Q21. Ishita invested a sum of Rs. 12000 at 5 % per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.
Sol. Lets assume no. of years be n.
Given P =Rs. 12000, R= 5 % p.a , A = Rs. 13230.
$$ \text {We know that } A = P(1+\frac{R}{100})^{n} $$ $$ \Rightarrow \quad 13230 = 12000(1+\frac{5}{100})^{n} $$ $$ \Rightarrow \quad 13230 = 12000(\frac{105}{100})^{n} $$ $$ \Rightarrow \quad (1.05)^{n} = \frac {13230}{12000} $$ $$ \Rightarrow \quad (1.05)^{n} = 1.1025 = 1.05^{2} $$ $$ \Rightarrow \quad n = 2 $$Hence, Ishita invested the money for 2 yrs.
Q22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?
Sol. Given P = Rs. 4000, n= 2 years.
Let’s assume that rate of interest be R %
We know that CI = A – P
$$ \Rightarrow \quad CI = P(1+\frac{R}{100})^{n} – P $$ $$ \Rightarrow \quad 410 = 4000(1+\frac{R}{100})^{2} – 4000 $$ $$ \Rightarrow \quad \frac {410 + 4000}{4000}= (1+\frac{R}{100})^{2} $$$$ \Rightarrow \quad \frac {4410}{4000}= (1+\frac{R}{100})^{2} $$ $$ \Rightarrow \quad (\frac {21}{20})^{2}= (1+\frac{R}{100})^{2} $$$$ \Rightarrow \quad \frac {21}{20}= 1+\frac{R}{100} $$ $$ \Rightarrow \quad \frac {21}{20} -1 = \frac{R}{100} $$ $$ \Rightarrow \quad \frac {1}{20} = \frac{R}{100} $$ $$ \Rightarrow \quad R = 5 $$Hence, the required rate of interest is 5 % per annum.
Q23. A sum of money deposited at 2 % per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.
Sol. Given A = Rs. 10404, R = 2 % p.a, n= 2 years
Let’s assume that sum of money be Rs. P.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 10404 = P(1+\frac{2}{100})^{2} $$ $$ \Rightarrow \quad 10404 = P(\frac{102}{100})^{2} $$ $$ \Rightarrow \quad P = 10404 \times (\frac{100}{102})^{2} $$ $$ \Rightarrow \quad P = 10404 \times \frac{10000}{10404} $$ $$ \Rightarrow \quad P = 10000 $$Hence, the required sum is Rs. 10000.
Q24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5 % per annum compound interest?
Sol. Given P = Rs. 1600, R = 5 % p.a, A = Rs. 1852.20
Let’s assume that time period be n years.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 1852.20 = 1600(1+\frac{5}{100})^{n} $$ $$ \Rightarrow \quad 1852.20 = 1600(\frac{105}{100})^{n} $$ $$ \Rightarrow \quad (\frac{105}{100})^{n} = \frac{1852.20}{1600} $$ $$ \Rightarrow \quad (\frac{21}{20})^{n} = \frac{9261}{8000} $$ $$ \Rightarrow \quad (\frac{21}{20})^{n} = (\frac{21}{20})^{3} $$ $$ \Rightarrow \quad n = 3 $$Hence, the required time is 3 years.
Q25. At what rate percent will a sum of Rs 1000 amount to Rs. 1102.50 in 2 years at compound interest?
Sol. Given P = Rs. 1000, n = 2 years, A = Rs. 1102.50
Let’s assume that rate of interest be R % p.a
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 1102.50 = 1000(1+\frac{R}{100})^{n} $$ $$ \Rightarrow \quad 1102.50 = 1000(\frac{100+R}{100})^{2} $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = \frac{1102.50}{1000} $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = 1.1025 $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = 1.05^{2} $$ $$ \Rightarrow \quad \frac{100+R}{100} = 1.05 $$ $$ \Rightarrow \quad 100+R = 105 $$ $$ \Rightarrow \quad R = 5 $$Hence, the required rate is 5 %.
Q26. The compound interest on Rs. 1800 at 10 % per annum for a certain period of time is Rs. 378. Find the time in years.
Sol. Given P = Rs. 1800, R = 10 % p.a, C.I =Rs. 378
Let’s assume that time period be n years.
$$ \text {We know that } CI =P(1+\frac{R}{100})^{n} – P $$$$ \Rightarrow \quad 378 = 1800(1+\frac{10}{100})^{n} – 1800 $$ $$ \Rightarrow \quad 378 = 1800(\frac{110}{100})^{n} – 1800 $$ $$ \Rightarrow \quad 378 = 1800[(\frac{110}{100})^{n} – 1] $$ $$ \Rightarrow \quad \frac {378}{1800} + 1 = (\frac{110}{100})^{n} $$ $$ \Rightarrow \quad \frac {2178}{1800} = (\frac{110}{100})^{n} $$ $$ \Rightarrow \quad 1.21 = (1.1)^{n} $$ $$ \Rightarrow \quad 1.1^{2} = (1.1)^{n} $$ $$ \Rightarrow \quad n = 2 $$Hence, the required time period is 2 years.
Q27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually?$$ \text {Sol. Given A = Rs. 45582.25, R = } 6 \frac{3}{4} \% \text { p.a, n = 2 years.} $$Let’s assume that sum of money be Rs x.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 45582.25 = x(1+\frac{27}{400})^{2} $$ $$ \Rightarrow \quad x = 40000 $$Hence, the required sum is Rs. 40000.
Q28. Sum of money amounts to Rs. 453690 in 2 years at 6.5 % per annum compounded annually. Find the sum.
Sol. Given A = Rs. 453690, R = 6.5 % p.a, n = 2 years.
Let’s assume that sum of money be Rs x.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 453690 = x(1+\frac{6.5}{100})^{2} $$ $$ \Rightarrow \quad x = 400000 $$Hence, the required sum is Rs. 400000.
Sol. Lets assume no. of years be n.
Given P =Rs. 12000, R= 5 % p.a , A = Rs. 13230.
$$ \text {We know that } A = P(1+\frac{R}{100})^{n} $$ $$ \Rightarrow \quad 13230 = 12000(1+\frac{5}{100})^{n} $$ $$ \Rightarrow \quad 13230 = 12000(\frac{105}{100})^{n} $$ $$ \Rightarrow \quad (1.05)^{n} = \frac {13230}{12000} $$ $$ \Rightarrow \quad (1.05)^{n} = 1.1025 = 1.05^{2} $$ $$ \Rightarrow \quad n = 2 $$Hence, Ishita invested the money for 2 yrs.
Q22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?
Sol. Given P = Rs. 4000, n= 2 years.
Let’s assume that rate of interest be R %
We know that CI = A – P
$$ \Rightarrow \quad CI = P(1+\frac{R}{100})^{n} – P $$ $$ \Rightarrow \quad 410 = 4000(1+\frac{R}{100})^{2} – 4000 $$ $$ \Rightarrow \quad \frac {410 + 4000}{4000}= (1+\frac{R}{100})^{2} $$$$ \Rightarrow \quad \frac {4410}{4000}= (1+\frac{R}{100})^{2} $$ $$ \Rightarrow \quad (\frac {21}{20})^{2}= (1+\frac{R}{100})^{2} $$$$ \Rightarrow \quad \frac {21}{20}= 1+\frac{R}{100} $$ $$ \Rightarrow \quad \frac {21}{20} -1 = \frac{R}{100} $$ $$ \Rightarrow \quad \frac {1}{20} = \frac{R}{100} $$ $$ \Rightarrow \quad R = 5 $$Hence, the required rate of interest is 5 % per annum.
Q23. A sum of money deposited at 2 % per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.
Sol. Given A = Rs. 10404, R = 2 % p.a, n= 2 years
Let’s assume that sum of money be Rs. P.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 10404 = P(1+\frac{2}{100})^{2} $$ $$ \Rightarrow \quad 10404 = P(\frac{102}{100})^{2} $$ $$ \Rightarrow \quad P = 10404 \times (\frac{100}{102})^{2} $$ $$ \Rightarrow \quad P = 10404 \times \frac{10000}{10404} $$ $$ \Rightarrow \quad P = 10000 $$Hence, the required sum is Rs. 10000.
Q24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5 % per annum compound interest?
Sol. Given P = Rs. 1600, R = 5 % p.a, A = Rs. 1852.20
Let’s assume that time period be n years.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 1852.20 = 1600(1+\frac{5}{100})^{n} $$ $$ \Rightarrow \quad 1852.20 = 1600(\frac{105}{100})^{n} $$ $$ \Rightarrow \quad (\frac{105}{100})^{n} = \frac{1852.20}{1600} $$ $$ \Rightarrow \quad (\frac{21}{20})^{n} = \frac{9261}{8000} $$ $$ \Rightarrow \quad (\frac{21}{20})^{n} = (\frac{21}{20})^{3} $$ $$ \Rightarrow \quad n = 3 $$Hence, the required time is 3 years.
Q25. At what rate percent will a sum of Rs 1000 amount to Rs. 1102.50 in 2 years at compound interest?
Sol. Given P = Rs. 1000, n = 2 years, A = Rs. 1102.50
Let’s assume that rate of interest be R % p.a
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 1102.50 = 1000(1+\frac{R}{100})^{n} $$ $$ \Rightarrow \quad 1102.50 = 1000(\frac{100+R}{100})^{2} $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = \frac{1102.50}{1000} $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = 1.1025 $$ $$ \Rightarrow \quad (\frac{100+R}{100})^{2} = 1.05^{2} $$ $$ \Rightarrow \quad \frac{100+R}{100} = 1.05 $$ $$ \Rightarrow \quad 100+R = 105 $$ $$ \Rightarrow \quad R = 5 $$Hence, the required rate is 5 %.
Q26. The compound interest on Rs. 1800 at 10 % per annum for a certain period of time is Rs. 378. Find the time in years.
Sol. Given P = Rs. 1800, R = 10 % p.a, C.I =Rs. 378
Let’s assume that time period be n years.
$$ \text {We know that } CI =P(1+\frac{R}{100})^{n} – P $$$$ \Rightarrow \quad 378 = 1800(1+\frac{10}{100})^{n} – 1800 $$ $$ \Rightarrow \quad 378 = 1800(\frac{110}{100})^{n} – 1800 $$ $$ \Rightarrow \quad 378 = 1800[(\frac{110}{100})^{n} – 1] $$ $$ \Rightarrow \quad \frac {378}{1800} + 1 = (\frac{110}{100})^{n} $$ $$ \Rightarrow \quad \frac {2178}{1800} = (\frac{110}{100})^{n} $$ $$ \Rightarrow \quad 1.21 = (1.1)^{n} $$ $$ \Rightarrow \quad 1.1^{2} = (1.1)^{n} $$ $$ \Rightarrow \quad n = 2 $$Hence, the required time period is 2 years.
Q27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually?$$ \text {Sol. Given A = Rs. 45582.25, R = } 6 \frac{3}{4} \% \text { p.a, n = 2 years.} $$Let’s assume that sum of money be Rs x.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 45582.25 = x(1+\frac{27}{400})^{2} $$ $$ \Rightarrow \quad x = 40000 $$Hence, the required sum is Rs. 40000.
Q28. Sum of money amounts to Rs. 453690 in 2 years at 6.5 % per annum compounded annually. Find the sum.
Sol. Given A = Rs. 453690, R = 6.5 % p.a, n = 2 years.
Let’s assume that sum of money be Rs x.
$$ \text {We know that } A =P(1+\frac{R}{100})^{n} $$$$ \Rightarrow \quad 453690 = x(1+\frac{6.5}{100})^{2} $$ $$ \Rightarrow \quad x = 400000 $$Hence, the required sum is Rs. 400000.