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SSC Selection Post 2022 PYQ — Question 67

SSC Selection Post - Matriculation Level 2022 - 2 Feb 2022 - Shift 2 • Quantitative Aptitude • Q67

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A women invested in a private company fund, a total sum of ₹10,25,000 in the names of her son and daughter aged 13 years and 15 years, respectively, such that they get an equal amount when they are of 18 years. If the company gives 20% compund interest compounded annually, then how much money (in ₹, to the nearest whole number) should be deposited in the name of her son?

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Correct answer: C

Explanation

The correct answer is option 3 i.e. 4,20,082.

A=P×(1+r100)tA = P \times \left(1 + \frac{r}{100} \right)^t

Let us assume the amount for the son is P
⇒ Amount for the daughter = 10,25,000 − P

Amount for the son at the end of 5 years = P×(1+20100)5P \times \left(1 + \frac{20}{100} \right)^5

Amount for the daughter at the end of 3 years = (10,25,000−P)×(1+20100)3(10,25,000 - P) \times \left(1 + \frac{20}{100} \right)^3

P×(1+20100)5=(10,25,000−P)×(1+20100)3P \times \left(1 + \frac{20}{100} \right)^5 = (10,25,000 - P) \times \left(1 + \frac{20}{100} \right)^3

P×(65)5=(10,25,000−P)×(65)3P \times \left( \frac{6}{5} \right)^5 = (10,25,000 - P) \times \left( \frac{6}{5} \right)^3

36P25=10,25,000−P\frac{36P}{25} = 10,25,000 - P

36P=25×10,25,000−25P36P = 25 \times 10,25,000 - 25P

36P+25P=2,56,25,000⇒61P=2,56,25,00036P + 25P = 2,56,25,000 \Rightarrow 61P = 2,56,25,000

P=2,56,25,00061=₹4,20,082P = \frac{2,56,25,000}{61} = ₹4,20,082