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

Compound Interest (Without using formula) Exercise Ex. 2(A)

Solution 1(a)

Correct option: (i) Rs. 880

For a particular year, R = 10%, Interest = Rs. 800

Amount = P + I = Rs. 8000 + Rs. 800  = Rs. 8800

For next year at compound interest,

P = Rs. 8800

Solution 1(b)

Correct option: (ii) Rs. 250

P = Rs. 5000, R = 10%, T = 6 months   year

Solution 1(c)

Correct option: (iii) Rs. 512.50

For 1st  year: P = Rs. 5000, R = 10%, T   year

And, A = P + I = Rs. 5000 + Rs. 250 = Rs. 5250

 

For 2nd  year: P = Rs. 5250, R = 10%, T   year

And, A = P + I = Rs. 5250 + Rs. 262.50 = Rs. 5512.50

Therefore, compound interest = A - P = Rs. 5512.50 - 5000 = Rs. 512.50

Solution 1(d)

Correct option: (iii) Rs. 2,688

For 1st year: P = Rs. 20,000, R = 12%, T = 1 year

And, A = P + I = Rs. 20,000 + Rs. 2,400 = Rs. 22,400

 

For 2nd year: P = Rs. 22,400, R = 12%, T = 1 year

Solution 1(e)

Correct option: (i) Rs. 1,312.50

For a year 2022, R = 5%, Interest = Rs. 1,250

Amount = P + I = Rs. 25000 + Rs. 1250  = Rs. 26,250

For next year 2023 at compound interest,

P = Rs. 26,250

Solution 1(f)

Correct option: (ii) 23.2%

Let the initial value = Rs. 100

After one year, increase % = 10% of Rs. 100

Therefore, increased value = Rs. 110

After second year, increase % = 12% of Rs. 110

Therefore, increased value = Rs. 123.2

Now,

Solution 2

 

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 3

Solution 4

(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

Solution 5

Solution 6

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 7

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 8

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 9

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 10

Solution 11

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 12

Compound Interest (Without using formula) Exercise Ex. 2(B)

Solution 1(e)

Correct option: (iii) Rs. 4,000

Let the Principal = Rs. 100

Then,

For S.I.: P = Rs. 100, R = 5% and T = 2 years

For C.I.:

Principal for 1st year = Rs. 100

And, amount = Rs. 100 + Rs. 5 = Rs. 105

Then, Principal for 2nd year = Rs. 105

Then, C.I. of 2 years = Rs. 5 + Rs. 5.25 = Rs. 10.25

Difference between C.I. and S.I. = Rs. 10.25 – Rs. 10 = Rs. 0.25

When difference between C.I. and S.I. for 2 years is Rs. 0.25, Principal = Rs. 100

Therefore, when the difference between C.I. and S.I. for 2 years is Rs. 100,

Principal

Solution 1(a)

Correct option: (i) Rs. 00

C.I. and S.I. are same for the first year on the same sum and at the same rate percent.

Solution 1(b)

Correct option: (ii) 572.80

For year 2023:

Since money deposited at the beginning of the year = Rs. 2,000

Principal for year 2023 = Rs. 2,000

And, amount = Rs. 2,000 + Rs. 160 = Rs. 2160

For year 2024:

Since money deposited at the beginning of the year = Rs. 3,000

Principal for year 2024 = Rs. 2,160 + Rs. 3,000 = Rs. 5,160

And, amount = Rs. 5,160 + Rs. 412.80 = Rs. 5572.80

Therefore, compound interest at the end of 2024 = Rs. 160 + Rs. 412.80 = Rs. 572.80

 

*Back answer is different.

Solution 1(c)

Correct option: (i) Rs. 880

For 1st year:

P = Rs. 1000, R = 10%, T = 1 year

And, amount = Rs. 1000 + Rs. 100 = Rs. 1100

Money repaid = Rs. 300

Balance = Rs. 1100 - Rs. 300 = Rs. 800

For 2nd year:

P = Rs. 800, R = 10%, T = 1 year

And, amount = Rs. 800 + Rs. 80 = Rs. 880

Solution 1(d)

Correct option: (iv) Rs. 40

For S.I.: P = Rs. 4000, R = 10% and T = 2 years

For C.I.:

Principal for 1st year = Rs. 4000

And, amount = Rs. 4000 + Rs. 400 = Rs. 4400

Then, Principal for 2nd year = Rs. 4400

Then, C.I. of 2 years = Rs. 400 + Rs. 440 = Rs. 840

Therefore, difference between C.I. and S.I. = Rs. 840 - Rs. 800 = Rs. 40

Solution 2

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 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

(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 5

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 6

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

straight I equals fraction numerator 14000 cross times 15 cross times 1 over denominator 100 end fraction equals 2100

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 7

Solution 8

For 1st six months:

P = Rs. 5,000, R = 12% and T = 1 half 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 = 1 half 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 = 1 half 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 9

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 10

Solution 11

Compound Interest (Without using formula) Exercise Ex. 2(C)

Solution 1(a)

Correct option: (iv) 20%

The difference between the compound interest for any two consecutive conversion periods is the interest of one period on the C.I. of the preceding conversion period.

Here,

Difference in compound interest for two consecutive years = Rs. 360 - Rs. 300 = Rs. 60

Now,   

Solution 1(b)

Correct option: (ii) 20%

The difference between the amounts for any two consecutive conversion periods is the interest of one period on the C.I. of the preceding period.

Here,

Difference in amounts for 6th year and 5th years = Rs. 6000 - Rs. 5000 = Rs. 1000

Now,   

Solution 1(c)

Correct option: (i) 3500

Let C.I. at the end of 2019 = Rs. x

Since, C.I. at the end of 2020 = C.I. of year 2019 + Interest on its for one year

Rs. 3850 = Rs. x + 10% of Rs. x

Solution 1(d)

Correct option: (ii) 5,040

Amount in 7 years = Amount in 6 years + Interest on it for 1 year

= Rs. 4500 + 12% of Rs. 4500 

Solution 1(e)

Correct option: (iii) 15%

Amount in 2nd year = Amount in 1st year + Interest on it for 1 year

Rs. 690 = Rs. 600 + R% of Rs. 600

Solution 1(f)

Correct option: (ii) 15%

Amount in 2nd year = Amount in 1st year + Interest on it for 1 year

Rs. 2760 = Rs. 2400 + R% of Rs. 2400

Solution 2

(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 3

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 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

Solution 6

(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 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

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 9

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 10

(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 11

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 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=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 13

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 Test Yourself

Solution 1

Let principal (p) = Rs. 100

For 1st year

P = Rs. 100

R = 10%

T = 1 year

straight I equals fraction numerator 100 cross times 10 cross times 1 over denominator 100 end fraction equals 10

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. 

 

S i m p l e space I n t e r e s t equals fraction numerator x cross times 8 cross times 2 over denominator 100 end fraction equals 0.16 x

 

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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