Income allocation and trust savings: A person gives 2/5 of his income to his elder son and 30% to his younger son. The rest is saved in three trusts A, B, and C in the ratio 3 : 5 : 2. If the difference between the amounts received by the two sons is ₹ 2000, how much is saved in trust C?
Correct Answer: ₹ 1200
Introduction / Context:This problem combines fractional allocations with a remainder split in a specified ratio. The given difference between sons’ shares identifies total income, after which the savings portion can be divided among the trusts.
Given Data / Assumptions:
- Elder son = 2/5 of income.
- Younger son = 30% = 3/10 of income.
- Difference between sons = ₹ 2000.
- Savings to A : B : C = 3 : 5 : 2.
Concept / Approach:Difference between sons = (2/5 − 3/10) * Income = 1/10 * Income. Thus income = 10 * difference. The remainder after giving to sons is 1 − (2/5 + 3/10) = 3/10 of income, to be split 3 : 5 : 2 among A, B, C.
Step-by-Step Solution:Income = 10 * 2000 = ₹ 20000.Savings portion = 3/10 of 20000 = ₹ 6000.Trust ratio total parts = 3 + 5 + 2 = 10.Trust C share = 2/10 of 6000 = ₹ 1200.
Verification / Alternative check:Sons receive ₹ 8000 and ₹ 6000 (difference ₹ 2000). Savings ₹ 6000 split as A = ₹ 1800, B = ₹ 3000, C = ₹ 1200, matching the ratio 3 : 5 : 2.
Why Other Options Are Wrong:
- ₹ 1000 and ₹ 1140 do not correspond to the 2/10 share of ₹ 6000.
- ₹ 1256 is not a clean part of the given ratio.
Common Pitfalls:
- Mistaking the difference as 1/5 instead of 1/10 of income.
- Dividing the total income in the 3 : 5 : 2 ratio instead of only the savings.
Final Answer:₹ 1200