a train travels at a certain average speed for a distance of 63 kilometer and then Travels at a distance of 72 km at an average speed of 6 km per hour more than its original speed. if it takes 3 hour to complete total journey what is the original average speed.
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let the average speed be x km/hr
63/x+72/x+6=3
63(x+6)+72(x)/x^2+6x=3
63x + 378+72x/x^2+6x=3
135x+378/x^2+6x=3
135x+378=3x^2+18x
3x^2+18x-135x-378=0
3x^2-117x-378=0
x^2-39x-126=0
x^2-42x+3x-126=0
x(x-42)+3(x-42)=0
(x-42)(x+3)=0
x=42 km/hr
63/x+72/x+6=3
63(x+6)+72(x)/x^2+6x=3
63x + 378+72x/x^2+6x=3
135x+378/x^2+6x=3
135x+378=3x^2+18x
3x^2+18x-135x-378=0
3x^2-117x-378=0
x^2-39x-126=0
x^2-42x+3x-126=0
x(x-42)+3(x-42)=0
(x-42)(x+3)=0
x=42 km/hr
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Answer is attached.
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Here's is Answer key for whole maths Leaked board paper 2018 =>
brainly.in/question/3121329
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