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Answers
Explanation:
Solution:-
Let the deposited money be Rs. X
Rate of Interest = 10%
Time = 2 years
Amount = P[1+(R/100)]^n
On Substituting these values in the above formula
=> A = X[1+(10/100)]²
=> A = X[1+(1/10)]²
=> A=X[(10+1)/10]²
=> A = X[11/10]²
=> A = X(121/100)
=> A = 121X/100
We know that
Amount = P+I
=> CI = Amount - Principle
=> CI = (121X/100)-X
=> CI = (121X-100X)/100
=> CI = 21X/100
Compound Interest for 2 years = Rs. 21X/100 --(1)
He withdrew after 2 years 2050 then
Remaining money = (121X/100)-(2050)
=> (121X-205000)/100
It will be the principle for the third year
I = PTR/100
=> [(121X-205000)/100]×1×10/100
=>((121X-205000)/100)×1/10
=> I = (121X-205000)/1000-----(2)
Ratio of the third year CI to the CI of the first two years
=> (121X-205000)/1000 : 21X/100
=> (121X-205000)/10 : 21X
According to the given problem
The given ratio = 8:21
=> (121X-205000)/10 : 21X = 8:21
=> [(121X-205000)/10] / 21X = 8/21
=> (121X-205000)/210X = 8/21
=> (121X-205000)/10X = 8
=> 121X-205000 = 8×10X
=> 121X-205000 = 80X
=> 121X-80X = 205000
=> 41X = 205000
=> X = 205000/41
=> X = 5000
Therefore , X =Rs. 5000
Answer:-
The initial deposited money in the bank by the person is Rs. 5000
Used formulae:-
- Amount = P[1+(R/100)]^n
- I = PTR/100
- Amount = P+I