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 :
∴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 :
∴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$=P\left(1+\frac{R}{100}\right)^n$
$=8000\left(1+\frac{1}{20}\right)^{3}$
$=8000 \times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}$
=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 :
∴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 :
Time (n) $=1\frac{1}{2}$ years, 3 half year
∴Amount$=P\left(1+\frac{R}{100}\right)^n$
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 :
$=\frac{2000 \times 10 \times 1}{100}$
=200...(i)
In case of Prasant
Principal (P)=2000
Rate=10% , 5 % half yearly
Time (n)=1 year, 2 half years
∴Amount$=P\left(1+\frac{R}{100}\right)^n$
$=2000\left(1+\frac{5}{100}\right)^{2}$
$=2000\left[1+\frac{1}{20}\right]^{2}$
$=2000 \times \left(\frac{21}{20}\right)^{2}$
$=2000 \times \frac{21}{20} \times \frac{21}{20}$
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 :
∴Total amount after 3 year
$=P\left[1+\frac{r_{1}}{1000}\right] *\left[1+\frac{r_{2}}{100}\right] \times\left[1+\frac{r_{3}}{100}\right]$
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.
Sol :
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 :
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 :
∴After 3 month no. of stray dog
$=P\left[1-\frac{R}{100}\right]^n$
$=1250\left[1-\frac{20}{100}\right]^3$
$=1250\left[\frac{4}{5}\right]^3$
$=1250 \times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}$
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 :
∴After 3 half year Increased donors
$=P\left[1+\frac{R}{100}\right]^n$
$=8000\left[1+\frac{20}{100}\right]^3$
$=8000 \times \left[1+\frac{1}{5}\right]^3$
$=8000 \times \frac{6}{5}\times \frac{6}{5}\times \frac{6}{5}$
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