\(\frac{1}{3}\)
1
3
9
Correct answer is B
\(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\) = \(\frac{(3^2)^{\frac{1}{2}} \times (3^3)^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\)
= \(\frac{3^{\frac{2}{3}} \times 3^{-\frac{3}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\)
= \(\frac{3^{-\frac{5}{6}}}{3^{-\frac{5}{6}}}\)
= 1
Find n if log\(_{2}\) 4 + log\(_{2}\) 7 - log\(_{2}\) n = 1
10
14
27
28
Correct answer is B
log\(_2\) 4 + log\(_2\) 7 - log\(_2\) n = 1
= log\(_2\) (4 x 7) - log\(_2\) n = 1
\(\therefore\) log\(_2\) 28 - log\(_2\) n = 1
= \(\frac{28}{n} = 2^1\)
\(\frac{28}{n}\) = 2
2n = 28
∴ n = 14
\(\frac{\sqrt{3}}{\sqrt{5}}\)
\(\frac{2 \sqrt{3}}{7}\)
-2
-1
Correct answer is D
\((\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}\)
\(\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}\)
\(\frac{(\sqrt{5} - \sqrt{3}) - (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})}\)
= \(\frac{\sqrt{5} - \sqrt{3} - \sqrt{5} - \sqrt{3}}{5 - \sqrt{15} + \sqrt{15} - 3}\)
= \(\frac{-2\sqrt{3}}{2}\)
= \(- \sqrt{3}\)
\(\therefore (\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}} = - \sqrt{3} \times \frac{1}{\sqrt{3}}\)
= \(-1\)
1981
1979
1982
1978
Correct answer is A
Let the no. of years be y
24 - y = \(\frac{1}{2}\)(45 - y)
45 - y = 2(24 - y)
45 - y = 48 - 2y
2y - y = 48 - 45
∴ y = 3
The exact year = 1984
1984 - 3 = 1981
(12, 6)
(23, 17)
(17, 11)
(18, 12)
Correct answer is C
x - y = 6.......(i)
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)