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Parin Sir > Quantitative Aptitude (Maths) > Surds & Indices > Examples(2) > Examples(2)

Examples(2)

May 30, 2020
Category: Examples(2)

(8) What is the least number which is divisible by 2, 5, 6 and 9 and at the same time a perfect square ?

Soln. [ 2, 5, 6, 9 ] = LCM of 2, 5, 6 and 9 = 2 $\times$ 5 $\times$ 3 $\times$ 3 = 2 $\times$ 3² $\times$ 5 = 90

Since it should be a square, we have to multiply it by 2 $\times$ 5 = 10

$\therefore$ The number is 90 $\times$ 10 = 900.

(9) Simplify : $\left (\frac{4}{25} \right ) ^{\frac{-3}{2}}$ $\times$ $\left [\left (\frac{125}{27} \right )^{\frac{1}{3}} \div \left (\frac{2}{25} \right )^{-1} \right ]$

Soln. Fraction

= $\left (\frac{4}{25} \right ) ^{\frac{-3}{2}}$ $\times$ $\left [\left (\frac{125}{27} \right )^{\frac{1}{3}} \div \left (\frac{2}{25} \right )^{-1} \right ]$

= $\left (\frac{25}{4} \right )^{\frac{3}{2}} \times$ $\left [\left (\frac{5^{3}}{3^{3}} \right )^{\frac{1}{3}} \div \left (\frac{25}{2} \right ) \right ]$

= $\left (\frac{5^{2}}{2^{2}} \right )^{\frac{3}{2}} \times \left [\frac{5}{3}\times \frac{2}{25} \right ]$

= $\frac{5^{3}}{2^{3}} \times \frac{2}{15}$

= $\frac{125}{8} \times \frac{2}{15} = \frac{25}{12}$

(10) Solve : $a^{2x}a^{3y+5} = a^{13}$ and $a^{5y-9}a^{4x}=a^{5}$

Soln. $a^{2x}a^{3y+5} = a^{13}$ and $a^{5y-9}a^{4x}=a^{5}$

$\therefore$ 2x + 3y + 5 = 13 and 4x + 5y – 9 = 5

$\therefore$ 2x + 3y = 8 and 4x + 5y = 14

solving, we get x = 1 and y = 2.

(11) Simplify : $\left (\frac{3^{\frac{1}{2}} \times 2^{\frac{1}{3}} \times 9^{\frac{1}{4}}}{15^{\frac{-1}{5}}\times 5^{\frac{3}{5}}} \div \frac{2^{\frac{4}{3}}\times 5^{\frac{3}{5}}}{9^{\frac{-3}{5}}\times 6} \right )$

Soln.

Given Exp. = $\left (\frac{3^{\frac{1}{2}} \times 2^{\frac{1}{3}} \times 9^{\frac{1}{4}}}{15^{\frac{-1}{5}}\times 5^{\frac{3}{5}}} \div \frac{2^{\frac{4}{3}}\times 5^{\frac{3}{5}}}{9^{\frac{-3}{5}}\times 6} \right )$

= $\left (\frac{3^{\frac{1}{2} } \times 2^{\frac{1}{3}} \times \left (3^{2} \right )^{\frac{1}{5}} \times \left (3\times5 \right )^{\frac{1}{5}}}{5^{\frac{3}{5}} } \div \frac{2^{\frac{4}{3}}\times\left (3^{2} \right )^{\frac{3}{5}}}{5^{\frac{7}{5}}\times2 \times 3} \right )$

= $\left (\frac{3^{\frac{1}{2}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{2}} \times 3^{\frac{1}{5}} \times 5^{\frac{1}{5}}}{5^{\frac{3}{5}}} \times \frac{5^{\frac{7}{5}} \times 2^{1} \times 3^{1}}{ 2^{\frac{4}{3}}\times 3^{\frac{6}{5}}} \right )$

= $2^{\left (\frac{1}{3}+1-\frac{4}{3} \right )}$ $\times$ $3^{\left (\frac{1}{2} +\frac{1}{2} +\frac{1}{5} + 1 -\frac{6}{5} \right ) }$ $\times$ $5^{\left (\frac{1}{5} +\frac{7}{5} -\frac{3}{5} \right ) }$

= $2^{0} \times 3^{1} \times 5^{1}$

= ( 3 $\times$ 5 )

= 15

(12) If $2^{x}$ = $3^{y}$ = $18^{z}$ , show that z = $\frac{xy}{2x+y}$

Soln. Let $2^{x}$ = $3^{y}$ = $18^{z}$ = k. Then, 2 = $k^{\frac{1}{x}}$ , 3 = $k^{\frac{1}{y}}$ & 18 = $k^{\frac{1}{z}}$

Now, 18 = 2 $\times 3^{2} \Leftrightarrow k^{\frac{1}{z}}=k^{\frac{1}{x}}.$ $\left (k^{\frac{1}{y}} \right )^{2} \Leftrightarrow k^{\frac{1}{z}}$ = $k^{\frac{1}{x}} \times k^{\frac{2}{y}} \Leftrightarrow k^{\frac{1}{z}}$ = $k^{\frac{1}{x}+ \frac{2}{y}}$