Question

A man borrowed some money at the rate of interest 6 % per annum for the first 2 years at the rate of interest 9 % per annum for the next 3 years. And at the rate of 14 % per annum for the period beyond 5 years. If he pays total interest of Rs11400 at the end of 9 years. How much money did he borrowed?​

Answer 1

here I have assumed interest as simple interest

Answer 2

Given :-

For 2 years rate of interest = 6%

For next 3 years rate of interest = 9%

For next 4 years rate of interest = 14%

Total rate of interest = ₹11,400

To Find :-

The sum borrowed i.e, the principal

Solution :-

${\sf{\huge{\boxed{\blue{Case\: I}}}}}$

Time = 2years

Rate = 6%

${\sf{S.I\: =\: {\dfrac{P\: \times\: R\: \times\: T}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{P\: \times\: 6\: \times\: 2}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{3P}{25}}}}$

${\sf{\huge{\boxed{\blue{Case\: II}}}}}$

Time = 3years

Rate = 9%

${\sf{S.I\: =\: {\dfrac{P\: \times\: R\: \times\: T}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{P\: \times\: 9\: \times\: 3}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{27P}{100}}}}$

${\sf{\huge{\boxed{\blue{Case\: III}}}}}$

Time = 4years

Rate = 14%

${\sf{S.I\: =\: {\dfrac{P\: \times\: R\: \times\: T}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{P\: \times\: 14\: \times\: 4}{100}}}}$

${\sf{\implies\: S.I\: =\: {\dfrac{14P}{25}}}}$

According to question,

${\sf{{\dfrac{3P}{25}}\: +\: {\dfrac{27P}{100}}\: +\: {\dfrac{14P}{25}}\: =\: 11400}}$

${\sf{(LCM\: =\: 100)}}$

${\sf{{\dfrac{12p\: +\: 27p\: +\: 56p}{100}}\: =\: 11400}}$

${\sf{{\dfrac{95p}{100}}\: =\: 11400}}$

${\sf{P\: =\: {\dfrac{11400\: \times\: 100}{95}}}}$

${\sf{P\: =\: 12000}}$

Therefore principal is ₹12,000