Question

A sum of money is borrowed at compound interest payable annually. The interests for the 2nd and 3rd year are Rs.330 and Rs.335 respectively. Find the sum.​

Answer 1

Given :

A sum of money is borrowed at compound interest payable annually

The interest for the second year = Rs 330

The interest for the third year = Rs 335

To Find :

The sum of money

Solution :

From compound Interest

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

For First year

Interest = Amount - Principal

Or,  Rs 330 =  P × $(1+\dfrac{r}{100})^{2}$  -  P

Or, Rs 330 = P [ 1 + 0.000 r² + 0.02 r - 1 ]

Or,  330 = P × (   0.000 r² + 0.02 r )       .............1

Again

For Second year

Interest = Amount - Principal

Or,  Rs 335 =  P × $(1+\dfrac{r}{100})^{3}$  -  P

Or, Rs 335 = P [ ( 1 + 0.01 r)³ - 1 ]  

Or, Rs 335 = P [ 1 +  0.000001 r³  + 0.03 r ( 1 + 0.01 r ) - 1 ]

Or, Rs 335 = P [  0.000001 r³  + 0.03 r  + 0.0003 r ]    

Or, Rs 330 = P [ 0.000001  r³  + 0.0303 r ]        ...........2

Solving eq 1 and 2

 $\dfrac{335}{330}$ = $\dfrac{0.000001 r^{3} + 0.0303 r }{0.000 r^{2} + 0.02 r}$

Or,  $\dfrac{67}{66}$ = $\dfrac{0.000001 r^{2} + 0.0303 }{0.000 r + 0.02 }$

Or, 67 × 0.0001 × r  + 1.34 = 66 × [  0.000001 r² + 1.9998]

Or,  0.0067 r + 1.34 = 0.000066 r² + 132

or,   0.000066 r² - 0.0067 r + 130.66

Solving this equation

  r = 50 + 1406 i   , 50- 14.06 i

i,e  r = 1406.8 % = 14

From eq 1

330 = P × (   0.000 r² + 0.02 r )    

330 =  P × (   0.0001 × ( 14)² + 0.02 × (14 ) )    

330 =  P × 0.2996

∴  principal = $\dfrac{330}{0.2996}$

i.e Principal = Rs 1101

Hence, The sum of money borrowed is Rs 1101 Answer

Answer 2

Compound Interest

Let P be the sum and r% be the rate of compound interest.

1st year:

Sum = P

Rate of interest = r%

Amount = P (1 + r/100)

2nd year:

Sum = P (1 + r/100)

Rate of interest = r%

Amount = P (1 + r/100)²

Interest = P (1 + r/100)² - P (1 + r/100)

= P (1 + r/100) (1 + r/100 - 1)

= P (1 + r/100) * r/100

==> P (1 + r/100) * r/100 = 330 .....(i)

3rd year:

Sum = P (1 + r/100)²

Rate of interest = r%

Amount = P (1 + r/100)³

Interest = P (1 + r/100)³ - P (1 + r/100)³

= P (1 + r/100)² (1 + r/100 - 1)

= P (1 + r/100)² * r/100

==> P (1 + r/100)² * r/100 = 335 .....(ii)

Dividing (ii) by (i), we get

1 + r/100 = 335/330

Or, r/100 = 5/330 = 1/66

Or, r = 100/66

Or, r = 50/33

Putting r = 55/33 in (i), we get

P {1 + (50/33)/100} * (50/33)/100 = 330

Or, P (1 + 1/66) * 1/66 = 330

Or, P * 67/66 * 1/66 = 330

Or, P = (330 * 66 * 66)/67

Or, P = 21454.93

Hence, the sum is Rs. 21454.93

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