Mr. Jayesh plans to save amount for higher studies of his daughter, required after 10 years. How much amount should he save at the beginning of each year to accumulate Rs.1,00,000 at the end of 10 years. If rate of interest is 12% compounded annually? [Given $(1.12)^{11} = 3.5$]

Mr. Jayesh wants to save Rs. 1,00,000 in 10 years by depositing money at the beginning of each year. Since payments are at the beginning of each year, this is an annuity due.

Given:

• Future Value = Rs. 1,00,000
• Time period = 10 years
• Interest rate = 12% per year
• $(1.12)^{11} = 3.5$

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For payments at the beginning of each year:

$FV = P \times \dfrac{(1+r)^n - 1}{r} \times (1+r)$

Where $P$ is the annual payment, $r = 0.12$, and $n = 10$.

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To find $(1.12)^{10}$:

$(1.12)^{10} = \dfrac{(1.12)^{11}}{1.12}$

$(1.12)^{10} = \dfrac{3.5}{1.12}$

$(1.12)^{10} = 3.125$

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Substituting into the formula:

$1,00,000 = P \times \dfrac{(3.125 - 1)}{0.12} \times 1.12$

$1,00,000 = P \times \dfrac{2.125}{0.12} \times 1.12$

$1,00,000 = P \times 17.708333 \times 1.12$

$1,00,000 = P \times 19.833333$

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Solving for $P$:

$P = \dfrac{1,00,000}{19.833333}$

$P = 5,042$ (approximately)

Therefore, Mr. Jayesh should save Rs. 5,042 at the beginning of each year.