Question

Find the compound interest for 2*1/2years on ₹10,000lent at 5%p.a.reckoned annually

Answer 1

Applying the rule A = P(1+R100)n on the given situations, we get:

(i) A = 3000(1+5100)2 = 3000(1.05)2 = Rs 3307.5

Now, CI = A – P = Rs 3307.50 – Rs 3000 = Rs. 307.50

(ii) A = 3000(1+18100)2 = 3000(1.18)2 = Rs 4177.2

Now, CI = A – P = Rs 4177.20 – Rs 3000 = Rs. 1177.20

(iii) A = 5000(1+10100)2 = 5000(1.10)2 = Rs 6050

Now, CI = A – P = Rs 6050 – Rs 5000 = Rs. 1050

(iv) A = 2000(1+4100)3 = 2000(1.04)3 = Rs 2249.68

Now, CI = A – P = Rs 2249.68 – Rs 2000 = Rs. 249.68

(v) A = 12800(1+7.5100)3 = 12800(1.075)3 = Rs 15901.40

Now, CI = A – P = Rs 15901.40 – Rs 12800 = Rs. 3101.40

(vi) A = 10000(1+20200)4 = 10000(1.1)4 = Rs 14641

Now, CI = A – P = Rs 14641 – Rs 10000 = Rs. 4641

(vii) A = 160000(1+10200)4 = 160000(1.05)4= Rs 194481

Now, CI = A – P = Rs 194481 – Rs 160000 = Rs. 34481

Answer 2

Answer:

interest for 1st yr=》 interest = 10000×1×5/100

=rs 500

amount =10000+500=10500

interest for 2nd yr=10500×1×5/100

=rs525

amount =10500+525=rs11025

interest for 1/2 year =11025×1/2×5/100

=55125/200

=27562.5/100

=rs275.625

=275.63(approx)

amount =11025+275.63=11300.63

required compound interest =amount -principal

=11300.63-10000

=rs 1300.63

Step-by-step explanation:

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