Ratio, Proportion, Indices, LogarithmMCQMTP June 24 Series IIQuestion 864 of 305
All Questions

What must be added to each term of the ratio 49:68\displaystyle 49:68, so that it becomes 3:4\displaystyle 3:4?

Options

A3\displaystyle 3
B5\displaystyle 5
C8\displaystyle 8
D4\displaystyle 4
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Correct Answer

✅ Option c — 8\displaystyle 8

All Options:

  • A3\displaystyle 3
  • B5\displaystyle 5
  • C8\displaystyle 8
  • D4\displaystyle 4

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Detailed Solution & Explanation

Let the number to be added to each term of the ratio be x\displaystyle x.
According to the problem, we can set up the following equation:
49+x68+x=34\frac{49 + x}{68 + x} = \frac{3}{4}
To solve for x\displaystyle x, we cross-multiply the terms:
4(49+x)=3(68+x)4(49 + x) = 3(68 + x)
Expand both sides:
196+4x=204+3x196 + 4x = 204 + 3x
Rearrange the terms to solve for x\displaystyle x:
4x−3x=204−1964x - 3x = 204 - 196
x=8x = 8
Thus, the number that must be added to each term is 8\displaystyle 8. This matches **Option C**.
Hence, **Option C** 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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If A:B=3:7\displaystyle A:B=3:7, then 3a+2b:4a+5b=?\displaystyle 3a+2b:4a+5b=?

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