Question

solve this problem 33
please

Answer 1

Let the speed of the stream be x km/hr
Speed of the boat in still water = 18 km/hr
Speed of the boat in upstream = (18 - x) km/hr
Speed of the boat in downstram = (18 + x) km/hr

Distance between the places is 24 km.

Time to travel in upstream = d / (18 - x) hr
Time to travel in downstream = d / (18 + x) hr
Difference between timings = 1 hr

[24 / (18 - x)] - [24 / (18 + x)] = 1

⇒ 24 [1 / (18 - x)] - [1 / (18 + x)] = 1

⇒ [1 / (18 - x)] - [1 / (18 + x)] = 1/24

⇒ (18 + x - 18 + x) / (324 - x2) = 1/24

⇒ (2x) / (324 - x2) = 1/24

⇒ 324 - x2 = 48x

⇒ x2 - 48x + 324 = 0

⇒ (x + 54)(x - 6) = 0

x = - 54 or 6

Speed of the stream can not be negative.

Therefore, speed of the stream is 6 km/hr.

Answer 2

Solution:

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Given:

Condition : A motor boat whose speed 18 kmph in still water takes 1 hour to go 24 km upstream and to return to the same spot .

∴ Speed of boat = 18kmph,.
to

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To find:

The speed of stream,.
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Let the speed of stream be x kmph,
then,.
The speed  when it goes upstream = 18-x kmph,
The speed when it goes down stream =18+x kmph,


$\frac{24}{18-x} - \frac{24}{18+x} =1$


$24( \frac{1}{18-x} - \frac{1}{18+x}) = 1$


$24( \frac{18+x-(18-x)}{(18+x)(18-x)} = 1$


[tex]24( \frac{2x}{324-x^2}) =1 [/tex]


$\frac{48x}{324-x^2} = 1$


$48x = 324-x^2$


$-x^2-48x+324=0$


$x^2+48x -324 =0$

By using quadratic formula ,
we get,
a = 1
b = 48
c = -324


$x = \frac{-b + \sqrt{b^2-4ac}}{2a}$


$x= \frac{-(48)+ \sqrt{(48)^2-4(1)(-324)}}{2(1)}$  (negative one is not taken here because,.the discriminant is neagtive and also -b,so the value of x will be negative,.In case of speed it is not possible.)


$x=\frac{-48+ \sqrt{2304+1296}}{2}$


$x= \frac{-48+60}{2}$


$x = \frac{12}{2}$


 ∴$x=6$

∴ The speed of stream is 6 kmph

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                                    Hope it Helps !!