Question

please solve the 10 sum quickly...it is easy . no spam plz

Answer 1

$<b><u><i>Hey mate!!! Here's your solution</i></u>$
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$\huge{To\: Find}$

Find the sum, invested at 10% compounded annually, on which the interest for third year exceeds the interest of first year by 252.

$\huge{Solution}$

Let the principal amount (P) be x.

Now, we consider two cases.

$\mathcal{\blue{Case\: 1}}$

P = x

Rate of interest (R) = 10%

Time (n) = 3 years

Therefore,

Compound interest (C.I) = {P (1+R/100)^n}-P

=> C.I.

= {x(1+10/100)³} - x

= {x (100+10/100)³} - x

= {x(110/100)³} - x

= {x(11/10)³} - x

= {1331x/1000} - x

= 1331x - 1000x/1000

= 331x/1000 [Equation 1]
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$\mathcal{\blue{Case\: 2}}$

P = x

R = 10%

n = 1 year

Therefore,

C.I.
= {P (1+ R/100)^n} - P

= {x(1+10/100)¹} - x

= {x(110/100)¹} - x

= {x(11/10)¹} - x

= 11x/10 - x

= 11x-10x/10

= x/10 [Equation 2]
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Now, according to question,

331x/1000 = x/10 + 252

=> 331x/1000 - x/10 = 252

=> 331x-100x/1000 = 252

=> 231x/1000 = 252

=> 231x = 252 × 1000

=> 231x = 252000

=> x = 252000/231

=> x = 1091 (approx)
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Hence, the sum is 1091 Rs.
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$\mathfrak{\blue{Finally,}}$

Your answer is

$\boxed{\bold{\green{\mathcal{1091 \: Rs}}}}$
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$\huge{\bold{\red{\mathfrak{Thank\: you}}}}$