Question:

An old woman decided to divide her gold among her daughter and daughters-in-law. She first kept exactly half of the gold for her daughter. Then she divided the rest of her gold among her daughters-in-law. The eldest one got 26 grams more than the youngest daughter-in-law. The middle one got twice as much as the youngest one. If the eldest daughter-in-law got 66 grams of gold, how much was received by the daughter.

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Always break down the problem into smaller variables and solve using logical equations.
Updated On: Aug 18, 2025
  • 198 grams
  • 172 grams
  • 186 grams
  • 194 grams
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The Correct Option is C

Solution and Explanation

Let the amount of gold the youngest daughter-in-law got be \( x \). Then the middle daughter-in-law received \( 2x \), and the eldest daughter-in-law received \( 66 \) grams. From the given information, we know: - The middle daughter-in-law got 26 grams more than the youngest: \( 2x = x + 26 \), which implies \( x = 26 \). - The total amount of gold for the daughters-in-law is \( 26 + 52 + 66 = 144 \). - Half of the total gold was given to the daughter, so the daughter received \( 186 \) grams of gold.

Option (A) 198 grams: Incorrect, this is more than half of the total gold.

Option (B) 172 grams: Incorrect, this is not the correct amount based on the total distribution.

Option (D) 194 grams: Incorrect, this does not align with the calculation of the remaining gold after distribution.
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