SChand CLASS 9 Chapter 2 Compound Interest Exercise 2(B)

 Exercise 2(B)

Question 1

Principal = ₹ 625, Rate of interest = 4 % p.a., Time (in years) = 2.

Sol :

Principal=625

Rate=4%

Period (n)=2 year

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=625\left(1+\frac{4}{100}\right)^{2}$

$=625\left(1+\frac{1}{25}\right)^{2}$

$=625\left(\frac{26}{25}\right)^2$

$=625 \times \frac{26}{25} \times \frac{26}{25}$

=676

Compound Interest=A-P

=676-625=51

Question 2

Principal = ₹ 8000, Rate of interest = 15 % p.a., Time (in years) = 3.

Sol :

Principal=8000

Rate=15%

Time (n)=3 years

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=8000 \times\left[1+\frac{15}{100}\right]^{3}$

$=8000\left(\frac{115}{100}\right)^{3}$

$=8000\left[\frac{23}{20}\right]^{3}$

$=8000 \times \frac{23}{20} \times \frac{23}{20} \times \frac{23}{20}$

=12167

∴Compound Interest=A-P

=12167-8000

=4161

Question 3

Principal = ₹ 1000, Rate of interest = 10 % p.a., Time (in years) = 3.

Sol :

Principal=1000

Rate=10%

Time (n)=3 years

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=1000\left(1+\frac{10}{100}\right)^3$

$=1000\left(1+\frac{1}{10}\right)^3$

$=1000\left(\frac{11}{10}\right)^3$

$=1000\left(\frac{11}{10}\right)^3$

$=1000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$

=1331

Compound Interest=A-P

=1331-100=331

Question 4

Principal = ₹ 8000, Rate of interest = 10 % half yearly, Time (in years) $=1\frac{1}{2}$

Sol :

Principal=8000

Rate=10% , 5% half

Time (n)$1 \frac{1}{2}$ years=3 half year

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=8000 \left[1+\frac{5}{100}\right]^3$

$=8000\left(1+\frac{1}{20}\right)^{3}$

$=8000 \times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}$

=9261

∴Compound Interest=A-P

=9261-8000=1261

Question 5

Principal = ₹ 700, Rate of interest = 20 % half yearly, Time (in years) $=1\frac{1}{2}$

Sol :

Principal=700

Rate=20%, 10% half yearly

Time (n) $=1 \frac{1}{2}$ year, 3 half year

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=700\left[1+\frac{10}{100 }\right)^{3}$

$=700\left(1+\frac{1}{10}\right)^{3}$

$=700\left\lfloor\frac{11}{10}\right]^3$

$=700 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$

$\frac{9817}{10}$

=981.7

Compound Interest=A-P

=981.7-700=231.70

Question 6

Sangeeta lent ₹ 40960 to Amar to purchase a shop at 12.5% per annum. If the interest is compounded semi-annually, find the interest paid by Amar after $1\frac{1}{2}$ year

Sol :

Amount of loan (P)=40960

Rate=12.5%$=\frac{125}{10}$ $=\frac{25}{2}\%$

$=\frac{25}{4}\%$ half yearly

Time (n) $=1\frac{1}{2}$ years, 3 half year

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=40960\left(1+\frac{25}{100}\right)^3$

$=40960\left(1+\frac{1}{16}\right)^3$

$=40960\left(\frac{17}{16}\right)^3$

$=40960 \times \frac{17}{16} \times \frac{17}{16}\times \frac{17}{16}$

=49130

Compound Interest=49130-40960

=8170

Question 7

Sudhir lent ₹ 2000 at compound interest at 10% payable yearly, while Prashant lent ₹ 2000 at compound interest at 10% payable half-yearly. Find the difference in the interest received by Sudhir and Prashant at the end of one year.

Sol :

In case of Sudhir

Amount lent (P)=2000

Rate=10%

Time = 1 year

∴Interest$=\frac{PRT}{100}$

$=\frac{2000 \times 10 \times 1}{100}$

=200...(i)

In case of Prasant

Principal (P)=2000

Rate=10% , 5 % half yearly

Time (n)=1 year, 2 half years

∴Amount$=P\left(1+\frac{R}{100}\right)^n$

$=2000\left(1+\frac{5}{100}\right)^{2}$

$=2000\left[1+\frac{1}{20}\right]^{2}$

$=2000 \times \left(\frac{21}{20}\right)^{2}$

$=2000 \times \frac{21}{20} \times \frac{21}{20}$

=50×21×21=2205

∴Compound Interest=A-P

=2205-2000=205...(ii)

equation (ii)-(i)

=205-200=5

$=40000 \times \frac{21}{20}\times \frac{11}{10}\times \frac{21}{20}$

=10×21×11×23

=13130

Question 8

Flow much will ₹ 25000 amount to in 2 years at compound interest, if the rates for the successive years be 4 and 5 per cent per year ?

Sol :

Principal=25000

Time=2 years

First year Rate (R₁)=4%

and Second year rate (R₂)=5%

∴Amount$=P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)$

$=25000 \times \left(1+\frac{4}{100}\right) \times \left(1+\frac{5}{100} \right)$

$=25000 \times \left(1+\frac{1}{25}\right) \times \left(1+\frac{1}{20} \right)$

$=25000 \times \left(\frac{26}{25}\right) \times \left(\frac{21}{20} \right)$

=50×26×21=27300

Question 9

Umesh set up a small factory by investing ₹ 40,000. During the first three successive years his profits were 5%, 10% and 15% respectively. If each year the profit was on previous year’s capital, calculate his total profit.

Sol :

Under Investment (P)=40000

Time (n)=3 years

First year rate (r₁)=5%

Second year rate (r₂)=10%

Third year rate (r₃)=15%

∴Total amount after 3 year

$=P\left[1+\frac{r_{1}}{1000}\right] *\left[1+\frac{r_{2}}{100}\right] \times\left[1+\frac{r_{3}}{100}\right]$

$=40000\left(1+\frac{5}{100}\right).\left(1+\frac{10}{100}\right) \cdot\left(1+\frac{15}{100}\right)$

Question 10

Himachal Pradesh State Electricity Board issued in July 1988, 20 year bonds worth ₹ 6.25 crore. The issue price of each bond is ₹ 100 and it carries an annual interest of 11.5%, compounded half-yearly. Jasbir invested ₹ 5000 in these bonds. Find the amount that he gets on maturity of the bonds in 2008.

[Given that (1.0575)^{40} = 9.35869]

Sol :

Given that $(1.0575)^40$=9.35869 is amount of total bond=6.25 

and Price of each bond=100

Rate=11.5%, 5.75 half yearly

Jasbir's Investment (P)=5000

Time = 2008-1988=20 years, 40 half yearly

∴After 40 half year amount

$=P\left[1+\frac{r}{100}\right]^n$

$=500\left[1+\frac{5.75}{100}\right]^{40}$

$=500(1+1.0575)^{40}$

=500×9.35861

=46793.45

Value of the maturity bond =46793.45

Question 11

A district contains 64000 inhabitants. If the population increases at the rate of $2\frac{1}{2}/%$ per annum , find the number of inhabitants at the end of 3 years.

Sol :

District Population (P)=6400

Rate=$2\frac{1}{2}\%=\frac{5}{2}\%$

Time (n)=3 year

∴After 3 year population $=P\left[1+\frac{R}{100}\right]^n$

$=64000\left[1+\frac{5}{2 \times 100}\right)^{3}$

$=64000 \left(1+\frac{1}{40}\right)^{3}$

$=64000\left(\frac{41}{40}\right)^{3}$

$=64000 \times \frac{41}{40}\times \frac{41}{40}\times \frac{41}{40}$

=41×41×41=68921

Question 12

The Nagar Palika of a certain city started campaign to kill stray dogs which numbered 1250 in the city. As a result, the population of stray dogs started decreasing at the rate of 20% per month. Calculate the number of stray dogs in the city three months after the campaign started.

Sol :

Principal=1250

Rate=20% / month

Time (n) =3  month

∴After 3 month no. of stray dog 

$=P\left[1-\frac{R}{100}\right]^n$

$=1250\left[1-\frac{20}{100}\right]^3$

$=1250\left[\frac{4}{5}\right]^3$

$=1250 \times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}$

=640

Question 13

8000 blood donors were registered with a charitable hospital. Some student organizations started mobilizing people for this noble cause. As a result, the number of donors registered, increased at the rate of 20% per half year. Find the total number of new registrants during $1\frac{1}{2}$ year

Sol :

No. of blood donors in the beginning (P)=8000

Rate=20% per half year

Period (n)$=1\frac{1}{2}$ year, 3 half year

∴After 3 half year Increased donors 

$=P\left[1+\frac{R}{100}\right]^n$

$=8000\left[1+\frac{20}{100}\right]^3$

$=8000 \times \left[1+\frac{1}{5}\right]^3$

$=8000 \times \frac{6}{5}\times \frac{6}{5}\times \frac{6}{5}$

=13824

∴Net Increase in donor=A-P

=13824-8000=5824