81
27
\(\frac{1}{9}\)
\(\frac{1}{27}\)
\(\frac{1}{81}\)
Correct answer is A
\(x \propto \frac{1}{y}\)
\(x = \frac{k}{y}\)
\(y \propto t^{2}\)
\(y = ct^{2}\)
k and c are constants.
\(x = \frac{k}{ct^{2}}\)
Let \(\frac{k}{c} = d\) (a constant)
\(x = \frac{d}{t^{2}}\)
\(1 = \frac{d}{3^{2}} \implies d = 9\)
\(\therefore x = \frac{9}{t^{2}}\)
\(x = 9 \div (\frac{1}{3})^{2} \)
= \( 9 \div \frac{1}{9} = 9 \times 9 = 81\)
30 years
6 years
9 years
3 years
1 year
Correct answer is D
1st 30 pupils at average age of 15.3 yrs. give total age of 15.3 x 30 = 459yrs 2nd group of 30 pupils at average age of 15.2 yrs give total age of 15.2 x 30 = 456yrs Difference in age = 459 - 456 = 3 yrs
10:3
1:100
3:1
5:2
Correct answer is A
The distinct sectors in the same circle substend 100o and 30o respectively at the centre of the circle. Their corresponding arcs are in ratio 100:30 = 10:3
A variable y is inversely proportional to x2, when y = 10, x = 2. What is y when x = 10?
2
4
100
0.4
0.1
Correct answer is D
y \(\alpha\) \(\frac{1}{x^2}\)
y = \(\frac{k}{x^2}\)
k = x2y
= (2)2 x 10
= 40
y = \(\frac{40}{x^2}\)
= \(\frac{40}{(10)^2}\)
= \(\frac{40}{100}\)
= 0.4
2cm
8cm
4cm
2\(\sqrt{3}\)cm
16cm
Correct answer is D
\(\frac{x}{2}\) = sin60° = cos30°
x = 2 sino
= 2 x \(\frac{\sqrt{3}}{2}\)
= \(\sqrt{3}\)
length of the diagonal = 2 x \(\sqrt{2}\)
= 2\(\sqrt{3}\)