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 :
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 :
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 :
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 :
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 :
=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 :
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 :
∴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 :
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 :
=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 :
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 :
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 :
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
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