Question

A certain sum of money was lent under the following repayment scheme based on simple interest:; 9% per annum for the initial 3 years; 10% per annum for the next 4 years; 11% per annum for the next 2 years; 12% per annum after the first 9 years; Find the amount which a sum of Rs. 15,000 taken 11 years taken for 11 years becomes at the end of 11 years?
A) 29000
B) 30000
C) 31950
D) 35000

Answer 1

Answer:

C) 31950

Step-by-step explanation:

A sum of money was taken for 11 years under certain schemes under Simple Interest.

For the first 3 year the rate of interest is 9% per annum.

So, if Rs 15000 is the sum of money,

S.I= (15000×9×3)/100

= Rs 4050.

For the next 4 years , rate of interest is 10% per annum. So ,

S.I= ( 15000×4×10)/100

= Rs 6000

For the next 2 years, rate of interest is 11% per annum. So,

S.I= (15000×11×2)/100

=Rs 3300

And for the next remaining 2 years , rate of interest is 12% per annum. So,

S.I = (15000×12×2)/100

=Rs 3600

Adding up all the amount for 11 years we get Rs 16950.

So the total amount receivable is Rs (16950+15000)= Rs 31950

Answer 2

Answer:

Option C is correct.

Step-by-step explanation:

Given:

Rate of interest  from 1st to 3rd year = 9%

Rate of interest from 4th to 7th year = 10%

Rate of interest from 8th to 9th year = 11%

Rate of interest 10th year = 12%

Principal = 15000

To find: Amount at the end of 11 years.

Using Simple Interest Formula,

$SI=\frac{P\times R\time T}{100}$

A = P + SI

According to the Question,

$SI=\frac{15000\times9\times3}{100}+\frac{15000\times10\times4}{100}+\frac{15000\times11\times2}{100}+\frac{15000\times12\times2}{100}$

$SI=150\times9\times3+150\times10\times4+150\times11\times2+150\times12\times2$

$SI=150(9\times3+10\times4+11\times2+12\times2)$

$SI=150(27+40+22+24)$

SI = 150 × 113

SI = 16950

A = 15000 + 16950 = 31950

Therefore, Option C is correct.