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A woman is 4 times as old as her daughter. After 6 years, the woman's age will be three times that of her daughter. The sum of their present ages is:
60 years
90 years
80 years
70 years
60 years
Use the ratio difference method: Initial ratio is 4:1. After 6 years, it becomes 3:1. The difference in ratio parts (4-3=1 and 1-1=0) suggests checking difference stability. 4:1 to 3:1 requires balancing; scale up to 8:2 and 9:3 (diff 1 unit=6 years). Daughter = 2 * 6 = 12, Woman = 8 * 6 = 48. Sum = 60.
Woman's age is 4 times her daughter's age. After 6 years, the woman's age is 3 times her daughter's age.
W=4D, W+6=3(D+6)
Use the ratio difference method: Initial ratio is 4:1. After 6 years, it becomes 3:1. The difference in ratio parts (4-3=1 and 1-1=0) suggests checking difference stability. 4:1 to 3:1 requires balancing; scale up to 8:2 and 9:3 (diff 1 unit=6 years). Daughter = 2 * 6 = 12, Woman = 8 * 6 = 48. Sum = 60.
Students often add 6 years only to the woman's age or forget that both ages increase by the same amount.
Define Variables
Let the daughter's present age be D and the woman's present age be W. Given W=4D.
W=4D
Set up the equation for 6 years later
After 6 years, both are 6 years older. The woman's age becomes W+6 and daughter's D+6. Given W+6=3(D+6).
W+6=3(D+6)
Solve for D
Substitute W=4D into the second equation: 4D+6=3D+18. Subtracting 3D from both sides: D=12.
D=12
Calculate W and Sum
Since D=12, then W=4├Ч12=48. The sum of their ages is 48+12=60.
48+12=60
A is correct because the daughter's current age is 12 and the mother's is 48, totaling 60 years.
This type of age problem is common in ratio-based arithmetic and can be solved using the 'Unit Method' (comparing gap difference) in other competitive exams like SSC or banking.
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