Question

An express train takes 1 hour less than a passenger train to travel 132 kilometre between mysore and bangalore if the average speed of the trains site:brainly.In

Answer 1

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Speed of passenger train is 33 km/hr.

Speed of express train is 44 km/hr.

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Appropriate question :

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore

If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

GIVEN :

• An express train takes 1 hour less than a passenger train
• Distance = 132 kilometre
• Average speed of the express train is 11 km/h more than the passenger train.

TO FIND :

• Average speed of the two trains.

SOLUTION :

Let the average speed of the passenger train = x km/hr.

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• Average speed of the express train is 11 km/h more than that of the passenger train.

•°• Average speed of express train is x + 11 km/hr.

Let's calculate the time for both the trains.

For passenger train :

Speed = x km/hr.

Distance = 132 km

We have the formula,

$\bold{\large{\boxed{\rm{\red{Speed\:=\:{\dfrac{Distance}{Time}}}}}}}$

So by blocking in the values in the formula we will be able to calculate time.

$\rightarrow$ $\bold{x\:=\:{\dfrac{132}{T}}}$

$\rightarrow$ $\bold{x\times\:T\:=\:132}$

$\rightarrow$ $\bold{T\:=\:{\dfrac{132}{x}}}$ ---> (1)

For express train :

Speed = x + 11 km/hr.

Distance = 132 km

Using the same formula,

$\rightarrow$ $\bold{x\:+\:11\:=\:{\dfrac{132}{T}}}$

$\rightarrow$ $\bold{x+11\times\:T\:=\:132}$

$\rightarrow$ $\bold{T\:=\:{\dfrac{132}{x+11}}}$ ---> (2)

Comparing equations 1 and 2,

$\bold{\dfrac{132}{x}}$ = $\bold{\dfrac{132}{x+11}}$

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• An express train takes 1 hour less than a passenger train

Constituting it mathematically,

$\rightarrow$ $\bold{\dfrac{132}{x+11}}$ = $\bold{\dfrac{132}{x}}$ - 1

$\rightarrow$ $\bold{\dfrac{132}{x+11}}$ + 1 = $\bold{\dfrac{132}{x}}$

$\rightarrow$ $\bold{\dfrac{132+x+11}{x+11}}$ = $\bold{\dfrac{132}{x}}$

Cross multiplying,

$\rightarrow$ $\bold{x(132+x+11)=132(x+11)}$

$\rightarrow$ $\bold{132x+x^2+11x=132x+1452}$

$\rightarrow$ $\bold{132x-132x+x^2+11x=1452}$

$\rightarrow$ $\bold{x^2+11x=1452}$

We will now further solve the question via factorization method.

$\rightarrow$ $\bold{x^2+11x-1452=0}$

$\rightarrow$$\bold{x^2+44x-33x-1452=0}$

$\rightarrow$ $\bold{x(x+44)-33(x+44)=0}$

$\rightarrow$$\bold{(x+44)\:\:\:(x-33)=0}$

$\rightarrow$ $\bold{x+44=0\:\:OR\:\:x-33=0}$

$\rightarrow$ $\bold{x=-44\:\:OR\:\:x=33}$

x = - 44 is not acceptable since speed cannot be negative.

•°• x = 33 is the speed of the passenger train.

Speed of passenger train = 33 km/hr.

Speed of express train :

Speed of express train = x + 11

Speed of express train = 33 + 11

Speed of express train = 44 km/hr.

Answer 2

Answer:

et the speed of passenger train be x km/hour.  Then speed of express train will be x + 11 km/hour

Time taken by the passenger train = 132 over straight x space hours.

Time taken by the express train = fraction numerator 132 over denominator straight x plus 11 end fraction space hours.  

According to question

132 over straight x minus fraction numerator 132 over denominator straight x plus 11 end fraction equals 1                        

Thus speed of passenger train is 33 km/hour and that of express train = 44 km/hour

Step-by-step explanation: