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KG1Question 1: Find the effective rate per cent per annum equivalent to a nominal rate of 10% per annum interest payable half-yearly?

Answer:

Compounded Half Yearly

P=100\  Rs.; \ r=10\%; Compounded \ half \ yearly \ n=1 \ year

A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 100(1+\frac{10}{2 \times 100})^{1 \times 2} =  110.25 \ Rs. 

\therefore Equivalent \ Nominal \ rate = 10.25\%

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Question 2: A property decreases, in value, every year at the rate of 6% at the beginning of that year. If its value at the end of 2 years is Rs. 2,25,000, what was its worth at the beginning of these 2 years?

Answer:

At the start of 1^{st} \ year 

P=x\  Rs.; \ r=6\%; \ n=1 \ Year

At the end of 1^{st} \ year 

A=P(1-\frac{r}{100})^n = x(1-\frac{6}{100})^1 =  0.94x \ Rs.  

P=0.94x\  Rs.

At the end of 2^{nd} year 

A=P(1+\frac{r}{100})^n = 0.94x(1-\frac{6}{100})^1 =  0.8836x \ Rs.  

Given 0.8836x=225000 \Rightarrow x = 254640.10 \ Rs.

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Question 3: A man wishes to accumulate capital worth Rs. 50,440 at the begining of 3 years from now. If he can invest his savings at 5% per annum compound interest, what equal sum must be put aside each year, beginning at the end of the first, to obtain the required amount.

Answer:

Start of 1st Year. Let us say that x Rs. are set aside each year.

P=x\  Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^n = x(1+\frac{5}{100})^1 =  1.05x \ Rs.  

Start of 2nd Year

P=(1.05x+x)=2.05x\  Rs.; \ r=5\%; Compounded \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^n = 2.05x(1+\frac{5}{100})^1 =  2.1525x \ Rs.  

Given 2.125x+x=50440 \Rightarrow = x= 16000 \ Rs.

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Question 4: Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for  1 year, if the interest on the same sum and at the same rate for  year, if the interest  is reckoned half-yearly. [1997]

Answer:

Simple Interest

P=x \ Rs.; T=2 \ Years; r=4\%

S.I=x \times \frac{4}{100} \times 2 = \frac{2}{25}x

450 = \frac{2}{25}x \Rightarrow x= 5625 \ Rs.

Compounded Half Yearly

P=5625\  Rs.; \ r=4\%; Compounded \ half \ yearly \ n=1 \ year

A=P(1+\frac{r}{2 \times 100})^{1 \times 2} = 5625(1+\frac{4}{2 \times 100})^{1 \times 2} =  5852.25 \ Rs. 

Compound \ Interest =5852.25-5625= 227.25 \ Rs.

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EG8Question 5: Find the compound interest to the nearest rupee on Rs.10,800 for 2 ½ years at 10% per annum.

Answer:

Compounded Yearly

P=10800\  Rs.; \ r=10\%; Compounded  \ yearly \ n=\frac{5}{2} \ year

A=P(1+\frac{r}{2 \times 100})^{\frac{5}{2} \times 2} 

A=10800(1+\frac{10}{2 \times 100})^{\frac{5}{2} \times 2} =  13783.84 \ Rs.

C.I. = 13783.84-10800 = 2983.85 \ Rs.

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Question 6: A bought a plot of land for Rs.70,000 and a car for Rs.32,000 on the same day. The value of the plot appreciates uniformly at the rate of 10% every year while the value of the car depreciates by 20% for the first year and by 10% for the second year. If A sells the plot of land as well as the car after 2 years, what will be the profit or loss on the whole?

Answer:

Value of Plot of Land at the end of 2 years

P=70000\  Rs.; \ r=10\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} = 7000(1+\frac{10}{100})^{2} =  84700 \ Rs.

Value of Car at the end of 2 years

P=32000\  Rs.; \ r=20\%; Compounded  \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^{1} = 32000(1-\frac{20}{100})^{1} =  25600 \ Rs.

After 2 years

A=P(1+\frac{r}{100})^{1}=25600(1-\frac{10}{100})^{1}=23040 \ Rs.

Total investment = 70000+32000 = 102000 \ Rs.

Total Return = 84700+23040 = 107740 \ Rs.

\Rightarrow gain = 107740-102000 = 5740 \ Rs.

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Question 7: The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs. 97200, find;

  1. Its value after 2 years.
  2. Its value when it was purchased.

Answer:

Value after 2 years

P=97200\  Rs.; \ r=10\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{1} = 97200(1-\frac{10}{100})^{2} =  78732 \ Rs.

Value of Machine at the end of 1st year

P=x\  Rs.; \ r=10\%; Compounded  \ yearly \ n=1 \ year

A=x(1+\frac{r}{100})^{1} = x(1-\frac{10}{100})^{1} =  0.9x \ Rs.

After 2 years

A=P(1+\frac{r}{100})^{1} = 0.9x(1-\frac{10}{100})^{1} =  0.81x \ Rs.

Given 0.81x = 97200 \Rightarrow x = 120000 \ Rs.

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Question 8: A man borrowed a sum of money and agrees to pay it off by paying Rs.9450 at the end of the first year and Rs. 13,230 at the end of the second year. If the rate of compound interest is 5% per annum, find the sum borrowed.

Answer:

1st Year. Let x be the sum borrowed.

P=x\  Rs.; \ r=5\%; Compounded  \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^{1} = x(1+\frac{5}{100})^{1} =  1.05x \ Rs.

2nd Year

P=(1.05x-9450)\  Rs.; \ r=5\%; Compounded  \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^{1} = (1.05x-9450)(1+\frac{5}{100})^{1} =  1.05(1.05x-9450) \ Rs.

Given 1.05(1.05x-9450) = 13230 \Rightarrow x = 21000 \ Rs.

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Question 9: A and B each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. B received Rs. 64 more than A. Find the money lent by each and interest received.

Answer:

Let A’s S.I. investment be x

S.I. = x \times \frac{8}{100} \times 2 = 0.16x

Let B’s C.I. investment be x

P=x\  Rs.; \ r=8\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} = x(1+\frac{8}{100})^{2} =  1.1664x \ Rs.

C.I. = 1.1664x-x=0.1664x

Given 0.1664x-0.16x = 64 \Rightarrow x = 10000 \ Rs.

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EG2Question 10: Calculate the sum of money on which the compound interest (Payable annually) for 2 years be four times the simple interest on Rs. 4715 for 5 years, both at the rate of 5% per annum.

Answer:

Simple Interest

S.I. = 4715 \times \frac{5}{100} \times 5 = 1178.75 \ Rs.  

Compound \ Interest = 4 \ times 1178.75 = 4715 \ Rs.

P=x\  Rs.; \ r=5\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} \Rightarrow 4715+x= x(1+\frac{5}{100})^{2} \Rightarrow x = 46000  \ Rs.

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Question 11: A sum of money was invested for 3 years. Interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to Rs. 4950, find the sum invested.

Answer:

P=x \ Rs. ; r_1=10\%; r_2=15\%; r_3=18\%; n=3 \  years

A = P(1+\frac{r_1}{100})^1 \times (1+\frac{r_2}{100})^1  

A = x(1+\frac{10}{100})^1 \times (1+\frac{15}{100})^1  = 1.265x

Given 1.265x-1.1x=4950 \Rightarrow x= 30000 \ Rs.

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Question 12: A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at the end of 6 months and 12 months is Rs.189, find the sum of money invested.

Answer:

Compounded Half Yearly

P=x\  Rs.; \ r=10\%; Compounded \ half \ yearly \ n=1 \ year

After first period

A=P(1+\frac{r}{2 \times 100})^{1 \times 1} = x(1+\frac{10}{2 \times 100})^{1 \times 1} =  1.05x \ Rs. 

After second period

A=P(1+\frac{r}{2 \times 100})^{1 \times 1} = 1.05x(1+\frac{10}{2 \times 100})^{1 \times 1} =  1.1025x \ Rs. 

Given 1.1025x-1.05x=189 \Rightarrow x=3600 \ Rs.  

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Question 13: Rohit borrows Rs. 86000 from Arun for 2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohits profit at the end of two years. [2010]

Answer:

Simple Interest for 2 years

S.I. = P \times \frac{r}{100} \times 2  \ Rs.  

S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \ Rs.  

Compound Interest for 2 years

P=8600\  Rs.; \ r=5\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} \Rightarrow A= 86000(1+\frac{5}{100})^{2} \Rightarrow A = 9481.50  \ Rs.

Gain =(94815-86000)-8600 = 215 \ Rs.

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Question 14: A person borrowed Rs.60,000 from a money lender at 5% simple interest for two year. The money lender deducts the interest that would be due at the end of the period and handed over the balance to the person. This person then  deposits this amount in a bank which pays 5% compound interest per annum. How much will the person will have to add to pay the money back to the money lender.

Answer:

Simple Interest for 2 years

S.I. = P \times \frac{r}{100} \times 2  \ Rs.  

S.I. = 60000 \times \frac{5}{100} \times 2 = 6000 \ Rs.  

Money given = 60000-6000 = 54000 \ Rs.

Compound Interest for 2 years

P=54000\  Rs.; \ r=5\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} \Rightarrow A= 54000(1+\frac{5}{100})^{2} \Rightarrow A = 59535  \ Rs.

Therefore money to be paid to square off the load = 60000-59535= 465 \ Rs.

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EG13Question 15: The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1200. Find the amount due and the compound interest on this sum of money at the same rate and after 2 years, interest is reckoned annually.

Answer:

Simple Interest for 2 years. Let P = x \ Rs.

S.I. = P \times \frac{r}{100} \times 3  \ Rs.  

S.I. = x \times \frac{5}{100} \times 3 = \frac{3}{20}x \ Rs.  

Given 1200 = \frac{3}{20}x  \Rightarrow x = 8000 \ Rs.

Compound Interest for 2 years

P=8000\  Rs.; \ r=5\%; Compounded  \ yearly \ n=2 \ year

A=P(1+\frac{r}{100})^{2} \Rightarrow A= 8000(1+\frac{5}{100})^{2} \Rightarrow A = 8820  \ Rs.

Compound Interest =(8820)-8000 = 820 \ Rs.

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Question 16: A person invests Rs.6000 for two years at a certain rate of interest compounded annually, At the end of first year it amounts to Rs.6720. Calculate;

  1. The rate of interest;
  2. The amount at the end of the second year. [2010]

Answer:

Compound Interest for 1 year

P=6000\  Rs.; \ r=x\%; Compounded  \ yearly \ n=1 \ year

A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{x}{100})^{1}

Given 6000(1+\frac{x}{100})^{1}=6720 \Rightarrow x= 12\%

Amount at the end of second year

A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{12}{100})^{2} = 7526.40 \ Rs.  

 

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