A train travels at a certain average speed for a distance of 54

A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32km upstream than to return downstream to the same spot. Find the speed of the stream.


Let the speed of the stream be s km/h.
Speed of the motor boat 24 km / h
Speed of the motor -boat upstream 24 s
Speed of the motor boat downstream 24 s
According to the given condition,

fraction numerator 32 over denominator 24 minus straight s end fraction minus fraction numerator 32 over denominator 24 plus s end fraction space equals 1
therefore space 32 space open parentheses fraction numerator 1 over denominator 24 minus s end fraction minus fraction numerator 1 over denominator 24 plus s end fraction close parentheses equals 1
therefore space 32 space open parentheses fraction numerator 24 plus straight s minus 24 plus straight s over denominator 576 minus straight s squared end fraction close parentheses space equals 1
therefore space 32 space straight x space 2 straight s space equals 576 minus straight s squared
therefore space straight s squared plus 64 straight s minus 576 space equals 0
therefore space left parenthesis straight s plus 72 right parenthesis left parenthesis straight s minus 8 right parenthesis space equals 0
Therefore comma space
straight s equals negative 72 space ors equals 8

Since, speed of the stream cannot be negative, the speed of the stream is 8 km/h.

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A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?


Let the average speed of train for the first 54 km be x km/h

⇒ Average speed for the next 63 km = ( x + 6) km/h

We know


Time space equals fraction numerator space Distance over denominator Speed end fraction

therefore space Time space taken space by space the space train space to space cover space 54 space km space equals space 54 over straight x space straight h
Also comma space time space taken space by space the space train space to space cover space the space next space 63 space km space equals space fraction numerator 63 over denominator straight x space plus space 6 end fraction space straight h

According space to space the space question comma

54 over straight x space plus space fraction numerator 63 over denominator straight x space plus space 6 end fraction space equals space 63

rightwards double arrow space fraction numerator 54 space left parenthesis straight x space plus space 6 space right parenthesis space plus 63 straight x over denominator straight x space left parenthesis straight x space plus space 6 right parenthesis end fraction space space equals space 3

⇒ 117x + 324 = 3 ( x2 + 6x)

⇒ 117x + 324 = 3x2 + 18x

⇒ 3x2 - 99x -324 = 0 ..(i)

Taking common from the above equation (i)

we have

x2 - 33x - 108 = 0

⇒ x2 - 36x + 3x -108 = 0

⇒ x (x - 36) + 3 (x - 36) = 0

⇒ (x - 36) (x +3 )=0

⇒ x -36 = 0 Or x + 3 =0

x = 36 or x = -3

 The speed of the train cannot be negative. Thus, x = 36

Hence, the speed of the train to cover 54 km or its first speed is 36 km/h.

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If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k.


Given points are A( k+1, 2k), B(3k, 2k+3), and C(5k-1, 5k)

These points will be collinear, if area of the triangle formed by them is zero.

We have,

              

i.e. 


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