Math, asked by simranthakral97, 9 months ago

Sol
2
Example 41
A customer, Lord Venkateswara, took a loan of
$10 million from Mr Kuber. In the second year of the
loan period, the interest that accrued to Mr Kuber was
$0.1 million more than that in the first year.
(Assume CI)
At what rate of interest did Kuber lend to Lord
Venkateswara?
b. When will the principal double itself
(approximately)?
a.
2 Exar

Answers

Answered by RvChaudharY50
2

Given :- A customer, Lord Venkateswara, took a loan of $10 million from Mr Kuber. In the second year of the loan period, the interest that accrued to Mr Kuber was

$0.1 million more than that in the first year.

To Find :-

A) At what rate of interest did Kuber lend to Lord Venkateswara ?

Answer :-

→ Principal = $ 10 million = 10 * 10 lakh = 100 lakh = 1 crore .

→ Rate = Let R% per annum .

so,

→ Interest of first year = (P * R * T)/100 = (10000000 * x)/100 = $ 100000x

then,

→ Amount after 1 year = 10000000 + 100000x = $ 100000(100 + x) = Principal for second year.

so,

→ Interest for second year = (P * R * T)/100 = 100000(100 + x) * x/100 = $ 1000x(100 + x)

A/q,

→ 1000x(100 + x) - 100000x = 0.1 million = 100000

→ 100000x - 100000x + 1000x² = 100000

→ 1000x² = 100000

→ x² = 100

→ x = 10% (Ans.)

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Answered by amitnrw
4

Given : A customer, Lord Venkateswara, took a loan of $10 million from Mr Kuber. In the second year of the loan period, the interest that accrued to Mr Kuber was $0.1 million more than that in the first year.  

(Assume CI)

To Find :  At what rate of interest did Kuber lend to Lord Venkateswara?

b. When will the principal double itself

Solution:

In the second year of the loan period, the interest that accrued to Mr Kuber was $0.1 million more than that in the first year.

this $0.1 million  excess interest  is basically interest of the interest earned in 1st year.

let say interest earned in 1st year  = x   million

Then  x   *  R * 1 / 100   =  0.1

=> xR  = 10   Eq1

interest earned in 1st year  = 10 *  R * 1  / 100    =   x

=>  x = R/10

substituting  value of x in Eq1

=> (R/10) R = 10

=> R² = 100

=> R = 10

A = P ( 1 + R/100)ⁿ

A = 2P

=> 2P  = P ( 1 + 10/100)ⁿ

=> 1.1ⁿ =  2

=> n = 7.28

Approx 7.3  years    principal double itself

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