ML Aggarwal Solutions Class 9 Math 2nd Chapter Compound Interest Exercise 2.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Second Chapter Compound Interest Exercise 2.1. APC Solution Class 9 Exercise 2.1.
(1) Find the amount and the compound interest on ₹8000 at 5% per annum for 2 years.
Solution:
Given, Principal = ₹8000
Rate of interest = 5%
Time = 2 years
∴ Interest after 1 year = PRT/100
= 8000×5×1/100
= ₹400
Amount after 1 year = 8000 + 400
= ₹8400
Interest after 2nd year = 8400×5×1/100
= ₹440
∴ Amount after 2 years = 8400 + 440
= ₹8840
∴ Total compound interest for 2 years = 8840 – 8000
= ₹840
(2) A man invests ₹46875 at 4% per annum compound interest for 3 years. Calculate:
(i) The interest for the first year
(ii) The amount standing to his credit at the end of the second year
(iii) The interest for the third year.
Solution:
Given principal = ₹46875
Rate of interest = 4%
Time = 3 years
(i) ∴ interest for 1 year = 46875×4×1/100
= ₹1875
Amount after 1year = 46875 + 1875
= ₹48750
Interest for 2nd year = 48750×4×1/100
= ₹1950
(ii) ∴ Amount after 2years = 48750 + 1950
= ₹50700
(iii) Interest for 3rd year = 50700×4×1/100
= ₹2028
(3) Calculate the compound interest for the second year on ₹8000 invested for years 10% p.a.
Also find the sum due at the end of the third year.
Solution:
Given principal = ₹8000
Rate of interest = 10%
Time = 3 years
∴ Interest for 1st year = 8000×10×1/100
= ₹800
∴ Amount at the end of 1 year = 8000 + 800
= ₹8800
Interest for 2nd year = 8800×10×1/100
= ₹880
∴ Amount at the end of 2nd year = ₹8800 + 880
= ₹9680
Interest for 3rd year = 9680×10×1/100
= ₹968
∴ Amount at the end of 3rd year = 9680+968
= ₹10648
∴ Compound interest for 2 years = 800+880 = ₹1680
Amount due at the end of 3 years = ₹10648
(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.
Solution:
Given Principal = ₹12800
Rate of interest = 10%
∴ Interest for 1st year = 12800×10×1/100
= ₹1280
(i) ∴ Sum due at the end of last year = 12800 + 1280
= ₹15080
(ii) Interest earned for 2nd year = 15080×10×1/100
= ₹1508
∴ Amount at the end of 2 years = 15080+1508
= ₹16588
∴ Interest for 3rd year = 16588×10×1/100
= ₹1658.8
(iii) ∴ Amount due at the end of 3rd year = 1658+1658.8
= ₹18246.8
(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.
Solution:
Given S.I on amount = ₹1360
rate of interest = 12%
Time = 2 years
(i) ∴ S.I = PRT/100
Or, 1380 = P×12×2/100
Or, P = 1380×100/12×2
= ₹5750
For compound interest half yearly on same amount, rate for 1/2 year = 12/2 = 6%
Interest for 1st year = 5750×6×1/100
= ₹345
Amount for 1st half year = 5750 + 345
= ₹6095
Interest for 2nd half year = 6095×6×1/100
= 36570/100
= ₹365.7
(ii) ∴ Total compound interest for 1 year = 345+365.7
= ₹710.7
(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
Solution:
Let, the rate of interest be x %
Given, Principal = ₹10000
Amount at the end of 1 year = ₹11200
Time = 2 years
(i) ∴ Interest for 1st year = 10000× x ×1/100
= ₹100 x
∴ Amount at the end of 1 year = 10000 + 100x
Or, 11200 = 10000 + 100x [Given]
Or, 11200 = 100 (100 + x)
Or, x = 112 – 100
= 12%
∴ Rate of interest is 12%
∴ Interest for 2nd year = 11200×12×1/100
= ₹1344
(ii) ∴ Amount at the end of 2 years = 11200 + 1344
= ₹12544
(7) Mr.Lalit invested ₹5000 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.
Solution:
Let, the rate of interest – be x %
Given, principal = ₹50000
Amount at the end of last year ₹5325
Time = 2 years
∴ Interest for last year = 5000×x×1/100
= ₹50x
(i) ∴ Amount at the end of 1st year = 5000 + 50x
Or, 5325 = 5000 + 50x [Given]
Or, 50x = 5325 – 5000
Or, x = 325/50
Or, x = 6.5%
∴The rate of interest is 6.5%
(ii) Interest for 2nd year = 5325×6.5×1/100
= 34612.5/100
= ₹346.125
∴ Amount at the end of 2nd year = 5325 + 346.125
= ₹5671.125
(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 the third year.
Solution:
Given, Principal = ₹5000
Amount at the end of 1st year = ₹5325
Let, the rate of interest be x %
∴ Interest for 1st year = 5000× x ×1/100
= ₹50x
(i) ∴ Amount at the end of 1st year = 5000 + 50x
Or, 5325 – 5000 = 50x [Given]
Or, x = 325/50
Or, x = 6.5%
∴ Rate of interest is 6.5%
(ii) Interest for 2nd year = 6325×6.5×1/100
= 34612.5/100
= ₹346.12
∴ Amount at the end of 2nd year = 5325 + 346.12
= ₹5671.12
∴ Interest for 3rd year = 5671.12×6.5×1/100
= 36862.28/100
= ₹368.62
(iii) ∴ Amount at the end of 3rd year = 5671.12 + 368.48
= ₹6039.4
(9) Find the amount and the compound interest on ₹2000 at 10% p.a. for 2 1/2 years, compounded annually.
Solution:
Given, Principal = ₹2000
Rate = 10%
Time = 2 1/2 years
∴ Interest for 1st year = 2000×10×1/100
= ₹200
∴ Amount at the end of 1st year = 2000+200
= ₹2200
Interest for 2nd year = 2200×10×1/100
= ₹220
∴ Amount at the end of the 2nd year = 2200+220
= ₹2420
∴ Interest for last half year = (2420×10×1/2)/100
= ₹121
∴ Amount at the end of 2 1/2 years = 2420+121
= ₹2541
∴ Total compound interest = 2541 – 2000
= ₹541
(10) Find the amount and the compound interest on ₹50000 for 1 1/2 years at 8% per annum, the interest being compounded semi-annually.
Solution:
Given principal = ₹50000
Rate of interest per annum = 8%
Rate of interest half yearly = 8/2 = 4%
Time = 1 1/2 years
∴ Interest for 1st half year = 5000×4×1/100
= ₹2000
∴ Amount for 1st half year = 50000 + 2000
= ₹52000
Interest for 2nd half year = 52000×4×1/100
= ₹2080
∴ Amount at the end of 1 year = 52000 + 2080
= ₹54080
Interest for 3rd half = 54080×4×1/100
= 21632/10
= ₹2163.2
∴ Amount at the end of 1 1/2 year = 54080+2163.2
= ₹56243.2
Total compound interior = 56243.2 – 5000
= ₹6243.2
(11) Calculate the amount and the compound interest on ₹5000 for 2 years when the rate of interest for successive years is 6% and 8% respectively.
Solution:
Given, Principal = ₹5000
Time = 2 years
Rate of interest for 1st year = 6%
Rate of interest for 2nd year = 8%
∴ Interest for 1st year = 5000×6×1/100
= ₹300
Amount at the end of 1st year = 5000 + 300
= ₹5300
Interest for 2nd year = 5300×8×1/100
= ₹424
∴ Amount at the end of 2nd year = 5300 + 424
= ₹5724
∴ Total compound interest = 5724 – 5000
= ₹724
(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.
Solution:
Given, principal = ₹17000
Time = 3 years
Rate of interests for successive years = 10%, 10%14%
∴ Interest for 1st year = 17000×10×1/100
= ₹1700
∴ Amount at the end of 1st year = 17000 + 1700
= ₹18700
Interest for 2nd year = 18700×10×1/100
= ₹1870
∴ Amount at the end of 2nd year = 18700 + 1870
= ₹20570
∴ Interest for 3rd year = 20570×14×1/100
= 28798/10
= ₹2879.8
Amount at the end of 3rd year = 20570 + 2879.8
= 23449.8
∴ Total compound interest = 23449.8 – 17000
= ₹6449.8
(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 first year?
(ii) What is the sum due at the end of the first 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 of one year.
(v) Hence, write down the compound interest for the third year.
Solution:
Given, Principal = ₹9600
Rate of interest = 10%
Time = 3 years
(i) ∴ Interest for 1st year = 9600×10×1/100
= ₹960
Amount at the end of 1st year = 9600+960
= ₹10560
(ii) Interest for 2nd year = 10560×10×1/100
= ₹1056
∴ Amount at the end of 2nd year = 10500+1056
= ₹11616
(iii) Total compound interest after 2 years = 11616 – 9600
= ₹2016
(iv) Difference between amount for 2nd year & 1st year.
= 11616 – 10560
= ₹1056
Interest on amount of diff = 1056×10×1/100
= ₹105.6
(v) ∴ interest for 3rd year = 11616×10×1/100
= ₹1161.6
(14) The simple interest on a certain sum money for 2 years at 10% per annum 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.
Solution:
Given, Simple interest = ₹1600
Rate of interest = 10%
Time = 2 years
∴ Simple interest = PRT /100
Or, 1600 = P×10×2/100
Or, P = ₹800
∴ Amount due after 2 years = 8000 + 1600
= ₹9600
∴ Interest for 1st year on same amount = 9600×10×1/100
= ₹960
Amount at the end of 1st year = 9600 + 960
=₹10560
∴ Interest for 2nd year = 10560×10×1/100
= ₹1056
∴ Amount at the end of 2nd year = 10560+1056
= ₹11616
∴ Interest for 3rd year = 11616×10×1/100
= ₹1161.6
∴ Amount at the end of 3rd year = 11616+1161.6
= 12777.6
∴ Total compound interest = 12777.6 – 9600
= ₹5177.6
(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 1/2 years.
Solution:
Given, Principal = ₹20000
Rate of interest = 10%
Time = 2 1/2 years = 5/2 years
∴ Simple interest on bank = 20000×10×5/100
= ₹5000
For compound interest,
Interest for 1st year = 20000×10×1/100
= ₹2000
∴ Amount at the end of 1st year = 20000 + 2000
= ₹22000
Interest for 2nd year = 22000×10×1/100
= ₹2200
∴ Amount at the end of 2nd year = 22000 + 2200
= ₹24200
∴ Interest for last 1/2 year = 24200×10×1/100
= ₹1210
∴ Amount at the end of 2 1/2 years = 24200+1210
= ₹25410
∴ Total compound interest = 25410-20000
= ₹5410
∴ Total profit made by vikram = 5410 – 5000
= ₹410
(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.
Solution:
Given, Principal = ₹6000
Rate of interest = 5%
Man repays ₹1200 each year.
∴ Interest for 1st year = 6000×5×1/100
= ₹300
Amount at the end of 1st year = 6000+300
= ₹6300
Amount at the end of 1st year after repayment
= 6300 – 1200
= ₹5100
Interest for 2nd year = 5100×5×1/100
= ₹255
Amount at the end of 2nd year = 5100+255
= ₹5355
Amount at the end of 2nd year after repayment
= 5355 – 1200
= ₹4155
∴ Amount due at the start of 3rd year = ₹4155.
(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.
Solution:
Given, Principal = ₹100000
Rate of interest = 11%
∴ Interest for last year = 100000×11×1/100
= ₹11000
Amount due at the end of 1st year = 100000 + 11000
= ₹111000
Amount repaid after 1 year = ₹41000
∴ Amount due at the end of 1st year after repayment
= 111000 – 41000
= ₹70000
∴ Interest for 2nd year = 70000×11×1/100
= ₹7700
∴ Amount due at the end of 2nd year = 70000 – 7700
= ₹77700
Amount repaid after 2nd year = ₹47700
∴ Amount due at the beginning of 3rd year = 77700 – 47700
= ₹30000
(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.
Solution:
Given, Principal = ₹50000
Successive rates of interest = 12%, 15%
∴ Interest for 1st year = 50000×12×1/100
= ₹6000
∴ Amount due at the end of 1st year = 50000+6000
= ₹56000
Amount repaid after 1 year = ₹33000
∴ Amount due at the end of 1st year after repayment = 56000 – 33000
= ₹23000
∴ Interest for 2nd year = 23000×15×1/100
= ₹3450
∴ Total amount due after 2 years = 23000+3450
= ₹26450