Neha borrowed INR 64000 from a bank for 11-2 years at the rate of 10

Compound Interest (CI) = 64000[(1+10200)^3 1] #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard VIII

Mathematics

Finding CI When Rate Compounded Anually or Semi - Anually

Neha borrowed...

Question

Neha borrowed INR 64000 from a bank for $1 \frac{1}{2}$ years at the rate of 10 % per annum . Compute the total compound interest payable by Neha after $1 \frac{1}{2}$ years , If the interest is compound half -yearly.

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Solution

Compound Interest (CI) = $64000 [(1 + \frac{10}{200})^{3} - 1]$
= $64000 [(1 + \frac{1}{20})^{3} - 1]$
= $64000 [(\frac{21}{20})^{3} - 1]$
= $64000 [(1.05)^{3} - 1]$
= $64000 [1.158 - 1]$
= $64000 \times 0.158$
= $10112$ INR

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Neha borrowed INR 64000 from a bank for 112 years at the rate of 10 % per annum . Compute the total compound interest payable by Neha after 112 years , If the interest is compound half -yearly.

Compound Interest (CI) = 64000[(1+10200)3−1] = 64000[(1+120)3−1] = 64000[(2120)3−1] = 64000[(1.05)3−1] = 64000[1.158−1] = 64000×0.158 = 10112 INR

Neha borrowed INR 64000 from a bank for $1 \frac{1}{2}$ years at the rate of 10 % per annum . Compute the total compound interest payable by Neha after $1 \frac{1}{2}$ years , If the interest is compound half -yearly.

Compound Interest (CI) = $64000 [(1 + \frac{10}{200})^{3} - 1]$
= $64000 [(1 + \frac{1}{20})^{3} - 1]$
= $64000 [(\frac{21}{20})^{3} - 1]$
= $64000 [(1.05)^{3} - 1]$
= $64000 [1.158 - 1]$
= $64000 \times 0.158$
= $10112$ INR