# SELINA Solutions for Class 9 Maths Chapter 2 - Compound Interest (Without using formula)

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## Chapter 2 - 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

## Chapter 2 - 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

## Chapter 2 - 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.

## Chapter 2 - 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.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

Solution 10

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