Question

A person deposited Rs. 4000 in a bank at 6% compounded continuously, after 3
years the rate of interest was increased to 7% and after 5 more years, the rate was
further increased to 8%, the money was withdrawn at the end of 10 years. Find
the amount​

Answer 1

Answer:

A person deposited 1000 Rs. in a bank. He got rate of interest for first 5 years 5 %. A = P ( 1+ ) A = future amount P = Principal amount = 1000 Rs. r = Rate of interest = 5 % = 0.05 n = number of compounding = 1 t = time = 5 years A = 1000 ( 1 + ( 0.05 ) ( 1x5) = 1000 ( 1 + 0.05 ) 5 = 1000 x ( 1.05 ) 5 = 1000 x 1.276281 = 1276.28 Rs. Now principal amount would be = 1276.28 Rs. T = 6 % = 0.06 t = 4 years A = 1276.28 ( 1 + 0.06 ) .1 = 1276.28 ( 1 + 0.06 ) 4 = 1276.28 x 1.26247 = 1611.27 Rs. The money was withdrawn at the end of 12 years. The remaining years of deposit = 12 - ( 5 + 4 ) = 3 years Now principal amount would be = 1611.27 Rs. r = 7 % = 0.07 t = 3 years A = 1611.27 ( 1 + 0.07) ( 1 × 3 )= 1611.27 ( 1 + 0.07) = 1611.27 x 1.2250 = 1973.8750 1973.88 Rs. He will received 1973.88 Rs. at the end of 12 years.

Step-by-step explanation:

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Answer 2

Answer:

The amount at the end of $10years$ is Rs.$5375.5$.

Step-by-step explanation:

Step-1:

The formula of compound interest is $A=P(1+\frac{r}{n})^{nt}$

Where A is the final amount, P is the initial principle, r is the rate of interest, n is the number of times interest is applied per period, and t is the time period.

Step-2:

The amount deposited $=Rs.4000$, Compound interest rate $=6$%, and time for which the money was deposited at this rate $=3years$.

Amount at the end of $3years$

$A=4000(1+\frac{0.03}{1} )^{1*3}\\A=4000*1.09\\A=4370.9$

Step-3:

Now this A is the principle for the following rate of interest $7$%, time $5years$

$A=4370.9(1+\frac{0.03}{1} )^{1*5}\\A=5067$

Step-4:

Now this A is the principle for the following rate of interest  $8$% and time $2years$

$A=5067(1+\frac{0.03}{1})^{1*2}\\ A=5375.5$

Hence the amount at the end of $10years$ is Rs.$5375.5$.

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