If P=((Q(R-T))/15 )〖^⅓〗, make T the subject of the formula.
A. T= (15R-Q)/P³
B. T=R– 15P³/Q
C. T= (R-15P³)/Q
D. T=(R+P³)/15Q
Correct Option: Answer is B
Solution:
P=((Q(R-T))/15 )〖^⅓〗
P³ = ((Q(R-T))/15 )
Q(R−T)=15P³
R−T=15P³/Q
T=R– 15P³/Q
If P=((Q(R-T))/15 )〖^⅓〗, make T the subject of the formula. Read More »
Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree.
A. 24°
B. 56°
C. 127°
D. 156°
Correct Option: Answer is D
Solution:
tanθ=9/20=0.45
θ=tan−1(0.45)
= 24.23°
∴ The bearing of Y from X = 180° – 24.23°
= 155.77°
= 156° (to the nearest degree)
Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30
A. x = 3 or 7, y = 12 or 8
B. x = 6 or 1, y = 11 or 5
C. x = 2 or 5, y = 15 or 6
D. x = 1 or 5, y = 10 or 7
Correct Option: Answer is C
Solution:
3x + y = 21 … (i);
xy = 30 … (ii)
From (ii), y=30/x. Putting the value of y in (i), we have
3x + 30/x = 21
⟹ 3x² + 30 = 21x
3x² – 21x + 30 = 0
3x² – 15x – 6x + 30 = 0
3x(x – 5) – 6(x – 5) = 0
(3x – 6)(x – 5) = 0
3x – 6 = 0 ⟹ x = 2.
x – 5 = 0 ⟹ x = 5.
If x = 2, y = 30/2 = 15;
If x = 5, y = 30/5 = 6.
Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30 Read More »
Simplify (0.0839 X 6.381)/5.44 to 2 significant figures.
A. 0.2809
B. 2.51
C. 3.5
D. 0.098
Correct Option: Answer is D
Simplify (0.0839 X 6.381)/5.44 to 2 significant figures. Read More »
If 2^(x+y )=16 and 4^(x-y )=1/32, find the values of x and y.
A. x = 3/4, y = 11/4
B. x = 3/4, y = 13/4
C. x = 2/3, y = 4/5
D. x = 2/3, y = 13/4
Correct Option: Answer is B
Solution:
2^(x+y )=16 ; 4^(x-y )=1/32
= 2^(x+y )=2^4
x +y = 4….(1)
= 2^(2(x-y) )=2^(-5)
2x – 2y = -5 ….(2)
Solve simultaneously,
x +y = 4
2x – 2y = -5
X = ¾ ; y = 13/4
If 2^(x+y )=16 and 4^(x-y )=1/32, find the values of x and y. Read More »
Evaluate ( 6/0.32 ÷ 2/0.084)¯¹ correct to 1 decimal place.
A. 1.3
B. 2.5
C. 4.6
D. 3.2
Correct Option: Answer is D
Evaluate ( 6/0.32 ÷ 2/0.084)¯¹ correct to 1 decimal place. Read More »
The weight of a day-old chick was measured to be 0.21g. If the actual weight of the chick is 0.18g, what was the percentage error in the measurement?
A. 15.5%
B. 18.2%
C. 14.8%
D. 16.7%
Correct Option: Answer is D
Solution:
Actual weight = 0.18g
Error = 0.21g – 0.18g
= 0.03g
% error = 0.03/0.18×100
= 16.7%
The angles of a polygon are given by 2x, 5x, x and 4x respectively. The value of x is
A. 31°
B. 30°
C. 26°
D. 48°
Correct Option: Answer is B
Solution:
Since there are 4 angles given, the polygon is a quadrilateral.
Sum of angle in a quadrilateral = 360°
∴ 2x + 5x + x + 4x = 360°
12x = 360°
x = 30°
The angles of a polygon are given by 2x, 5x, x and 4x respectively. The value of x is Read More »
If S = (4t + 3)(t – 2), find ds/dt when t = 5 secs.
A. 50 units per sec
B. 35 units per sec
C. 22 units per sec
D. 13 units per sec
Correct Option: Answer is B
Solution:
s=(4t+3)(t−2)
ds/dt=(4t+3)(1)+(t−2)(4)
= 4t+3+4t−8
= 8t – 5
ds/dt (t=5secs)=8(5)−5
= 40 – 5
= 35 units per sec
If S = (4t + 3)(t – 2), find ds/dt when t = 5 secs. Read More »
Find the value of ( (0.5436)²)/(0.017 X 0.219) to 3 significant figures.
A. 46.2
B. 43.1
C. 534
D. 431
Correct Option: Answer is B
Solution:
= ( (0.5436)²)/(0.017 X 0.219)
= ( 0.16063)/(0.017 X 0.219) = 43.1 (3 s.f)
Find the value of ( (0.5436)²)/(0.017 X 0.219) to 3 significant figures. Read More »
