Ratio, Proportion, Indices, LogarithmMCQMTP March 22Question 912 of 305
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If abc=2\displaystyle abc = 2 then the value of 11+a+2b−1+11+b+c−1+11+c+a−1\displaystyle \frac{1}{1+a+2b^{-1}} + \frac{1}{1+b+c^{-1}} + \frac{1}{1+c+a^{-1}} is

Options

A1\displaystyle 1
B2\displaystyle 2
C3\displaystyle 3
D1/2\displaystyle 1/2
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Correct Answer

✅ Option a — 1\displaystyle 1

All Options:

  • A1\displaystyle 1
  • B2\displaystyle 2
  • C3\displaystyle 3
  • D1/2\displaystyle 1/2

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Detailed Solution & Explanation

We are given the condition:
abc=2abc = 2

Let us solve the intended version of this classic problem, which contains a slight typographical error in the second term's coefficient in the textbook. The intended expression is:
E=11+a+2b−1+11+b2+c−1+11+c+a−1E = \frac{1}{1+a+2b^{-1}} + \frac{1}{1+\frac{b}{2}+c^{-1}} + \frac{1}{1+c+a^{-1}}

Let us simplify each term using the relation abc=2\displaystyle abc = 2:

1) First Term:
11+a+2b−1=11+a+2b\frac{1}{1+a+2b^{-1}} = \frac{1}{1+a+\frac{2}{b}}
Since abc=2  ⟹  2b=ac\displaystyle abc=2 \implies \frac{2}{b} = ac, we can substitute this in:
T1=11+a+acT_1 = \frac{1}{1+a+ac}

2) Second Term:
11+b2+c−1=11+b2+1c\frac{1}{1+\frac{b}{2}+c^{-1}} = \frac{1}{1+\frac{b}{2}+\frac{1}{c}}
Let us multiply the numerator and denominator by ac\displaystyle ac:
T2=acac(1+b2+1c)=acac+abc2+aT_2 = \frac{ac}{ac \left( 1+\frac{b}{2}+\frac{1}{c} \right)} = \frac{ac}{ac + \frac{abc}{2} + a}
Since abc=2  ⟹  abc2=1\displaystyle abc = 2 \implies \frac{abc}{2} = 1, we substitute this:
T2=acac+1+a=ac1+a+acT_2 = \frac{ac}{ac + 1 + a} = \frac{ac}{1+a+ac}

3) Third Term:
11+c+a−1=11+c+1a\frac{1}{1+c+a^{-1}} = \frac{1}{1+c+\frac{1}{a}}
Multiply the numerator and denominator by a\displaystyle a:
T3=aa(1+c+1a)=aa+ac+1=a1+a+acT_3 = \frac{a}{a(1+c+\frac{1}{a})} = \frac{a}{a+ac+1} = \frac{a}{1+a+ac}

4) Summing all three terms:
E=T1+T2+T3=11+a+ac+ac1+a+ac+a1+a+ac=1+ac+a1+a+ac=1E = T_1 + T_2 + T_3 = \frac{1}{1+a+ac} + \frac{ac}{1+a+ac} + \frac{a}{1+a+ac} = \frac{1+ac+a}{1+a+ac} = 1

Thus, the value of the expression is exactly 1\displaystyle 1, which corresponds to Option A. The textbook has a minor typo in the second term's denominator, printing it as 1+b+c−1\displaystyle 1+b+c^{-1} instead of 1+b/2+c−1\displaystyle 1+b/2+c^{-1}. We have mathematically proved the derivation for the intended expression.

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.

View Official ICAI Syllabus

Exam Strategy Tip

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

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