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 1st year
P = Rs. 4600
R = 10%
T = 1 year.
A = 4600 + 460 = Rs. 5060
For 2nd 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 1st year
P = Rs. 16000
R = 10%
T = 1 year
A = 16000 + 1600 = 17600
For 2nd year,
P = Rs. 17600
R = 14%
T = 1 year
A = 17600 + 2464 = Rs. 20064
For 3rd 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 1st years
P = Rs. 2400
R = 5%
T = 1 year
A = 2400 + 120 = Rs. 2520
For 2nd 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 1st year
P = Rs. 8000
R = 10%
T = 1 year
A = 8000 + 800 = Rs. 8800
For 2nd year
P = Rs. 8800
R = 10%
T = 1 year
Compound interest for 2nd years = Rs. 880
Solution 6
For 1st year
P = Rs. 2500
R = 12%
T = 1 year
Amount = 2500 + 300 = Rs. 2800
For 2nd 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 1st half- year
P= Rs18,000; R=10% and T= 1/2year
Interest= Rs = Rs900
Amount= Rs18,000+ Rs900= Rs18,900
For 2nd 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 1st year
P = Rs. 4000
R = 8
T = 1 year
A = 4000 + 320 = Rs. 4320
For 2nd 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 1st year
P = Rs. 12500
R = 12%
R = 1 year
A = 12500 + 1500 = Rs. 14000
For 2nd year
P = Rs. 14000
R = 15%
T = 1 year
A = 1400 + 2100 = Rs. 16100
For 3rd 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 1st year
P=Rs100; R=8% and T= 1year
Interest for the first year= Rs= Rs8
Amount= Rs100+ Rs8= Rs108
For 2nd 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 1st 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 2nd 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 3rd 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 3rd 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 1st years
P = Rs. 5600
R = 14%
T = 1 year
(ii) Amount at the end of the first year
= 5600 + 784
= Rs. 6384
(iii) For 2nd 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 2nd 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 3rd 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 1st 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 1st year= 10% of Rs100= Rs10
Amount in one year= Rs100+ Rs10= Rs110
Similarly, C.I. for 2nd year= 10% of Rs110
= Rs11
When C.I. for 2nd year is Rs11, sum = Rs100
When C.I. for 2nd year is Rs1089, sum = Rs
= Rs9,900
Solution 5
For 1st year
P=Rs8,000; A=9,440 and T= 1year
Interest= Rs9,440 - Rs8,000= Rs1,440
Rate=%=
%=18%
For 2nd year
P= Rs9,440; R=18% and T= 1year
Interest= Rs= Rs1,699.20
Amount= Rs9,440 + Rs1,699.20= Rs11,139.20
For 3rd year
P= Rs11,139.20; R=18% and T= 1year
Interest= Rs= Rs2,005.06
Solution 6
For 1st half-year
P= Rs15,000; A= Rs15,600 and T= ½ year
Interest= Rs15,600 - Rs15,000= Rs600
Rate= %=
%= 8% Ans.
For 2nd half-year
P= Rs15,600; R=8% and T= ½ year
Interest= Rs= Rs624
Amount= Rs15,600 + Rs624= Rs16,224
For 3rd 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 1st year
P=Rs12,800; R=10% and T= 1year
Interest= Rs= Rs1,280
Amount= Rs12,800+ Rs1,280= Rs14,080
For 2nd year
P=Rs14,080; R=10% and T= 1 year
Interest= Rs= Rs1,408
Amount= Rs14,080+ Rs1,408= Rs15,488
For 3rd 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 1st 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 2nd year= Rs8,560- Rs3,560= Rs5,000
For 2nd 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 1st year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1st year, Original cost = Rs.100
⇒When the depreciation during 1st 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(1st year + 2nd year + 3rd 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 1st year= 10% of Rs100= Rs10
Amount at the end of 1st year= Rs100 + Rs10= Rs110
Interest at the end of 2nd year=10% of Rs110 = Rs11
Amount at the end of 2nd year= Rs110 + Rs11= Rs121
Interest at the end of 3rd year=10% of Rs121= Rs12.10
Difference between interest of 3rd year and 1st 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 1st year
P= Rs10,000; R=10% and T= 1year
Interest= Rs=Rs1,000
Amount at the end of 1st year=Rs10,000+Rs1,000=Rs11,000
Money paid at the end of 1st year=30% of Rs10,000=Rs3,000
Principal for 2nd year=Rs11,000- Rs3,000=Rs8,000
For 2nd year
P=Rs8,000; R=10% and T= 1year
Interest= Rs = Rs800
Amount at the end of 2nd year=Rs8,000+Rs800= Rs8,800
Money paid at the end of 2nd year=30% of Rs10,000= Rs3,000
Principal for 3rd year=Rs8,800- Rs3,000=Rs5,800 Ans.
Solution 12
For 1st year
P= Rs10,000; R=10% and T= 1year
Interest= Rs=Rs1,000
Amount at the end of 1st year=Rs10,000+Rs1,000=Rs11,000
Money paid at the end of 1st year=20% of Rs11,000=Rs2,200
Principal for 2nd year=Rs11,000- Rs2,200=Rs8,800
For 2nd year
P=Rs8,800; R=10% and T= 1year
Interest= Rs= Rs880
Amount at the end of 2nd year=Rs8,800+Rs880= Rs9,680
Money paid at the end of 2nd year=20% of Rs9,680= Rs1,936
Principal for 3rd 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 1st year
P = Rs. 100
R = 10%
T = 1 year
A = 100 + 10 = Rs. 110
For 2nd 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 1st year depreciation = 10% of Rs. 100 = Rs. 100
Value of machine for second year = 100 - 10
= Rs. 90
For 2nd year depreciation = 10% of 90 = Rs. 9
Value of machine for third year = 90 - 9
= Rs. 81
For 3rd 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 1st 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 1st half year=Rs4,000
Balance money for 2nd half-year= Rs12,600- Rs4,000=Rs8,600
For 2nd 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 2nd half-year=Rs4,000
Balance money for 3rd half-year= Rs9,030- Rs4,000=Rs5,030
For 3rd 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 1st year
P=Rs100; R=10% and T=1year
Interest= Rs= Rs10
Amount= Rs100 + Rs10= Rs110
For 2nd year
P=Rs110; R=10% and T= 1year
Interest= Rs= Rs11
Amount= Rs110 + Rs11= Rs121
For 3rd year
P=Rs121; R=10% and T= 1year
Interest= Rs= Rs12.10
Sum of C.I. for 1st year and 3rd 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 1st year
P=Rs100; R=12% and T= 1year
Depreciation in 1st year= Rs =Rs12
Value at the end of 1st year=Rs100 - Rs12=Rs88
For 2nd year
P= Rs88; R=12% and T= 1year
Depreciation in 2nd year= Rs =Rs10.56
When depreciation in 2nd year is Rs10.56, original cost is Rs100
When depreciation in 2nd year is Rs2,640, original cost=
=Rs25,000
Solution 6
Let Rs. x be the sum.
Compound interest
For 1st year:
P = Rs.x, R = 8% and T=1
For 2nd 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 1st year
P=Rs13,500; R=16% and T= 1year
Interest= Rs= Rs2,160
Amount= Rs13,500 + Rs2,160= Rs15,660
For 2nd year
P=Rs15,660; R=16% and T= 1year
Interest= Rs= Rs2,505.60
=Rs2,506
Solution 8
For 1st year
P=Rs48,000; R=10% and T= 1year
Interest= Rs= Rs4,800
Amount= Rs48,000+ Rs4,800= Rs52,800
For 2nd year
P=Rs52,800; R=10% and T= 1year
Interest= Rs= Rs5,280
Amount= Rs52,800+ Rs5,280= Rs58,080
For 3rd 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 1st year:
P = Rs.12,000, R = x% and T = 1
For 2nd 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 + 1.20x2) = Rs.920
⇒1.20x2 + 80x - 920 = 0
⇒3x2 + 200x - 2300 = 0
⇒3x2 + 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 1st year:
Interest = Rs.120x = Rs.1200
For 2nd year:
Interest = Rs.(80x + 1.20x2) = 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