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Surds and Indices Formulas for Bank Exams and SSC

Vikram Singh3 months ago 3.9K Views

Surds and indices questions are a challenging part of quantitative aptitude. So, you should try to use surds and indices formulas for bank exams and other competitive exams and should learn how to use surds and indices formulas in different-2 equations.

Here in this blog, you can learn easily with examples, how to use formulas of surds and indices to solve the questions. Obtain a high score in the competitive exam with these formulas which will help you in practice.

After learning to use formulas, you can learn how to solve surds and indices problems with solutions and can practice more with surds and indices questions and answers to obtain a better rank.

Surds and Indices Formulas for Competitive Exams

Surds:

Let a and n are rational numbers and a positive integer, respectively. If a1/n is an irrational number, then a1/n is known as surd of power n.

a1/n = n√a = nth root of a.

Sign n√ is known as radical sign and n and a are known as radical power and radicand, respectively.

e.g. √2, √5, a+√b etc. are surds

Laws of Surds -

$$(A)\ ^n\sqrt {a } \ = a^{1\over n}$$

$$(B) \ \ ^n\sqrt {ab }=\ ^n\sqrt {a }\ × \ ^n\sqrt {b }$$

$$(C) \ \ ^n\sqrt {{a\over b}}= {\ ^n\sqrt {a }\over \ ^n\sqrt {b }}$$

$$(D) \ \left(^n\sqrt {a }^ \ \right)^n = a$$

$$(E) \ \left(^m\sqrt {^n\sqrt {a } \ } \ \right)= \ ^{mn}\sqrt {a } \$$

$$(F) \ \left(^n\sqrt {a }^ \ \right)^n = \ ^n\sqrt {a^m }^ \$$

Indices:

When a number P is multiplied by itself n times, then the product is called n th power of P and is written as Pn. Here, P is called the base and n is known as the index of the power.

Laws of Indices -

$$(A) \ a^m×a^n=a^{m+n}$$

$$(B) \ {a^m\over a^n}=a^{m-n}$$

$$(C) \ (a^m)^n= a^{mn}$$

$$(D) \ (ab)^n= a^nb^n$$

$$(E) \ \left({a\over b} \right)^n = {a^n\over b^n}$$

$$(F) \ a^0=1$$

$$EX.1.\ The \ value \ of \ (256)^{5\over4} \ is :$$

$$(A) \ 512 \ (B) \ 984 \ (C) \ 1024 \ (D)1032$$

$$Explanation:$$

$$(256)^{5\over4} =(4^4)^{5\over4}= 4^\left(4×{5\over4} \right)= 4^5=1024$$

$$EX.2.\ The \ value \ of \ \left(\sqrt { 8} \ \right)^{1\over3} \ is :$$

$$(A) \ 2 \ (B) \ 4 \ (C) \ \sqrt { 2} \ \ (D)8$$

$$Explanation:$$

$$\left(\sqrt { 8} \ \right)^{1\over3} \ = \left(8^{{1\over2}} \right)^{1\over3}= 8^\left({1\over2}×{1\over3} \right)$$

$$=8^{{1\over6}}=(2^3)^{1\over6}=2^\left(3×{1\over6} \right)=2^{1\over2}=\sqrt { 2} \$$

$$EX.3.\ The \ value \ of \ \left((10)^{150}÷(10)^{146} \right)\ is :$$

$$(A) \ 1000 \ (B) \ 10000 \ (C) \ 100000 \ \ (D)10^6$$

$$Explanation:$$

$$(10)^{150}÷(10)^{146}={(10)^{150}\over(10)^{146}}=(10)^{(150-146) }=10^4=10000.$$

$$EX.4.(2.4×10^3)÷(8×10^{-2})=?$$

$$(A)\ 3×10^{-5} (B) \ 3×10^4 (C)\ 3×10^5 (D) 30$$

$$Explanation:$$

$$(2.4×10^3)÷(8×10^{-2})= {2.4×10^3\over 8×10^{-2}}=(3×10^4)$$

EX.5. If 5= 3125, then the value of 5(a-3) is :

(A) 25                    (B) 125                  (C) 625                  (D) 1625

$$Explanation:$$

If 5= 3125 ⟺ 5a = 55 ⟺ a = 5

∴ 5(a-3) = 5(5-3) = 52 = 25.

$$Ex.6.\ The \ value \ of \ \left({32\over243} \right)^{{-} 4\over5 } is :$$

$$(A) \ {4\over9}$$

$$(B) \ {9\over4}$$

$$(C) \ {16\over81}$$

$$(D) \ {81\over16}$$

$$Explantion:$$

$$\left({32\over243} \right)^{{-} 4\over5 }= \left(\left( {2\over3}\right)^5 \right)^{{-} 4\over5 }$$

$$=\left({2\over3} \right)^\left(5×{-4\over5} \right)$$

$$= \left({2\over3} \right)^{(-4)} = \left({3\over2} \right)^{(4)}$$

$$= {3^4\over2^4} = {81\over16}$$

$$Ex.7.\ The \ value \ of \ \left(-{1\over216} \right)^{{-} 2\over3 } is :$$

$$(A) \ 36$$

$$(B) \ -36$$

$$(C) \ {1\over36}$$

$$(D) \ -{1\over36}$$

$$Explantion:$$

$$\left(-{1\over216} \right)^{{-2} \over3 }= \left(\left( -{1\over6}\right)^3 \right)^{{-} 2\over3 }$$

$$=\left(-{1\over6} \right)^\left(3×{-2\over3} \right)$$

$$= \left(-{1\over6} \right)^{(-2)}$$

$$= {1\over\left(-{1\over6} \right)^2}$$

$$= {1\over\left({1\over36} \right)}= 36$$

$$Ex.8. \ The \ value \ of \ {1\over\left( 216\right)^{-{2\over 3}}}+{1\over\left( 256\right)^{-{3\over 4}}}+{1\over\left( 32\right)^{-{1\over 5}}}is :$$

(A) 102

(B) 105

(C) 107

(D) 109

$$Explantion:$$

$${1\over\left( 216\right)^{-{2\over 3}}}+{1\over\left( 256\right)^{-{3\over 4}}}+{1\over\left( 32\right)^{-{1\over 5}}}$$

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

$$= {1\over\left( 6\right)^\left(3× {-2\over 3}\right)}+{1\over\left( 4\right)^\left(4× {-3\over 4}\right)}+{1\over\left( 2\right)^\left(5× {-1\over 5}\right)}$$

$$={1\over6^{-2}}+{1\over4^{-3}}+{1\over2^{-1}} =(6^2+4^3+2^1)=36+64+2=102$$

Q.9. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:

(A) 1.45

(B) 1.88

(C) 2.9

(D) 3.7

Explain:

Xz = y2 ⟺ (100.48)z = (100.70)2 ⟺ 10(0.48z) = 10(2×0.70) = 101.40

⟺ 0.48z = 1.40 ⟺ z = 140/48 = 35/12 = 2.9 (approx).

Q.10. If m and n are whole numbers such that mn = 121, then the value of (m-1)n+1 is :

(A) 1

(B) 10

(C) 121

(D) 1000

Explain:

We know that 112 = 121. Putting m = 11 and n = 2, we get:

(m – 1)n+1 = (11 – 1)(2+1) = 103 = 1000.

Keep practice these surds and indices formulas for bank or other competitive exams. If you face any difficulty or problem using formulas and want to ask anything regarding surds and indices, you can ask me in the comment section.