Question

An ordinary train takes 1 hour less for journey of 150km if its speed is increased 5 km per hour from its visual speed. find the visual speed of train

Answer 1

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հίί ʍαtε_______✯◡✯
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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯ ♦Let the usual speed of train be x km/hr

Then, increased speed of the train  = (x + 5)km/hr

▶Time taken by the train under usual speed speed to cover  150 km = 150/x hr

Time taken by the train under increased speed to cover 150km = 150(x +5)hr

▶So,

150 x - 150 (x + 5) = 1

$\frac{150 } {x} - \frac{150}{(x + 5)} = 1$

150( x + 5) - 150xx (x + 5 ) = 1

=》

$\frac{150(x + 5) - 150x}{x(x + 5)} = 1$
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$150x + 750 - 150x {x}^{2} + 5x = 1$
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=》
$750 = {x}^{2} + 5 x$

${x}^{2} + 5x - 750 = 0$
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${x}^{2} - 25x + 30x - 750 = 0$
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x(x – 25) + 30(x – 25) = 0

(x – 25) (x + 30) = 0

x = 25  or x = -30

▶Since, the speed of the train can never be negative

♦Therefore, the usual speed of the train is x = 25km/hr

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$hope \: it \: helps$


★BRAINLY★

Answer 2

Let usual speed be u kmph. Then

u=St=150tt=150u.……[1]

When u is increased by 5 kmph

u + 5=St - 1=150t - 1u + 5=150150u−1...[∵t=150u]u + 5=150u150 - u(u+5)(150−u)=150u150u - u2+750−5u=150uu2+5u−750=0u2+30u−25u−750=0u(u + 30) - 25(u + 30) = 0(u - 25)(u + 30) = 0u = 25 (or) u = -30

Speed can’t be negative. Therefore, usual speed of train is 25 kmph