Class 9 SELINA Solutions Maths Chapter 2 - Compound Interest (Without using formula)
Compound Interest (Without using formula) Exercise Ex. 2(A)
Solution 1
Year ↓ |
Initial amount (Rs.) |
Interest (Rs.) |
Final amount (Rs.) |
1 st |
16,000 |
800 |
16,800 |
2 nd |
16,800 |
840 |
17,640 |
3 rd |
17,640 |
882 |
18,522 |
4 th |
18,522 |
926.10 |
19448.10 |
5 th |
19448.10 |
972.405 |
20420.505 |
Thus, the amount in 4 years is Rs. 19448.10.
Solution 2(i)
Solution 2(ii)
Solution 3
(i)
For 1^{st} year
P = Rs. 4600
R = 10%
T = 1 year.
A = 4600 + 460 = Rs. 5060
For 2^{nd} year
P = Rs. 5060
R = 12%
T = 1 year.
A= 5060 + 607.20 = Rs. 5667.20
Compound interest = 5667.20 - 4600
= Rs. 1067.20
Amount after 2 years = Rs. 5667.20
(ii)
For 1^{st} year
P = Rs. 16000
R = 10%
T = 1 year
A = 16000 + 1600 = 17600
For 2^{nd} year,
P = Rs. 17600
R = 14%
T = 1 year
A = 17600 + 2464 = Rs. 20064
For 3^{rd} year,
P = Rs. 20064
R = 15%
T = 1 year
Amount after 3 years = 20064 + 3009.60
= Rs. 23073.60
Compound interest = 23073.60 - 16000
= Rs. 7073.60
Solution 4
For 1^{st} years
P = Rs. 2400
R = 5%
T = 1 year
A = 2400 + 120 = Rs. 2520
For 2^{nd} year
P = Rs. 2520
R = 5%
T = 1 year
A = 2520 + 126 = Rs. 2646
For final year,
P = Rs. 2646
R = 5%
T = year
Amount after years = 2646 + 66.15
= Rs. 2712.15
Compound interest = 2712.15 - 2400
= Rs. 312.15
Solution 5
For 1^{st} year
P = Rs. 8000
R = 10%
T = 1 year
A = 8000 + 800 = Rs. 8800
For 2^{nd} year
P = Rs. 8800
R = 10%
T = 1 year
Compound interest for 2^{nd} years = Rs. 880
Solution 6
For 1^{st} year
P = Rs. 2500
R = 12%
T = 1 year
Amount = 2500 + 300 = Rs. 2800
For 2^{nd} year
P = Rs. 2800
R = 12%
T = 1 year
Amount = 2800 + 336 = Rs. 3136
Amount repaid by A to B = Rs. 2936
The amount of watch =Rs. 3136 - Rs. 2936 = Rs. 200
Solution 7
Solution 8
Solution 9
To calculate S.I.
P=Rs18,000; R=10% and T=1year
S.I.= Rs = Rs1,800
To calculate C.I.
For 1^{st} half- year
P= Rs18,000; R=10% and T= 1/2year
Interest= Rs = Rs900
Amount= Rs18,000+ Rs900= Rs18,900
For 2^{nd} year
P= Rs18,900; R= 10% and T= 1/2year
Interest= Rs = Rs945
Amount= Rs18,900+ Rs945= Rs19,845
Compound interest= Rs19,845- Rs18,000= Rs1,845
His gain= Rs1,845 - Rs1,800= Rs45
Solution 10
Compound Interest (Without using formula) Exercise Ex. 2(B)
Solution 1
For 1^{st} year
P = Rs. 4000
R = 8
T = 1 year
A = 4000 + 320 = Rs. 4320
For 2^{nd} year
P = Rs. 4320
R=8%
T = 1 year
A = 4320 + 345.60 = 4665.60
Compound interest = Rs. 4665.60 - Rs. 4000
= Rs. 665.60
Simple interest for 2 years =
= Rs. 640
Difference of CI and SI = 665.60 - 640
= Rs 25.60
Solution 2
For 1^{st} year
P = Rs. 12500
R = 12%
R = 1 year
A = 12500 + 1500 = Rs. 14000
For 2^{nd} year
P = Rs. 14000
R = 15%
T = 1 year
A = 1400 + 2100 = Rs. 16100
For 3^{rd} year
P = Rs. 16100
R = 18%
T = 1 year
A = 16100 + 2898 = Rs. 18998
Difference between the compound interest of the third year and first year
= Rs. 2898 - Rs. 1500
= Rs. 1398
Solution 3
Let money be Rs100
For 1^{st} year
P=Rs100; R=8% and T= 1year
Interest for the first year= Rs= Rs8
Amount= Rs100+ Rs8= Rs108
For 2^{nd} year
P=Rs108; R=8% and T= 1year
Interest for the second year= Rs= Rs8.64
Difference between the interests for the second and first year = Rs8.64 - Rs8 = Rs0.64
Given that interest for the second year exceeds the first year by Rs.96
When the difference between the interests is Rs0.64, principal is Rs100
When the difference between the interests is Rs96, principal=Rs=Rs15,000
Solution 4
Solution 5
For 1^{st} six months:
P = Rs. 5,000, R = 12% and T = year
∴ Interest = = Rs. 300
And, Amount = Rs. 5,000 + Rs. 300 = Rs. 5,300
Since money repaid = Rs. 1,800
Balance = Rs. 5,300 - Rs. 1,800 = Rs. 3,500
For 2^{nd} six months:
P = Rs. 3,500, R = 12% and T = year
∴ Interest = = Rs. 210
And, Amount = Rs. 3,500 + Rs. 210 = Rs. 3,710
Again money repaid = Rs. 1,800
Balance = Rs. 3,710 - Rs. 1,800 = Rs. 1,910
For 3^{rd} six months:
P = Rs. 1,910, R = 12% and T = year
∴ Interest = = Rs. 114.60
And, Amount = Rs. 1,910 + Rs. 114.60 = Rs. 2,024.60
Thus, the 3^{rd} payment to be made to clear the entire loan is 2,024.60.
Solution 6
Let principal (p = Rs. 100
R = 10%
T = 1 year
SI =
Compound interest payable half yearly
R = 5% half yearly
T = year = 1 half year
For first year
I =
A = 100 + 5 = Rs. 105
For second year
P = Rs. 105
Total compound interest = 5 + 5.25
= Rs. 10.25
Difference of CI and SI = 10.25- 10
= Rs. 0.25
When difference in interest is Rs. 10.25, sum = Rs. 100
If the difference is Rs. 1 ,sum =
If the difference is Rs. = 180,sum =
= Rs. 72000
Solution 7
Solution 8
(i) For 1^{st} years
P = Rs. 5600
R = 14%
T = 1 year
(ii) Amount at the end of the first year
= 5600 + 784
= Rs. 6384
(iii) For 2^{nd} year
P = 6384
R = 14%
R = 1 year
= Rs. 893.76
= Rs. 894 (nearly)
Solution 9
Savings at the end of every year = Rs. 3000
For 2^{nd} year
P = Rs. 3000
R = 10%
T = 1 year
A = 3000 + 300 = Rs. 3300
For third year, savings = 3000
P = 3000 + 3300 = Rs. 6300
R = 10%
T = 1 year
A = 6300 + 630 = Rs. 6930
Amount at the end of 3^{rd} year
= 6930 + 3000
= Rs. 9930
Solution 10
Compound Interest (Without using formula) Exercise Ex. 2(C)
Solution 1
Solution 2
Difference between the C.I. of two successive half-years
= Rs760.50 - Rs650= Rs110.50
Rs110.50 is the interest of one half-year on Rs650
Rate of interest= Rs%= %= 34%
Solution 3
(i)Amount in two years= Rs5,292
Amount in three years= Rs5,556.60
Difference between the amounts of two successive years
= Rs5,556.60 - Rs5,292= Rs264.60
Rs264.60 is the interest of one year on Rs5,292
Rate of interest= Rs%= %= 5%
(ii) Let the sum of money= Rs100
Interest on it for 1^{st} year= 5% of Rs100= Rs5
Amount in one year= Rs100+ Rs5= Rs105
Similarly, amount in two years= Rs105+ 5% of Rs105
= Rs105+ Rs5.25
= Rs110.25
When amount in two years is Rs110.25, sum = Rs100
When amount in two years is Rs5,292, sum = Rs
= Rs4,800
Solution 4
(i)C.I. for second year = Rs1,089
C.I. for third year = Rs 1,197.90
Difference between the C.I. of two successive years
= Rs1,197.90 - Rs1089= Rs108.90
Rs108.90 is the interest of one year on Rs1089
Rate of interest= Rs%= %= 10%
(ii) Let the sum of money= Rs100
Interest on it for 1^{st} year= 10% of Rs100= Rs10
Amount in one year= Rs100+ Rs10= Rs110
Similarly, C.I. for 2^{nd} year= 10% of Rs110
= Rs11
When C.I. for 2^{nd} year is Rs11, sum = Rs100
When C.I. for 2^{nd} year is Rs1089, sum = Rs = Rs9,900
Solution 5
For 1^{st} year
P=Rs8,000; A=9,440 and T= 1year
Interest= Rs9,440 - Rs8,000= Rs1,440
Rate=%=%=18%
For 2^{nd} year
P= Rs9,440; R=18% and T= 1year
Interest= Rs= Rs1,699.20
Amount= Rs9,440 + Rs1,699.20= Rs11,139.20
For 3^{rd} year
P= Rs11,139.20; R=18% and T= 1year
Interest= Rs= Rs2,005.06
Solution 6
For 1^{st} half-year
P= Rs15,000; A= Rs15,600 and T= ½ year
Interest= Rs15,600 - Rs15,000= Rs600
Rate= %=%= 8% Ans.
For 2^{nd} half-year
P= Rs15,600; R=8% and T= ½ year
Interest= Rs= Rs624
Amount= Rs15,600 + Rs624= Rs16,224
For 3^{rd} half-year
P= Rs16,224; R=8% and T= ½ year
Interest= Rs= Rs648.96
Amount= Rs16,224+ Rs648.96= Rs16,872.96 Ans.
Solution 7
For 1^{st} year
P=Rs12,800; R=10% and T= 1year
Interest= Rs= Rs1,280
Amount= Rs12,800+ Rs1,280= Rs14,080
For 2^{nd} year
P=Rs14,080; R=10% and T= 1 year
Interest= Rs= Rs1,408
Amount= Rs14,080+ Rs1,408= Rs15,488
For 3^{rd} year
P=Rs15,488; R=10% and T= 1year
Interest= Rs= Rs1,548.80
Amount= Rs15,488+ Rs1,548.80= Rs17,036.80
Solution 8
(i) For 1^{st} year
P= Rs8,000; R=7% and T=1year
Interest= Rs= Rs560
Amount= Rs8,000+ Rs560= Rs8,560
Money returned= Rs3,560
Balance money for 2^{nd} year= Rs8,560- Rs3,560= Rs5,000
For 2^{nd} year
P= Rs5,000; R=7% and T=1year
Interest paid for the second year= Rs= Rs350 Ans.
(ii)The total interest paid in two years= Rs350 + Rs560
= Rs910 Ans.
(iii) The total amount of money paid in two years to clear the debt
= Rs8,000+ Rs910
= Rs8,910 Ans.
Solution 9
(i)
Difference between depreciation in value between the first and second years
Rs.4,000 - Rs.3,600 = Rs.400
⇒ Depreciation of one year on Rs.4,000 = Rs.400
(ii)
Let Rs.100 be the original cost of the machine.
Depreciation during the 1^{st} year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1^{st} year, Original cost = Rs.100
⇒When the depreciation during 1^{st} year = Rs.4,000,
The original cost of the machine is Rs.40,000.
(iii)
Total depreciation during all the three years
= Depreciation in value during(1^{st} year + 2^{nd} year + 3^{rd} year)
= Rs.4,000 + Rs.3,600 + 10% of (Rs.40,000 - Rs.7,600)
= Rs.4,000 + Rs.3,600 + Rs.3,240
= Rs.10,840
The cost of the machine at the end of the third year
= Rs.40,000 - Rs.10,840 = Rs.29,160
Solution 10
Let the sum of money be Rs 100
Rate of interest= 10%p.a.
Interest at the end of 1^{st} year= 10% of Rs100= Rs10
Amount at the end of 1^{st} year= Rs100 + Rs10= Rs110
Interest at the end of 2^{nd} year=10% of Rs110 = Rs11
Amount at the end of 2^{nd} year= Rs110 + Rs11= Rs121
Interest at the end of 3^{rd} year=10% of Rs121= Rs12.10
Difference between interest of 3^{rd} year and 1^{st} year
=Rs12.10- Rs10=Rs2.10
When difference is Rs2.10, principal is Rs100
When difference is Rs252, principal = =Rs12,000 Ans.
Solution 11
For 1^{st} year
P= Rs10,000; R=10% and T= 1year
Interest= Rs=Rs1,000
Amount at the end of 1^{st} year=Rs10,000+Rs1,000=Rs11,000
Money paid at the end of 1^{st} year=30% of Rs10,000=Rs3,000
Principal for 2^{nd} year=Rs11,000- Rs3,000=Rs8,000
For 2^{nd} year
P=Rs8,000; R=10% and T= 1year
Interest= Rs = Rs800
Amount at the end of 2^{nd} year=Rs8,000+Rs800= Rs8,800
Money paid at the end of 2^{nd} year=30% of Rs10,000= Rs3,000
Principal for 3^{rd} year=Rs8,800- Rs3,000=Rs5,800 Ans.
Solution 12
For 1^{st} year
P= Rs10,000; R=10% and T= 1year
Interest= Rs=Rs1,000
Amount at the end of 1^{st} year=Rs10,000+Rs1,000=Rs11,000
Money paid at the end of 1^{st} year=20% of Rs11,000=Rs2,200
Principal for 2^{nd} year=Rs11,000- Rs2,200=Rs8,800
For 2^{nd} year
P=Rs8,800; R=10% and T= 1year
Interest= Rs= Rs880
Amount at the end of 2^{nd} year=Rs8,800+Rs880= Rs9,680
Money paid at the end of 2^{nd} year=20% of Rs9,680= Rs1,936
Principal for 3^{rd} year=Rs9,680- Rs1,936=Rs7,744 Ans.
Compound Interest (Without using formula) Exercise Ex. 2(D)
Solution 1
Let principal (p) = Rs. 100
For 1^{st} year
P = Rs. 100
R = 10%
T = 1 year
A = 100 + 10 = Rs. 110
For 2^{nd} year
P = Rs. 110
R = 11%
T = 1 year
A = 110 + 12.10 = Rs. 122.10
If Amount is Rs. 122.10 on a sum of Rs. = 100
If amount is Rs. 1, sum =
If amount is Rs. 6593.40, sum =
= Rs. 5400
Solution 2
Let the value of machine in the beginning= Rs. 100
For 1^{st} year depreciation = 10% of Rs. 100 = Rs. 100
Value of machine for second year = 100 - 10
= Rs. 90
For 2^{nd} year depreciation = 10% of 90 = Rs. 9
Value of machine for third year = 90 - 9
= Rs. 81
For 3^{rd} year depreciation = 15% of 81
= Rs. 12.15
Value of machine at the end of third year = 81 - 12.15
= Rs. 68.85
Net depreciation = Rs. 100 - Rs. 68.85
= Rs. 31.15
Or 31.15%
Solution 3
For 1^{st} half-year
P=Rs12,000; R=10% and T=1/2 year
Interest= Rs= Rs600
Amount= RS12,000 + Rs600= Rs12,600
Money paid at the end of 1^{st} half year=Rs4,000
Balance money for 2^{nd} half-year= Rs12,600- Rs4,000=Rs8,600
For 2^{nd} half-year
P=Rs8,600; R=10% and T=1/2 year
Interest=Rs=Rs430
Amount= Rs8,600+ Rs430= Rs9,030
Money paid at the end of 2^{nd} half-year=Rs4,000
Balance money for 3^{rd} half-year= Rs9,030- Rs4,000=Rs5,030
For 3^{rd} half-year
P=Rs5,030; R=10% and T=1/2 year
Interest = Rs= Rs251.50
Amount= Rs5,030 + Rs251.50= Rs5,281.50
Solution 4
Let Principal= Rs 100
For 1^{st} year
P=Rs100; R=10% and T=1year
Interest= Rs= Rs10
Amount= Rs100 + Rs10= Rs110
For 2^{nd} year
P=Rs110; R=10% and T= 1year
Interest= Rs= Rs11
Amount= Rs110 + Rs11= Rs121
For 3^{rd} year
P=Rs121; R=10% and T= 1year
Interest= Rs= Rs12.10
Sum of C.I. for 1^{st} year and 3^{rd} year=Rs10+Rs12.10=Rs22.10
When sum is Rs22.10, principal is Rs100
When sum is Rs2,652, principal =Rs=Rs12,000 Ans.
Solution 5
Let original value of machine=Rs100
For 1^{st} year
P=Rs100; R=12% and T= 1year
Depreciation in 1^{st} year= Rs =Rs12
Value at the end of 1^{st} year=Rs100 - Rs12=Rs88
For 2^{nd} year
P= Rs88; R=12% and T= 1year
Depreciation in 2^{nd} year= Rs =Rs10.56
When depreciation in 2^{nd} year is Rs10.56, original cost is Rs100
When depreciation in 2^{nd} year is Rs2,640, original cost=
=Rs25,000
Solution 6
Let Rs. x be the sum.
Compound interest
For 1^{st} year:
P = Rs.x, R = 8% and T=1
For 2^{nd} year:
P = Rs.x + Rs. 0.08x = Rs. 1.08x
Amount = 1.08x + 0.0864x = Rs. 1.1664x
CI = Amount - P = 1.1664x - x = Rs. 0.1664x
The difference between the simple interest and compound interest at the rate of 8% per annum compounded annually should be Rs. 64 in 2 years.
⇒ Rs. 0.1664x - Rs. 0.16x = Rs.64
⇒ Rs. 0.0064x = Rs.64
⇒ x = 10000
Hence the sum is Rs. 10000.
Solution 7
For 1^{st} year
P=Rs13,500; R=16% and T= 1year
Interest= Rs= Rs2,160
Amount= Rs13,500 + Rs2,160= Rs15,660
For 2^{nd} year
P=Rs15,660; R=16% and T= 1year
Interest= Rs= Rs2,505.60
=Rs2,506
Solution 8
For 1^{st} year
P=Rs48,000; R=10% and T= 1year
Interest= Rs= Rs4,800
Amount= Rs48,000+ Rs4,800= Rs52,800
For 2^{nd} year
P=Rs52,800; R=10% and T= 1year
Interest= Rs= Rs5,280
Amount= Rs52,800+ Rs5,280= Rs58,080
For 3^{rd} year
P=Rs58,080; R=10% and T= 1year
Interest= Rs= Rs5,808
Solution 9
(i)
Let x% be the rate of interest charged.
For 1^{st} year:
P = Rs.12,000, R = x% and T = 1
For 2^{nd} year:
After a year, Ashok paid back Rs.4,000.
P = Rs.12,000 + Rs.120x - Rs.4,000 = Rs.8,000 + Rs.120x
The compound interest for the second year is Rs.920
Rs. (80x + 1.20x^{2}) = Rs.920
⇒1.20x^{2} + 80x - 920 = 0
⇒3x^{2} + 200x - 2300 = 0
⇒3x^{2} + 230x - 30x - 2300 = 0
⇒x(3x + 230) -10(3x + 230) = 0
⇒(3x + 230)(x - 10) = 0
⇒x = -230/3 or x = 10
As rate of interest cannot be negative so x = 10.
Therefore the rate of interest charged is 10%.
(ii)
For 1^{st} year:
Interest = Rs.120x = Rs.1200
For 2^{nd} year:
Interest = Rs.(80x + 1.20x^{2}) = Rs.920
The amount of debt at the end of the second year is equal to the addition of principal of the second year and interest for the two years.
Debt = Rs.8,000 + Rs.1200 + Rs.920 = Rs.10,120
Solution 10