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Mathematics

Premlata deposits ₹ 5,000 in an R.D. account at 8% p.a. rate of interest. How much per month must she deposit to get the same interest when the rate of interest is increased by 2%. Time in both the cases is same.

Banking

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Answer

Let the time in both the cases be n months.

In first case :

P = ₹ 5,000, r = 8%

I=P×n(n+1)2×12×r100I1=5000×n(n+1)24×8100I = P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] I_{1} = 5000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100}

In second case:

New rate of interest (r) = 8% + 2% = 10%

Let new monthly deposit = x

I2=x×n(n+1)24×10100I_{2} = x \times \dfrac{n(n + 1)}{24} \times \dfrac{10}{100}

Since interest is same,

⇒5000×n(n+1)24×8100=x×n(n+1)24×10100⇒5000×8100=x×10100⇒5000×8=10x⇒40000=10x⇒x=4000.\Rightarrow 5000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} = x \times \dfrac{n(n + 1)}{24} \times \dfrac{10}{100} \\[1em] \Rightarrow 5000 \times \dfrac{8}{100} = x \times \dfrac{10}{100} \\[1em] \Rightarrow 5000 \times 8 = 10x \\[1em] \Rightarrow 40000 = 10x \\[1em] \Rightarrow x = 4000.

Hence, Premlata must deposit ₹ 4,000 per month to get the same interest when the rate of interest is increased by 2%.

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