Question

At what rate percent will a sum of 64,000 be compounded to ₹68,921 in three years

Answer 1

Given :-
Principal (P) = ₹ 64000
Time (n) = 3 years
Amount = ₹ 68921
Let Rate of interest = R

$Amount = P {(1 + \frac{R}{100} )}^{n} \\ \\ 68921 = 64000 {(1 + \frac{R}{100} )}^{3} \\ \\ \frac{68921}{64000} = {(1 + \frac{R}{100} )}^{3} \\ \\ {( \frac{41}{40}) }^{3} = {(1 + \frac{R}{100} )}^{3} \\ \\ \frac{41}{40} = 1 + \frac{R}{100} \\ \\ \frac{41}{40} - 1 = \frac{R}{100} \\ \\ \frac{41 - 40}{40} = \frac{R}{100} \\ \\ \frac{1}{40} = \frac{R}{100} \\ \\ \frac{1}{40} \times 100 =R \\ \\ 2.5 = R \\ \\ \\ R\% = 2.5\%$

Answer 2

7(y+3)-2(x+2)=14 4(y-2)+3(x-3)=2