Question

the compound interest on certain sum for 2 years is rupees 41.60 and simple interest is rupees 40 find the sum​

Answer 1

Answer:

RS 500.00

Step-by-step explanation:

we know that the SI for every year is same

hence SI for 1st yr is 20 and for 2nd year is 20

again for 1st year SI = CI so the difference in CI-SI = 0.8 due to 2nd year

and this difference is due to the interest on the interest of 1st year

so Rate = (0.8/20)*100 = 4%

now we can calculate the sum by formula SI = (p*r*i)/100

p = (20*100)/4 = 500 (we have calculated for 1 year )

1st yr--->> SI:20

CI:20 (1st yr CI alwz same as SI)

2nd yr--->> SI:20 (same for every year)

CI = 20 + (20*r*1)/100 {note: (1st yr interest)}

so 2nd yr CI = 2nd year CI - 1st year CI

20 + (20*r*1)/100 = 20.8

r = 4%

for 1st year,

p = (100*20)/4*1 = 500

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Answer 2

The sum is Rs. 250

Given:

The compound interest on certain sum for 2 years is Rs. 41.60 and simple interest is Rs. 40

To find:

The sum

Step-by-step explanation:

Let the sum be P, rate of both compound interest and simple interest be r per annum.

Compound interest on P for 2 years is Rs. 41.60

$\Rightarrow P(1+\dfrac{r}{100})^{2}-P=41.60$

$\Rightarrow P(1+\dfrac{2r}{100}+\dfrac{r^{2}}{10000}-1)=41.60$

$\Rightarrow P(\dfrac{2r}{100}+\dfrac{r^{2}}{10000})=\dfrac{4160}{100}$

$\Rightarrow \dfrac{Pr}{10000}(200+r)=\dfrac{4160}{100}$

$\Rightarrow \dfrac{Pr}{100}(200+r)=4160$ ... ... (1)

Simple interest on P for 2 years is Rs. 40

$\Rightarrow \dfrac{P\times 2\times r}{100}=40$

$\Rightarrow 2\dfrac{Pr}{100}=40$ ... ... (2)

Dividing (1) by (2), we get

$\dfrac{200+r}{2}=\dfrac{4160}{40}$

$\Rightarrow 8000+40r=8320$

$\Rightarrow 40r=320$

$\Rightarrow r=8$

Putting r = 8 in $\dfrac{P\times 2\times r}{100}=40$, we get

$\dfrac{P\times 2\times 8}{100}=40$

$\Rightarrow P=\dfrac{40\times 100}{2\times 8}=\dfrac{4000}{16}$

$\Rightarrow P=250$

Thus the sum of money is Rs. 250

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