Question

(a) Bharathi borrows an amount of rupees 12500 at 12% per annum for 3 years at a simple interest and Madhuri
borrows the same amount for the same time period at 10%
per annum , compounded annually. Who pays more interest and by how much
(b) The numerator of a fraction is 3 a less than its denominator. If the denominator is increased by 5 and the numerator by 2 then the fraction becomes 1/2. Find
the fraction.
​

Answer 1

For Bharathi

S.I. = P×R×T/100

= 12500×12×3/100

= 4500

For Madhuri

S.I = P×R×T/100

= 12500×10× 3/100

= 3750

Therefore, Bharathi pays more interest by ₹750.

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Answer 2

Answer (a):-

Given:

Bharathi borrows with

• Principal = Rs. 12500
• Rate = 12 %
• Time = 3 years

Madhuri borrows with

• Principal = Rs. 12500
• Rate = 10 %
• Time = 3 years

What To Find:

We have to find

• Who paid more interest and by how much.

How To Find:

To find it we have to,

• First, find the simple interest (SI) for Bharathi.
• Next, find the compound interest (CI) for Madhuri.
• Then, compare who got more interest.
• Finally, subtract the interest to find how much the person got more.

Formulas Needed:

• For SI -

$\bf \longrightarrow SI = \dfrac{P \times R \times T}{100}$

• For CI -

$\bf i. \longrightarrow A = P \bigg( 1 + \dfrac{R}{100} \bigg) ^T$

$\bf ii. \longrightarrow CI = A - P$

Solution:

• Finding the SI for Bharathi.

Using the formula,

$\sf \Longrightarrow SI = \dfrac{P \times R \times T}{100}$

Substitute the values,

$\sf \Longrightarrow SI = \dfrac{12500 \times 12 \times 3}{100}$

Cancel the zeros,

$\sf \Longrightarrow SI = 125 \times 12 \times 3$

Multiply the values,

$\sf \Longrightarrow SI = Rs. \: 4500$

• Finding the CI for Madhuri.

First, amount (A) -

$\sf \Longrightarrow A = P \bigg( 1 + \dfrac{R}{100} \bigg) ^T$

Substitute the values,

$\sf \Longrightarrow A = 12500 \bigg( 1 + \dfrac{10}{100} \bigg) ^3$

Solve and remove the brackets,

$\sf \Longrightarrow A = 12500 \times \dfrac{110}{100} \times \dfrac{110}{100}\times \dfrac{110}{100}$

Cancel the zeros,

$\sf \Longrightarrow A = 125 \times 11 \times 11 \times \dfrac{11}{10}$

Multiply them,

$\sf \Longrightarrow A = \dfrac{166375}{10}$

Divide 166375 by 10,

$\sf \Longrightarrow A = Rs. \: 16637.5$

Next, the CI -

Using the formula,

$\sf \Longrightarrow CI = A - P$

Substitute the values,

$\sf \Longrightarrow CI = Rs. \: 16637.5 - Rs. \: 12500$

Subtract the amounts,

$\sf \Longrightarrow CI = Rs. \: 4137.5$

• Finding who paid more interest and by how much.

Who paid more interest -

Here,

• Bharathi's Interest = Rs. 4500
• Madhuri's Interest = Rs. 4137.5

Since,

Rs. 4500 > Rs. 4137.5

∴ Therefore, Bharathi paid more interest.

By how much -

Subtract the amount,

$\sf \Longrightarrow Rs. \: 4500 - Rs. \: 4137.5$

That is,

$\sf \Longrightarrow Rs. \: 362.5$

Final Answer:

∴ Thus, Bharathi paid more interest by Rs. 362.5.

$\rule{300}{3}$

Answer (b):-

Given:

A fraction with

• Numerator = 3 less than the denominator
• Denominator

If in the fraction

• Numerator = Increased by 2
• Denominator = Increased by 5

Then it becomes,

• $\bf \dfrac{1}{2}$

What To Find:

We have to

• The fraction.

How To Find:

To find it, we have to,

• First, form the linear equation.
• Next, solve the equation formed and find the value.
• Then, substitute the value.
• Finally, the fraction.

Solution:

• Forming the equation.

Let,

• The denominator is x.

Then,

• The numerator is x - 3.

If the denominator is

• Increased by 5

Then it will be,

• Denominator = x + 5

If the numerator is,

• Increased by 2

Then it will be,

• Numerator = x - 3 + 2 = x - 1

And the fraction is

• $\bf \dfrac{1}{2}$

Hence, the equation is,

• $\bf \dfrac{x - 1}{x + 5} = \dfrac{1}{2}$

• Solving the equation.

$\sf \implies \dfrac{x - 1}{x + 5} = \dfrac{1}{2}$

Use cross multiplication,

$\sf \implies 2(x - 1) = 1(x + 5)$

Solve the brackets on both sides,

$\sf \implies 2x - 2 = x + 5$

Take x to LHS, and -2 to RHS,

$\sf \implies 2x - x = 5 + 2$

Solve the LHS and the RHS,

$\sf \implies x = 7$

• Substituting the values,

The fraction,

$\sf \implies \dfrac{x - 1}{x + 5}$

Substitute the values,

$\sf \implies \dfrac{7 - 1}{7 + 5}$

Solve,

$\sf \implies \dfrac{6}{12}$

Final Answer:

$\underline{\underline{\sf \therefore The \: fraction \: is \: \dfrac{6}{12} \: .}}$

$\rule{300}{3}$