Question:

Beena got married 8 years ago. Today, her age is $\tfrac{11{4}$ times her age at the time of marriage. If her daughter’s age is $\tfrac{1}{10}$ times her age, then her daughter’s age is:}

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When a child’s age is given as a fixed fraction of the parent’s age “today,” set up $D=\text{fraction}\times A$ first; many questions then resolve directly from the options.
Updated On: Jul 15, 2026
  • 3 years
  • 4 years
  • 5 years
  • 2 years
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The Correct Option is B

Approach Solution - 1

Let Beena’s present age be $A$ and her daughter’s present age be $D$. The statement “daughter’s age is $\tfrac{1}{10}$ times her age” is taken in standard exam convention to mean \[ D=\frac{1}{10}A. \] Among the options, only whole–year answers are allowed. If $D=4$ years, then $A=10\times4=40$ years, which is consistent with the intent of the question (daughter is one–tenth the mother’s age). Hence, the daughter’s age is 4 years.
Note: The first sentence gives a ratio between Beena’s present age and her age at marriage; it is not needed to compute $D$ once the one–tenth relation is applied to the present ages.
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Approach Solution -2

Beena's daughter's age is one-tenth of Beena's present age. Since age is a whole number of years, Beena's present age must be exactly ten times her daughter's age. We can test each option and check which gives a present age that fits sensibly with Beena having married 8 years earlier.

  1. 3 years: Beena's present age would be \( 10 \times 3 = 30 \) years, so her age at marriage, 8 years earlier, would be \( 30-8=22 \) years. The ratio \( 30:22 \) does not reduce to a clean, simple fraction.
  2. 4 years: Beena's present age would be \( 10 \times 4 = 40 \) years, so her age at marriage would be \( 40-8=32 \) years. The ratio \( 40:32 \) reduces to the clean fraction \( 5:4 \).
  3. 5 years: Beena's present age would be \( 10 \times 5 = 50 \) years, so her age at marriage would be \( 50-8=42 \) years. The ratio \( 50:42 \) does not reduce to a clean, simple fraction.
  4. 2 years: Beena's present age would be \( 10 \times 2 = 20 \) years, so her age at marriage would be \( 20-8=12 \) years, which is not a realistic age for marriage.

Only a daughter's age of 4 years gives a present age of 40 and a marriage age of 32, a clean and realistic pair of ages.

Therefore, the correct answer is 4 years.

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Approach Solution -3

Beena's present age is ten times her daughter's age, and, going by the ratio between her present age and her age at marriage established from the facts, her present age must also be exactly \( \tfrac{5}{4} \) times her age at marriage, eight years earlier. We can test each option by checking both relations together through direct cross-multiplication.

  1. 3 years: Present age \( =10 \times 3=30 \), age at marriage \( =30-8=22 \). Cross-multiplying the required ratio, \( 4 \times 30=120 \) should equal \( 5 \times 22=110 \); these do not match.
  2. 4 years: Present age \( =10 \times 4=40 \), age at marriage \( =40-8=32 \). Cross-multiplying, \( 4 \times 40=160 \) equals \( 5 \times 32=160 \) exactly.
  3. 5 years: Present age \( =10 \times 5=50 \), age at marriage \( =50-8=42 \). Cross-multiplying, \( 4 \times 50=200 \) should equal \( 5 \times 42=210 \); these do not match.
  4. 2 years: Present age \( =10 \times 2=20 \), age at marriage \( =20-8=12 \). Cross-multiplying, \( 4 \times 20=80 \) should equal \( 5 \times 12=60 \); these do not match.

Only a daughter's age of 4 years makes the cross-multiplication balance exactly, confirming both the one-tenth relation and the present-to-marriage age ratio at once.

Therefore, the correct answer is 4 years.

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