If the surface area of a sphere increased by 44%, find the percentage increase in diameter
44
30
22
20
Correct answer is D
Surface Area of Sphere A = 4\(\pi r^2\)
∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\)
= \(\frac{(D)^2}{2}\)
= \(\pi\)D2
When increased by 44% A = \(\frac{144 \pi D^2}{100}\)
\(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\)
Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D
Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)%
= 20%
2\(\frac{2}{3}\)
4
5\(\frac{1}{3}\)
8
Correct answer is B
P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125
After sometimes, A = 1.5 x 800 = N1,200
A = P(1 + Tr) = 1200
= 800(1 + 0.125T)
= 1200
= 800(1 + 0.125T)
1200 = 800(1 + 0.125T)
0.125T = 0.5
T = \(\frac{0.5}{0.125}\)
= 4
If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\)
125 : 128
3 : 5
3 : 4
2 : 5
Correct answer is D
a : b = 5 : 8 = 2.5 : 40
x : y = 25 : 16
\(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\)
= 1 : \(\frac{40}{16}\)
= 16 : 40
= 2 : 5
10-5
7 x 10-4
8 x 10-5
10-6
Correct answer is A
0.007685 = 0.00769 (three significant figures)
0.007685 = 0.0077(4d.p)
the difference = 0.0077 - 0.00769
= 0.00001
= 1.0 x 10-5
Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
\(\frac{r}{2}\)
2r
\(\frac{1}{2r}\)
\(\frac{2}{r}\)
Correct answer is B
\(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
= \(((64r^{-6})^{1/_2})^{1/_3}\)
=\((64r^{-6})^{1/_6}\)
=\(64)^{1/_6}(r^{-6})^{1/_6}\)
=2/r