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Raju and Lalitha originally had marbles in the ratio 4:94:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:65:6. What fraction of her original number of marbles was given by Lalitha to Raju?

Solution

✅ Correct Option: 2

Raju and Lalitha start with marbles in the ratio 4:9. After Lalitha gives some marbles to Raju, their new ratio becomes 5:6. We need to find what fraction of her original marbles Lalitha gave away.


Since the initial ratio is 4:9, we can write:

Raju's initial marbles = 4x

Lalitha's initial marbles = 9x

When we have a ratio like 4:9, we use a common factor 'x' to represent the actual quantities. This ensures the ratio remains 4:9 regardless of the actual numbers.

Let's say Lalitha gives y marbles to Raju.


After the transfer:

Raju's marbles = 4x + y (he receives y marbles)

Lalitha's marbles = 9x - y (she gives away y marbles)


The new ratio is 5:6, so:

4x+y9x−y=56\dfrac{4x + y}{9x - y} = \dfrac{5}{6}


Cross-multiplying:

6(4x+y)=5(9x−y)6(4x + y) = 5(9x - y)

24x+6y=45x−5y24x + 6y = 45x - 5y

This eliminates fractions and gives us a linear equation that's easier to solve.


Moving all terms with y to one side and all terms with x to the other:

6y+5y=45x−24x6y + 5y = 45x - 24x

11y=21x11y = 21x

Therefore: y=21x11y = \dfrac{21x}{11}


The fraction of original marbles that Lalitha gave to Raju is:

y9x=21x119x\dfrac{y}{9x} = \dfrac{\frac{21x}{11}}{9x}

21x11×19x=21x11×9x=2199\frac{21x}{11} \times \dfrac{1}{9x} = \dfrac{21x}{11 \times 9x} = \dfrac{21}{99}

Reducing to lowest terms:

2199=733\dfrac{21}{99} = \dfrac{7}{33}


Answer: 733\dfrac{7}{33}


When solving ratio problems with transfers, always:

Use variables that maintain the original ratio

Account for what each person gains/loses

Set up the new ratio equation

Verify your answer with concrete numbers

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