ML Aggarwal Solution Class 9 Chapter 2 Compound Interest Exercise 2.1

 Exercise 2.1

Question 1

Sol :

Principal for the first year=8000

Interest for the first year 8,000×5×1100

=400

Amount after one year=8000+400=8400

Interest for the second year 8400×5×1100

=420

Amount after second year=8400+420

=8820

Compound interest after two years=final amount-principal

=8820-8000

=820


Question 2

Sol :

Principal for the first year =₹ 46,875

Rate of interest=4%

(i) Interest for the first year46,875×4×1100

=1875

Amount after one year=46875+1875

=48750


(ii) Interest for the second year 48,750×4×1100

=1950

The amount standing to his credit at the end of

Second year=48750+1950

=50700


(iii) Interest for the third year50,700×4×1100

=5007×4

=2028


Question 3

Sol :

Principal for the first year =28,000

Rate of Interest =10 %

Interest for the first year 8,000×10×1100

=800

Amount after one year=8000+800

=8800

Interest for the second year8800×10×1100

=880

Amount after second year=8800+880

=9680

Compound interest for the second year=800+880

=1680

Interest for the third year9,680×10×1100

=968

Sum due at the end of the third year

=9680+968

=10648


Question 4

Sol :

Principal for the first year =₹ 12,800

Rate of interest =10 %

Interest for the first year

12,800×10×1100

=₹ 1,280

(i) Amount at the end of first year=12800+1280

=14080

(ii) Interest for the second year14,080×10×1100

=1408

Amount at the end of second year=14080+1408

=15488

Interest for the third year15,488×10×1100

=1548.8

(iii) The total amount due to him at the end of three years

=15488+1548.8

=17036.5


Question 5

Sol :

Interest for 2 years =1380

x×2×12100=1380

24x=13800

Sum of money1,38,00024

=5,750

Principal for the first half year=5750

Interest for the first half year5,750×6×1100

=57.5×6

=345

Amount after the first half year=5750+345

=6095

Interest for the second half year6095×6×1100

=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

Sol :

Principal for the first year =10,000

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=10,000×x×1100

1200×10010000=x

x=12%

Interest for the second year=11200×12×1100

=112×12

=1344

Amount at the end of second year=11200+1344

=12544


Question 7

Sol :

Principal for the first year=5000

Amount after one year=5600

Interest for first year=5600-5000

5000×x×1100=600

x=12

(i) Rate of Interest=12%

(ii) Interest acured in the second year5,600×12×1100

=672

Amount after second year=5600+672

=6272

(iii) Interest for the first third year=6272×12×1100

=62.72×12

=752.64

Amount after third year=62.72+752.64

=7024.64


Question 8

Sol :

Principal for the first year =₹ 2,000

Interest for the first year2,000×10100
=2,000×220
=200

Amount after first year=2000+200
=2200

Interest for the second year=2200×10×1100

=220

Amount after second year=2200+220

=2420

Interest for the second half 212 year

=2420×5×1100

=121

∴Compound Interest after 212 year=200+220+121

=541

Amount to be paid after 212 year=2000+541

=2541


Question 9

Sol :

Principal for the first half year =50,000

Interest for the first half year 50,000×4×1100

=2000

Amount after the first half year

=50000+2000

=52000

Principal after the second half year=52000

Interest for the second half year52000×4×1100

=520×4

=2080

Amount after the second half year=52000+2080

=54080

Principal for third year=54080


Question 10

Sol :

Given that A=9261, n=3 and R=5

A=P(1+r100)n

9261=P(1+5100)3

9261=P(1+120)3

9261=P(2120)3

9261×20×20×2021×21×21=P

P=9261×80009261

∴Principal=8000


Question 11

Sol :

Given that A=7803, n=2, r=2

A=P(1+r100)n

7803=P(1+2100)2

7803=P×51×5150×50

P=19507,5002500

∴P=7500

Interest for the third half year54080×4×1100

=540.8×4

=2163.2

Amount after third half year=54080+2163.2

=56.243.20

∴Compound Interest for 112 years 

=56,243-50,000

=6243


Question 10

Sol :

Principal amount for the first year=5000

Interest for the first year5,000×6×1100

=50×6

=300

Amount after first year=5000+300

=5300

Interest for the second year5,300×8×1100

=53×8

=424

∴Amount after second year=5300+424

=5724

Compound interest for 2 years=5724-5000

=724


Question 11

Sol :

Principal for the first half year=17000

Interest for the first year at 10%=17,000×10×1100

=1700

Amount after first year=17000+1700

=18700

Interest for the second year at 10% 18,700×10×1100

=1870

Amount after second year

=18700+1870

=20570

Interest for the third year at 14%

=20,570×14×1100

=2879.8

Amount after third year=20570+2879.8

=23449.8

Compound Interest after three year=6449.80


Question 12

Sol :

Principal for the first year=9600

Interest for the first year9,600×10×1100

=960

(i) Sum due at the end of first years=9600+960

=10560

Interest for the second year10,560×10×1100

=1056

(ii) Sum due at the end of second year=10560+1056

=11626

(iii) Compound Interest Earned n 2 years

=11626-9600

=2026

(iv) Difference of (ii) and (i)=11626-10560

=1056

Interest for difference of amount for one year

=1,056×10×1100

=105.60

compound Interest for third year=1056+105.6

=1161.60


Question 13

Sol :

Simple interest on certain money for 2 years at 10% is 1600

x×10×2100=1,600

x=8000

Principal=8000

Interest for the first year8,000×10×1100

=800

Amount to be paid after one year=8000+800

=8800

Interest for the second year8800×10×1100

=880010

=9680

Interest for the third year9,680×10×1100

=968

Amount after three year =9680+968

=10648

Compound Interest for three year=800+880+968

=2648


Question 14

Sol :

Principal for the first year =6000

Interest for the first year6,000×5×1100

=300

Amount after first year=6300

Money repaid=1200

Principal for the second year=6300-1200

=5100

Interest for the Second year5,100×5×1100

=255

Principal amount after second year=5100+255

=5355

Amount outstanding at the beginning of third year

=5355-1200

=4155


Question 15

Sol :

Principal for the first year=100000

Interest for the first year=1000000×11×1100

=11000

Amount after one year=100000+11000

=111000

Money repaid after one year=41000

Principal for second year=111000-41000

=70000

Interest for the second year70,000×11×1100

=7700

Amount after second year=70000+7700

=77700

Principal for third year=77700-47700

=30000

Amount=30000


Question 16

Sol :

Simple Interest =PNR100

=20000×2.5×10100

=200×2.5

=5000

Interest for the first year=20000×1×10100

=2000

Amount after one year=20000+2000

=22000

Interest after the second year22,000×10×1100

=2200

Amount after second year=22000+2200

=24200

Interest for second year 6 months=24200×5100

=1210

∴Total amount=24200+1210

=25410

∴Gain=25410-25000

=410

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2