Mathematics 2010 | Study Mode

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Question 61
A
\(\frac{6}{35}\)
B
\(\frac{10}{21}\)
C
\(\frac{21}{10}\)
D
\(\frac{35}{6}\)
Explanation
W \(\alpha\) U

W = ku

u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\)

= \(\frac{6}{35}\)

Correct option: A
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Question 62
A
x < -1 or x > 1
B
-1 < x < 1
C
x > 0
D
x < -1
Explanation
x(x - 1) > 0

x < -1 or x > 1

Correct option: A
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Question 63
A
75o
B
65o
C
55o
D
50o
Explanation
In the diagram above, < STU = < TRS = 25o

(< between tangent to circle and a circle and a chord through the point of contact = < in the alternate segment)

< RTS = 90o

(< in a semicircle = 90o)
xo + < RTS + < STU = 180o

xo + 90o + 25o = 180o

xo + 115o = 180o

xo = 180o - 115o = 65oMaths-explained-image

Correct option: B
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Question 64
A
20o
B
23o
C
24o
D
26o
Explanation
In the diagram above 88o + (3x + 20)o = 180o

(oppsite < s of circlic quad. are supplementary)

88o + 20o + 3x = 180o

108o + 3x = 180o

3x = 180o - 108o

= 72oMaths-explained-image

Correct option: C
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Question 65
A
7cm
B
8cm
C
5cm
D
6cm
Explanation
Let A denote the area of \(\bigtriangleup\)PQR, then A = \(\frac{1}{2} bh\)

Using Sin 60o = \(\frac{h}{q}\)

h = q sin 60o

So A = \(\frac{1}{2}b(q \sin 60^o)\)

12\(\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}\)

12\(\sqrt{3}\) - 2q\(\sqrt{3}\)

q = \(\frac{12}{2} = 6\)cmMaths-explained-image

Correct option: D

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