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Mathematics

Assertion (A) : Mr. Sharma deposited ₹ 80 per month in a cumulative deposit account for 6 years at the rate of 6% p.a. Qualifying sum of his whole deposit = ₹ 6811.20

Reason (R) : Let a sum ₹ P be deposited every month in a bank for n months. If the rate of interest be r% p.a., then interest on the whole deposit (I) = P×n(n+1)12×r100P \times \dfrac{n(n + 1)}{12} \times \dfrac{r}{100}.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Banking

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Answer

If sum ₹ P is deposited every month in a bank for n months and the rate of interest is r% p.a., then interest on the whole deposit (I) = P×n(n+1)2×12×r100P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}.

∴ Reason (R) is false.

Given,

Mr. Sharma deposited ₹ 80 per month in a cumulative deposit account for 6 years at the rate of 6% p.a.

By formula,

Qualifying sum = P×n+P×n(n+1)2×12×r100P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

P = ₹ 80/month

r = 6%

n = (6 × 12) = 72

Substituting values we get :

Qualifying sum=80×72+80×72(72+1)2×12×6100=5760+80×72×7324×6100=5760+20×72×73100=5760+52565=5760+1051.20=₹ 6811.20\text{Qualifying sum} = 80 \times 72 + 80 \times \dfrac{72(72 + 1)}{2 \times 12} \times \dfrac{6}{100} \\[1em] = 5760 + 80 \times \dfrac{72 \times 73}{24} \times \dfrac{6}{100} \\[1em] = 5760 + \dfrac{20 \times 72 \times 73}{100} \\[1em] = 5760 + \dfrac{5256}{5} \\[1em] = 5760 + 1051.20 \\[1em] = \text{₹ 6811.20}

∴ Assertion (A) is true.

Hence, Option 1 is the correct option.

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