# Class 9 SELINA Solutions Maths Chapter 3 - Compound Interest (Using Formula)

## Compound Interest (Using Formula) Exercise Ex. 3(A)

### Solution 1

Given : P= Rs12,000; n=3years and r=5%

Amount= =

=

=Rs13,891.50 Ans.

C.I. =RS13,891.50 - Rs12,000

= Rs1,891.50 Ans.

### Solution 2

Given : P= Rs15,000; n=2years; r_{1} =8% and r_{2} =10%

Amount==

=

=Rs17,820 Ans.

### Solution 3

Given : P=Rs6,000; n= 3years; r_{1}= 5%; r_{2}= 8% and r_{3} =10%

Amount=

=

=

=Rs7,484.40

C.I. = Rs7,484.40 - Rs6,000 = Rs1,484.40 Ans.

### Solution 4

Given : Amount= Rs5,445; n= 2years and r = 10%

A=

5,445=

5,445=

P==Rs4,500 Ans.

### Solution 5

Given : C.I.= Rs768.75; n= 2years and r = 5%

A=

A=

A==

A - P =C.I

- P=Rs768.75

=Rs768.75

P=Rs Ans.

### Solution 6

Given : C.I.= Rs1,655; n= 3years and r = 10%

A=

A=

A=

A - P =C.I

- P=Rs1,655

=Rs1,655

P=Rs Ans.

### Solution 7

Given : Amount =Rs9,856; n=2years; r_{1} =10% and r_{2} =12%

Ans.

### Solution 8

The sum is Rs.16,000

### Solution 9

At 5% per annum the sum of Rs.6,000 amounts to Rs.6,615 in 2 years when the interest is compounded annually.

### Solution 10

Let Principal = Rs y

Then Amount= Rs 1.44y

n= 2years

### Solution 11

### Solution 12

Given: P=Rs5,000; A=Rs6,272 and n= 2years

(i)

(ii) Amount at the third year

### Solution 13

Given : P=Rs7,000; A=Rs9,317 and r= 10%

### Solution 14

Given : P=Rs4,000; C.I.=Rs630.50 and r=5%

### Solution 15

Let share of A = Rs y

share of B = Rs (28,730 - y)

rate of interest= 10%

According to question

Amount of A in 3years= Amount of B in 5years

Therefore share of A=Rs15,730

Share of B=Rs28,730 - Rs 15,730=Rs13,000

### Solution 16

(i)Let share of John = Rs y

share of Smith = Rs (44,200 - y)

rate of interest= 10%

According to question

Amount of John in 4years= Amount of Smith in 2years

Therefore share of John=Rs20,000

Share of Smith=Rs44,200- Rs 20,000=Rs24,200

(ii)Amount that each will receive

### Solution 17

### Solution 18

## Compound Interest (Using Formula) Exercise Ex. 3(B)

### Solution 1

Let principal (P) = x

R = 8%

T = 2 years

_{}

Given, CI = SI = 54.40

_{}

_{}

Thus, principal sum = Rs. 8500

### Solution 2

(for 2 years) A = Rs. 19360

T = 2 years

Let P = X

_{}...(1)

A (for 4 years) = Rs. 23425.60

_{}...(2)

(2) _{}(1)

_{}

_{}

Form (1), we have

_{}

_{}

Thus, sum = Rs. 16000

### Solution 3

Let principal = x, A = 3x, T = 8 years, R = ?

Case I,

Case II,

P = x, A = 27x, T = ?

_{}

From (1) and (2)

_{}

Hence, time = 24 years.

### Solution 4

P = Rs. 9430

R = 5%

R = 10 years

SI = _{}

Let sum = x

CI = 4715, T = 2 years, Rs= 5%

CI = A - P

_{}

Thus principal from = Rs. 46,000

### Solution 5

Let principal = Rs. 100, R = 5% T = 2 years

For Kamal, SI = _{}

For Anand,

_{}

CI =

Difference of CI and SI = _{}

_{}

When difference is Rs. _{}, then principal = Rs. 100

If difference is 1, principal = 100 _{}4

If difference is Rs, 15, principal = 100 _{}4 _{}15 = Rs. 6000

For kamal, interest = _{}

For Anand, interest = _{}

_{}

### Solution 6

SI = Rs. 450

R = 4%

T = 2 years

P = ?

_{}

Now, P = 5625, R = 4%, T = 2 years

A = _{}

_{}

CI = A - P = 6084 - 5625

= Rs. 459

### Solution 7

Let principal (P), R = 4%, T = 4 years

_{}

Given: SI - CI = Rs. 228

_{}

Thus, Principal = Rs. 96000

### Solution 8

CI = Rs. 246, R = 5%, T = 2 years

CI = A - P

Now, P = Rs. 2400, R = 6%, T = 3 years

_{}

### Solution 9

Let the sum (principle) = *x*

Given Amount = 23400, R = 10% and T = 3 years

Amount = Principle + Interest

23400 = *x *+

*x *= 18000

Principle = 18000

Now,

Principle = `18000, r = 10% and n = 2 years

The amount of the same sum in 2 years and at 10% p.a. compound interest is 21780.

### Solution 10

## Compound Interest (Using Formula) Exercise Ex. 3(C)

### Solution 1

Given: P=Rs7,400; r=5% p.a. and n= 1year

Since the interest is compounded half-yearly,

Then

### Solution 2

(i)When interest is compounded yearly

Given: P=Rs10,000; n=18months=year and r=10%p.a.

For 1year

For1/2 year

P=Rs11,000;n= 1/2 year and r=10%

C.I.= Rs11,550 - Rs10,000= Rs1,550

(ii)When interest is compounded half-yearly

P=Rs10,000; n= year and r=10%p.a.

C.I.= Rs11,576.25 - Rs10,000=Rs1,576.25

Difference between both C.I.= Rs1,576.25 - Rs1,550

= Rs26.25 Ans.

### Solution 3

For the first 2 years

Amount in the account at the end of the two years is Rs.22,400.

For the remaining one year

The total amount to be paid at the end of the three years is Rs.27,104.

### Solution 4

The sum of Rs.24,000 amount Rs.27,783 in one and a half years at 10% per annum compounded half yearly.

### Solution 5

(i)For Ashok(interest is compounded yearly)

Let P=Rs y; n=18months=year and r=20%p.a.

For 1year

For1/2 year

P;n= ½ year and r=20%

(ii)For Geeta(interest is compounded half-yearly)

P=Rs y; n= year and r=20%p.a.

According to question

Money invested by each person=Rs3,000 Ans.

### Solution 6

The rate of interest is 8%.

### Solution 7

Given: P=Rs1,500; C.I.=Rs496.50 and r=20%

Since interest is compounded semi-annually

Then

Ans.

### Solution 8

Given: P=Rs 3,500; r=6% and n= 3years

Since interest is being compounded half-yearly

Then

Ans.

### Solution 9

Given: P=Rs12,000; n= years and r= 10%

To calculate C.I.

For 1 year

P=Rs 12,000; n=1 year and r=10%

For next 1/2 year

P=Rs 13,200; n= 1/2 year and r=10%

C.I. = Rs 13,860 - Rs 12,000= Rs 1,860

Difference between C.I. and S.I

= Rs 1,860 - Rs 1,800=Rs 60 Ans.

### Solution 10

Given: P=Rs12,000; n= years and r= 10%

To calculate C.I.(compounded half-yearly)

P=Rs12,000;n= and r=10%

C.I. - Rs12,000= Rs1,891.50

Difference between C.I. and S.I

=Rs1,891.50 - Rs1,800=Rs91.50 Ans.

## Compound Interest (Using Formula) Exercise Ex. 3(D)

### Solution 1

Cost of machine in 2008 = Rs44,000

Depreciation rate=12%

(i) Cost of machine at the end of 2009

(ii) Cost of machine at the beginning of 2007(P)

### Solution 2

Let x be the value of the article.

The value of an article decreases for two years at the rate of 10% per year.

The value of the article at the end of the 1^{st} year is

X - 10% of x = 0.90x

The value of the article at the end of the 2^{nd} year is

0.90x - 10% of (0.90x) = 0.81x

The value of the article increases in the 3^{rd} year by 10%.

The value of the article at the end of 3^{rd } year is

0.81x + 10% of (0.81x) = 0.891x

The value of the article at the end of 3 years is Rs.40,095.

_{ }

The original value of the article is Rs.45,000.

### Solution 3

Population in 2009 (P) = 64,000

Let after n years its population be 74,088(A)

Growth rate= 5% per annum

Ans.

### Solution 4

Let the population in the beginning of 1998 = P

The population at the end of 1999 = 2,85,120(A)

r_{1} = - 12% and r_{2} = +8%

Ans.

### Solution 5

Let sum of money be Rs P and rate of interest= r%

Money after 1year= Rs16,500

Money after 3years=Rs19,965

For 1year

For 3years

Divide eq^{n} (2) by eq^{n} (1)

On comparing, we get

r= 10% Ans.

Put value of r in eq^{n} (1)

Ans

### Solution 6

Given: P = Rs 7,500 and Time(n)= 2 years

Let rate of interest = y%

Given: C.I. - S.I. = Rs 12

Ans.

### Solution 7

Let Principal be Rs y and rate= r%

According to 1^{st} condition

Amount in 10 years = Rs 3y

According to 2^{nd} condition

Let after n years amount will be Rs 27y

### Solution 8

At the end of the two years the amount is

_{ }

Mr. Sharma paid Rs.19,360 at the end of the second year.

So for the third year the principal is A_{1} - 19,360.

Also he cleared the debt by paying Rs.31,944 at the end of the third year.

_{ }

Mr. Sharma borrowed Rs.40,000.

### Solution 9

Let sum of money be RS y

To calculate S.I.

To calculate C.I.(compounded half-yearly)

### Solution 10

Let Rs.x and Rs.y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.

Pramod will attain the age of 25 years after 25 - 16 = 9 years

Rohit will attain the age of 25 years after 25 -18 = 7 years

_{}

Pramod and Rohit should invest in 400:441 ratio respectively such that they will get the same sum on attaining the age of 25 years.

## Compound Interest (Using Formula) Exercise Ex. 3(E)

### Solution 1

1^{st} case

Given: S.I. = Rs 450;Time= 2 years and Rate = 4%

2^{nd} case(compounded half-yearly)

P = Rs.5,625;n= 1 year and r = 4%

### Solution 2

For 2years

For ½ year

_{ }

C.I. = A - P = Rs.13,721 - Rs.10,800 = Rs.2,921

### Solution 3

(i) Present value of machine(P) = Rs.97,200

Depreciation rate = 10%

=Rs.78732

(ii) Present value of machine(A) = Rs.97,200

Depreciation rate = 10% and time = 2 years

To calculate the cost 2 years ago

### Solution 4

Let the sum of money lent by both Rs.y

For Anuj

P = Rs.y ;rate = 8% and time = 2 years

For Rajesh

P = Rs.y ;rate = 8% and time = 2 years

Given : C.I. - S.I. = Rs.64

### Solution 5

Given: Principal = Rs.4,715;time = 5 years and rate= 5% p.a.

Then C.I. = Rs.1,178.75 x 4 = Rs.4,715

Time = 2 years and rate = 5%

### Solution 6

Given: C.I. for the 2^{nd} year = Rs.4,950 and rate = 15%

Then amount at the end of 2^{nd} year= Rs.33,000

For first 2years

A = Rs.33,000; r_{1} =10%

The sum invested is Rs.30,000.

### Solution 7

Let the sum of money be Rs.y

and rate = 10% p.a. compounded half yearly

For first 6months

For first 12 months

Given: The difference between the above amounts = Rs.189

y = 3600

### Solution 8

P = Rs.86,000;time = 2 years and rate = 5% p.a.

To calculate S.I.

To calculate C.I.

Profit = C.I. - S.I. = Rs.8,815 - Rs.8,600 = Rs.215

### Solution 9

Let Rs.x be the sum of money.

Rate = 5 % p.a. Simple interest = Rs.1,200, n = 3years.

The amount due and the compound interest on this sum of money at the same rate and after 2 years.

P = Rs.8,000;rate = 5% p.a., n = 3 years

The amount due after 2 years is Rs.8,820 and the compound interest is Rs.820.

### Solution 10

Let x% be the rate of interest.

P = Rs.6,000, n = 2 years, A = Rs.6,720

- For the first year

The rate of interest is x% = 12%.

- The amount at the end of the second year.

The amount at the end of the second year = Rs.7,526.40