Find the compound interest on rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly.

Solution:

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time(n) = 1\ \frac{1}{2} years = 3 years (compounded half yearly)

Amount (A) = P\left(1+\frac{R}{100}\right)^n

= 10000\left(1+\frac{5}{100}\right)^3

= 10000\left(1+\frac{1}{20}\right)^3

= 10000\left(\frac{21}{20}\right)^3

= 10000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n) = 1\ \frac{1}{2} years.

Amount (A) for 1 year = P\left(1+\frac{R}{100}\right)^n

= 10000\left(1+\frac{10}{100}\right)^1

= 10000\left(1+\frac{1}{10}\right)^1

= 10000\left(\frac{11}{10}\right)^1

= 10000\times\frac{11}{10}

= Rs. 11,000

Interest for \frac{1}{2} year = \frac{11000\times1\times10}{2\times100}=RS.\ 550

\therefore Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

Solution

Here,

Principal (P)=Rs. 10000

Rate of interest (R)=10% per annum compounded half yearly.

⇒R=5% per half yearly
Time (n)=112 years =3 half years
Amount (A)=P(1+R100)n

=10000(1+5100)3

=10000(1+120)3
=10000(2120)3

=10000×2120×2120×2120

=Rs11,576.25
Compound Interest (C.I)=A−P

=Rs 11,576.25−Rs 10,000

⇒C.P=Rs. 1576.25

So, Compound interest half yearly is Rs.1576.25---(1)

Rate of interest (R)=10%

Time (n)=1 years
Amount (A) for 1 year

A=P(1+R100)n

=10000(1+10100)1

=10000(1+110)1
=10000(1110)1

=10000×1110
=Rs.11,000
Interest for half year 12 year,

=11000×1×102×100 [ ∵S.I=PTR100]

=Rs.550
Therefore, total amount =Rs. 11,000+Rs. 550=Rs.11,550
Now,

C.I=A−P

=Rs. 11,550−Rs.10000

=Rs.1550

So, Compound interest per annum is Rs.1550. ---(2)
From (1) and (2),

the interest Rs.1576.25 is more than Rs.1550.

Hence, the answer is 'Yes'.

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