Simplify \(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)
\(\frac{1}{2}\)
p - 2r
\(\frac{1}{p - 2r}\)
\(\frac{2p}{p - 2r}\)
Correct answer is A
\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)
= \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\
= \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\)
= \(\frac{p^2 - 2pr}{2p(p - 2r)}\)
= \(\frac{p(p - 2r)}{2p(p - 2r)}\)
= \(\frac{1}{2}\)
If 20(mod 9) is equivalent to y(mod 6), find y.
1
2
3
4
Correct answer is B
First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9)
If 2(mod 9) y(mod 6), then y = 2 by comparism
Examine M' \(\cap\) N from the venn diagram
{f, g}
{e}
{e, f, g}
{e, f, g}
Correct answer is A
From the venn diagram given,
M = (a, b, c), N = (c, f, g)
U = (a, b, c, d, e, f, g)
Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g)
= (f, g)
70o
60o
55o
42o
Correct answer is B
Perimeter of a sector
= 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21
64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21
64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3
64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3
22 = \(\frac{33\theta}{90}\)
\(\theta = \frac{22 \times 30}{11}\)
= 60o
96cm2
90cm2
81cm2
27cm2
Correct answer is C
Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2
Area of \(\Delta\) QMN = Area of \(\Delta\) QNP
= Area of \(\Delta\) PNO (triangles between the same parallels)
Hence, area of the trapezium
3 x area of \(\Delta\) QNP
= 3 x 27
= 81cm2
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