Arithmetic aptitude :: Surds and Indices :: General Questions

• Surds and Indices - General Questions
• Surds and Indices - Important Formulas and Facts

Laws of Indices :

1.  $a^{m} \times a^{n}= a^{m+n}$
2.  $\frac{a^{m}}{a^{n}}= a^{m-n}$
3.  $(a^{m})^{n}=a^{mn}$
4.  $(ab)^{n} = a^{n}b^{n}$
5.  $\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}$
6.  $a^{0}=1$

Surds:

When we can't simplify a number to remove a square root (or cube root etc) then it is called a surd.

Rational Number:

A real number that can be written as a simple fraction then it is called rational number

e.g. 1.5 = $\frac{3}{2}$

Irrational Number:

A real number that can't be written as a simple fraction is called Irrational number.

e.g. $\pi$ = 3.1415926535897932384626433832795

Let a be rational number and n be a positive integer such that $a^{\frac{1}{n}}=\sqrt[n]{a}$ is irrational. Then $\sqrt[n]{a}$ is called surd of order n.

Laws of Surds

1. $\sqrt[n]{a}=a^{\frac{1}{n}}$
2. $\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$
3. $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$
4. $(\sqrt[n]{a})^{n}= a$
5. $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$
6. $(\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}$

Number of prime numbers in $\frac{6^{12}\times 35^{28}\times 15^{16}}{14^{12}\times 21^{11}}is:$

Answer:

Explanation:

$\frac{6^{12}\times 35^{28}\times 15^{16}}{14^{12}\times 21^{11}}=$ $=\frac{\left(2\times 3\right)^{12}\times \left(5\times7\right)^{28}\times \left(3\times 5\right)^{16}}{\left(2\times 7\right)^{12}\times \left(3 \times 7\right)^{11}}$ $=\frac{2^{12}\times 3^{12}\times 5^{28}\times 7^{28}\times 3^{16}\times 5^{16}}{2^{12}\times 7^{12}\times 3^{11}\times 7^{11}}$ $=2^{\left(12-12\right)}\times 3^{\left(12+16-11\right)}\times5^{\left(28+16\right)}\times7^{\left(28-12-11\right)}$ $=2^{0} \times3^{17} \times5^{44} \times7^{-5}$ $=\frac{3^{17}\times 5^{44} }{7^{5}}$ Number of prime factors = 17 + 44 + 5 = 66.

Name of prime factors in $\left(216\right)^{\frac{3}{5}}\times \left(2500\right)^{\frac{2}{5}}\times \left(300\right)^{\frac{1}{5}}$

Answer:

Explanation:

$\left(216\right)^{\frac{3}{5}}\times \left(2500\right)^{\frac{2}{5}}\times \left(300\right)^{\frac{1}{5}}=$ $\left(3^{3}\times 2^{3}\right)^{\frac{3}{5}}\times \left(5^{4}\times 2^{2}\right)^{\frac{2}{5}}\times \left(5^{2} \times 2^{2} \times 3\right)^{\frac{1}{5}}$ $=3^{\left[3\times\frac{3}{5}\right]}\times2^{\left[3\times\frac{3}{5}\right]}\times5^{\left[4\times\frac{2}{5}\right]}\times2^{\left[2\times\frac{2}{5}\right]}\times5^{\left[2\times\frac{1}{5}\right]}\times2^{\left[2\times\frac{1}{5}\right]}\times 3^{\frac{1}{5}}$ $=3^{\frac{9}{5}} \times 2^{\frac{9}{5}} \times 5^{\frac{8}{5}} \times 2^{\frac{4}{5}} \times 5^{\frac{2}{5}} \times 2^{\frac{2}{5}} \times 3^{\frac{1}{5}}$ $=3^{\left[\frac{9}{5}+\frac{1}{5}\right]}\times 2^{\left[\frac{9}{5}+\frac{4}{5}+\frac{2}{5}\right]}\times 5^{\left[\frac{8}{5}+\frac{2}{5}\right]}$ $=3^{2}\times 2^{3}\times 5^{2}$ Hence, the number of prime factors = $\left(2+3+2\right)=7$

$\frac{243^{\frac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}=?$

Answer:

Explanation:

Given Expression $=\frac{243^{\frac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$ $=\frac{\left(3^{5}\right)^{\frac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$ $=\frac{3^{\left(5 \times \frac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$ $=\frac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$ $=\frac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$ $=\frac{ 3^{3n+1}}{ 3^{3n-1}}$ $=3^{\left(3n+1-3n+1\right)}$ $=3^{2}=9$

Given that $10^{0.48}=x,10^{0.70}=y$ and $x^{2}=y^{2}$, then the value of z is close to:

Answer:

Explanation:

$x^{2}=y^{2}$ $\Leftrightarrow \left(10^{0.48}\right)^{z}$ $\Leftrightarrow \left( 10^{0.70}\right)^{2}$ $\Leftrightarrow 10^{0.48z}$ $\Leftrightarrow 10^{2\times 0.70 }$ $\Leftrightarrow 10^{1.40 }$ $\Leftrightarrow 10^{1.40 }$ $\Leftrightarrow z =\frac{140}{48}$ $=\frac{35}{12}$ = 2.9 (approx.).

If $2^{n-1}+2^{n+1}=320$ then n is equal to:

Answer:

Explanation:

$2^{n-1}+2^{n+1}=320$ $\Leftrightarrow 2^{n-1} \left( 1+ 2^{2}\right)=320$ $\Leftrightarrow 5 \times 2^{n-1} = 320$ $\Leftrightarrow 2^{n-1} = \frac{320}{5}$ $=64=2^{6}$ $\Leftrightarrow n-1=6$ $\Leftrightarrow n=7$

• Surds and Indices - General Questions