Question

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1.)A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

2.)Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12,600 at the end of first year and ₹ 17,640 at the end of the second year. Find the sum borrowed.
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Answer 1

(1).

Let the principal be p and rate of interest be r%.

Given that the sum of money becomes three times of itself in 10 years.

A = 3P  ----- (1)


We know that A = P(1 + r/100)^10

                        3P = P(1 + r/100)^10 

                           (1 + r/100)^10 = 3 ------ (2)


Now,

Given that it becomes 27-times of itself at the same rate per annum.

A = 27P

A = P(1 + r/100)^n

27P = P(1 + r/100)^n

(1 + r/100)^n = 27 

(1 + r/100)^n = (3)^3 ------ (3)


On solving (2) & (3), we get

(1 + r/100)^n = 3^3

(1 + r/100)^n = ((1 + r/100)^10)^3

(1 + r/100)^n = (1 + r/100)^30

n = 30.

Therefore After 30 years, the money becomes twenty-seven times of itself.


(2)

Let the principal be P, R = 5%.

Given Amount at the end of first year = 12600.

We know that A = P(1 + r/100)^n

= > 12600 = P(1 + 5/100)^1

= > 12600 = P(21/20)

= > 12600 * 20 = 21P

= > 252000 = 21P

= > 252000/21

= > 12000.


Given Amount at the end of 2nd year = 17640

= > A = P(1 + r/100)^n

= > 17640 = P(1 + 5/100)^2

= > 17640 = P(21/20)^2

= > 17640 = P(441/400)

= > 17640 * 400 = 441P

= > 7056000 = 441P

= > P = 7056000/441

= > P = 16000.



The sum borrowed = 16000 + 12000

                                  = 28000.




Therefore the sum of money borrowed = 28000.


Hope this helps!

Answer 2

Solution 1
Let's assume that the principal and rate of interest be P and r% respectively.
Now according to the question.
Money after 10 years
A =3P  ......(i)

 A = P(1 + r/100)^10
3P = P(1 + r/100)^10  
(1 + r/100)^10 = 3 ........(ii)


Now, for A = 27P
A = P(1 + r/100)^n
27P = P(1 + r/100)^n
(1 + r/100)^n = 3^3
(1 + r/100)^n = ((1 + r/100)^10)^3
(1 + r/100)^n = (1 + r/100)^30
n = 30.

So, it will take 30 years to become 27 times

Solution 2
At the end of first year
A = P(1 + r/100)^n
12600 = P(1 + 5/100)^1
12600 = P(21/20)
12600 x 20 = 21P
252000 = 21P
P=252000/21
P=12000.


At the end of 2nd year:
A = P(1 + r/100)^n
17640 = P(1 + 5/100)^2
17640 = P(21/20)^2
17640 = P(441/400)
17640 * 400 = 441P
7056000 = 441P
P = 7056000/441
P = 16000.

Net borrowing
=16000+12000
=28000.