If the ratio of two numbers is $7:11$. If $7$ is added to each number then the new ratio will be $2:3$ then the numbers are.

The correct answer is Option a: $49,77$. • Let the two numbers be $7x$ and $11x$, since their ratio is $7:11$. • According to the problem, if $7$ is added to each number, the new ratio becomes $2:3$. So, we can write the equation: $\frac{7x + 7}{11x + 7} = \frac{2}{3}$. • Now, we solve this equation for $x$: $3(7x + 7) = 2(11x + 7)$ $21x + 21 = 22x + 14$ • To find $x$, rearrange the terms: $21 - 14 = 22x - 21x$ $7 = x$ • So, the value of $x$ is $7$. • Now, substitute the value of $x$ back into our original expressions for the numbers: First number $= 7x = 7 \times 7 = 49$ Second number $= 11x = 11 \times 7 = 77$ • Therefore, the two numbers are $49$ and $77$. • Why Option (A) is correct: The numbers $49$ and $77$ have a ratio of $49:77 = (7 \times 7):(11 \times 7) = 7:11$. If $7$ is added to each, they become $49+7=56$ and $77+7=84$. The new ratio is $56:84$. Dividing both by $28$, we get $2:3$. This matches the conditions given in the question. • Why Option (B) is incorrect: The ratio of $42:45$ is $14:15$, not $7:11$. So, this option is immediately incorrect.

Ratio, Proportion, Indices, LogarithmPYQ Jun 19Question 790 of 211

All Questions

If the ratio of two numbers is $\displaystyle 7:11$ . If $\displaystyle 7$ is added to each number then the new ratio will be $\displaystyle 2:3$ then the numbers are.

Options

\displaystyle 49,77

\displaystyle 42,45

\displaystyle 43,42

\displaystyle 39,40

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Correct Answer

✅ Option a — $\displaystyle 49,77$

All Options:

A $\displaystyle 49,77$
B $\displaystyle 42,45$
C $\displaystyle 43,42$
D $\displaystyle 39,40$

Detailed Solution & Explanation

• Let the two numbers be

\displaystyle 7x

and

\displaystyle 11x

, since their ratio is

\displaystyle 7:11

. • According to the problem, if

\displaystyle 7

is added to each number, the new ratio becomes

\displaystyle 2:3

. So, we can write the equation:

\displaystyle \frac{7x + 7}{11x + 7} = \frac{2}{3}

. • Now, we solve this equation for

\displaystyle x

\displaystyle 3(7x + 7) = 2(11x + 7)

\displaystyle 21x + 21 = 22x + 14

• To find

\displaystyle x

, rearrange the terms:

\displaystyle 21 - 14 = 22x - 21x

\displaystyle 7 = x

• So, the value of

\displaystyle x

\displaystyle 7

. • Now, substitute the value of

\displaystyle x

back into our original expressions for the numbers: First number

\displaystyle = 7x = 7 \times 7 = 49

Second number

\displaystyle = 11x = 11 \times 7 = 77

• Therefore, the two numbers are

\displaystyle 49

and

\displaystyle 77

. • Why Option (A) is correct: The numbers

\displaystyle 49

and

\displaystyle 77

have a ratio of

\displaystyle 49:77 = (7 \times 7):(11 \times 7) = 7:11

. If

\displaystyle 7

is added to each, they become

\displaystyle 49+7=56

and

\displaystyle 77+7=84

. The new ratio is

\displaystyle 56:84

. Dividing both by

\displaystyle 28

, we get

\displaystyle 2:3

. This matches the conditions given in the question. • Why Option (B) is incorrect: The ratio of

\displaystyle 42:45

\displaystyle 14:15

, not

\displaystyle 7:11

. So, this option is immediately incorrect.

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 the ratio of two numbers is 7:11\displaystyle 7:117:11. If 7\displaystyle 77 is added to each number then the new ratio will be 2:3\displaystyle 2:32:3 then the numbers are.