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

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

Correct option: (ii) Rs. 250

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

Correct option: (iii) Rs. 512.50

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

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

For 2^nd 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

Correct option: (iii) Rs. 2,688

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

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

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

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

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,

Thus, the amount in 4 years is Rs. 19448.10.

(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

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

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

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

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

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

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 1^st year = Rs. 100

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

Then, Principal for 2^nd 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

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.

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.

Correct option: (i) Rs. 880

For 1^st 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 2^nd year:

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

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

Correct option: (iv) Rs. 40

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

For C.I.:

Principal for 1^st year = Rs. 4000

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

Then, Principal for 2^nd 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

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

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

(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)

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

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

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.

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

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,  

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 6^th year and 5^th years = Rs. 6000 - Rs. 5000 = Rs. 1000

Now,  

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

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 

Correct option: (iii) 15%

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

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

Correct option: (ii) 15%

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

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

(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

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

(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

(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

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

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%

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.

(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.

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.

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.

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.

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

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%

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

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.

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

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.

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

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

(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 + 1.20x²) =  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 1^st year:

Interest = Rs.120x = Rs.1200

For 2^nd 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