Question

Harish invested rupees 80000 in two different parts 1 @ 20% CI per annum and another one at the rate of 15% SI per annum if he interchanges the rate of interest he got rupees 2875 less at the end of 2 years find the difference of two sums

Answer 1

150

Step-by-step explanation:

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Answer 2

Given :

Total investment amount = Rs 80000

The rate of interest at compound interest = 20%

The rate of interest at simple interest = 15 %

The rate of interest interchange then amount of Rs 2875 less at the end of 2 years

Time period = 2 years

To Find :

The difference of two sum

Solution :

Let the investment amount at compound interest = Rs x

So, The investment amount at simple interest = Rs ( 80000 - x )

From compound interest

Amount = Principal × $(1+\dfrac{rate}{100})^{time}$

Or,  A = x × $(1+\dfrac{20}{100})^{2}$

Or,  A = 1.44 x

Again

From simple interest

 S.I = $\dfrac{principal\times rate\times time}{100}$

Or, S.I = $\dfrac{(80000-x)\times 15\times 2}{100}$

Or,  S.I = $\dfrac{(80000-x)\times 30}{100}$

So, Amount = Interest + Principal

                   = $\dfrac{(80000-x)\times 30}{100}$ + 80000

So, Sum of Amount from C.I and S.I =

1.44 x + $\dfrac{(80000-x)\times 30}{100}$ + 80000                     ............1

Again

When rate of interest changes

From compound interest

Amount = Principal × $(1+\dfrac{rate}{100})^{time}$

Or,  A = x × $(1+\dfrac{15}{100})^{2}$

Or,  A = 1.3225 x

Again

From simple interest

 S.I = $\dfrac{principal\times rate\times time}{100}$

Or, S.I = $\dfrac{(80000-x)\times 20\times 2}{100}$

Or,  S.I = $\dfrac{(80000-x)\times 40}{100}$

So, Amount = Interest + Principal

                   = $\dfrac{(80000-x)\times 40}{100}$ + 80000

So, Sum of Amount from C.I and S.I  =

1.3225 x + $\dfrac{(80000-x)\times 40}{100}$ + 80000                ...............2

∵  After interchanging the rate of interest , He got Rs 2875 less

i.,e   1.3225 x + $\dfrac{(80000-x)\times 40}{100}$ + 80000 -2875 =  1.44 x + $\dfrac{(80000-x)\times 30}{100}$ + 80000

Or,    $\dfrac{(80000-x)\times 40}{100}$  - $\dfrac{(80000-x)\times 30}{100}$  = 1.4 - 1.3 + 2875

Or,   $\dfrac{(80000-x)}{100}$ × ( 40 - 30 ) = 2875.1

Or,    $\dfrac{(80000-x)}{100}$  = 287.51

Or,  80000 - x = 28751

∴    x = 80000 - 28751

        = 51249

Now,

Difference of two sums =  [ 1.44 x + $\dfrac{(80000-x)\times 30}{100}$ + 80000   ]  -  [ 1.3225 x + $\dfrac{(80000-x)\times 40}{100}$ + 80000  ]

                                       = $\dfrac{(80000-x)}{100}$ × ( 30 - 40 ) -  ( 1.4 - 1.3 ) x

                                       = -  $\dfrac{(80000-x)}{100}$ × 10 + 0.1 x

                                      = -  $\dfrac{(80000-51249)}{100}$ × 10 + 0.1 × 51249

                                      = -  2875.1 + 5124.9

                                      = Rs 2249.8

Hence, The  difference of two sums is Rs 2249.8  Answer