ICSE Board: Suggested Books        ICSE Solved Board Papers  
SAT Preparation Books       ICSE Kindle Books

KG3Question 1: Find the money, invested at 10\%   compounded annually, on which the sum of interest for the first year and the third year is  Rs. \ 1768  .

Answer:

For 1st year: P = Rs. \ x; \ R=10\% \ and \ T=1 \ year

Therefore Interest = \frac{x \times 10 \times 1}{100} = Rs. \ 0.1x 

and, Amount = x+0.1x = Rs. \ 1.1x 

For 2nd year: P = Rs. \ 1.1x; \ R=10\% \ and \ T=1 \ year

Therefore Interest = \frac{1.1x \times 10 \times 1}{100} = Rs. \ 0.11x 

and, Amount = 1.10x + 0.11x = Rs. \ 1.21x 

For 3rd year: P = Rs. \ 1.21x; \ R=10\% \ and \ T=1 \ year

Therefore Interest = \frac{1.21x \times 10 \times 1}{100} = Rs. \ 0.121x 

and, Amount = 1.21x + 0.121x = Rs. \ 1.331x 

Given 0.1x+0.121x= 1768 \Rightarrow x= Rs. \ 8000

\\

Question 2: A sum of money is invested at compound interest payable annually. The interest in first two successive years is  Rs. \ 1350   and  Rs. \ 1440   respectively. Find;

  1. The rate of interest
  2. The original Sum
  3. The interest earned in the third year.

Answer:

Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years

= 1440-1350 = Rs. \ 90 

\Rightarrow Rs. \ 90   \ is \ the \ interest \ on \ Rs. \ 1350 

\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 90}{1350 \times 1} \% = 6\frac{2}{3}\%

For 1st year: P = Rs. \ P; \ R=6\frac{2}{3}\% \ and \ T=1 \ year

Interest = Rs. 1350 = P \times \frac{6\frac{2}{3}}{100} \times 1 \Rightarrow P= 20250

and, Amount = 20250 + 1350x = Rs. \ 21600 

For 2nd year: P = Rs. \ 21600; \ R=6\frac{2}{3}\% \ and \ T=1 \ year

Therefore Interest = \frac{21600 \times6\frac{2}{3} \times 1}{100} = Rs. \ 1440 

and, Amount =21600 +1440 = Rs. \ 23040 

For 3rd year: P=Rs. \ 23040; \ R=6\frac{2}{3}\% \ and \ T=1 \ year

Therefore Interest = \frac{23040 \times 6\frac{2}{3} \times 1}{100} = Rs. \ 1536 

and, Amount =23040 +1536 = Rs. \ 24576 

\\

EG9Question 3: A sum of money amounts to  Rs. \ 46305   in 1   year and to  Rs. \ 48620.25   in 1\frac{1}{2} years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.

Answer:

Interest \ earned \ in \ 3rd \ half\ year=48620.25-46305 = Rs. \ 2315.25 

P=Rs. \ 46305; I = Rs. \ 2315.25 T = \frac{1}{2}; R=x\%

Interest = Rs. \ 2315.25=46305 \times \frac{x}{100} \times \frac{1}{2} \Rightarrow x= 10\% 

For 1st half year: P = Rs. \ P; \ R=10\% \ and \ T=\frac{1}{2} \ year

Interest =  P \times \frac{10}{100} \times\frac{1}{2} = 0.05P

and, Amount = P+0.05P = 1.05P 

For 2nd half year: P = Rs. \ 1.05P; \ R=10\% \ and \ T=\frac{1}{2} \ year

Interest =  1.05 \times \frac{10}{100} \times\frac{1}{2} = 0.0525P 

\therefore Amount = 1.05P+0.0525P = 1.1025P = 46305 \Rightarrow P = Rs. \ 42000 

\\

Question 4: The cost of machine in Rs. \ 32,000  . Its value depreciates at the rate of 5\%   every year. Find the total depreciation in its value by the end of   2     years.

Answer:

For 1st year: P = Rs. \ 32000; \ R=5\% \ and \ T=1 \ year

Depreciation =  32000 \times \frac{5}{100} \times 1 = Rs. \ 1600

and, Value = 32000-1600  = Rs. \ 30400 

For 2nd  year: P = Rs. \ 30400; \ R=5\% \ and \ T=1 \ year

Depreciation =  30400 \times \frac{5}{100} \times 1 = Rs. \ 1520 

and, Value = 30400-1520  = Rs. \ 28880 

\therefore Total \ Depreciation = 1600+1520 = Rs. \ 3120

\\

Question 5: Find the sum, invested at  10\%   compounded annually, on which the interest for the third year exceeds the interest of the first year by  Rs. \ 252  .

Answer:

For 1st year: P = Rs. \ x; \ R=10\% \ and \ T=1 \ year

Interest =  x \times \frac{10}{100} \times 1 = Rs. \ 0.1x 

and, Amount = x+0.1x  = Rs. \ 1.1x 

For 2nd year: P = Rs. \ 1.1x; \ R=10\% \ and \ T=1 \ year

Interest =  1.1x \times \frac{10}{100} \times 1 = Rs. \ 0.11x 

and, Amount = 1.1x+0.11x  = Rs. \ 1.21x 

For 3rd year: P = Rs. \ 1.21x; \ R=10\% \ and \ T=1 \ year

Interest =  1.21x \times \frac{10}{100} \times 1 = Rs. \ 0.121x 

and, Amount = 1.21x+0.121x  = Rs. \ 1.331x 

Given 0.121x-0.1x=252 \Rightarrow x= \frac{252}{0.021} = Rs. \ 12000

\\

Question 6: The compound interest, compounded annually, on a certain sum is  Rs. \ 9680   in second year and is  Rs. \ 10648   in third year. Calculate;

  1. The rate of interest, sum borrowed
  2. The interest of 1^{st}    year.

Answer:

Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years

= 10648-9680 = Rs. \ 968 

\Rightarrow Rs. \ 968  is the interest on Rs. \ 9680 

\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 968}{9680 \times 1} \% = 10\%

Let the sum of money  = Rs. \ 100

Therefore Interest on it for 1st Year  = 10\% \ of  \ Rs. \ 100 = Rs. \ 10 

 \Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110

 \therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 10\% \ of \ 110 = Rs. \ 11 

 \Rightarrow Amount \ in \ 2nd \ year = 110 + 11 = Rs. \ 121 

When interest of 2nd year  = Rs. \ 11, sum \ is \ Rs. \ 100

 \Rightarrow \ When \ interest \ of \ 2nd \ year= Rs. \ 9680 , then

 sum= \frac{100}{11} \times 9680 = Rs. \ 88000 

Interest \ of \ 1st \ year = 88000 \times \frac{10}{100} \times 1 = Rs. \ 8800

\\

\\

EG2Question 7: A man borrows  Rs. \ 10,000   at  10\%   compound interest compounded yearly. At the end of each year, he pays back  30\%   of the sum borrowed. How much money is left unpaid just after the second year?

Answer:

For 1st year: P = Rs. \ 10000; \ R=10\% \ and \ T=1 \ year

Interest =  10000 \times \frac{10}{100} \times 1 = Rs. \ 1000 

and, Amount = 10000+1000  = Rs. \ 11000 

For 2nd year: P = Rs. \ (11000-3000)=8000; \ R=10\% \ and \ T=1 \ year

Interest =  8000 \times \frac{10}{100} \times 1 = Rs. \ 800 

and, Amount = 8000+800  = Rs. \ 8800 

Amount left = 8800-3000 = Rs. \ 5800

\\

Question 8: A man borrows Rs. \ 10000   at  10\%   compound interest compounded yearly. At the end of each year, he pays back  20\%   of the amount for that year. How much money is left unpaid just after the second year?

Answer:

For 1st year: P = Rs. \ 10000; \ R=10\% \ and \ T=1 \ year

Interest =  10000 \times \frac{10}{100} \times 1 = Rs. \ 1000 

and, Amount = 10000+1000  = Rs. \ 11000 

For 2nd year: P = Rs. \ (11000-2000)=9000; \ R=10\% \ and \ T=1 \ year

Interest =  9000 \times \frac{10}{100} \times 1 = Rs. \ 900 

and, Amount = 9000+900  = Rs. \ 9900 

Amount left = 9900-2000 = Rs. \ 7900

\\

Question 9: The population of a town increases  10\%    every  3   years. If the present population of the town is  72600  , calculate:

  1. Its population after  2   years;
  2. Its population  2   years ago;

Answer:

For 2 years ago: P = Rs. \ x; \ R=10\% \ and \ T=1 \ year

Increase = x \times \frac{10}{100} \times 1 = Rs. \ 0.1x 

and, Population = x+0.1x  = Rs. \ 1.1x 

For 1 year ago: P = Rs. \ 1.1x; \ R=10\% \ and \ T=1 \ year

Increase = 1.1x \times \frac{10}{100} \times 1 = Rs. \ 0.11x 

and, Population = 1.1x+0.11x  = Rs. \ 1.21x 

Given 1.21x= 72600 \Rightarrow x= 60000

For 1st year: P = Rs. \ 72600; \ R=10\% \ and \ T=1 \ year

Increase = 72600 \times \frac{10}{100} \times 1 = Rs. \ 7260 

and, Population = 72600+7260  = Rs. \ 79860 

For 2nd year: P = Rs. \ 79860; \ R=10\% \ and \ T=1 \ year

Increase = 79860 \times \frac{10}{100} \times 1 = Rs. \ 7986

and, Population = 79860+7986  = Rs. \ 87846 

\\

EG12Question 10: The compound interest, calculated yearly, on a certain sum of money for the second year is Rs. \ 1320    and for the third year is Rs. \ 1452  . Calculate the rate of interest and the original sum of money. [2014]

Answer:

Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years

= 1452-1320 = Rs. \ 132 

\Rightarrow Rs. \ 132  is the interest on Rs. \ 1320 

\therefore Rate \ of \ Interest = \frac{100 \times I}{P \times T} \% = \frac{100 \times 132}{1320 \times 1} \% = 10\%

Let the sum of money  = Rs. \ 100

Therefore Interest on it for 1st Year  = 10\% \ of  \ Rs. \ 100 = Rs. \ 10 

 \Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110

 \therefore Interest \ on \ it \ for \ 2^{nd} \ Year = 10\% \ of \ 110 = Rs. \ 11 

 \Rightarrow Amount \ in \ 2nd \ year = 110 + 11 = Rs. \ 121 

When interest of 2nd year  = Rs. \ 11, sum \ is \ Rs. \ 100

 \Rightarrow \ When \ interest \ of \ 2nd \ year= Rs. \ 1320 , then

 sum= \frac{100}{11} \times 1320 = Rs. \ 12000 

\\

 

 

Advertisements