Compound Interest Using Formula Concise ICSE Class-9th

Compound Interest Using Formula Concise ICSE Class-9th Mathematics Selina solutions Chapter-3 . We provide step by step Solutions of Exercise / lesson-3 Compound Interest (Using Formula)   for ICSE Class-9 Concise Selina Mathematics by RK Bansal.

Our Solutions contain all type Questions with Exe-3 A, Exe-3 B, Exe-3 C, Exe-3 D and Exe-3 E to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

Compound Interest Using Formula Concise ICSE Class-9th Mathematics Selina solutions Chapter-3

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–: Select Topics :–

Concept Part-1

Concept Part-2

Exe-3 A,

Exe-3 B,

Exe-3 C,

Exe-3 D,

Exe-3 E,

 

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Exercise – 3(A) Compound Interest (Using Formula) Concise Selina Mathematics ICSE Class-9th 

Question 1

Find the amount and the compound interest on Rs. 12,000 in 3 years at 5% compounded annually.

Answer

Given : P= Rs. 12,000; n = 3 years and r = 5%

= Rs. 13,891.50

C.I. = Rs. 13,891.50 – Rs. 12,000 = Rs. 1,891.50.

Question 2

Calculate the amount of Rs. 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% and 10% respectively.

Answer

Given : P = Rs. 15,000; n = 2 years;  r₁ = 8 % and r₂ = 10%

Answer 

Given : P = Rs. 6,000; n = 3 years; r₁ = 5%; r₂ = 8% and r₃ = 10%

= Rs. 7,484.40
∴ C.I. = Rs. 7,484.40 – Rs. 6,000 = Rs. 1,484.40.

Answer 

Given : P= Rs. 5,445 ; n = 2 years and r = 10%

Answer 

Given : C.I.= Rs. 768.75; n= 2 years and r = 5%

Answer 

Given : C.I. = Rs. 1,655; n = 3 years and r = 10%

Answer 

Given : A = Rs. 9,856 ; n = 2 years ;  r₁ = 10 % and r₂ = 12%

Question 8

Answer 

‘

⇒ ( P + 4240 ) = $P\left(1.265\right)$

⇒ P = Rs. 16000

The sum is Rs.16,000.

Answer 

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

Answer 

Let Principal = Rs. y
Then Amount= Rs 1.44y
n= 2 years

On solving, we get
r = 20 %

Answer 

Given : P = Rs. 4,000, C.I. = Rs. 1,324 and n = 3 years
Now, A = P + I
⇒ A = Rs. ( 4,000 + 1,324 ) = Rs. 5,324

Answer 

Given: P = Rs. 5,000; A = Rs. 6,272 and n = 2 years.

Answer 

Given : P = Rs. 7,000; A = Rs. 9,317 and r = 10%.

On comparing,
n = 3 years

Answer 

Given : P= Rs. 4,000; C.I.= Rs. 630.50 and r = 5%

On comparing,

n = 3 years

Answer  

Let share of A = Rs. y
share of B = Rs (28,730 – y)
rate of interest= 10%

According to question,
Amount of A in 3 years= Amount of B in 5 years

Therefore share of A = Rs. 15,730

Share of B = Rs. 28,730 – Rs.15,730 = Rs. 13,000

Answer 

(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 4 years = Amount of Smith in 2 years

Answer 

(i) I = Rs. 6,000, T = 2 years and R = 10%

= Rs. 39,930

(iii) C.I. earned in 3 years = A – P = Rs. (39,930 – 30,000) = Rs. 930.

Question 18

Answer

Given : P = Rs. 8,000, R = 5%, T = 2 years
For simple interest,

C.I. = A – P
= Rs. (8,820 – 8,000)
= Rs. 820

Now, C.I. – S.I. = Rs. ( 820 – 800 ) = Rs. 20.
Thus, the difference between the compound interest and the simple interest is Rs. 20.

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Compound Interest (Using Formula) Exe-3 B, Concise Selina Mathematics ICSE Class-9th  

Question 1

The difference between simple interest and compound interest on a certain sum is Rs. 54.40 for 2 years at 8 per cent per annum. Find the sum.

Answer 

Let principal (P) = x
R = 8%
T = 2 years

 

 

X = Rs. 8500
Thus, principal sum = Rs. 8500

Question 2

Answer 

(for 2 years) A = Rs. 19360
T = 2 years
Let P = X

 

X = Rs. 16,000

Thus, sum = Rs. 16000

Question 3

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money becomes twenty-seven times of itself at the same rate of interest p.a.

Answer 

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

Case I,

T = 24

Time = 24 years.

Question 4

Answer 

P = Rs. 9430
R = 5%
T = 10 years

Question 5

Answer 

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

Question 6

Answer 

SI = Rs. 450
R = 4%
T = 2 years
P = ?

Question 7

Answer 

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

P = Rs. 96000

Thus, Principal = Rs. 96000

Question 8

Answer 

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

CI = A – P

Question 9

Answer 

Let the sum (principle) = X
Given Amount = 23400, R = 10% and T = 3 years

A = 21780

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

 Question 10

Answer 

For the payment of Rs. 12,600 at the end of first year :
A = Rs. 12,600; n = 1 year and r = 5%

Concise Selina Mathematics ICSE Class-9th Exe-3 C, Compound Interest (Using Formula)

 Question 1

Answer 

Given: P = Rs. 7,400; r = 5% p.a. and n = 1 year

Since the interest is compounded half-yearly,

Question 2

Answer 

(i) When interest is compounded yearly :
Given : P = Rs. 10,000 ; n = 18 months = $1.5$ year and r = 10% p.a.
For 1 year

 

= Rs. 11,576.25

C.I.= Rs.11,576.25 – Rs.10,000 = Rs. 1,576.25

Difference between both C.I. = Rs. 1,576.25 – Rs. 1,550 = Rs. 26.25

Question 3

A man borrowed Rs.16,000 for 3 years under the following terms:
20% simple interest for the first 2 years.
20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.
Find the total amount to be paid at the end of the three years.

Answer 

For the first 2 years

⇒ A = 27,104

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

Question 4

Answer 

 

⇒ P = 24,000

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

Question 5

Answer 

(i) For Ashok (interest is compounded yearly) : 

Let P = Rs. y; n = 18 months = $1.5$ year and r = 20% p.a.

For 1 year

Question 6

Answer 

⇒ r = 8

The rate of interest is 8%.

Question 7

Answer 

Given: P=Rs. 1,500; C.I.= Rs. 496.50 and r = 20%
Since interest is compounded semi-annually

 Question 8

Calculate the C.I. on Rs. 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly.
Do not use mathematical tables. Use the necessary information from the following:
(1.06)³ =1.191016; (1.03)³ = 1.092727
(1.06)⁶ =1.418519; (1.03)⁶ = 1.194052

Answer 

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

Since interest is being compounded half-yearly

= 3,500[(1.03)⁶ – 1 ]

= 3,500[ 1.194052 – 1 ]

= 3,500 x 0.194052

= Rs. 679.18

Question 9

Find the difference between compound interest and simple interest on Rs.12,000 and in  $1.5$ years at 10% compounded yearly.

Answer 

= Rs. 13,860

∴ 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.

Question 10

Find the difference between compound interest and simple interest on Rs. 12,000 and in $1.5$ years at 10% compounded half-yearly.

Answer 10

A = Rs. 13,891.50

C.I. = Rs. 13,891.50 – Rs. 12,000 = Rs. 1,891.50

∴ Difference between C.I. and S.I = Rs. 1,891.50 – Rs. 1,800 = Rs. 91.50.

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EXERCISE- 3(D), Compound Interest (Using Formula)Selina solutions for ICSE Class -9th

Question 1

The cost of a machine is supposed to depreciate each year at 12% of its value at the beginning of the year. If the machine is valued at Rs. 44,000 at the beginning of 2008, find its value :
(i) at the end of 2009.
(ii) at the beginning of 2007.

Answer  

Cost of machine in 2008 = Rs. 44,000
Depreciation rate = 12%

(i)  ∴ Cost of machine at the end of 2009

Question 2

Answer  

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.
0.891X = 40,095

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

Question 3

Answer  

Population in 2009 (P) = 64,000
Let after n years its population be 74,088(A)
Growth rate= 5% per annum

On comparing, we get,
n = 3 years

Question 4

Answer  

Let the population in the beginning of 1998 = P
The population at the end of 1999 = 2,85,120(A)
r₁ = – 12% and r₂ = + 8%

Question 5

Answer  

Let sum of money be Rs P and rate of interest = r %
Money after 1 year = Rs. 16,500
Money after 3 years = Rs. 19,965

For 1 year

Question 6

Answer  

Given : P = Rs. 7,500 and Time(n) = 2 years
Let rate of interest = y%

⇒ y² = 16
⇒ y = 4 %

Question 7

Answer  

Let Principal be Rs y and rate= r%
According to 1^st condition
Amount in 10 years = Rs 3

On comparing, we get
n = 10 x 3 = 30 years

Question 8

Answer  

At the end of the two years the amount is

⇒  P = Rs. 40,000

Mr. Sharma borrowed Rs. 40,000.

Question 9

Answer  

Let sum of money be Rs. y
To calculate S.I.

Question 10

Answer  

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.

Simple interest on a sum of money for 2 years at 4% is Rs .450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half yearly.

Answer  

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

= Rs. 5852.25

∴ C.I. = 5,852.25 – 5,625 = Rs. 227.25

Question 2

Find the compound interest to the nearest rupee on Rs. 10,800 for $2.5$ years at 10% per annum.

Answer  

Given : P = Rs. 10,800 ; Time = $2.5$ years and Rate = 10% p.a.

For 2 years

Question 3

The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs.97,200, find:

1.   Its value after 2 years.
2.   Its value when it was purchased.

Answer  

(i) Present value of machine(P) =  Rs.97,200
Depreciation rate = 10%

 

Question 4

Answer  

Let the sum of money lent by both Rs. y
For Anuj
P = Rs.y ; rate = 8% and time = 2 years

Question 5

Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on Rs. 4,715 for 5 years, both at the rate of 5% per annum.

Answer  

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

Question 6

Answer  

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

 

 

Question 7 Compound Interest Using Formula

Answer  

Let the sum of money be Rs. y
and rate = 10% p.a. compounded half yearly

Question 8

Answer  

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

Question 9

Answer  

Let Rs.X be the sum of money.
Rate = 5 % p.a. Simple interest = Rs.1,200, n = 3 years.

 

⇒ A = 8,000( 1.1025 )

⇒ A = 8,820

C.I. = A – P
⇒ C.I. = 8,820 – 8,000
⇒ C.I. = 820.
The amount due after 2 years is Rs. 8,820 and the compound interest is Rs. 820.

Question 10

Nikita invests Rs.6,000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts to Rs.6,720. Calculate:
(a) The rate of interest.
(b) The amount at the end of the second year.

Answer  

Let X % be the rate of interest.
P = Rs. 6,000, n = 2 years, A = Rs.6,720
For the first year

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

 End of Compound Interest Using Formula Solutions :–

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