Age problems involve finding current ages or ages at specific times when given ratios between different people's ages. The key insight is that while individual ages change over time, age differences remain constant
First, age difference between two people never changes. If A is 5 years older than B today, A will always be 5 years older. Second, when we add or subtract the same number of years to different ages, their ratio changes. Third, present age problems often give ratios at two different time points.
• If ratio of ages is a:b, then ages are ax and bx where x is common factor
• After n years: (current age + n)
• Before n years: (current age - n)
SSC CGL typically asks three types - current age ratios with future/past conditions, age ratios at two different time points, and problems involving sum of ages with ratios. Questions often involve 2-3 people with time shifts of 2-10 years.
For two-time-point problems, use the 'difference method'. If ratio changes from a:b to c:d after n years, then (cx-ax) = (dx-bx) where x is the time difference. This eliminates one variable immediately.
Let current ages be 3x and 4x
After 6 years, ages become (3x+6) and (4x+6)
New ratio = (3x+6):(4x+6) = 4:5
Cross multiply: 5(3x+6) = 4(4x+6)
15x + 30 = 16x + 24
30 - 24 = 16x - 15x
6 = x
Current ages are 3×6 = 18 years and 4×6 = 24 years Verification: After 6 years, ages are 24 and 30, ratio = 24:30 = 4:5 ✓
Students often forget to add/subtract years from both ages equally. Another error is setting up wrong equations when dealing with 'before' scenarios - remember to subtract years, not add them.
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