On Simplification 1/1+za-b+z^a-c + 1/1+zb-c+z^b-a + 1/1+zc-a+z^c-b would reduces to

Option a: 1. Solution: We are given the expression: E = 1/1+za-b+z^a-c + 1/1+zb-c+z^b-a + 1/1+zc-a+z^c-b Let us rewrite the exponents using the division rule zu-v = zu/zv : T1 = 1/1 + + za/zc T2 = 1/1 + + zb/za T3 = 1/1 + + zc/zb To make the denominators common, let us multiply the numerator and denominator of each term by a factor: 1) For T1 , multiply numerator and denominator by z-a : T1 = -az-a ( 1 + za-b + za-c ) = -az-a + z-b + z-c 2) For T2 , multiply numerator and denominator by z-b : T2 = -bz-b ( 1 + zb-c + zb-a ) = -bz-b + z-c + z-a 3) For T3 , multiply numerator and denominator by z-c : T3 = -cz-c ( 1 + zc-a + zc-b ) = -cz-c + z-a + z-b Now, all three terms have the same denominator, z-a + z-b + z-c . Adding them together: E = T1 + T2 + T3 = -a + z-b + z-cz-a + z-b + z-c = 1 Hence, Option A is the correct answer.

On Simplification $\frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}$ would reduces to

The correct answer is Option a: 1. We are given the expression: $$E = \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}$$ Let us rewrite the exponents using the division rule $z^{u-v} = \frac{z^u}{z^v}$: $$T_1 = \frac{1}{1 + \frac{z^a}{z^b} + \frac{z^a}{z^c}}$$ $$T_2 = \frac{1}{1 + \frac{z^b}{z^c} + \frac{z^b}{z^a}}$$ $$T_3 = \frac{1}{1 + \frac{z^c}{z^a} + \frac{z^c}{z^b}}$$ To make the denominators common, let us multiply the numerator and denominator of each term by a factor: 1) For $T_1$, multiply numerator and denominator by $z^{-a}$: $$T_1 = \frac{z^{-a}}{z^{-a} \left( 1 + z^{a-b} + z^{a-c} \right)} = \frac{z^{-a}}{z^{-a} + z^{-b} + z^{-c}}$$ 2) For $T_2$, multiply numerator and denominator by $z^{-b}$: $$T_2 = \frac{z^{-b}}{z^{-b} \left( 1 + z^{b-c} + z^{b-a} \right)} = \frac{z^{-b}}{z^{-b} + z^{-c} + z^{-a}}$$ 3) For $T_3$, multiply numerator and denominator by $z^{-c}$: $$T_3 = \frac{z^{-c}}{z^{-c} \left( 1 + z^{c-a} + z^{c-b} \right)} = \frac{z^{-c}}{z^{-c} + z^{-a} + z^{-b}}$$ Now, all three terms have the same denominator, $z^{-a} + z^{-b} + z^{-c}$. Adding them together: $$E = T_1 + T_2 + T_3 = \frac{z^{-a} + z^{-b} + z^{-c}}{z^{-a} + z^{-b} + z^{-c}} = 1$$ Hence, **Option A** is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQMTP Sep 24 Series IIQuestion 925 of 305

All Questions

On Simplification $\displaystyle \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}$ would reduces to

Options

B-1

\displaystyle z^{a}(a+b+c)

\displaystyle z^{a}(a+b+c)

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Correct Answer

✅ Option a — 1

All Options:

A1
B-1
C $\displaystyle z^{a}(a+b+c)$
D $\displaystyle z^{a}(a+b+c)$

Detailed Solution & Explanation

We are given the expression:

E = \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}

Let us rewrite the exponents using the division rule

\displaystyle z^{u-v} = \frac{z^u}{z^v}

T_1 = \frac{1}{1 + \frac{z^a}{z^b} + \frac{z^a}{z^c}}

T_2 = \frac{1}{1 + \frac{z^b}{z^c} + \frac{z^b}{z^a}}

T_3 = \frac{1}{1 + \frac{z^c}{z^a} + \frac{z^c}{z^b}}

To make the denominators common, let us multiply the numerator and denominator of each term by a factor:
1) For

\displaystyle T_1

, multiply numerator and denominator by

\displaystyle z^{-a}

T_1 = \frac{z^{-a}}{z^{-a} \left( 1 + z^{a-b} + z^{a-c} \right)} = \frac{z^{-a}}{z^{-a} + z^{-b} + z^{-c}}

2) For

\displaystyle T_2

, multiply numerator and denominator by

\displaystyle z^{-b}

T_2 = \frac{z^{-b}}{z^{-b} \left( 1 + z^{b-c} + z^{b-a} \right)} = \frac{z^{-b}}{z^{-b} + z^{-c} + z^{-a}}

3) For

\displaystyle T_3

, multiply numerator and denominator by

\displaystyle z^{-c}

T_3 = \frac{z^{-c}}{z^{-c} \left( 1 + z^{c-a} + z^{c-b} \right)} = \frac{z^{-c}}{z^{-c} + z^{-a} + z^{-b}}

Now, all three terms have the same denominator,

\displaystyle z^{-a} + z^{-b} + z^{-c}

. Adding them together:

E = T_1 + T_2 + T_3 = \frac{z^{-a} + z^{-b} + z^{-c}}{z^{-a} + z^{-b} + z^{-c}} = 1

Hence, **Option A** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

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This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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On Simplification 11+za−b+za−c+11+zb−c+zb−a+11+zc−a+zc−b\displaystyle \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}1+za−b+za−c1​+1+zb−c+zb−a1​+1+zc−a+zc−b1​ would reduces to

Options

Correct Answer

All Options:

Detailed Solution & Explanation

About This Chapter: Ratio, Proportion, Indices, Logarithm

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On Simplification $\displaystyle \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}}$ would reduces to