Question

A bus covers a distance of 90 km at a uniform speed had the speed been 15 km/hr more it would have taken 30 minutes less for the journey. find the original speed of the bus​

Answer 1

Let original speed =xkm/hr

⇒ Time taken to travel =90/xhr

⇒ New speed =(x+15)km/hr

⇒ Time taken to travel = 90/x+15hr

∴ 90/x=90xx+15+1/2

=>90/x+90/x-15=1/2

=>90[1/x-1/x+15]=1/2

=>x+15-x/x(x+15)=1/180

⇒ 15X180=x(x+15)

⇒ 15×180=x(x+15)

⇒ 2700=x^2+15x

⇒ x^2+15x−2700=0

⇒ x^2 +60x−45x−2700=0

⇒ x(x+60)−45(x+60)=0

⇒ (x+60)(x−45)=0

Speed can not be negative.

∴ The original speed of the train is 45km/hr

Answer 2

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• A bus covers a distance of 90 km at a uniform speed

• If the speed is 15 km/hr more it would have taken 30 minutes less for the journey

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$\sf\large\bold{\orange{\underline{\blue{ To \: Find :-}}}}$

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• Original speed of the bus

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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• Let the orginal speed of bus be "b"

• Let the time taken with original speed be "T1"

• Let the time taken with new speed be "T2"

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We know that ,

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➠ $\sf T = \dfrac { D } { S }$ ⚊⚊⚊⚊ ⓵

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Where ,

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• T = Time

• D = Distance

• S = Speed

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$\underline{\bold{\texttt{With original speed :}}}$

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• T = T1

• D = 90

• S = b

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf T = \dfrac { D } { S }$

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➜ $\sf T1 = \dfrac { 90} { b}$ ⚊⚊⚊⚊ ⓶

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$\underline{\bold{\texttt{With new speed :}}}$

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• T = T2

• D = 90

• S = b + 15

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf T = \dfrac { D } { S }$

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➜ $\sf T2 = \dfrac { 90} { b + 15}$ ⚊⚊⚊⚊ ⓷

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Given that , If the speed is 15 km/hr more it would have taken 30 minutes less for the journey

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$\underline{\bold{\texttt{Converting minutes to hours to match the units :}}}$

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➻ $\sf 1 \: Minute = \dfrac { 1 } { 60 } \: Hour$

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➻ $\sf 30 \: Minutes = \dfrac { 30 } { 60 } \: Hour$

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So,

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➜ $\sf T1 = T2 + \dfrac { 30 } { 60 }$

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➜ $\sf T1 = T2 + \dfrac { 1 } { 2 }$ ⚊⚊⚊⚊ ⓸

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⟮ Putting the values of T1 & T2 from ⓶ & ⓷ to ⓸ ⟯

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➜ $\sf T1 = T2 + \dfrac { 1 } { 2 }$

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➜ $\sf \dfrac { 90} { b} = \dfrac { 90} { b + 15} + \dfrac { 1 } { 2 }$

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➜ $\sf \dfrac { 90} { b} = \dfrac { 90 \times 2 + b + 15} {2(b + 15)}$

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➜ $\sf \dfrac { 90} { b} = \dfrac { 180 + b + 15} { 2b + 30}$

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➜ $\sf \dfrac { 90} { b} = \dfrac { 195 + b} { 2b + 30}$

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Cross multiplication

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➜ 90(2b + 30) = b(195 + b)

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➜ 180b + 2700 = 195b + b²

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➜ b² + 195b - 180b - 2700 = 0

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➜ b² + 15b - 2700 = 0

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Splitting the middle term

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➜ b² + 60b - 45b - 2700 = 0

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➜ b(b + 60)-45(b + 60) = 0

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➜ (b + 60)(b - 45) = 0

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• b = - 60
• b = 45

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As speed can't be negative hence ,

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➨ b = 45

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• Hence the original speed of the bus is 45 km/hr