MATHEMATICS IN EVERYDAY LIFE–8

[Pages:20]MATHEMATICS IN EVERYDAY LIFE?8

Chapter 10 : Compound Interest

EXERCISE 10.1

1. Principal for 1st year = `7600 Rate of interest = 5% per annum

Interest

for

1st

year

=

`

7600 5 100

1

= `380 Amount at the end of 1st year = `7600 + `380

= `7980 Principal for 2nd year = ` 7980

Interest

for

2nd

year

=

`

7980 5 100

1

= `399 Amount at the end of 2nd year = `7980 + `399

= `8379 Compound interest = Amount ? Principal

= `8379 ? `7600

= `779

Hence, compound interest = `779. 2. Principal for 1st year = `5000

Rate of interest = 15% per annum

5000 15 1 Interest for 1st year = ` 100

= `750 Amount at the end of 1st year = `5000 + `750

= `5750 Principal for 2nd year = `5750

Interest

for

2nd

year

=

`

5750 15 100

1

= `862.50 Amount at the end of 2nd year = `5750 + `862.50

= `6612.50 Compound interest = Amount ? Principal

= `6612.50 ? `5000 = `1612.50 Hence, compound interest = `1612.50.

Mathematics In Everyday Life-8

ANSWER KEYS

3. Principal for 1st year = `16000 Rate of interest = 5% per annum

16000 ? 5 ? 1 Interest for 1st year = ` 100

= `800 Amount at the end of 1st year = `16000 + `800

= `16800 Principal for 2nd year = `16800

Interest

for

2nd

year

=

`

16800 5 100

1

= `840

Amount at the end of 2nd year = `(16800 + 840)

= `17640

Compound interest = Amount ? Principal

= `(17640 ? 16000)

= `1640

Hence, she will pay `17640 at the end of 2 years and interest is `1640.

4. Principal for 1st year = `8000

Rate of interest = 18% per annum

Interest

for

1st

year

=

`

8000 18 100

1

=

`1440

Amount at the end of 1st year = `(8000 + 1440)

= `9440

Principal for 2nd year = `9440

Interest

for

2nd

year

=

`

9440 18 100

1

= `1699.20

Amount at the end of 2nd year = `(9440 + 1699.20)

= `11139.20

Principal for 3rd year = `11139.20

Interest

for

3rd

year

=

`

11139.20 100

18

1

= `2005.06

Amount at the end of 3rd year

= `(11139.20 + 2005.06)

= `13144.26

1

 Compound interest = `(13144.26 ? 8000) Compound interest = `5144.26. 5. Principal for 1st year = `3000 Rate of interest = 4% per annum

Interest

for

1st

year

=

`

3000 4 100

1

= `120

Amount at the end of 1st year = `(3000 + 120) = `3120

Principal for 2nd year = `3120

Interest

for

2nd

year

=

`

3120 4 100

1

= ` 124.80

Amount at the end of 2nd year = `(3120 + 124.80) = `3244.80

Compound interest = Amount ? Principal

= `(3244.80 ? `3000)

= `244.80 6. Principal for 1st year = `4000

Rate of interest = 9 1 % 19 % per annum 22

4000 19 1

Interest for 1st year = `

100 2

= `380

Amount at the end of 1st year = `(4000 + 380) = `4380

Principal for 2nd year = `4380

4380 19 1

Interest for 2nd year = `

100 2

= `416.10 Amount at the end of 2nd year = `(4380 + 416.10)

= `4796.10 Compound interest = Amount ? Principal

= `(4796.10 ? 4000) Compound interest = `796.10. 7. Principal for 1st year = `2000 Rate of interest = 10% per annum

Interest

for

1st

year

=

`

2000 10 100

1

= `200

Amount at the end of 1st year = `(2000 + 200)

= `2200

Principal for 2nd year = `2200

2

Interest

for

2nd

year

=

`

2200 10 100

1

= `220

Amount at the end of 2nd year = `(2200 + 220) = `2420

Principal for 3rd year = `2420

Interest

for

3rd

year

=

`

2420 10 100

1

= `242 Amount at the end of 3rd year = `(2420 + 242)

= `2662 Hence, he will repay `2662 after 3 years. 8. Principal for 1st year = `2500 Rate of interest = 8% per annum

Interest

for

1st

year

=

`

2500 8 100

1

= `200

Amount at the end of 1st year = `(2500 + 200) = `2700

Principal for 2nd year = `2700

Interest

for

2nd

year

=

`

2700 8 100

1

= `216

Amount at the end of 2nd year = `(2700 + 216)

= `2916 Compound interest = Amount ? Principal

= `(2916 ? 2500) = ` 416.

9. Principal for 1st year = `3500

Rate of interest = 20% per annum

3500 20 1

Interest for 1st year = `

100

= `700

Amount at the end of 1st year = `(3500 + 700)

= `4200

Principal for 2nd year = `4200

Interest

for

2nd

year

=

`

4200 20 100

1

= `840

Amount at the end of 2nd year = `(4200 + 840)

= `5040 Compound interest = Amount ? Principal

= `(5040 ? 3500) = ` 1540 Hence, compound interest = `1540.

Answer Keys

10. Principal for 1st year = `20,000 Rate of interest = 12% per annum

Interest

for

1st

year

=

`

20,000 12 100

1

= `2400

Amount at the end of 1st year = `(20,000 + 2400)

= `22,400

Principal for 2nd year = `22,400

22400 12 1

Interest for 2nd year = `

100

= `2688

Amount at the end of 2nd year = `(22400 + 2688)

= `25088

Hence, Rahim has to pay `25088 after 2 years.

EXERCISE 10.2

1. Principal (P) = `6400, Rate (R) = 5% per annum

Time (T) = 2 years

We have,

A =

P

1

R 100

T

=

`6400

1

5 100

2

=

`6400

1

1 20

2

=

`6400

21 2 20

= `6400

21 21 20 20

=

`

64

21 4

21

=

`

7056

Compound interest = Amount ? Principal

= `(7056 ? 6400)

Compound interest = ` 656

2. P = ` 2700, R = 6 2 % per annum = 20 % per annum,

3

3

T = 2 years

We have,

A=

P

1

R 100

T

=

`2700

1

3

20 100

2

=

`2700

1

20 300

2

=

`2700

1

1 15

2

=

`2700

16 2 15

=

`2700

16 16 15 15

=

`

2700 16 15 15

16

= `3072

Compound interest = A ? P = `(3072 ? 2700) = `372

3. P = `12800, R = 7 1 % per annum = 15 % per annum,

2

2

T = 1 year

If interest compounded semi-annually. Then

We have,

R=

15 22

=

15 % per 4

half

yearly

T = 1 2 = 2 half years.

A=

P

1

R 100

T

=

`12800

1

4

15 100

2

=

`12800

1

15 400

2

=

`12800

1

3 80

2

=

`12800

83 2 80

= `12800

83 83 80 80

128 83 83

= `

88

= `13778

Hence, Arun paid `13778 after one year.

4.

P =

` 6400,

T

=

2

years,

R

=

11 2

%

per

annum

We have,

A=

P

1

R 100

T

Mathematics In Everyday Life-8

3

=

`6400

1

2

11 100

2

=

`6400

1

11 200

2

=

`6400

211 2 200

= `6400 211 211 200 200

=

`

64

211 400

211

= `7123.36

Compound interest = A ? P

= `(7123.36 ? 6400)

= `723.36

5. P = `30,000, R = 7% p.a., T = 2 years

We have,

A=

P

1

R 100

T

=

`30,000

1

7 100

2

=

`30,000

107 2 100

= `30,000

107 107 100 100

= `(3 107 107)

= `34347

Hence, the amount is `34347.

6. P = `12000, R = 15% p.a., T = 3 years

We have,

A=

P

1

R 100

T

=

`12,000

1

15 100

3

=

`12,000

1

3 20

3

=

`12,000

23 3 20

= `12,000 23 23 23 20 20 20

=

`

12

23

8

23

23

A = `18250.50

4

Compound interest = A ? P = `(18250.50 ? 12000) = `6250.50

Compound interest = `6250.50

7. P = `2,50,000, R = 8% p.a., T = 1 1 = 3 years 22

If interest compounded half yearly. Then,

R=

8% 2

=

4% per

half year

3 T = 2 2 = 3 half years

Now, we have

A=

P

1

R 100

T

=

`2,50,000

1

4 100

3

=

`2,50,000

1

1 25

3

=

`2,50,000

26 3 25

=

`2,50,000

26 26 26 25 25 25

= `(16 26 26 26)

A = `2,81,216

Compound interest = A ? P

= `(2,81,216 ? 2,50,000)

= `31,216.

8. P = ` 25,000, R = 15% p.a., T = 2 years

Simple

interest

for

2

years

=

`

25000 15 100

2

S.I.

P

R 100

T

= `7500 Simple interest for 2 years = `7500

Amount to be paid by Ravi to Shishir after two years = `(25000 + 7500) = `32500

But, Ravi lent same money to Rakesh at the same rate of interest but compounded annually for 2 years. Then,

A=

P

1

R 100

T

=

`

25000

1

15 100

2

Answer Keys

=

`

25000

1

3 20

2

=

`

25000

23 2 20

= ` 25000 23 23 20 20

A

=

`

250

4

529

= ` 33062.50

Now, gain in this transaction

= `(33062.50 ? 32500)

= ` 562.50

Hence, he gain `562.50 in this transaction.

9. P = `9000, R = 10% p.a., T = 3 years

A=

P

1

R 100

T

=

`9000

1

10 100

3

=

`9000

1

1 10

3

=

`9000

11 3 10

= `9000 11 11 11 10 10 10

= `(9 1331) = `11979 Hence, Saurabh paid ` 11979 after three years. 10. Let principal be `P. T = 2 years, R = 12% p.a., S.I. = `1920

P RT S.I. = 100

`1920

=

`

P

12 100

2

Now,

1920 100

P=

12 2

P = `8000

A=

P

1

R 100

T

=

`8000

1

12 100

2

=

`8000

1

3 25

2

Mathematics In Everyday Life-8

=

`

8000

28 2 25

= ` 8000 28 28 25 25

=

`

64

28 5

28

= ` 10035.20

Compound interest = A ? P

= `(10035.20 ? `8000)

= `2035.20

Compound interest = `2035.20

11. P = ` 15000, R = 20% p.a., T = 2 years

If interest compounded half yearly. Then,

R = 20 % = 10% per half year, T = 2 ? 2 = 4 half-years. 2

We have,

A=

P

1

R 100

T

=

`15000

1

10 100

4

=

`15000

1

1 10

4

11 4 = `15000 10

= `15000 11 11 11 11 10 10 10 10

=

`

15000 14641 10000

= `21961.50 Compound interest = A ? P

= `(21961.50 ? 15000) = `6961.50 Compound interest = `6961.50.

12.

P = `

60000, R

= 10% p.a.,

T =

1 1 years 2

=

3 2

years

If interest compounded half yearly. Then,

10 R = 2 = 5% per half-year

T=

3 2

2 = 3 half-years

5

We have,

A=

P

1

R 100

T

=

`60000

1

5 100

3

=

`60000

1

1 20

3

=

`60000

21 3 20

= `60000

21 21 21 20 20 20

A

=

`

15

9261 2

Amount = `69457.50

Compound interest = A ? P

= `(69457.50 ? 60000)

= `9457.50. 13. P = ` 24000, R = 10% p.a., T = 2 years

If interest compounded semi-annually. Then,

We have,

10 R = 2 = 5% per semi-annually T = 2 2 = 4 half-years

A=

P

1

R 100

T

=

`24000

1

5 100

4

=

`24000

1

1 20

4

=

`24000

21 4 20

= `24000

21 21 21 21 20 20 20 20

3 194481 = ` 20 = ` 29172.15

Amount = `29172.15.

14. P = `8000, R = 12% per annum

If interest compounded quarterly. Then,

T = 9 months = 3 quarters ( 1 quarter = 3 months)

R=

12 % 4

=

3%

per

quarter

We have,

A=

P

1

R 100

T

=

`8000

1

3 100

3

= `8000 103 103 103 100 100 100

8741816 = ` 1000 = `8741.82 Amount = `8741.82.

15. P = ` 16000, R1 = 5% p.a. for first year, R2 = 10% per annum for second year We have,

A=

P

1

R1 100

1

R2 100

=

`16000

1

5 100

1

10 100

=

`16000

1

1 20

1

1 10

= `16000 21 11 20 10

= `(80 21 11) = `18480 Hence, amount to be paid for machine is `18480.

16.

P = `18000, R = 10% p.a., T =

31 3

years

When interest is compounded annually but time is a

fraction. Therefore,

Amount after

31 3

years = `18000

10 3 1 100

1

1310010

=

`18000

1

1 10

3

1

10 300

=

`

18000

11 3 10

31 30

= ` 18000 11 11 11 31 10 10 10 30

18 1331 31

= `

30

= ` 24756.60

6

Answer Keys

Compound interest = A ? P = `24756.60 ? `18000

Compound interest = `6756.60.

17.

P = ` 16,00,000,

R =

12 1 % = 2

25 % 2

per annum,

T = 1 year

If interest compounded half yearly. Then,

R=

25 25 % 22 4

per half year

T = 1 2 = 2 half years

A=

P

1

R 100

T

=

`16,00,000

1

25 400

2

=

`16,00,000

1

1 16

2

=

`16,00,000

17 2 16

= `16,00,000 17 17 16 16

= `6250 289

A = ` 18,06,250 Hence, he paid ` 18,06,250 after 1 year. 18. Let Principal be `P. T = 2 years, R = 5% per annum, S.I. = `800

We have,

P RT S.I. = 100

S.I. 100

P = RT

800 100

= `

52

P = `8000

Now, we have

A=

P

1

R 100

T

=

`8000

1

5 100

2

=

`8000

1

1 20

2

=

`8000

21 2 20

Mathematics In Everyday Life-8

= `8000 21 21 20 20

= `(20 441) A = `8820 Compound interest = A ? P

= `8820 ? `8000 Compound interest = `820. 19. P = ` 16000, R = 20% p.a., T = 1 year If interest compounded quarterly. Then,

We have,

R=

20 % 4

=

5%

per

quarter.

T = 1 4 = 4 quarters

A=

P

1

R 100

T

=

`16000

1

5 100

4

=

`16000

1

1 20

4

=

`16000

21 4 20

= `16000 21 21 21 21 20 20 20 20

A = `19448.10 Compound interest = `(19448.10 ? 16000)

= `3448.10 20. P = ` 18000, R = 20% p.a., T = 2 years

A=

P

1

R 100

T

=

`18000

1

20 100

2

=

`18000

1

1 5

2

=

`18000

6 2 5

A = `18000 6 6 55

= `25920 Compound interest = `(25920 ? 18000)

= `7920 If interest compounded half-yearly. Then,

R=

20 % 2

= 10% per half year.

T = 2 2 = 4 half-year

7

A=

P

1

R 100

T

=

`18000

1

10 100

4

=

`18000

1

1 10

4

=

`18000

11 4 10

= `18000 11 11 11 11 10 10 10 10

A = `26353.80. New compound interest = `(26353.80 ? 18000)

= `8353.80 Hence, Indu earns = `(8353.80 ? 7920)

= `433.80 more money.

EXERCISE 10.3

1. Let sum be `P.

R = 5% p.a., T = 2 years

S.I.

=

PRT P52

100

100

S.I. = P

10

A =

P

1

R 100

T

=

P

1

5 100

2

P

1

1 20

2

=

P

21 20

21 20

A = 441P 400

Compound interest = A ? P =

441P 400

? P =

41P 400

C.I. ? S.I.

=

41P P 400 10

=

41P 40P 400

=

P 400

But the difference is `60.

P

400 = `60

P = `24000

Hence, the sum is `24000.

2. R = 10% p.a., T = 3 years

For 1st finance company:

P R T P 10 3 S.I. = 100 100

S.I.

=

3P 10

8

For 2nd finance company:

A=

P

1

R 100

T

=

P

1

10 100

3

=

P

1

1 10

3

=

P

11 10

3

1331P A = 1000

1331P Compound interest = 1000 ? P

331P = 1000

331P 1000

=

3P 10

+ `186

331P 1000

?

3P 10

= `186

331P 300P

1000

= `186

31P = `186

1000

186000 P = ` 31

P = `6000

Hence, the sum is `6000.

3. P = `8000, A = `9261, T = 3 years, R = ?

Now,

A =

P

1

R 100

T

`9261

=

`8000

1

R 100

3

` 9261 ` 8000

=

1

R 100

3

9261 8000

=

1

R 100

3

21 3 20

=

1

R 100

3

21 20

=

1

R 100

[If am = bm, then a = b]

R 100

=

21 1 20

R 100

21 20 = 20

Answer Keys

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