Inheritance sharing with fractional conditions: A man left 1/7 of his property to his daughter and the remainder to be shared equally among his sons. If each son’s share equals twice the daughter’s share, how many sons are there?

Introduction / Context:This is an algebraic sharing problem using fractions. The property is split such that the daughter receives a fixed fraction, and the rest is divided equally among sons. A proportional condition links the son’s share to the daughter’s share, allowing us to determine the number of sons.

Given Data / Assumptions:

• Daughter’s share = 1/7 of the total property.
• Remaining property = 1 − 1/7 = 6/7.
• Each son’s share is twice the daughter’s share, i.e., 2 × (1/7) = 2/7.
• There are n sons sharing 6/7 equally.

Concept / Approach:Set up the equality: (each son’s share) = (total remainder)/(number of sons). Because each son’s share is given to be 2/7, we can solve for n directly. This is a single-step proportion problem once the remainder is computed.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 2 or 4 or 6: Do not satisfy (6/7)/n = 2/7.

Common Pitfalls:

• Forgetting to compute the remainder 6/7 before dividing among sons.
• Misreading “twice the daughter’s share” as “two times the number of daughters” or similar.

Final Answer: