Question

Answered what is the difference of the compound interest and the simple interest on rs 14,000 at the interest rate of 10% p.a for 2 year​

Answer 1

P = Rs 42000 T=2 years

Let rate = r %

S.I=

100

PRT

= 100

42000×r×2

=840 Rs

For C.I

A=P(1+

100

r ) n

=42000(1+

100

r ) 2=42000(1+

10000r 2 + 50r )=42000+ 1042r 2 +840r

C.I = A - P

=42000+ 1042r 2 +840r−42000= 1042r 2 +840r

Given C.I−S.I=105= 1042r 2 +840r−840r=105= 1042r 2

=105

=42r 2

=1050

=25

r=5%

C.I at 2nd year = 42000(1+

100 ) 2 −42000=42000(1.1025−1)=4305 Rs

Answer 2

Solution

Given Here :-

• Principal (p) = 14,000 Rs.
• Rate (r) = 10%
• Time(t) = 2 years

Find :-

• The difference of the compound interest and the simple interest.

Explanation

Using Formula

★ C.I. = p [ (1 + r/100)^t - p]

★ S.I. = (p × r × t)/100

First Calculate Compound Interest

➡ C.I. = p [ (1 + r/100)^t ] - p

Keep all above values

➡ C.I. = 14,000 [ (1 + 10/100)² ] - 14,000

➡ C.I. = 14,000[ ( 1 + 1/10)² ] - 14,000

➡ C.I. = 14,000[ (11/10)² ] - 14,000

➡ C.I. = 14,000 [ ( 11/10 × 11/10 )] - 14,000

➡ C.I. = 14,000 [ 121/100 ] - 14,000

➡ C.I. = 14,000 × 121/100 - 14,000

➡ C.I. = 140 × 121 - 14,000

➡ C.I. = 16,940 - 14,000

➡ C.I. = 2,940 Rs.

_______________________

Now, Calculate Simple Interest

➡ S.I. = (p × r × t)/100

Keep all above values

➡ S.I. = (14,000 × 10 × 2)/100

➡ S.I. = 140 × 10 × 2

➡ S.I. = 2,800 Rs.

________________________

Now, Calculate difference between both of them,

➡ C.I. - S.I.

Keep Value of C.I. & S.I.

➡ (C.I. - S.I.) = 2,940 - 2,800

➡ ( C.I. - S.I. ) = 140 Rs.

Hence

• Difference between C.I. & S.I. will be = 140 Rs.

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