Question

Find amount and C.I on ₹20000 and 10% p.a for
$1\frac{1}{12}$
year when interest is compounded half yearly .​

Answer 1

Answer:

23152.50

Step-by-step explanation:

Principal (P) = ₹ 20,000

Time (t) = 1. ½ Years = 3/2 years

Rate (r) = 10%

Amount = P × (1 + (r/2 × 100)) n × 2

= ₹ 20,000 × (1 + (10/200) 3/2 × 2

= ₹ 20,000 × (210/200)3

= ₹ 20,000 × 21/20 × 21/20 × 21/20

= ₹ 23,152.50

C.I. Amount – Principal

= ₹ 23,152.50 - ₹20,000 = 23,152.50

Answer 2

Answer

• Amount = ₹ 23,152.5.
• C.I. = ₹ 3152.5.

Given

• Principal = ₹ 20,000.
• Rate = 10 % p.a.
• Time = 1½.
• Compounded half-yearly.

To Find

• Amount and Compound Interest.

Step By Step Explanation

Formula Used :

$\underline{\boxed{ \bold{ \purple{Half-yearly = P {\bigg(1 + \cfrac{ \cfrac{R}{2}}{100} \bigg)}^{2n}}}}} \: \: \: \: \: \: \red\bigstar$

Where, P = Principal, R = Rate and N = Time.

By substituting the values :

$\longmapsto \sf P {\bigg(1 + \cfrac{ \cfrac{R}{2}}{100} \bigg)}^{2n} \\ \\ \longmapsto \sf 20000{\bigg(1 + \cfrac{ \cfrac{ \cancel{10}}{ \cancel2}}{100} \bigg)}^{ \cancel2 \times\cfrac{3}{ \cancel2} } \\ \\ \longmapsto \sf 20000 {\bigg(1 + \cfrac{ \cancel5}{ \cancel{100}} \bigg)}^{3} \\ \\ \longmapsto \sf 20000 {\bigg(1 + \cfrac{1}{20} \bigg)}^{3} \\ \\ \longmapsto \sf 20000 {\bigg(\cfrac{21}{20} \bigg)}^{3} \\ \\ \longmapsto \sf 20 \: \cancel{000} \times \cfrac{21}{2 \cancel0} \times \cfrac{21}{2 \cancel0} \times \cfrac{21}{2 \cancel0} \\ \\ \longmapsto \sf20 \times \cfrac{9261}{8} \\ \\ \longmapsto \sf \cancel\cfrac{185220}{8} \\ \\ \longmapsto \bold{23152.5}$

Therefore, Amount = ₹ 23,152.5 and C.I. → 23,152.5 - 20,000 = ₹ 3,152.5.

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