Math, asked by krono01, 5 months ago

A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹96, find the sum of money.​

Answers

Answered by Aashi6552
21

Answer:

Let P=Rs x,R=8% then

Then interest for the first year= x×8×1/100 = 8x/100

Then Principal for the second year=x+ 8/100x=Rs 108/100x

hence interest for the second year=

(108/100x×8×1)/100 =Rs. 864x10000

According to the question

⇒ 864x10000 = 8x/100 +96

⇒ 864x10000 − 8x/100 =96

⇒ 864x−800x/10000 =96

⇒ 64x/10000 =96

⇒64x=96×10000

⇒x= 96×10000/64

⇒x=15000

Hence the sum=Rs. 15000$$

Answered by Mɪʀᴀᴄʟᴇʀʙ
52

Solution:-

Let us assume that Principal = 100x

For I year:

Rate = 8% p.a.

Principal = ₹100x

Time = 1 year

Interest = \sf\dfrac{P\times R\times T}{100}

= \dfrac{100x\times 8\times 1}{100}

= ₹8x

Amount = Principal + Interest

= 100x + 8x

= 108x

For II year:

Principal = ₹108x

Rate = 8% p.a.

Time = 1 year

Interest = \sf\dfrac{108x\times 8\times 1}{100}

= \sf\dfrac{216x}{25}

= ₹8.64x

According to Question:

₹8x + ₹96 = ₹8.64x

⟹ 96 = 8.64x - 8x

⟹ 96 = 0.64x

⟹ x = \sf\dfrac{96}{0.64}

⟹ x = \sf\cancel \dfrac{9600}{64}

⟹ x = 150

So:

Principal = ₹100x

= 100 × 150

= ₹15,000

∴ Principal = ₹15,000

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