First half of Exercise 4(a)…

Question 26: The compound interest on a sum of money for 2 years is Rs. \ 1050 and the simple interest on the same sum for the same period and at the same rate is Rs. \ 1000 . Find: (i) the rate of interest (ii) the sum.

Answer:

Compound Interest:

Principal = P , Rate of Interest = r\% per annum, T = 2 \ years

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

Compound Interest 1050 = P \Big( 1+ \frac{r}{100} \Big)^2  - P

\Rightarrow 1050 = P \Big[ (1+ \frac{r}{100})^2 - 1 \Big]

\Rightarrow 1050 = P (2+ \frac{r}{100} )( \frac{r}{100} ) … … … … … (i)

Simple Interest:

Principal = P , Rate of Interest :  R=r\%, annually, T=  2 years

Interest = \frac{P \times R \times T}{100}

1000 = \frac{P \times r \times 2}{100} \Rightarrow 50000 = Pr

\Rightarrow P = \frac{50000}{r} … … … … … (ii)

Now solving (i) and (ii) we get

1050 =  \frac{50000}{r} (2+ \frac{r}{100} )( \frac{r}{100} $latex )

\Rightarrow 1050 = 1000 + 5r

\Rightarrow r = 10\%

\Rightarrow P = \frac{50000}{10} = 5000 \ Rs.

\\

Question 27: Find the rate percent per annum if Rs. \ 2000 amounts to Rs. \ 2662 in 1\frac{1}{2} years, interest being compounded half-yearly.

Answer:

Principal = 2000 \ Rs. , Rate of Interest :  R=r\%, semi-annually, T=  1.5 \ years year  Amount = 2662 \ Rs.

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

26620 = 2000 \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}

1.331 = \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}

1.10 = 1 + \frac{r}{200} \Rightarrow r = 20\% per annum

\\

Question 28: Find the rate at which a sum of money will double itself in 3 years if the interest is compounded annually.

Answer:

Principal = x \ Rs. , Rate of Interest :  R=r\%, annually, n=  3 year  Amount = 2x \ Rs.

A= P \Big( 1+ \frac{R}{100} \Big)^{n}

2x = x \Big( 1+ \frac{r}{100 } \Big)^{3}

2 = \Big( 1+ \frac{r}{100 \times 2} \Big)^3

1.2599 = 1 + \frac{r}{100} \Rightarrow r = 25.99\% per annum

\\

Question 29: Find the rate at which a sum of money will become four amount times the original in 2 years if the interest is compounded half yearly.

Answer:

Principal = x \ Rs. , Rate of Interest :  R=r\%, annually, n=  2 year  Amount = 4x \ Rs.

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

4x = x \Big( 1+ \frac{r}{100 \times 2 } \Big)^{2 \times 2}

4 = \Big( 1+ \frac{r}{100 \times 2} \Big)^4

1.4142 = 1 + \frac{r}{200} \Rightarrow r = 82.84\% per annum

\\

Question 30: A sum compounded annually becomes \frac{25}{16} times of itself in 2 years. Determine the rate of interest.

Answer:

Principal = x \ Rs. , Rate of Interest :  R=r\%, annually, n=  2 year  Amount = \frac{25}{16}x \ Rs.

A= P \Big( 1+ \frac{R}{100} \Big)^{n}

\frac{25}{16} x = x \Big( 1+ \frac{r}{100} \Big)^{2}

\frac{25}{16} = \Big( 1+ \frac{r}{100 \times 2} \Big)^2

\frac{5}{4} = 1 + \frac{r}{100} \Rightarrow r = 25\% per annum

\\

Question 31: Rishi invested Rs. \ 30000 in a finance company and received Rs. \ 39930 after 1\frac{1}{2} years. Find the rate of interest per annum compounded semi-annually.

Answer:

Principal = 30000 \ Rs. , Rate of Interest :  R=r\%, semi-annually, T=  1.5 \ years year  Amount = 39930 \ Rs.

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

39930 = 30000 \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}

1.331 = \Big( 1+ \frac{r}{100 \times 2} \Big)^{1.5 \times 2}

1.10 = 1 + \frac{r}{200} \Rightarrow r = 20\% per annum

\\

Question 32: In how much time would Rs. \ 5000 amounts to Rs. \ 6655 at 10\% per annum compound interest?

Answer:

Principal = 5000 \ Rs. , Rate of Interest :  R=10\%, annually, T=  n \ years year  Amount = 6655 \ Rs.

A= P \Big( 1+ \frac{R}{100} \Big)^{n}

6655 = 5000 \Big( 1+ \frac{10}{100} \Big)^{n}

1.331 = \Big( 1+ \frac{10}{100} \Big)^{n}

1.1^3 = \Big( 1+ \frac{10}{100} \Big)^{n}

\Rightarrow n = 3 \ years

\\

Question 33: In what time will Rs. \ 4400 become Rs. \ 4576 at 8\% per annum interest compounded half-yearly?

Answer:

Principal = 4400 \ Rs. , Rate of Interest :  R=8\%, annually, T= n \ years year  Amount = 4576 \ Rs.

A= P \Big( 1+ \frac{R}{100k} \Big)^{nk}

4576 = 4400 \Big(1+ \frac{8}{100 \times 2} \Big)^{n \times 2}

1.04 = \Big( 1+ \frac{4}{100} \Big)^{2n}

1.04 = ( 1.04)^{2n}

\Rightarrow 2n = 1 \ year  \Rightarrow  n = \frac{1}{2} \ year

\\

Question 34: 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.

Answer:

1^{st} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_1

A_1 = P \Big( 1+ \frac{r}{100} \Big)^1 … … … … (i)

2^{nd} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_2

A_2 = P \Big( 1+ \frac{r}{100} \Big)^2 … … … … (ii)

3^{rd} \ Year

Principal = P , Rate of Interest = r\% per annum, n = 1 Amount = A_3

A_3 = P \Big( 1+ \frac{r}{100} \Big)^3 … … … … (iii)

Given: A_2 - A_1 = 1320   and A_3 - A_2 = 1452

Therefore

P \Big( 1+ \frac{r}{100} \Big)^2 - P \Big( 1+ \frac{r}{100} \Big)^1 = 1320

\Rightarrow P (1+ \frac{r}{100} )( \frac{r}{100} ) = 1320 … … … … (iv)

P \Big( 1+ \frac{r}{100} \Big)^3 - P \Big( 1+ \frac{r}{100} \Big)^1 = 1452

\Rightarrow P (1+ \frac{r}{100} )^2( \frac{r}{100} ) = 1452 … … … … (v)

Dividing (v) by (iv) we get

\frac{1452}{1320} = 1 + \frac{r}{100}

\Rightarrow r = 1.1 - 1 = 10\%

Substituting in (iv) P = \frac{1320 \times 100}{10 \times 1.1} = 12000 \ Rs.

\\

Question 35: On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to Rs. \ 25 , if the rate of interest charged for both is 5\% per annum?

Answer:

Compound Interest:

Principal = P , Rate of Interest = 5\% per annum, T = 2 \ years

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

A = P \Big( 1+ \frac{5}{100} \Big)^2 = 1.1025 P

Compound Interest = 1.1025 P  - P = 0.1025P

Simple Interest:

Principal = P , Rate of Interest :  R=5\%, annually, T=  2 years

Interest = \frac{P \times R \times T}{100}

= \frac{P \times 5 \times 2}{100} \Rightarrow = 0.1P

Given: 25 = 0.1025 P - 0.1P \Rightarrow P = 10000 \ Rs.

\\

Question 36: Mr Kumar borrowed Rs. \ 15000 for two years. The rate of interest for the two successive years are 8\% and 10\% respectively. If he repays Rs. \ 6200 at the end of the first year, find the outstanding amount at the end of second year.

Answer:

For the 1^{st} \ Year : We have,

Principal = Rs. 15000 , Rate of Interest = 8\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{15000 \times 8 \times 1}{100} = 1200  \ Rs.

Therefore, the Amount at the end of 1^{st} year =15000 + 1200 = 16200 \ Rs.

For the 2^{nd} \ Year : We have,

Principal = Rs. 16200 - 6200 = 10000 , Rate of Interest = 10\% per annum, T = 1

Therefore Interest = \frac{P \times R \times T}{100} = \frac{10000 \times 10 \times 1}{100} =1000  \ Rs.

Therefore, the Amount at the end of 2^{nd} year = 10000 + 1000 = 11000 \ Rs.

\\  

Depreciation Problems

Question 37: The value of a machine depreciates at the rate of 10\% per annum. what will be its value 2 years hence if the present value is Rs. \ 100000 ? Also, find the total depreciation during this period.

Answer:

Present Value V_0 = 100000 \ Rs.   Rate of Depreciation = 10\% No of Years = 2

V_2 = V_0 \Big( 1- \frac{R}{100} \Big)^n

V_2 = 100000 \Big( 1- \frac{10}{100} \Big)^2

V_2= 81000 \  Rs.

Total depreciation = 100000 - 81000 = 19000 \ Rs.

\\

Question 38: Pritam bought a plot of land for Rs. \ 640000 . Its value is increasing by 5\% of its previous value after every six months. What will be the value of the plot after 2 years?

Answer:

Present Value = 640000 \ Rs. , Rate of Interest = 5\% half yearly , T = 2 \ years

V_2 = P \Big( 1+ \frac{r}{100k} \Big)^{nk}

V_2 = 640000 \Big( 1+ \frac{10}{100 \times 2} \Big)^{2 \times 2}

V_2 = 777924 \  Rs.

\\

Question 39: The value of a machine depreciates at the rate of 10\% per annum. It was purchased 3 years ago. If its present value is Rs. \ 43740 , find its purchase price.

Answer:

Present Value V_0 = 43740 \ Rs.   Rate of Depreciation = 10\% No of Years = 3  Value 3 years back  V_{-3} = x 

V_0 = V_{-3} \Big( 1- \frac{R}{100} \Big)^3

43740 = V_{-3} \Big( 1- \frac{10}{100} \Big)^3

V_{-3} = 60000 \  Rs.

Total depreciation = 60000 - 43740 = 16260 \ Rs.

\\

Question 40: The cost of a T.V. set was quoted Rs. \ 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5\% . Because of decrease in demand the cost was reduced by 4\% in the beginning of 2001. What was the cost of the T.V. set in 2001?

Answer:

Cost of TV at the beginning of 1999 = 17000 \ Rs.

Cost of TV at the beginning of 2000 = 17000 \times 1.05 = 17850 \ Rs.

Cost of TV at the beginning of 2001 = 17850 \times 0.96 = 17136 \ Rs.

\\

Question 41: Ashish staffed the business with an initial investment of Rs. \ 500000 . In the first he incurred a loss of 4\% . However, during the second year he earned a profit of 5\% which in third year rose to 10\% . Calculate the net profit for the entire period of 3 years.

Answer:

Initial Investment at the beginning of 1st Year = 500000 \ Rs.

Capital at the beginning of 2nd Year = 500000 \times 0.96 = 480000 \ Rs.

Capital at the beginning of 3rd Year = 480000 \times 1.05 = 504000 \ Rs.

Capital at the beginning of 4th Year = 504000 \times 1.10 = 554400 \ Rs.

Net profit = 554400 - 500000 = 54400 \ Rs.

\\

Population Questions

Question 42: The present population of a town is 28000 . If it increases at the rate of 5\% per annum, what will be its population after 2 years?

Answer:

Population P = 28000 , Rate R = 5\% per annum, n = 2 years

Population after n years = P \Big(1+ \frac{R}{100} \Big)^{n}

Population after 2 years = 28000 \Big(1+ \frac{5}{100} \Big)^{2}  = 30870

\\

Question 43: The present population of a town is 25000 . It grows at first 4\%, 5\% and 8\% during year, second year and third year respectively. Find lts population after 3 years.

Answer:

Current Population P = 25000 , Rate R_1 = 4\%, R_2=5%, R_3 = 8\% per annum, n = 3 years

Population after 3^{rd} \ Year = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)

Population after 3^{rd} \ Year = 25000 \Big(1+ \frac{4}{100} \Big).\Big(1+ \frac{5}{100} \Big).\Big(1+ \frac{8}{100} \Big)

= 25000 \times 1.04 \times 1.05 \times 1.08 = 29484

\\

Question 44: There is a continuous growth in population of a village at the rate of 5\% per annum. If its present population is 9261 , what was it 3 years ago?

Answer:

Population 3 years ago = P_{-3} Current Population P_0 = 9261 , Rate R = 5\% per annum, n = 3 years

Therefore P_{0} = P_{-3} \Big(1+ \frac{R}{100} \Big)^{n}

9261 =P_{-3} \Big(1+ \frac{5}{100} \Big)^{3}

\Rightarrow P_{-3} = 8000

\\

Question 45: In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.

Answer:

P = 40000 \ A = 46305 \ n = 3 \ years, Rate = R\%

46305 = 40000 \Big(1+ \frac{R}{100} \Big)^{3}

1.05 = 1 + \frac{R}{100}

\Rightarrow R = 5\%

\\

Question 46: The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050 . Find its present population.

Answer:

Rate of Increase = 5% \ \ A (population in 2 years) = 22050, \ P (Initial population) = P

22050 = P \Big(1+ \frac{5}{100} \Big)^{2}

\Rightarrow P = 20000

\\

Question 47: The count of bacteria in a culture grows by 10\% in the first hour, decreases by 8\% in the second hour and again increase by 12\% in third hour. If the count of bacteria in the sample 13125000 , what will be the count of bacteria after 3 hour.

Answer:

Current bacteria Population P = 13125000 , Rate R_1 = 10\%, R_2=-8%, R_3 = 12\% per annum, n = 3 hours

Bacteria Population after 3^{rd} \ hours = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)

Bacteria Population after 3^{rd} \ hours = 13125000 \Big(1+ \frac{10}{100} \Big).\Big(1+ \frac{-8}{100} \Big).\Big(1+ \frac{12}{100} \Big)

= 13125000 \times 1.10 \times 0.92 \times 1.12 = 14876400

\\

Question 48: 6400 workers were employed to construct a river bridge in four years. At the end First year, 25\% workers were retrenched. At the second year 25\% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25\% the end of the third year. How many workers were working during the fourth year?

Answer:

Current worker Population P = 6400 , Rate R_1 = -25\%, R_2=-25%, R_3 = 25\% per annum, n = 3 years

Worker Population after 3^{rd} \ years = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)

Worker Population after 3^{rd} \ years = 6400 \Big(1+ \frac{-25}{100} \Big).\Big(1+ \frac{-25}{100} \Big).\Big(1+ \frac{25}{100} \Big)

= 64000 \times 0.75 \times 0.75 \times 1.25 = 4500

\\

Question 49: A man started a factory with an initial investment of Rs. \ 100000 . In the first year, he incurred a loss of 5\% . However, during the second year; he earned a profit of 10\% which in the third year rose to 12\% . Calculate the profit for the entire period of three years.

Answer:

Initial Investment P = 100000 , Rate R_1 = -5\%, R_2=10%, R_3 = 12\% per annum, n = 3 years

Worker Population after 3^{rd} \ years = P \Big(1+ \frac{R_1}{100} \Big).\Big(1+ \frac{R_2}{100} \Big).\Big(1+ \frac{R_3}{100} \Big)

Worker Population after 3^{rd} \ years = 100000 \Big(1+ \frac{-5}{100} \Big).\Big(1+ \frac{10}{100} \Big).\Big(1+ \frac{12}{100} \Big)

= 100000 \times 0.95 \times 1.10 \times 1.12 = 117040 \ Rs.

Therefore net profit = 117040 - 100000 = 17040 \ Rs.

\\

Question 50: The population of a city increases each year by 4\% of what it had been at the beginning of each year. If the population in 1999 had been 6760000 , find the population of the city in (i) 2001 (ii) 1997.

Answer:

P (1999) = 6760000 \ , Rate = 4\%

P_{2001} = 6760000 \Big(1+ \frac{4}{100} \Big)^{2} = 7311616

6760000= P_{1997} \Big(1+ \frac{4}{100} \Big)^{2}

\Rightarrow P_{1997} = 6250000

\\

Advertisements