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Mathematics

Asit saves ₹ 50 daily and deposits ₹ 1,500 every month in a recurring deposit account for 2122\dfrac{1}{2} years at r% per annum. If he gets ₹ 50,812.50 on maturity, find the value of r.

Banking

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Answer

Given,

P = ₹ 1500/month

n = 2.5 years = 2.5 × 12 = 30 months

M.V. = ₹ 50,812.50

By formula,

M.V. = P×n+P×n(n+1)2×12×r100P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Substituting values we get :

⇒50812.50=1500×30+1500×30×(30+1)2×12×r100⇒50812.50=45000+15×30×31×r24⇒50812.50−45000=15×5×31×r4⇒5812.50=2325r4⇒r=5812.50×42325⇒r=2.5×4=10%.\Rightarrow 50812.50 = 1500 \times 30 + 1500 \times \dfrac{30 \times (30 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 50812.50 = 45000 + 15 \times \dfrac{30 \times 31 \times r}{24} \\[1em] \Rightarrow 50812.50 - 45000 = \dfrac{15 \times 5 \times 31 \times r}{4} \\[1em] \Rightarrow 5812.50 = \dfrac{2325r}{4} \\[1em] \Rightarrow r = \dfrac{5812.50 \times 4}{2325} \\[1em] \Rightarrow r = 2.5 \times 4 = 10\%.

Hence, r = 10%.

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