Exercise 2.1
Question 1
Find the amount and the compound interest on ₹ 8000 at 5% per annum for 2 years.
Sol :
Principal for the first year=8000
Interest for the first year $\frac{₹ 8,000 \times 5 \times 1}{100}$
=400
Amount after one year=8000+400=8400
Interest for the second year $\frac{8400 \times 5 \times 1}{100}$
=420
Amount after second year=8400+420
=8820
Compound interest after two years=final amount-principal
=8820-8000
=820
Question 2
A man invests ₹ 46875 at 4% per annum compound interest for 3 years. Calculate:
Sol :
Principal for the first year =₹ 46,875
Rate of interest=4%
(i) Interest for the first year$\frac{46,875 \times 4 \times 1}{100}$
=1875
Amount after one year=46875+1875
=48750
(ii) Interest for the second year $\frac{48,750 \times 4 \times 1}{100}$
=1950
The amount standing to his credit at the end of
Second year=48750+1950
=50700
(iii) Interest for the third year$\frac{50,700 \times 4 \times 1}{100}$
=5007×4
=2028
Question 3
Calculate the compound interest for the second year on ₹ 8000 for three years at 10% p.a.Also find the sum due at the end of third year.
Sol :
Principal for the first year =28,000
Rate of Interest =10 %
Interest for the first year $\frac{8,000 \times 10 \times 1}{100}$
=800
Amount after one year=8000+800
=8800
Interest for the second year$\frac{8800 \times 10 \times 1}{100}$
=880
Amount after second year=8800+880
=9680
Compound interest for the second year=800+880
=1680
Interest for the third year$\frac{9,680 \times 10 \times 1}{100}$
=968
Sum due at the end of the third year
=9680+968
=10648
Question 4
Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest.
Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of three years.
Sol :
Rate of interest =10 %
Interest for the first year
$\frac{12,800 \times 10 \times 1}{100}$
=₹ 1,280
(i) Amount at the end of first year=12800+1280
=14080
(ii) Interest for the second year$\frac{14,080 \times 10 \times 1}{100}$
=1408
Amount at the end of second year=14080+1408
=15488
Interest for the third year$\frac{15,488 \times 10 \times 1}{100}$
=1548.8
(iii) The total amount due to him at the end of three years
=15488+1548.8
=17036.5
Question 5
The simple interest on a sum of money for 2 years at 12% per annum is ₹ 1380. Find:
(i) the sum of money.
(ii) the compound interest on this sum for one year payable half-yearly at the same rate.
Sol :
$\frac{x \times 2 \times 12}{100}=1380$
24x=13800
Sum of money$\frac{1,38,000}{24}$
=5,750
Principal for the first half year=5750
Interest for the first half year$\frac{5,750 \times 6 \times 1}{100}$
=57.5×6
=345
Amount after the first half year=5750+345
=6095
Interest for the second half year$\frac{6095 \times 6 \times 1}{100}$
=60.95×6
=365.7
∴Compound Interest on this Sum for one year payable half yearly at the same rate
=345+365.7
=710.70
Question 6
A person invests ₹ 10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹ 11200. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of second year.
Sol :
Amount after one year=11200
Interest for the first year=11200-1000
=10200
Rate of Interest for the first year=?
Interest for the first year$=\frac{10,000 \times x \times 1}{100}$
$\frac{1200\times 100}{10000}=x$
x=12%
Interest for the second year$=\frac{11200\times 12\times 1}{100}$
=112×12
=1344
Amount at the end of second year=11200+1344
=12544
Question 7
Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate
(i) the rate of interest.
(ii) the amount at the end of second year, to the nearest rupee.
Sol :
Amount after one year=5600
Interest for first year=5600-5000
$\frac{5000\times x\times 1}{100}=600$
x=12
(i) Rate of Interest=12%
(ii) Interest acured in the second year$\frac{5,600 \times 12 \times 1}{100}$
=672
Amount after second year=5600+672
=6272
(iii) Interest for the first third year$=\frac{6272 \times 12 \times 1}{100}$
=62.72×12
=752.64
Amount after third year=62.72+752.64
=7024.64
Question 8
A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 5600. Calculate:
(i) the rate of interest per annum.
(ii) the interest accrued in the second year.
(iii) the amount at the end of the third year.
Sol :
=220
Amount after second year=2200+220
=2420
Interest for the second half $2\frac{1}{2}$ year
$=\frac{2420\times 5\times 1}{100}$
=121
∴Compound Interest after $2\frac{1}{2}$ year=200+220+121
=541
Amount to be paid after $2\frac{1}{2}$ year=2000+541
=2541
PAGE-49
Question 9
Find the amount and the compound interest on ₹ 2000 at 10% p.a. for 2 years, compounded annually.
Sol :
Interest for the first half year $\frac{50,000 \times 4 \times 1}{100}$
=2000
Amount after the first half year
=50000+2000
=52000
Principal after the second half year=52000
Interest for the second half year$\frac{52000 \times 4 \times 1}{100}$
=520×4
=2080
Amount after the second half year=52000+2080
=54080
Principal for third year=54080
Question 10
Find the amount and the compound interest on ₹ 50000 for $1\dfrac{1}{2}$ years at 8% per annum, the interest being compounded semi-annually.
Sol :
Given that A=9261, n=3 and R=5
$A=P\left(1+\frac{r}{100}\right)^{n}$
$9261=P\left(1+\frac{5}{100}\right)^{3}$
$9261=P\left(1+\frac{1}{20}\right)^{3}$
$9261=P\left(\frac{21}{20}\right)^{3}$
$\frac{9261 \times 20 \times 20 \times 20}{21 \times 21 \times 21}=P$
$\therefore P=\frac{9261 \times 8000}{9261}$
∴Principal=8000
Question 11
Calculate the amount and the compound interest on ₹ 5000 in 2 years when the rate of interest for successive years is 6% and 8%, respectively.
Sol :
∴$A=P\left(1+\frac{r}{100}\right)^{n}$
$7803=P\left(1+\frac{2}{100}\right)^{2}$
$7803=\frac{P \times 51 \times 51}{50 \times 50}$
$P=\frac{19507,500}{2500}$
∴P=7500
Interest for the third half year$\frac{54080 \times 4 \times 1}{100}$
=540.8×4
=2163.2
Amount after third half year=54080+2163.2
=56.243.20
∴Compound Interest for $1 \frac{1}{2}$ years
=56,243-50,000
=6243
Question 12
Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively.
Sol :
Interest for the first year$\frac{5,000 \times 6 \times 1}{100}$
=50×6
=300
Amount after first year=5000+300
=5300
Interest for the second year$\frac{5,300 \times 8 \times 1}{100}$
=53×8
=424
∴Amount after second year=5300+424
=5724
Compound interest for 2 years=5724-5000
=724
Question 13
A sum of ₹ 9600 is invested for 3 years at 10% per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in 2 years.
(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.
(v) Hence, write down the compound interest for the third year.
Sol :
It is given that
Principal = ₹ 9600
Rate of interest = 10% p.a.
Period = 3 years
We know that
Interest for the first year $=\dfrac{\text{P R T }}{100}$
Substituting the values
$=\dfrac{ (9600 × 10 × 1)}{100}$
= ₹ 960
(i) Amount after one year = 9600 – 960 = ₹ 10560
So the principal for the second year = ₹ 10560
Here the interest for the second year $=\dfrac{ (10560 × 10 × 1)}{100}$
= ₹ 1056
(ii) Amount after two years = 10560 + 1056 = ₹ 11616
(iii) Compound interest earned in 2 years = 960 + 10560 = ₹ 2016
(iv) Difference between the answers in (ii) and (i) = 11616 – 10560 = ₹ 1056
We know that
Interest on ₹ 1056 for 1 year at the rate of 10% p.a. $= \dfrac{(1056 × 10 × 1)}{100}$
= ₹ 105.60
(v) Here
Principal for the third year = ₹ 11616
So the interest for the third year $= \dfrac{(11616 × 10 × 1)}{ 100}$
= ₹ 1161.60
Question 14
The simple interest on a certain sum of money for 2 years at 10% p.a. is ₹ 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.
Sol :
It is given that
Period = 2 years
Rate = 10% p.a.
We know that
Sum $= \dfrac{(SI × 100)}{(r × n)}$
Substituting the values
$= \dfrac{(1600 × 100)}{(10 × 2)}$
= ₹ 8000
Here
Amount after 3 years $= P \left(1 + \dfrac{r}{100\times n}\right)^{nT}$
Substituting the values
$= 8000 \left(1 + \dfrac{10}{100}\right)^3$
By further calculation
$= 8000 × \dfrac{11}{10} × \dfrac{11}{10}× \dfrac{11}{10}$
= ₹ 10648
So the compound interest = A – P
Substituting the values
= 10648 – 8000
= ₹ 2648
Question 15
Vikram borrowed ₹ 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after $2\dfrac{1}{2}$ years.
Sol:
First case-
Principal = ₹ 20000
Rate = 10% p.a.
Period $= 2\dfrac{1}{2}= \dfrac{5}{2}$ years
We know that
Simple interest $= \dfrac{PRT}{100}$
Substituting the values
$= \dfrac{(20000 × 10 × 5)}{ (100 × 2)}$
= ₹ 5000
Second case-
Principal = ₹ 20000
Rate = 10% p.a.
Period $= 2\dfrac{1}{2}$ years at compound interest
We know that
Amount $=P \left(1 + \dfrac{r}{100n}\right)^{nT}$
Substituting the values
$= 20000 \left(1 + \dfrac{10}{100}\right)2 \left(1 + \dfrac{10}{(2 × 100)}\right)2$
By further calculation
$= 20000 × \dfrac{11}{10} × \dfrac{11}{10}× \dfrac{21}{20}$
= ₹ 25410
Here
Compound Interest = A – P
Substituting the values
= 25410 – 20000
= ₹ 5410
So his gain after 2 years = CI – SI
We get
= 5410 – 5000
= ₹ 410
Question 16
A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the amount outstanding at the beginning of the third year.
Sol :
It is given that
Principal = ₹ 6000
Rate of interest = 5% p.a.
We know that
Interest for the first year = Prt/100
Substituting the values
= (6000 × 5 × 1)/ 100
= ₹ 300
So the amount after one year = 6000 + 300 = ₹ 6300
Principal for the second year = ₹ 6300
Amount paid = ₹ 1200
So the balance = 6300 – 1200 = ₹ 5100
Here
Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255
Amount for the second year = 5100 + 255 = ₹ 5355
Amount paid = ₹ 1200
So the balance = 5355 – 1200 = ₹ 4155
Question 17
Mr. Dubey borrows ₹ 100000 from State Bank of India at 11% per annum compound interest. He repays ₹ 41000 at the end of first year and ₹ 47700 at the end of second year. Find the amount outstanding at the beginning of the third year.
Sol :
It is given that
Borrowed money (P) = ₹ 100000
Rate = 11% p.a.
Time = 1 year
We know that
Amount after first year = Prt/100
Substituting the values
= (100000 × 11 × 1)/ 100
By further calculation
= 100000 + 11000
= ₹ 111000
Amount paid at the end of the first year = ₹ 41000
So the principal for the second year = 111000 – 41000
= ₹ 70000
We know that
Amount after second year = P + (70000 × 11)/ 100
By further calculation
= 70000 + 700
= ₹ 77700
So the amount paid at the end of the second year = ₹ 47700
Here the amount outstanding at the beginning year = 77700 – 47700
= ₹ 30000
Question 18
Jaya borrowed ₹ 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.
Sol :
It is given that
Amount borrowed by Jaya = ₹ 50000
Period (n) = 2 years
Rate of interest for two successive years are 12% and 15%, respectively
We know that
Interest for the first year = Prt/100
Substituting the values
= (50000 × 12 × 1)/ 100
= ₹ 6000
So the amount after first year = 50000 + 6000 = ₹ 56000
Amount repaid = ₹ 33000
Here
Balance amount for the second year = 56000 – 33000 = ₹ 23000
Rate = 15%
So the interest for the second year = (230000 × 15 × 1)/ 100
= ₹ 3450
Amount paid after second year = 23000 + 3450 = ₹ 26450
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