Question

1. Calculate the compound interest on 10,000 for 2 years at 7% p.a.
2. Calculate the compound interest on 15,000 for 2 years at 6% p.a.
3. Calculate the amount and the compound interest on 8000 for 3 years at 15% p.a., interest
compounded annually.
4. Harish invests * 12,500 for 2 years at 12% p.a., calculate the amount and the compound interest
that Harish will get after 2 years.
5. Mahesh invests 3000 for 3 years at the rate of 10% p.a. compound interest. Find the amount and
the compound interest that Ramesh will get after 3 years.​

Answer 1

Solution 1

Given

• Principal = ₹10000.
• Time = 2 years.
• Rate = 7% p.a.

To Find

• Compound Interest.

Solution

A = P(1+ R/100)^n

⇒ A = 10000(1+ 7/100)^n

⇒ A = 10000(107/100)²

⇒ A = 10000 × 107/100 × 107/100

⇒ A = 107 × 107

∴ Amount = ₹11449.

C.I. = A - P

⇒ C.I. = 11449 - 10000

∴ Compound Interest = ₹1449.

Final Answer

• Compound Interest = ₹1449.

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Solution 2

Given

• Principal = ₹15000.
• Time = 2 years.
• Rate = 6% p.a.

To Find

• Compound Interest.

Solution

A = P(1+ R/100)^n

⇒ A = 15000(1+ 6/100)²

⇒ A = 15000(1+ 3/50)²

⇒ A = 15000(53/50)²

⇒ A = 15000 × 53/50 × 53/50

⇒ A = 6 × 53 × 53

∴ Amount = ₹16854.

C.I. = A - P

⇒ C.I. = 16854 - 15000

∴ Compound Interest = ₹1854.

Final Answer

• Compound Interest = ₹1854.

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Solution 3

Given

• Principal = ₹8000.
• Time = 3 years.
• Rate = 15% p.a.

To Find

• Amount.
• Compound Interest.

Solution

A = P(1+ R/100)^n

⇒ A = 8000(1+ 15/100)³

⇒ A = 8000(1+ 3/20)³

⇒ A = 8000(23/20)³

⇒ A = 8000 × 23/20 × 23/20 × 23/20

⇒ A = 23 × 23 × 23

∴ Amount = ₹12167.

C.I. = A - P

⇒ C.I. = 12167 - 8000

∴ Compound Interest = ₹4167.

Final Answer

• Amount = ₹12167.
• Compound Interest = ₹4167.

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Solution 4

Given

• Principal = ₹12500.
• Time = 2 years.
• Rate = 12% p.a.

To Find

• Amount.
• Compound Interest.

Solution

A = P(1+ R/100)^n

⇒ A = 12500(1+ 12/100)²

⇒ A = 12500(1+ 3/25)²

⇒ A = 12500(28/25)²

⇒ A = 12500 × 28/25 × 28/25

⇒ A = 20 × 28 × 28

∴ Amount = ₹15680.

C.I. = A - P

⇒ C.I. = 15680 - 12500

∴ Compound Interest = ₹3180.

Final Answer

• Amount = ₹15680.
• Compound Interest = ₹3180.

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Solution 5

Given

• Principal = ₹3000.
• Time = 3 years.
• Rate = 10% p.a.

To Find

• Amount.
• Compound Interest.

Solution

A = P(1+ R/100)^n

⇒ A = 3000(1+ 10/100)³

⇒ A = 3000(1+ 1/10)³

⇒ A = 3000(11/10)³

⇒ A = 3000 × 11/10 × 11/10 × 11/10

⇒ A = 3 × 11 × 11 × 11

∴ Amount = ₹3993.

C.I. = A - P

⇒ C.I. = 3993 - 3000

∴ Compound Interest = ₹993.

Final Answer

• Amount = ₹3993.
• Compound Interest = ₹993.

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Answer 2

Question ¹

GiveN:-

• Principal = Rs.10000
• Rate = 7% per annum
• Time = 2 years

To FinD:-

• The Compound Interest.

SolutioN:-

We know that,

$\large{\pink{\underline{\boxed{\bf{Amount=P\left(1+\dfrac{R}{100}\right)^n}}}}}$

where,

P is principal = Rs.10000

R is rate = 7%

n is time = 2 years

Putting the values,

$\sf :\implies{Amount = 10000 (1 + \dfrac{7}{100})^2 }$

$\sf :\implies{Amount = 10000 ( \dfrac{100 + 7}{100})^2 }$

$\sf :\implies{Amount = 10000 × (\dfrac{107}{100})^2 }$

$\sf :\implies{Amount = 10000 × (\dfrac{107}{100} × \dfrac{107}{100}) }$

$\sf :\implies{Amount = 10000 ×\dfrac{107}{100}\dfrac{107}{100} }$

$\sf :\implies{Amount = 1\cancel{00}\cancel{00} \dfrac{107}{\cancel{100}} × \dfrac{107}{\cancel{100}}}$

$\sf :\implies{Amount = 1 × 107 × 107 }$

$\sf\color{lime}\fbox{\bf\color{purple}{Amount = 11,449.}}$

Now,

We know that,

$\large{\pink{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

where,

Amount = Rs.11,449

Principal = Rs.10000

Putting the values,

$\sf :\implies{Compound \: Interest = Rs,(11,449 - 10, 000)}$

$\sf\color{lime}\fbox{\bf\color{purple}{Compound \: Interest = 1,449.}}$

The Compound Interest is Rs.1,449

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Question ²

GiveN:

Calculate the compound interest on ` 15,000 for 2 years at 6% p.a.

SolutioN:

We Know that,

$\rm{A = P(1 + \dfrac{r}{100})ⁿ}$

Here, A is the amount after $n$ year at a rate of $r\%$ on a principal P

$:\implies\sf { A = 15000(1 + \dfrac{6}{100})²}$

$:\implies\sf {A = 15000(\cancel{\dfrac{53}{50}}) × (\cancel{\dfrac{53}{50}})}$

$:\implies\sf {A = 6 × 53 × 53 }$

$:\implies\sf { A = Rs.16,854 }$

We know,

$\large{\pink{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

$\sf :\implies{Compound \: Interest = Rs,(16,854 - 15, 000)}$

$\sf\color{lime}\fbox{\bf\color{purple}{Compound \: Interest = Rs.854.}}$

The Compound Interest is Rs.854

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Question ³

GiveN:-

• Principal = ₹8000.
• Time = 3 years.
• Rate = 15% p.a.

To FinD:-

• Amount.
• Compound Interest.

SolutioN-

$\sf{A = P(1 + \dfrac{r}{100})ⁿ}$

$\sf :\implies{A = 8000(1 + \dfrac{15}{100})³ }$

$\sf :\implies{ A = 8000(1 + \dfrac{3}{20})³}$

$\sf :\implies{A = 8000(\dfrac{23}{20})³ }$

$\sf :\implies{A = 8\cancel{000} × \dfrac{23}{\cancel{20}} × \dfrac{23}{\cancel{20}} × \dfrac{23}{\cancel{20}} }$

$\sf :\implies{A = 23× 23 × 23}$

$\sf\color{lime}\fbox{\bf\color{purple}{Rs. 12,161.}}$

The Amount is Rs.12,161

We know,

$\large{\pink{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

$\sf :\implies{Compound \: Interest = Rs,(12,167 - 8,000)}$

$\sf\color{lime}\fbox{\bf\color{purple}{Compound \: Interest = Rs.4,167}}$

The Compound Intrest is Rs, 4,167.

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Question ⁴

GiveN:-

• Principal = ₹12500.
• Time = 2 years.
• Rate = 12% p.a.

To FinD:-

• Amount.
• Compound Interest.

SolutioN:-

$\sf{A = P(1+ \dfrac{r}{100})^n}$

$\sf :\implies{A = 12500(1 + \dfrac{12}{100})² }$

$\sf :\implies{A = 12500(1 + \dfrac{3}{25})^2}$

$\sf :\implies{ A = 12500(\dfrac{28}{25})²}$

$\sf :\implies{A = 125\cancel{00} × \dfrac{28}{\cancel{25}} × \dfrac{28}{\cancel{25}} }$

$\sf :\implies{A = 20 × 28 × 28 }$

$\sf\color{lime}\fbox{\bf\color{purple}{Rs.15,680}}$

The Amount is Rs.15,680

We know,

$\large{\pink{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

$\sf :\implies{Compound \: Interest = Rs,(15,680 - 12,500)}$

$\sf\color{lime}\fbox{\bf\color{purple}{Compound \: Interest = Rs.3180}}$

The Compund Interest is Rs.3180

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Question ⁵

GiveN:-

• Principal = ₹3000.
• Time = 3 years.
• Rate = 10% p.a.

To FinD:-

• Amount.
• Compound Interest.

SolutioN:-

$\sf{A = P(1+ \dfrac{r}{100})^n}$

$\sf :\implies{A = 3000(1 + \dfrac{10}{100})³ }$

$\sf :\implies{A = 3000(1 + \dfrac{1}{10})³ }$

$\sf :\implies{A = 3000(\dfrac{11}{10})³ }$

$\sf :\implies{A = 3\cancel{000} × \dfrac{11}{\cancel{10}} × \dfrac{11}{\cancel{10}} × \dfrac{11}{\cancel{10}}}$

$\sf :\implies{A = 3 × 11 × 11 × 11 }$

$\sf\color{lime}\fbox{\bf\color{purple}{Rs.3,993}}$

The Amount is Rs.3,993.

We Know,

$\large{\pink{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

$\sf :\implies{Compound \: Interest = Rs,(3,993 - 3,000)}$

$\sf\color{lime}\fbox{\bf\color{purple}{Compound \: Interest = Rs.993}}$

The Compund Interest is Rs.993.

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HOPES IT HELPS!:)