Answer:

25700

you can find by using Simple intrest trick because compound interest and simple interest for 1 year is same .

Principal = 100% or 100X let

Rate = 3%

Time = 1 year

SI = time × rate mean 3× 1 = 3 given

Amount = 26471

but we have assumed our principal is 100% and intrest is 3 and we know

Amount is Principal + intrest so

our amount is 103% and given Amount is 26471

it's mean

103%= 26471

value of 1% = 26471÷ 103 = 257

now we have Principal 100% putting value of % in 100% or 100x it becomes 100× 257 = 25700

Verified answer

Answer:

Given :-

• A sum of money becomes Rs 26,471 at an interest rate of 3% per annum compounded annually after a period of one year.

To Find :

• What is the sum or the principal.

Formula Used :-

[tex]\clubsuit[/tex] Amount formula when the interest is compounded annually :

[tex]\bigstar \: \: \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}\: \: \: \bigstar\\[/tex]

where,

• A = Amount
• P = Principal
• r = Rate of Interest
• n = Time Period

Solution :-

Given :

• Amount = Rs 26,471
• Rate of Interest = 3% per annum
• Time Period = 1 year

According to the question by using the formula we get,

[tex]\implies \bf A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n\\[/tex]

[tex]\implies \sf 26471 =\: P\bigg(1 + \dfrac{3}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 26471 =\: P\bigg(\dfrac{100 \times 1 + 3}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 26471 =\: P\bigg(\dfrac{100 + 3}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 26471 =\: P\bigg(\dfrac{103}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 26471 =\: P \times \dfrac{103}{100}\\[/tex]

[tex]\implies \sf 26471 =\: \dfrac{103P}{100}\\[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 103P =\: 26471 \times 100[/tex]

[tex]\implies \sf 103P =\: 2647100[/tex]

[tex]\implies \sf P =\: \dfrac{\cancel{2647100}}{\cancel{103}}[/tex]

[tex]\implies \sf P =\: \dfrac{25700}{1}[/tex]

[tex]\implies \sf\bold{\red{P =\: Rs\: 25700}}\\[/tex]

[tex]\sf\bold{\underline{\purple{\therefore\: The\: sum\: is\: Rs\: 25700\: .}}}\\[/tex]