A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km /hours more than its original speed. If it takes 3 hours to complete total journey, what is the original speed?
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Answered by
1
Heya
Here goes your answer
Let the initial speed of the train be x
Speed = Distance / time.
Time = Distance / Speed
Time taken to cover 63km = 63/x
Time taken to cover 72km = 72/(x+6)
According to the question,
63/x + 72/(x+6) = 3
[ 63(x+6) + 72x ] / x(x+6) = 3
(63x + 378 +72x) = 3x(x+6)
135x+378= 3x²+18x
3x²+18x-135x-378 = 0
3x²-117x-378=0
3(x²-39x-126)=0
x²-39x-126=0
x²-42x+3x-126=0
x(x-42) 3(x-42)=0
(X+3) (x-42) = 0
x+3=0
=> x=-3. But x can't be negative
x-42=0
=> x=42
Speed of train is 42 kmph
Here goes your answer
Let the initial speed of the train be x
Speed = Distance / time.
Time = Distance / Speed
Time taken to cover 63km = 63/x
Time taken to cover 72km = 72/(x+6)
According to the question,
63/x + 72/(x+6) = 3
[ 63(x+6) + 72x ] / x(x+6) = 3
(63x + 378 +72x) = 3x(x+6)
135x+378= 3x²+18x
3x²+18x-135x-378 = 0
3x²-117x-378=0
3(x²-39x-126)=0
x²-39x-126=0
x²-42x+3x-126=0
x(x-42) 3(x-42)=0
(X+3) (x-42) = 0
x+3=0
=> x=-3. But x can't be negative
x-42=0
=> x=42
Speed of train is 42 kmph
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0
Answer is attached.
Here's is Answer key for whole maths Leaked board paper 2018 =>
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Here's is Answer key for whole maths Leaked board paper 2018 =>
brainly.in/question/3121329
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