## Surds & Indices Problems To Prepare For IBPS Exam

Dear Reader,

Below are four problems based on surds and indices.

**Question 1**

The value of [(11)^{148} / (11)^{145}] is:

a) 14641 b) 1331 c) 11^293 d) none of these

**Answer : **b) 1331

Solution :

Using the formula, [(a)^{m} /(a)^{n}] = a^{(m-n)}.

Given expression is [(11)^{148} / (11)^{145}]

Here, a = 11, m = 148 and n = 145,

= [11^{(148-145)}]

= [(11)^{3}]

= 1331

**Question 2**

The value of (2.7 x 10^{4}) / (9 x 10^{-3}) is:

a) .3 x 10^{6} b) 3 x 10^{-6} c) .03 x 10^{6} d) 3 x 10^{6}

**Answer : **d) 3 x 10^{6}

Solution :

Given expression is (2.7 x 10^{4}) / (9 x 10^{-3})

= (27 x 10^{3}) / (9 x 10^{-3})

= (3 x 10^{3}) / (10^{-3})

= [3 x 10^{(3-(-3))}]

= [3 x 10^{(3+3)}]

= 3 x 10^{6}

**Question 3**

Find the value of (10000)^{6} / 10^{21}.

a) 1000 b) 10^{-27} c) 100 d) none of these

**Answer : **a) 1000

Solution :

Given expression is (10000)^{6} / 10^{21}

= [(10^{4})]^{6} / 10^{21}.

By using the formula, (a^{m})^{n} = a^{(mn)}, we have

= (10)^{24} / 10^{21}.

= 10^{(24-21)}

= 10^{3}

= 1000

**Question 4**

Find the value of [(3125)^{0.11}] x [(3125)^{0.09}].

a) 25 b) 5 c) 125 d) none of these

**Answer : **b) 5

Solution :

Given expression is [(3125)^{0.11}] x [(3125)^{0.09}]

By using the formula, (a^{m})x(a^{n}) = a^{(m+n)}.

Here, a = 3125, m = 0.11 and n = 0.09, then we have,

= [3125^{(0.11+0.09)}]

= 3125^{(0.20)}

= 3125^{(20/100)}

= 3125^{(1/5)}

= (5^{5})^{(1/5)} (since 5^{5} = 3125)

= (5)^{(5 x 1/5)}

= 5.