Exercise 2(A)
Question 1
Principal = ₹ 10000, Rate % p.a = 12 %, Number of years = 2.
Sol :
Principal=1000
Rate=12%
First year Interest$=\frac{PRT}{100}$
$\Rightarrow \frac{10000 \times 12 \times 1}{100}$
=100×12×1=1200
After first year Amount=P+S.I
=10000+1200=11200
∴Second year principal=11200
Second year Interest$=\frac{\text{Second PRT}}{100}$
$=\frac{11200\times 12 \times 1}{100}$
=112×12=1334
∴Interest for 2 year=1200+1344=2544
Hence Compound Interest=2544
Question 2
Principal = ₹ 50000, Rate % p.a = 10 %, Number of years = 2.
Sol :
Principal (P)=5000
Rate=10%
Time=2 year
First Year Interest$=\frac{PRT}{100}$
$=\frac{5000 \times 10 \times 1}{100}$
=50×10×1=500
∴After First year amount=P+S.I
=5000+500=5500
∴Second year Principal=5500
Second year Interest$=\frac{\text{Second year PRT}}{100}$
$=\frac{5500 \times 10 \times 1}{100}$
=550
∴Interest for 2 year =500+550
=1050
Hence Compound Interest for 2 year
=1050
Question 3
Principal = ₹ 2800, Rate % p.a = 10 %, Number of years $=1\frac{1}{2}$
Sol :
Principal=2800
Rate=10%
Time=$1 \frac{1}{2}$ years
First Year Interest$=\frac{PRT}{100}$
$=\frac{2800 \times 10 \times 1}{100}$
=28×10×1=280
∴After first year Amount=P+S.I.
=2800+280
=3080
Second Year Principal=3080
Next 6 month Interest $\left(\frac{1}{2}\right)$ year
$=\frac{\text{Second year }P \times R \times T}{100}$
$\frac{3080 \times 10 \times 1}{100 \times 2}$
=154
∴Total Interest for $1 \frac{1}{2}$ years
=280+154=434
Hence Compound Interest for $1\frac{1}{2}$ years
=434
Question 4
Principal = ₹ 2000, Rate % p.a = 20 %, Number of years = 2.
Sol :
Principal=2000
Rate=20%
Time=2 years
∴First Year Interest$=\frac{PRT}{100}$
$=\frac{2000 \times 20 \times 1}{100}$
=20×20×1=400
After One year amount=P+S.I
=2000+400=2400
Second year Principal=2400
Second year Interest$=\frac{\text{Second year } P\times R\times T}{100}$
$=\frac{2400 \times 20 \times 1}{100}$
=24×20×1=480
∴Total Interest for 2 year
=400+480=880
Compound Interest for 2 year=880
Question 5
Principal = ₹ 20480, Rate % p.a $=6\frac{1}{4}/%$ , Number of years = 2 years 73 days.
Sol :
Principal=20480
Rate$=6\frac{1}{4}%=\frac{25}{4}\%$ p.a.
Time=2 years 73 days
=2 years $\frac{73}{365}$
$=2\frac{1}{5}$ years
First Year Interest$\frac{PRT}{100}$
$=\frac{20480\times 25 \times 1}{100 \times 4}$
=1280
After One year amount=P+S.I.
=20480+1280
=21760
Second Year Principal=21760
Second Year Interest$=\frac{\text{Second Year}P\times R\times T}{100}$
$=\frac{21760 \times 25 \times 1}{4\times 100}$
=1280
After One Year amount=P+S.I
=20480+1280
=21760
Second year Principal=21760
Second Year Interest$=\frac{\text{Second year}P\times R\times T}{100}$
$=\frac{21760 \times 25 \times 1}{4 \times 100}$
=1360
After 2 year Amount=21760+1360
=23120
Next $\frac{1}{5}$ year principal=23120
Interest for $\frac{1}{5}$ year$=\frac{23120 \times 25 \times 1}{100 \times 4 \times 5}$
=289
Total Interest for $2\frac{1}{5}$ year
=1280+1360+289=2929
Question 6
Find the amount and compound interest on a sum of ₹ 15625 at 4% per annum for 3 years compounded annually.
Sol :
Principal=15625
Rate=4%
Time=3 years
First Year Interest$=\frac{PRT}{100}$
$=\frac{15625 \times 4 \times 1}{100}$
=625
∴After One Year Amount=P+S.I
=15625+625=16250
Second Year Principal=16250
Second Year Interest$=\frac{\text{P of the second year} \times R \times T}{100}$
$=\frac{10250 \times 4 \times 1}{100}$
=650
After Second year amount=16250+650
=16900
Third Year Principal$=\frac{16900 \times 4 \times 1}{100}$
=169×4×1=676
∴After Third year amount=16900+676
=17576
Compound Interest=A-P
=17576-15625=1951
Question 7
To renovate his shop, Anurag obtained a loan of T8000 from a bank. If the rate of interest at 5% per annum is compounded annually, calculate the compound interest that Anurag will have to pay after 3 years.
Sol :
Amount of loan (P)=8000
Rate=5%
Time (n)=3 years
∴First Year Interest$=\frac{PRT}{100}$
$=\frac{8000 \times 5 \times 1}{100}$
=80×5×1=400
∴After One Year amount=P+S.I
=8000+400=8400
Second Year Principal=8400
Second Year Interest$=\frac{\text{Second Year}P\times R \times T}{100}$
$=\frac{8400 \times 5 \times 1}{100}$
=420
After two year amount=8400+420=8820
Third year Principal=8820
Third Year Interest$=\frac{8820 \times 5 \times 1}{100}$
=441
∴After 3 year amount=8820+441=9261
Compound Interest=A-P
=9261-8000
=1261
Question 8
Maria invests ₹ 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate.
(i) the amount standing to her credit at the end of the second year.
(ii) the interest for the 3rd year.
Sol :
Maria's investment (P)=93750
Rate=9.6%
Time (n)= 3 year
First Year Interest$\frac{P R T}{150}$
$=\frac{93750 \times 9.6 \times 1}{1000}$
$=\frac{93750 \times 96 \times 1}{1000}$
=9000
∴After One Year Amount=P+S.I
=93750+9000=102750
Second Year Principal=102750
Second Year Interest$=\frac{\text{Second Year}P \times R \times T}{100}$
$=\frac{102750 \times 9.6 \times 1}{1000}$
$=\frac{102750 \times 96}{100}$
=9864
(i) After 2 year Amount=102750+9864
=112614
Third Year Principal=112614
Third Year Interest$=\frac{112614 \times 9.8 \times 1}{1000}$
$=\frac{112614 \times 98 \times 1}{1000}$
=10810.94
(ii) ∴After Third Year Amount
=112614+10810.94
=123424.94
Question 9
A sum of ₹ 9,600 is invested for 3 years at 10% p.a. 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 the first 2 years.
(iv) Find the compound interest at the end of 3 years.
Sol :
Principal=9600
Rate=10%
Time=3 year
First Year Interest$=\frac{ P \times R\times T}{100}$
$=\frac{9600 \times 10 \times 1}{100}$
=960
(i) ∴After First Year Amount=P+S.I
=9600+960=10560
(ii) Second Year Principal=10560
Second Year Interest$=\frac{10560 \times 10 \times 1}{100}$
=1056
After Second year Amount=10560+1056
=11616
(iii) Two Year Compound Interest
=960+1056
=2016
Principal of Third Year=11616
Third Year Interest$=\frac{11616 \times 10 \times 1}{100}$
=1161.60
(iv) After the end of third year Compound Interest
=2016+1161.60=3177.60
Question 10
Shanker takes a loan of ₹ 10,000 at a compound interest rate of 10% per annum (p.a.)
(i) Find the compound interest after one year.
(ii) Find the compound interest for 2 years.
(iii) Find the sum of money required to clear the debt at the end of 2 years.
(iv) Find the difference between the compound interest and the simple interest at the same rate for 2 years.
Sol :
Amount of loan (P)=10000
Rate=10%
Time (n)=2 year
(i) First Year Interest$=\frac{P \times R \times T}{100}$
$=\frac{10000 \times 10 \times 1}{100}$
=1000
After One Year Amount=P+S.I.
=10000+1000=11000
Second Year Principal=11000
Second Year Interest$=\frac{\text{Second year} P \times R \times T}{100}$
$=\frac{11000 \times 10 \times 1}{100}$
=1100
(ii) ∴2 Year Compound Interest=1000+1100
=2100
(iii) After 2 Year Amount=11000+1100
=12100
(iv) Simple Interest For 2 Year
$=\frac{1000 \times 10 \times 2}{100}$
=2000
2 year difference between C.I. and S.I
=2100-2000
=100
Question 11
Find compound interest on ₹ 5000 at 12% p.a. for 1 year, compounded half yearly.
Sol :
Principal=5000
Rate=12% or 6% half yearly
Time=1 year or 2 half year
First Half Year Amount=P+S.I.
$=\frac{5000 \times 6 \times 1}{100}$
=300
After One Year Principal=5300
=5000+300=5300
Second Half Year Principal=5300
Second Half Year Interest
$=\frac{5300\times 6 \times 1}{100}$
=318
∴After 2 Half Year Amount
=5300+318=5618
Compound Interest For 2 Half Year
=A-P
=5618-5000
=618
Question 12
Find the amount and the compound interest on ₹ 16000 for $1\frac{1}{2}$ years at 10% p.a. the interest being compounded half-yearly.
Sol :
Principal=1600
Rate=10% or 5% half yearly
Time=$1 \frac{1}{2}$ year or 3 half years
First Half Year Interest$=\frac{P \times R \times T}{100}$
$=\frac{16000 \times 5 \times 1}{100}$
=800
After First Half Year Amount=P+S.I
=16000+800
=16800
Second Half Year Principal=16800
Second Half Year Interest$\frac{16000 \times 5 \times 1}{100}$
=840
After 2 Half Year Amount=16800+840=17640
Third Half Year Principal=17640
Third Half Year Interest
$=\frac{17640 \times 5 \times 1}{100}$
=882
After Third Year Compound Interest=A-P
=18522-16000=2522
Question 13
Calculate the amount due and the compound interest on ₹ 40000 for 2 years when the rate of interest successive years is 7% and 8% respectively.
Sol :
Principal=40000
Time=2 year
First Year Rate of Interest (R1)=7%
Second Year Rate of Interest (R2)=8%
∴First Year Interest$=\frac{P R T}{100}$
$=\frac{4000 \times 7 \times 1}{100}$
=2800
After First Year Amount=P+S.I.
=40000+2800=42800
Second Year Principal =42800
Second Year Interest$=\frac{42800 \times 8 \times 1}{100}$
=3424
∴After Two Year Amount=42800+3424
=46224
2 Year Compound Interest=A-P
=46224-40000
=6224
Question 14
If the simple interest on a sum of money for 2 years at 5% per annum is ₹ 50, what will be the compound interest on the same sum at the same rate for the same time.
Sol :
Simple Interest For 2 Year
T=2 Year
R=5%
∵We know that S.I. and C.I. is same for the first year
∴First year simple interest$=\frac{50}{2}$
=25
Second year simple interest=25
∴Principal$=\frac{S.I. \times 100}{R\times T}$
$=\frac{50 \times 100}{5 \times 2}$
=500
After First Year Amount=P+S.I.
=500+25
=525
Second Year Principal=525
Second Year Interest$=\frac{525 \times 5 \times 1}{100}$
$=\frac{2625}{100}$
=26.25
2 Year Compound Interest
=25+26.25
=51.25
Question 15
A man invests ₹ 46,875 at 4% per annum compound interest for 3 years. Calculate :
(i) the interest for the 1st year;
(ii) the amount standing to his credit at the end of the 2nd year;
(iii) the interest for the 3rd year.
Sol :
Investment (P)=46875
Rate=4%
Time=3 year
(i) First Year Interest$=\frac{PRT}{100}$
$=\frac{46875 \times 4 \times 1}{100}$
=1875
(ii) After One Year Amount=P+S.I.
=46875+1875=48750
Second Year Principal=48750
Second Year Interest$=\frac{42750 \times 4 \times 1}{100}$
=1950
After Second Year Amount=48750+1950
=50700
(iii) Third Year Principal=50700
Third Year Interest$=\frac{50700 \times 4 \times 1}{100}$
=2028
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