Question

Please solve this question and give me the correct solution

DON'T de.lete my questions
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Answer 1

Answer:

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Step-by-step explanation:

here we know Principal, Time Period, and Rate of Interest but the Amount and Compound Interest (C.I.) is unknown which we have to find.

A = P[1 + (r/100)]n

P = ₹ 10,000

n = 1 1/2 years

R = 10% p.a. compounded annually and half-yearly

where , A = Amount, P = Principal, n = Time period and R = Rate percent

For calculation of C.I. compounded half-yearly, we will take the Interest rate as 5% and n = 3

A = P[1 + (r/100)]n

A = 10000[1 + (5/100)]3

A = 10000[1 + (1/20)]3

A = 10000 × (21/20)3

A = 10000 × (21/20) × (21/20) × (21/20)

A = 10000 × (9261/8000)

A = 5 × (9261/4)

A = 11576.25

Interest earned at 10% p.a. compounded half-yearly = A - P

= ₹ 11576.25 - ₹ 10000 = ₹ 1576.25

Now, let's find the interest when compounded annually at the same rate of interest.

Hence, for 1 year R = 10% and n = 1

A = P[1 + (r/100)]n

A = 10000[1 + (10/100)]1

A = 10000[1 + (1/10)]

A = 10000 × (11/10)

A = 11000

Now, for the remaining 1/2 year P = 11000, R = 5%

A = P[1 + (r/100)]n

A = 11000[1 + (5/100)]

A = 11000[(105/100)]

A = 11000 × 1.05

A = 11550

Thus, amount at the end of 112112when compounded annually = ₹ 11550

Thus, compound interest = ₹ 11550 - ₹ 10000 = ₹ 1550

Therefore, the interest will be less when compounded annually at the same rate.

Answer 2

Step-by-step explanation:

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

$\sf \: Time = 1\ \frac{1}{2} years = 3 years (compounded \: half \: yearly)$

$\sf \: Amount \: (A) = P\left(1+\frac{R}{100}\right)^n \\ \sf= 10000\left(1+\frac{5}{100}\right)^3 \\ \sf= 10000\left(1+\frac{1}{20}\right)^3 \\ \sf= 10000\left(\frac{21}{20}\right)^3$

$\sf= 10000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20} \\ \sf= Rs. 11,576.25$

Compound Interest (C.I.) = A – P

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

If it is compounded annually, then

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

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

$\sf= 10000\left(1+\frac{10}{100}\right)^1 \\ \sf= 10000\left(1+\frac{1}{10}\right)^1 \\ \sf= 10000\left(\frac{11}{10}\right)^1 \\ \sf= 10000\times\frac{11}{10}\\ \sf= Rs. 11,000$

$\sf \: Interest \: for \frac{1}{2} year = \frac{11000\times1\times10}{2\times100}=RS.\ 550 \\ \sf \: \therefore Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550</p><p>$

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.