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The age of a mother six years ago was six times the age of her daughter, at that time. Three years hence, the mother will be thrice as old as her daughter. What is the difference between the present age of the mother and her daughter?
24 years
42 years
36 years
30 years
30 years
Use the difference method: Let ages be M and D. M-6=6(D-6) => M-6D = -30. M+3=3(D+3) => M-3D = 6. Subtracting gives 3D = 36, D = 12. Then M = 42. Difference = 42 - 12 = 30.
Six years ago, mother's age (M-6) was 6 times daughter's age (D-6). Three years hence, mother's age (M+3) will be 3 times daughter's age (D+3).
MтИТ6=6(DтИТ6), M+3=3(D+3)
Use the difference method: Let ages be M and D. M-6=6(D-6) => M-6D = -30. M+3=3(D+3) => M-3D = 6. Subtracting gives 3D = 36, D = 12. Then M = 42. Difference = 42 - 12 = 30.
Students often forget to adjust both ages for the time elapsed (6 years ago vs 3 years hence) or calculate only the daughter's age instead of the difference.
Define variables and establish equations
Let the present age of the mother be M and the daughter be D. Six years ago, the ages were (MтИТ6) and (DтИТ6). Given MтИТ6=6(DтИТ6), which simplifies to MтИТ6D=тИТ30.
MтИТ6D=тИТ30
Form the second equation
Three years hence, the ages will be (M+3) and (D+3). Given M+3=3(D+3), which simplifies to M+3=3D+9, or MтИТ3D=6.
MтИТ3D=6
Solve the system of equations
Subtract the first equation from the second: (MтИТ3D)тИТ(MтИТ6D)=6тИТ(тИТ30), which gives 3D=36, so D=12. Substituting back, MтИТ3(12)=6, so M=42.
M=42,D=12
Calculate the difference
The difference between their present ages is MтИТD=42тИТ12=30.
42тИТ12=30
D is correct because the mother is 42 years old and the daughter is 12 years old, making their difference 30 years.
This age-difference logic is identical to speed-time problems where the gap between two moving objects remains constant regardless of time.
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