A point X is on the bearing 342° from a point Y. What is the bearing of Y from X?
342o
252o
198o
162o
Correct answer is D
If x is the bearing 342° from a point y
= 360° - 342° = 18°
The bearing of y from x = 180° - 18° = 162°
Find the mean deviation of 6, 7, 8, 9, 10
1.2
1.5
2
8
Correct answer is A
\(\begin{matrix}
x & x-m & |x-m| \\
6 & -2 & 2 \\
7 & -1 & 1 \\
8 & 0 & 0 \\
9 & 1 & 1 \\
10 & 2 & 2 \\
& & 6
\end{matrix}
\)
Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)
In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80°. Find ∠QRS.
90o
80o
50o
40o
Correct answer is D
QOP = 180 - 100(straight line angle) → 80°
OPQ and OQP are isosceles of 40° each.
∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 40°
∠QRS = 40°
Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q
{2}
{1,9}
{315.7}
{4, 6, 8, 10}
Correct answer is B
ξ = {1, 2, 3, . . . . . . ,10}, P= (2, 3, 5, 7) and Q = {1, 3, 5, 9},
Pl∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}
Find the sum of the roots of the equation 2x2 + 3x - 9 = 0
-18
-6
\(-\frac{9}{2}\)
\(-\frac{3}{2}\)
Correct answer is D
The general quadratic equation
x2 - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\\
Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)
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