Ten years ago, a mother was four times as old as her daughter. Ten years from now, the mother will be twice as old as the daughter. What is the daughter's present age?

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Age problems with “times as old” at two different times usually form a pair of linear equations in present ages. Solving gives the required current ages. Given Data / Assumptions: 10 years ago: Mother = 4 × Daughter In 10 years: Mother = 2 × Daughter Concept / Approach:Let present ages be M (mother) and D (daughter). Translate each time-based relation to an equation and solve simultaneously. Step-by-Step Solution: M − 10 = 4(D − 10) … (1)M + 10 = 2(D + 10) … (2)From (2): M = 2D + 20 − 10 = 2D + 10Plug into (1): 2D + 10 − 10 = 4D − 40 ⇒ 2D = 4D − 4040 = 2D ⇒ D = 20Then M = 2(20) + 10 = 50 Verification / Alternative check: 10 years ago: M = 40, D = 10 ⇒ 40 = 4×10 ✓In 10 years: M = 60, D = 30 ⇒ 60 = 2×30 ✓ Why Other Options Are Wrong: 5, 10, 30 do not satisfy both time-based multiplicative conditions. None of these: 20 works. Common Pitfalls:Using 4 times and 2 times on present ages instead of the specified past/future ages; or making arithmetic slips when substituting between equations. Final Answer:20 years

Ten years ago, a mother was four times as old as her daughter. Ten years from now, the mother will be twice as old as the daughter. What is the daughter's present age?

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