Question

A sum of money is invested for 3 years at 11% p.a. If it is invested for 5 years the intrest increases by ₹5500. Find the sum of money?
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Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large\underline{\green{\bf Given :-}}$

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• A sum of money is invested for 3 years at 11% P.A

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• If it is invested for 5 years the intrest increases by ₹5500

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$\large\underline{\red{\bf To \: Find :-}}$

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• Sum of money

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$\large\underline{\orange{\bf Solution :-}}$

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• Let "x" be the interest for 3 years

• Let the principal amount be "P"

• Let "T" be the rate

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We know that,

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➠ $\sf x = \dfrac { PTR } { 100 }$ ⚊⚊⚊⚊ ⓵

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$\underline{\bold{\texttt{For 3 years :}}}$

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• P = P

• T = 3

• R = 11

• x = x

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⟮ Putting these values in ⓵ ⟯

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➜ $\sf x = \dfrac { PTR } { 100 }$

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➜ $\sf x = \dfrac { P × 3 × 11 } { 100 }$

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➜ $\sf x = \dfrac { P × 33} { 100 }$

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➜ 100x = P × 33

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➜ $\sf P = \dfrac { 100x} { 33 }$ ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{For 5 years :}}}$

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• P = P

• T = 5

• R = 11

• x' = x + 5500

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⟮ Putting these values in ⓵ ⟯

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➜ $\sf x' = \dfrac { PTR } { 100 }$

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➜ $\sf x + 5500 = \dfrac { P × 5 × 11 } { 100 }$

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➜ $\sf x + 5500 = \dfrac { P × 55} { 100 }$

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➜ 100x + 550000 = P × 55

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➜ $\sf \cancel 5(20x + 110000) = P ×\cancel { 55 }$

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➜ 20x + 110000 = P × 11

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➜ $\sf P = \dfrac { 20x + 110000 } { 11 }$ ⚊⚊⚊⚊ ⓷

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⟮ Here equation ⓶ = ⓷ ⟯

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➜ $\sf \dfrac { 20x + 110000 } { \cancel { 11} } = \dfrac { 100x} {\cancel { 33 } }$

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➜ $\sf \dfrac { 20x + 110000 } { 1} = \dfrac { 100x} {3}$

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➜ 100x = 60x + 330000

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➜ 100x - 60x = 330000

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➜ 40x = 330000

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➜ $\sf x = \dfrac { 330000 } {40 }$

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➨ x = 8250 ⚊⚊⚊⚊ ⓸

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• Hence interest for 3 years is Rs 8250

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⟮ Putting x = 8250 from ⓸ to ⓶ ⟯

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➜ $\sf P = \dfrac { 100x} { 33 }$

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➜ $\sf P = \dfrac { 100 × \cancel { 8250} } { \cancel { 33 } }$

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➜ $\sf P = \dfrac { 100 × 250} { 1}$

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➨ P = 25000

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• Hence the sum of money is Rs 25000

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