Question

A invests ₹8,000 and B invests 11000 at the same rate of interest per annum. It at the end of 3 years, B gets ₹720 more Interest than A find the rate of interest.
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Answer 1

Answer:

S.I. = \frac{PRT}{100}

100

PRT

A invested = Rs. 8000

B invested = Rs. 11000

Time period of investment for both A & B, T = 3 years

Let the rate of interest in both the investment be “R”% and the interest earned by A and B be denoted as “Ia” & “Ib” respectively.

Interest earned by A after 3 years:

Ia = \frac{8000 * R * 3}{100}

100

8000∗R∗3

= 240R …… (i)

And,

Interest earned by B after 3 years:

Ib = \frac{11000 * 3 * R}{100}

100

11000∗3∗R

= 330R …… (ii)

It is given that at the end of 3 years the interest earned by B is Rs. 720 more than the interest earned by A, so we can write the eq. as,

Ib = Ia + 720

⇒ 330R = 240R + 720 ……. [substituting values from (i) & (ii)]

⇒ 90R = 720

⇒ R = 720/90

⇒ R = 8%

Answer 2

Required formula:

$\boxed{ \tt {\pink{S.I. = \frac{PRT}{100} } }}$

$: \implies \sf \red{A \: invested = Rs. 8000} \\ : \implies \sf \red{B \: invested = Rs. 11000}$

Time period of investment for both A & B, T = 3 years

Let the rate of interest in both the investment be “R”% and the interest earned by A and B be denoted as “Ia” & “Ib” respectively.

Interest earned by A after 3 years:

$\tt { : \implies \pink { Ia = \frac{8000 \times R \times 3 }{100} = 240R - - - - - (i)}}$

And,

Interest earned by B after 3 years:

$\tt { : \implies\pink{Ib = \frac{11000 \times 3 \times R}{100} = 330R - - - - (ii)}}$

It is given that at the end of 3 years the interest earned by B is Rs. 720 more than the interest earned by A, so we can write the equation as,

$\sf{Ib = Ia + 720} \\ \sf{⇒ 330R = 240R + 720 - - [substituting \: values \: from \: (i) \: and \: (ii)]} \\ \sf{⇒ 90R = 720} \\ \sf{⇒ R = \sf{\frac{720}{90}}} \sf{⇒ R = 8\%}$

Thus, the rate of interest is 8%.