Math, asked by aditiraj78434, 1 month ago

A man invests some money continuously for 3 years
at the rate
of 11% per annum. After 3 years man got
Rs. 30012 from the bank. Then find what
money he invested every year:
1) 9000, 4 yrs.
2) 7500, 4 yrs.
3) 8200, 3 yrs
4) 8000, 3 yrs.
lv for 3 years

Answers

Answered by orangesquirrel
0

The correct answer to this question is option 3) 8200, 3 years

Given:

Rate of interest = 11% p.a.

The amount that man got after 3 years = Rs 30012

To Find:

The money the man invested every year for three years =?

Solution:
Let the money invested by the man each year for three years be Rs 100 per year.

The interest on Rs 100 when invested in the first year = Rs 33

The interest on Rs 100 when invested in the second year = Rs 22

The interest on Rs 100 when invested in the third year = Rs 11

The total amount he will get after 3 years will be 3(100) + 33 + 22 + 11

= 300 + 66 = Rs 366

But the amount that man got after 3 years is given as Rs 30012.

So 366 corresponds to 30012

1 will correspond to 30012/366 = 82

100 will correspond to 82 × 100 = 8200

Hence, the money the man invested every year for three years is Rs 8200.

The correct answer to this question is option 3) 8200, 3 years

#SPJ1

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