the values of x and y
A. x = 22.5° and y = 33.75° B. x = 15° and y = 52.5° C. x = 22.5° and y = 45.0° D. x = 56.25° and y = 11.5° E. x = 18° and y = 56.5°
Solution:
Since PQR is a straight line, that means;
(3x/2 + 3y) + 45 = 180 (Sum of angles on a straight line)
(5x + y) + y = 180 (Sum of angles on a straight line)
3x/2 + 3y = 180 – 45 = 135
3x + 6y = 270
x + 2y = 90. …………….i
5x + 2y = 180…………….ii
Substitute x = 90 – 2y in equation ii
5(90 – 2y) + 2y = 180
450 – 10y + 2y = 180
-8y = 180 – 450 = -270
y = 270/8 = 33.75°
Substitute y = 33.75 in equation i
x + 2(33.75) = 90
x + 67.5 = 90
x = 90 – 67.5 = 22.5°
45. PQR is the diameter of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = QRT = 60°. Find the perimeter of the figure (PTRS p = 22/7)
A. 25cm B. 18ccm C. 36cm D. 29cm E. 255cm
Solution:
Radius = PQ = QR = 3.5cm
PR = PQ x 2 = 7cm
PR = PT = RT = 7cm (Sides of equilateral triangle)
PT + RT = 7 + 7 = 14cm
Perimeter of a circle = 2πr
Perimeter of a semicircle = 2πr/2 = πr
= 22/7 x 3.5 = 11cm
Therefore the perimeter of PTRS = 14cm + 11cm
= 25cm
46. In a triangle PQR, QR = √3cm, PR = 3cm, PQ = 2√3cm and PQR = 30°. Find angles P and R
A. P = 60° and R = 90° B. P = 30° and R = 120° C. P = 90° and R = 60° D. P = 60° and R = 60° E. P = 45° and R = 105°
Solution:
Using cosine rule; R² = P² + Q² – 2PQcosR
Cos R = (P² + Q² – R²)/2PQ
= [(√3)² + 3² – (2√3)²]/(2 x √3 x 3)
= (3 + 9 – 12)/10.39 = 0/10.39
Cos R = 0
P = Cos–¹0 = 90°
P = 180 – (30 + 90) = 60°
47. In the above diagram if PS = SR and PQ//SR. What is the size of PQR?
A. 25° B. 50° C. 55° D. 65° E. 75°
Solution:
Since PS = SR , Triangle PSR is an isosceles triangle which has equal bases
SPR = SRP = (180 – 130)/2
= 50/2 = 25°
SRP = RPQ = 25° (Alternate angles)
Therefore PQR = 180 – (100 + 25)
= 180 – 125 = 55°
48. Find the mean of the following 24.57, 25.63, 25.32, 26.01, 25.77
A. 25.12 B. 25.30 C. 25.26 D. 25.50 E. 25.73
Solution:
Mean = (24.57 + 25.63 + 25.32 + 26.01 + 25.77)/5
= 127.3/5 = 25.46 = 25.50
49. In the figure above PT is a tangent to the circle with centre O. If PQT = 30°. Find the value of PTO
A. 30° B. 15° C. 24° D. 12° E. 60°
Solution:
QPT = x + 90
PQT = 30°
QTP = x + 2x = 3x
(x + 90) + 30 + 3x = 180 (Total angles in a triangle)
4x + 120 = 180
4x = 180 – 120 = 60
x = 60/4 = 15°
Thus, PTO = 2x = 2 x 15 = 30°
50. A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Altogether he covered 600km. At what speed did he drive for the last 3 hours?
A. 120km/hr B. 60km/hr C. 600/7km/hr D. 50km/hr E. 100km/hr.
Solution:
Let the First speed be “y”, Time = 4 hours
Distance = y x 4 = 4y
Second speed is 2y, Time = 3 hours
Distance = 2y x 3 = 6y
4y + 6y = 600
10y = 600
y = 600/10 = 60km/hr
His speed for the last 3 hours is 2y
= 2 x 60 = 120km/hr
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