Question

एक कार 840 km की दूरी को एक नियत समय मे तय करती है यदि चाल में 10km /घण्टा की बढोत्तरी कर दी जाए तो यह 2 घंटा कम समय लेती है कार की वास्तविक चाल क्या है it's urgent

Answer 1

Heya mate,

Here is your answer,. ⬇️⬇️

${ANSWER :-}$

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$<i>$ Let, the speed of the car be x.

We know, time = distance/speed
So, time = 840/x. .......... (1)

Now, when the speed of the car is increased by 10 km/h, we have
Speed = x + 10

Thus, the car will take time
= 840/(x+10). ............. (2)
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It is given that with 10km/h increase in speed, the car will take 2 hours less.

Therefore, from (1) and (2),

We have, 840/x - 840/(x+10) = 2

=> [ 840(x+10) - 840(x)] / x(x+10) = 2

=> 840x + 8400 - 840x = 2 * x(x+10)

=> 8400 = 2 * (x^2 + 10x)

=> x^2 + 10x = 8400/2 = 4200

=> x^2 + 10x - 4200 = 0

=> x^2 + 70x - 60x - 4200 = 0

=> x(x+70) - 60(x+70) = 0

=> (x-60)(x+70) = 0

=> x = 60,(-70)

Now, we have that, speed cannot be negative.

Thus, x = 60.
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➡️ Hence, the speed of the car was 60 km/h.

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$<b>$ Hope this helps,.

If helps, please mark ' thanks'.

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REGARDS, ARNAB ✌️

THANK YOU

Answer 2

दूरी=840 km
असली समय मान लेते हैं=X hours
तब चाल=दूरी÷समय
=840/X km/h

अगर चाल 10 km/h बढ़ती है तब समय 2 घंटे कम लेता है
मतलब चाल जब=840/x+10
समय तब =x-2 hours
मतलब
=
$\frac{840}{ \frac{840 }{x} + 10} = x - 2 \\ = \frac{840}{ \frac{840 + 10x}{x} } = x - 2 \\ \frac{840x}{840 + 10x} = x - 2 \\ = \frac{840}{10(84 + x)} = x - 2 \: (common = 10) \\ = 84x = (84 + x)(x - 2) \\ = 84x = 84x - 168 + {x}^{2} - 2x \\ = 84x = {x}^{2} + 82x - 168 \\ 84x - ( {x}^{2} + 82x - 168) = 0 \: \: (moving \: to \: left) \\ = 84x - {x}^{2} - 82x + 168 = 0 \\ = { - x}^{2} + 2x + 168 \\ = 0 = 0 - ( { - x}^{2} + 2x + 168) \: moving \: to \: right \\ = {x}^{2} - 2x - 168 \\ = {x}^{2} - 14x + 12x - 168 \: ( \: split \: middle \: term) \\ = x(x - 14) - 12(x + 14)$
=(x-12)(x+14
हमें मिला
X=-12
X=14

समय कभी घाटे में नहीं होता इसलिए
समय=14 hours
चाल=840/X
=840/14
=60 km/h

$\boxed{\mathbf{चाल=60 km/h}}$