Question

find the difference between SI and CI on rs5000 at 4% p.a. for three years​

Answer 1

Solution:-

S.I=100P×R×n=1005000×10×3=1500

C.I=P(1+100R)n−P=5000[(1+10010)3−1]=1655

∴C.I−S.I=1655−1500=155

Answer 2

Difference between S.I. and C.I. on Rs.5000 at 4% p.a. for 3 years is 42.32

Explanation:

\rm \bullet \: \: Simple interest = \dfrac{p \times r \times t}{100}∙Simple interest= 100 p×r×t

Where,

p(principal) = Rs.5,000

r(rate) = 4% p.a.

t(time) = 3 years.

\rm = \dfrac{prt}{100}= 100prt

= \dfrac{50 \cancel{00}\times 4 \times 3}{1 \cancel{00} }= 1 0050 00×4×3

= 50 \times 4 \times 3=50×4×3

= 600=600

∴ Simple interest: Rs.600

Now,

Calculating compound interest,

We know,

\rm \: \bullet \: \:Compound \: interest = Amount - Principal∙Compoundinterest=Amount−Principal

Using amount formula,

\rm = p \bigg( 1 + \dfrac{r}{100} \bigg)^{t}=p(1+ 100r) t

Where,

p(principal) = Rs.5000

r(rate) = 4% p.a.

t(time) = 3 years

So, C. I. :-

\rm = p \bigg( 1 + \dfrac{r}{100} \bigg)^{t} - p=p(1+ 100r) t−p

\rm = 5000\bigg( 1 + \dfrac{4}{100} \bigg)^{3} - 5000=5000(1+ 71004) 3−5000

\rm = 5000\bigg( 1 + \dfrac{1}{25} \bigg)^{3} - 5000=5000(1+ 251 ) 3−5000

\rm = 5000\bigg( \dfrac{26}{25} \bigg)^{3} - 5000=5000( 2526 ) 3−5000

\rm = 5000 \times \dfrac{17576}{15625} - 5000=5000×

1562517576and 7−5000

\rm = 5624.32 - 5000=5624.32−5000

= 624.32=624.32

Hence, the difference between S.I. and C.I. is :-

= 642.32 - 600=642.32−600

= 42.32=42.32

∴ Required answer: Rs. 42.32