Question

(10) Raman deposited 7600 in a bank. He withdrew
3000 after 2 years. At the end of 7 years, her
an amount of 7656. Find the rate of interest.​

Answer 1

$\huge\underline\mathfrak\pink{Question-}$

Raman deposited 7600 in a bank. He withdrew  3000 after 2 years. At the end of 7 years, her  an amount of 7656. Find the rate of interest.​

$\huge\underline\bold\red{SolutiOn-}$

Money deposit, Principal = Rs 7600  

Time for interest = 2 years

We know that

$\sf\pink{So,}}}$ = $S.I = \frac{7600\times R \times 2}{100}$

After 2 years, he withdraw Rs 3000. So, new principal is 'P₂'.

End of year is 7 years. So, 5 years are there after new principal and for new simple interest.

Now, new simple interest =

Now, amount is A = Rs 7656

                             ⇒ A = P₂ + SI₂

                             ⇒ P₂ + SI₂

Now according to question = P₂ = 7600 + SI - 3000

                                              = P₂ = 4600 + SI

From above = 3056 = SI + SI₂

From above = $\frac{R}{100} \times (7600 \times 2\:+ 5P_2 = 3056)$............................( 1 )

Putting the equation in ( 1 ) = $\frac{R}{100} [7600\times R+5\:\frac{4600+7600\times R\times 2}{100}] = 3056$

After solving above,

$\large{\sf{\pink{R = 7.0196 %.}}}$

$\large{\sf{\purple{What\:is\:compound\:interest?}}}$

                        Compound Interest  

In the compound interest falls over the period of time is again inverted to further interest and if the principal does not remain the same for the entire loan period , due to addition of interest to the principal after a certain interest of time to form the new principal, then the interest obtained is called compound interest.

Here,

P = Principal

A = Amount

T = Time

R = Rate

n = Number of year, month, day.

${\fbox{\purple{Formula}}}}$

CI = A - P

1.e., P = $\large[(1 + \frac{R}{100})^n - 1]$

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