ML Aggarwal Solutions Class 9 Math 2nd Chapter Compound Interest Exercise 2.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Second Chapter Compound Interest Exercise 2.2. APC Solution Class 9 Exercise 2.2.
(1) Find the amount and the compound interest on ₹5000 for 2 years at 6% per annum, interest payable yearly.
Solution:
Given, principal = ₹5000
Time = 2 years
Rate of interest = 6%
∴ Amount = P (1 + r/100)n
= 5000 (1 + 6/100)2
= 5000 × 106×106/100×100
= ₹5618
∴ C-I = A – P = 5618 – 5000
= ₹618
(2) Find the amount and the compound interest on ₹8000 for 4 years at 10% per annum interest reckoned yearly.
Solution:
Given, Principal = ₹8000
Time = 4 years
Rate of interest = 10%
∴ A = 8000 (1 + 10/100)4
= 8000 × 11/10 × 11/10 × 11/10 × 11/10
= 117128/10
= ₹11712.8
∴ C.I = A – P
= 11712.8 – 8000
= ₹3712.8
(3) If the interest is compounded half yearly, calculate the amount when the principal is ₹7400, the rate of interest is 5% and he the duration is one year.
Solution:
Given, Principal = ₹7400
Rate of interest = 5%
Time = 1 year/half yearly
= 7400 × 41/40 × 41/40
= 124394/16 = ₹7774.625
∴ C.I = A – P = 7774.625 – 7400
= ₹374.625
(4) Find the amount and the compound interest on ₹5000 at 10% p.a. for 1 1/2 years, compound interest reckoned semi-annually.
Solution:
Given, Principal = ₹5000
Rate of interest = 10%/half yearly
Time = 1 1/2 years = 10/2 = 5% 1 year
= 3/2 years half yearly
= 3/2 × 2 = 3 years/yearly
∴ A = 5000 × (1 + 5/100)3
= 5000 × 21/20 × 21/20 × 21/20
= 46305/8
= ₹5788.12
∴ C.I = A – P = 5788.12 – 5000
= ₹788.12
(5) Find the amount and the compound interest on ₹100000 compounded quarterly for 9 months at the rate of 4% p.a.
(Hint. r = 1/4 of 4% = 1% and n = 9/3 = 3)
Solution:
Given, principal = ₹100000
Rate of interest = 4% quarterly.
= 4/4 = 1% yearly
n = 9 months = 4/12 × 9 years
= 3 years
∴ A = 100000 (1 + 1/100)3
= 100000 × 101/100 × 101/100 × 101/100
= 1030301/10 = ₹103030.1
∴ C.I = 103030.1 = 100000 = ₹3030.1
(6) Find the difference between C.I and S.I on sum of ₹4800 for 2 years at 5% per annum payable yearly.
Solution:
Given, Principal = ₹4800
n = 2 years
Rate of interest = 5%
∴ S.I = PRT/100
= 4800×5×2/100
= ₹480
C.I = P [(1 + r/100)n – 1]
= 4800 [(1 + 5/100)2 – 1]
= 4800 (441/400 – 1)
= 4800 × 41/400
= ₹492
∴ Difference = CI-SI
= 492 – 480
= ₹12
(7) Find the difference between the simple interest and compound interest on ₹2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually.
Solution:
Given, principal = ₹2500
n = 2 years/annum = 2×2 = 4 years/half yearly
Rate of interest = 4%/annum
= 4/2 = 2% half yearly
∴ S.I = PRT/100 = 2500×4×2/400
= ₹200
C.I = 2500 [(1 + 2/100)4 – 1]
= 2500 (6765201/6250000 -1)
= 2500 × 515201/6250000
= 20608.04/100
= ₹206.08
∴ Difference = C.I – S.I
= 206.08 – 200
=₹6.08
(8) Find the amount and the compound interest on ₹2000 in 2 years if the rate is 4% for the first year and 3% for the second year.
Solution:
Given, principal = ₹2000
n = 2 years
Rates = 4%, 3% successive years.
∴ A = 2000 (1 +4/100) (1 + 3/100)
= 2000 × 26/25 × 103/100
= 10712/5
= ₹2142.4
∴ CI = 2142.4 – 2000
= ₹142.4
(9) Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.
Solution:
Given, principal = ₹3125
n = 3 years
Rates = 4%, 5%, 6% successive years.
∴ A = 3125 (1 + 4/100) × (1 + 5/100) (1 + 6/100)
= 3125 × 26/25 × 21/20 × 53/50
= 14469/4
= 3617.25
∴ C.I = 367.25 – 3125
= ₹492.25
(10) What sum of money will amount to ₹9261 in 3 years at 5% per annum compound interest?
Solution:
Let, principal be x
Amount = ₹9261
n = 3 years
Rate = 5%
∴ 9261 = x (1 + 5/100)3
Or, x = 9261 × 20/21 × 20/21 × 20/21
= ₹8000
∴ Principal is ₹8000.
(11) What sum invested at 4% per annum compounded semi-annually amounts to ₹7803 at the end of one year?
Solution:
Given, rate = 4%/annum
= 4/2 = 2% half yearly
Amount = ₹7000
n = 1 year/annum
= 1 × 2 = 2year half yearly
∴ Let, principal be x.
∴ 7803 = x (1 + 2/100)2
Or, x = 7803 × 50/51 ×50/51
= ₹7500
∴Principal is ₹7500
(12) What sum invested for 1 1/2 years compounded half-yearly at the rate of 4% p.a. will amount ₹132651?
Solution:
Given, rate = 4% p.a.
= 4/2 = 2% half yearly
Amount = ₹132651
n = 1 1/2 years p.a.
= 3/2 × 2 = 3 years/half yearly.
∴ Let, principal be x
132651 = x (1 + 2/100)3
Or, x = 132651 × 50/51 × 50/51 × 50/51
Or, x = ₹125000
∴Principal is ₹125000
(13) On what sum will the compound interest for 2 years at 4% per annum be ₹5712?
Solution:
Given, C.I = ₹5712
Rate = 4%
n = 2 years
Let, principal be x
∴ 5712 = x [(1 + 4/100)2 – 1]
Or, 5712 = x (676/625 – 1)
Or, x = 5712 × 625/51
Or, x = ₹70000
∴Principal is ₹70000
(14) A man invests ₹1200 for two years at compound interest. After one year the money amounts to ₹1275. Find the interest for the second year correct to the nearest rupee.
Solution:
Given, Principal = ₹1200
Amount after 1 year = ₹1275
∴ Let, the rate of interest be r
∴ or, 1275 = 1200 (1+r/100)
or, 1275/1200 = 100+r/100
Or, r = 106.25 – 100
Or, r = 6.25%
∴ C.I for 2nd year = 1275 × 6.25/100
= 318.75/4
= 79.6875
= ₹80
(15) At what rate percent per annum compound interest will ₹2304 amount to ₹2500 in 2 years?
Solution:
Given, Principal = ₹2304
Amount = ₹2500
n = 2 years
Let, rate of interest be r
∴ 2500 = 2304 (1 + 2/100)2
Or, 2500 = 2304 × (100+r/100)2
Or, 2500/2304 = (100+r/10000)2
Or, 2500 × 10000/2304 = (100 + r)2
Or, 100+r = √2500×10000/2304
Or, 100+r = 50×100/48
Or, 100+r = 625/6
Or, r = 104.16 – 100
Or, r = 4.16%
(16) A sum compounded annually becomes 25/16 times of itself in two years. Determine the rate of interest per annum.
Solution:
Let, the principal be 100
∴ (ATQ) amount = 100 × 25/16
= 625/4
n = 2 years [Given]
∴ 625/4 = 100 (1+ r/100)2
Or, 25/16 = (100+r/10002
Or, 100+r/100 = √25/16
Or, 100+r = 5/4 × 100
Let, rate be r
Or, r = 125 – 100
= 25%
∴ rate = 25%
(17) At what rate percent will ₹2000 amount to ₹2315-25 in 3 years at compound interest?
Solution:
Given, Principal = ₹2000
Amount = ₹2315.25
n = years
Let, rate of interest be r
∴ 2315.25 = 2000 (1 + r/100)3
Or, 2315.25/2000 = (100+r/100)3
Or, 100+r/100 = √2315.25/2000
Or, 100+r/100 = 3√92.61/80
Or 100+r/100 = 3√92.61/80×100
Or, 100+r/100 = 3√9261/8000
Or, 100+r/100 = 21/20
Or, r = 105-100
= 5%
∴ The rate of interest is 5%
(18) If ₹40000 amounts to ₹48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.
Solution:
Given, Principal = ₹40000
Amount = ₹48620.25
n = 2 years p.a.
= 2×2 = 4 years/half yearly
Let, rate of interest be r p.a.
Rate of interest be r/2 half yearly.
∴ 48620.25 = 40000 (1 + (r/2)/100)4
Or, 48620.25/40000 = (200+r/200)4
Or, (200+r/200)4 = 4862025/4000000
Or, (200+r/200)2 = √194481/160000
Or, (200+r/200)2 = 441/400
Or, 200+r/200 = √441/400
Or, 200+r/200 = 21/20
Or, r = 210 – 200
Or, r = 10%
∴ The rate of interest is 10% p.a.
(19) Determine the rate of interest for a sum that becomes 216/125 times of itself in 1 1/2 years, compounded semi-annually.
Solution:
Given, n = 1 1/2 years p.a.
= 3/2 years p.a.
= 3/2 × 2 = 3 years/half yearly
Let principal be ₹100
∴ ATQ, Amount = 100 × 216/125
Let, rate of interest be = ₹ 4×216/5
r p.a.
∴ Rate of interest = r/2/half yearly
∴ 4×216/5 = 100 (1+ (r/2)/100)3
Or, 4×216/500 = (200+r/200)3
Or, (200+r/200)3 = 216/125
Or, 200+r/200 = 3√216/125
Or, 200+r/200 = 6/5
Or, r = 240 – 200
Or, r = 40%
∴ The rate of interest is 40% p.a.
(20) At what rate percent p.a. compound interest would ₹80000 amount to ₹88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.
Solution:
Given, Principal = ₹80000
Amount = ₹88200
n = 2 years
Let, the rate of interest be r
∴ 88200 = 80000 (1 + r/100)2
Or, 88200/80000 = (100+r/100)2
Or, 100+r/100 = √441/400
Or, 100+r/100 = 21/20
Or, r = 105-100
= 5%
∴ The rate of interest is 5%
Now, when n = 3 years.
Principal = ₹80000
Rate = 5%
∴ A = 80000 (1 + 5/100)3
= 80000 × 21/20 × 21/20 × 21/20
=₹92610
∴ The amount at 3 years is ₹92610
(21) A certain sum amounts to ₹5292 in 2 years and to ₹5556.60 in 3 years at compound interest. Find the rate and the sum.
Solution:
Given, Amount at 2 years = ₹5292
Amount at 3 years = ₹5556.60
Let, the principal be x
The rate of interest be r
∴ at 2 years
5292 = x (1 + r/100)2
Or, (10+r/100)2 = 5292/x —- (i)
Now, at 3 years,
5556.60 = x (1 + r/100)3
Or, (100+r/100)3 = 5556.60/x —- (ii)
Dividing equation (i) from equation (ii) we get,
Or, 100+r/100 = 5556.60/5292
Or, 100+r = 555660/5292
Or, r = 555660/5292 – 100
Or, r = 555660-529200/5292
Or, r = 26460/5292
Or, r = 6615/1323
Or, r = 5%
∴ r = 5%
∴ The rate of interest is 5%
∴ Putting the value of r on equation (i) we get.
5292 = x (1 + 5/100)2
Or, 5292 = x × (21/20)2
Or, x = 5292 × 20/21 × 20/21
Or, x = ₹4800
∴ The principal is ₹4800
(22) A certain sum amounts to ₹798.60 after 3 years and ₹878.46 after 4 years. Find the interest rate and the sum.
Solution:
Given, Amount after 3 years = ₹748.60
Amount after 4 years = ₹878.46
Let, the principal be x the rate of interest be r.
∴ At 3 years,
708.60 = x (1 + r/100)3
Or, 798.60/x = (100+r/100) —- (i)
At 4 years,
878.46 = x (1 + r/100)4
Or, (100+r/100)4 = 878.46/x —- (ii)
Now, dividing eqn (i) from eqn (ii) we get,
The rate of interest is 10%
Now, putting the value of r in equation (i) we get,
798.60 = x(100+10/100)3
Or, x = 798.60×100×100 x 100/110×110×110
Or, 798600/11×11v11
Or, x = ₹600
∴ The principal is ₹600
(23) In what time will ₹15625 amount to ₹17576 at 4% per annum compound interest?
Solution:
Given, Principal = ₹15625
Amount = ₹17576
rate of interest = 4%
Let, the time be n.
∴ 17576 = 15625 (1 + 4/100)2
Or, 17576/15625 = (26/25)n
Or, (26/25)n = (26/25)3
Equating both sides we get,
n = 3
∴ Time taken is 3 years
(24) (i) In what time will ₹1500 yield ₹496.50 as completed annually?
(ii) Find the time (in years) in which ₹12500 will produce ₹3246-40 as compound interest at 8% per annum, interest compounded annually.
Solution:
(i) Given, Principal = ₹1500
C.I = ₹496.50
Rate % = 10%
∴ Amount = 1500 + 496.50
= ₹1996.5
Let, the time taken be n years.
∴ 1996.5 = 1500 (1 + 10/100)n
Or, 1996.5/1500 = (11/10)n
Or, (11/10)n = 19965/15000
Or, (11/10)n = 1331/1000
Or, (11/10)n= (11/10)3
Equating both side we get,
n = 3 years
∴ Time taken = 3 years p.a.
(ii) Given principal = ₹12500
C.I = ₹3246.40
Rate% = 8% p.a.
∴ Amount = 12500 + 3246.40
= ₹15746.4
∴ 15746.4 = 12500 (1 + 8/100)n
Or, 15746.4/12500 = (27/25)n
Let, time be n year.
Equating both sides we get,
n = 3 years
∴ Time taken = 3 years
(25)₹16000 invested at 10% p.a. compounded semi-annually, amounts to ₹18522. Find the time period of investment.
Solution:
Given, Principal = ₹16000
Amount = ₹18522
Rate% = 10% p.a.
= 10/2 = 5%/half yearly
Let, the time be n p.a.
∴ Time = n × 2/half yearly
= 2n year
∴ 18522 = 16000 (1 + 5/100)2n
Or, 13522/16000 = (21/20)2n
Or, (21/20)2n = 9261/8000
Or, (21/20)2n = (x/2003
Now, equating both sides we get,
2n = 3
Or, n = 3/2 = 1 1/2 years
∴ Time taken = 1 1/2 years
(26) What sum will amount to ₹2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?
Solution:
Given, amount = ₹2782.50
Time = 2 years
Rates % = 5%, 6% successive years
∴ 2782.50 = x (1 + 5/100) (1 + 6/100)
Or, 2782.50 = x × 21/20 × 53/50
Or, x = 2782.50×20×50/21×53
Or, x = 278250×10/21×53
Or, x = ₹2500
∴ The principal is ₹2500.
(27) A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹225 and ₹240. Find:
(i) The rate of interest
(ii) The original sum
(iii) The interest earned in the third year
Solution:
Let, the principal be x
The rate of interest be r.
(ATQ)∴ Amount for 1st year = x + 225
Amount for 2nd year = x + 225 + 240
= x + 465
∴For 1st year,
x + 225 = x (1 + r/100)
Or, x+225/x = (100+r/100) —- (i)
For 2nd year,
x + 465 = (100+r/100)2
Putting the value of 100+r/100 from equation (i) we get
x + 465 = x (x+225/x)2
Or, x+465/x = (x+225)2/x
Or, x2 + 465x = x2 + 450x + 2252
Or, 465x – 450x = 225 × 225
Or, 15x = 225×225
Or, x = 225×225/15
Or, x = ₹3375
(ii) ∴ The Principal is ₹3375
(i) Putting the value of x in equation (i) we get,
x + 225 = x (1 + r/100)1
Or, 3375+225 = 3375 (100+r/100)
Or, 3600/3375 = 100+r/100
Or, 320/3 = 100+r /100
Or, r = 320/3 – 100
Or, r = 20/3
Or, r = 6.66%
(iii) C.I after 3rd year = 3375 [(1 + 6.66/100)3 – 1]
∴ C.I for 3rd year = 721 – (225 + 240)
= 721 – 465
= 256
(28) On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a.?
Solution:
Given, rate% = 5% p.a.
n = 2 years
Difference in C.I & S.I = ₹25
∴ Let, the principal be ₹x
∴ S.I for 2 years = x × 5 × x/100
= ₹ x/10
∴ C.I for 2 years = x [(1+ 5/100)2 – 1]
= x [(21/20)2 – 1]
= x [441-400/400]
= 41x/400
(ATQ) 41x/400 – x/10 = 25
Or, 41x-40x/400 = 25
Or, x = 25 × 400
Or, x = ₹10000
∴ The Principal is ₹10000
(29) The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹15. Find the sum of money lent out.
Solution:
Let, the principal be ₹ x.
Given, rate % = 10% p.a.
= 10/2 = 5% half yearly.
n = 1 year
= 1 × 2 = 2 years half yearly
For S.I in 1 year = x × 10 ×1/100
= ₹ x/10
For C.I in 1 year half yearly = x [(1+5/100)2– 1]
= x [(21/20)2 – 1]
= x (441-400/400)
= ₹ 41x/400
(ATQ) difference in C.I & S.I is = ₹15
Or, 41x/400 – x/10 = 15
Or, 41x-40x/400 = 15
Or, x = 15×400
Or, x = ₹6000
∴ The principal is ₹6000
(30) The amount at compound interest which is calculated yearly on a certain sum of money is ₹1250 in one year and ₹1375 in two years. Calculate the rate of interest.
Solution:
Given, Amount at the end of 1 year = ₹1250
Amount at the end of 2 years = ₹1375
Let, the principal be x
The rate of interest be r
∴ 1250 = x (1+r/100)1
Or, x = 1250/(100+r/100) —- (i)
∴ For 2nd year,
1375 = x (1 + r/100)2
Putting the value of a form equation (i) we get,
1375 = 1250/(100+r/100) × (100+r/100)1
Or, 1345/1250 = 100+r/100
Or, r = 110 – 100
Or, r = 10%
∴The rate of interest is 10%
(31) The simple interest on a certain sum for 3 years is ₹225 and the compound interest on the same sum at the same rate for 2 years is ₹153. Find the rate of interest and the principal.
Solution:
Given, S.I = ₹225
n = 3 years
C.I = ₹153
n = 2 years
Let, the principal be x, the rate of interest be x
For simple interest,
225 = x × r × 3/100
Or, x = 22500/3×r
Or, x = 7500/r —- (i)
Now, For compound interest,
153 = x [(1 + r/100)2 – 1]
Putting the value of x from equation (i) we get,
153 = 7500/r [(100+r/100)2 – 1]
Or, 153 = 7500/r × [(100+r)2/10000 – 1]
Or, 153 = 7500/r [(100+r)2-10000/10000]
Or, 51×4×r = 10000+200r+r2-10000
Or, r2 + 200r – 204r = 0
Or, r2 – 4r = 0
Or, r (r – 4) = 0
Or, r – 4 = 0
Or, r = 4%
∴ The rate of interest is 4%
Now, Putting the value of r in equation we get,
x = 7500/r
Or, x = 7500/r
Or, x = 7500/4
∴ The principal is ₹1875
(32) Find the difference between compound interest on ₹8000 for 1 1/2 years at 10% p.a. when compounded annually and semi-annually.
Solution:
Given, principal = ₹8000
n = 1 1/2 years p.a.
= 3/2 × 2 half yearly
= 3 years
Rate of interest = 10% p.a.
= 10/2 = 5% half yearly
∴ Amount when compounded yearly
∴ C.I when compounded yearly = 9240 – 8000
= ₹1240
Amount when compounded half yearly,
= 8000 (1 + 5/100)3
= 8000 × 21/20 × 21/20 × 21/20
= ₹9261
∴ C.I when compounded half yearly
= 9261 – 8000
= ₹1261
∴ Difference between C.I when compounded yearly and half yearly = 1261 – 1240
= ₹21
(33) A sum of money is lent out at compound interest for two years at 20% p.a, C.I being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate percent per annum. C.I being reckoned half yearly, it would have fetched ₹482 more by way of interest. Calculate the sum of money lent out.
Solution:
Given, Difference in C.I half yearly & C.I yearly = ₹482.
n = 2 years, p.a. = 2 ×2 = 4 years half yearly.
Rate of interest = 20%, p.a.
= 20/2 = 10% half yearly
Let, the principal be ₹x
∴ Amount when compounded yearly,
= x (1 + 20/100)2
= x × 36/25
= ₹ 36x/25
∴ C.I when compounded yearly = 36x/25 – x
= 36x-25x/25
= ₹ 11x/25
Now, Amount when compounded half yearly,
Now, (ATQ), 4641x/10000 – 11x/25 = 482
Or, 4641x – 4400x/10000 = 482
Or, 241x = 492 × 10000
Or, x = 482×10000/241
Or, x = ₹20000
∴ The principal is ₹20000
(34) A sum of money amounts to ₹13230 in one year and to ₹13891.50 in 1 1/2 years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.
Solution:
Given, Amount for 1 year = ₹13230
Amount for 1 1/2 years = ₹13891.5
Now, for half yearly,
n for 1st year = 1×2 = 2 years
n for 11/2 year = 3/2 × 2 = 3 years
Let, the rate of interest be r % p.a.
The rate of interest r/2 % half yearly
Let, the principal be ₹ x.
For 1st year,