Question

.............can anyone ans this que.....​

Answer 1

Answer:42km/hr

Step-by-step explanation:

Let x be the original speed.

A.G.C

(63/x)+(72/x+6)=3

(63x+378+72x)=3x^2+6x

X^2-39x-126=0

X=42.

Answer 2

Heya mate,

Here is your answer,. ⬇️⬇️

${ANSWER :-}$

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$<i>$ Let the original speed of the train be x km/hr.

Now, Time = distance /speed

Thus, time = 63/x h.

Then, it travels at a speed of 6 km/hr more than its original speed.

Thus, new speed = (x + 6) km/h

Now, Time = 72/(x+6) h.

By condition, we have,

63/x + 72/(x+6) = 3.

=> [63(x+6) + 72x] / x(x+6) = 3.

=> (63x + 378 +72x) / x^2 + 6x = 3

=> 135x + 378 = 3x^2 + 18x

=> 3x^2 - 117x - 378 = 0

=> x^2 - 39x - 126 = 0

=> x^2 - 42x + 3x - 126 = 0

=> x(x-42) + 3(x-42) = 0

=> (x + 3) ( x - 42) = 0

=> x = ( - 3) , 42

Speed can't be negative.

Thus, speed = 42 km/h.

➡️ Hence, the original average speed is 42 km/h.

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$<b>$ Hope this helps,.

If helps, please mark 'thanks'

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REGARDS, ARNAB ✌️

THANK YOU