Math, asked by nombikshashas, 1 year ago

The difference between C.I. and S.I . on rs 20000 for 2 yrs is rs 128 at a same rate of interest per annum find the rate of interest

Answers

Answered by tejasmba
6

CI = P[1+r/100]^n-P

 

SI = pnr/100

 

From the problem,

 

P = 20,000

N = 2

CI – SI = 128

 

CI – SI = P[1+r/100]^n-P-Pnr/100

 

=P[(1+r/100)^n-1-nr/100]

 

Substituting the values of P and n, we get

 

CI – SI = 128 = 20000[(1+r/100)^2-1-2r/100]

 

Let us consider the expression within the square brackets, (1+r/100)^2-1-2r/100.

 

Expanding the square expression and subtracting the remaining terms, we get

 

1+2r/100+r^2/10000-1-2r/100 = r^2/10000

 

Thus, 128 = 20,000*r^2/10,000 = 2r^2

 

Or r^2 = 128/2 = 64.

 

Thus, r = 8%.

 

Therefore, the rate of interest is 8%.

Answered by anushkachauhani30jun
2

Answer:

Rate = 8%

Step-by-step explanation:

For ,

N=2years

P=Rs20,000

We have S.I.=

PNR/100

=

20,000×2×R/100

=400R

And on interest being compounded for 2 years, Amount=P(1+ R/100) ²

=20,000×(1+ R/100) ²

=20,000×( 100+R/100) ²

=2R²+400R+20,000

So, C.I.=A−P=2R +400R+20,000−20,000=2R ² +400R

Given, C.I.−S.I=Rs128

=>2R ² +400R−400R=128

=>2R ² =128

=64

=>R=8 %

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