Question

a man moves to a place from Jorhat by car the distance between Jorhat and the place is 200 km if he drives 10 kilometre per hour less than the previous speed then he would take 1 hour more find the previous speed. ( class 10 )​

Answer 1

200 ÷ 10= 20

1 hour= 60 min

60 × 60 = 3600

20 × 1000= 20000

3600/20000

= 18 by 100

= 1800 km/ hr...

Answer 2

Previous speed of the man is 58.54 km per hour.

Step-by-step explanation:

Let the previous speed of the man is = x km per hour

and the current speed of the man is = y km per hour

If the man drives 10 km per hour less than the previous previous speed then

the current speed y = (x - 10) km per hour

Time ($t_{1}$) to cover the distance with speed x km per hour = $\frac{\text{Distance}}{\text{Speed}}=\frac{200}{x}$

Time ($t_{2}$) to cover the same distance with speed y km per hour = $\frac{\text{Distance}}{\text{Speed}}=\frac{200}{y}$

Now $t_{1}-t_{2}=1$

$\frac{200}{x}-\frac{200}{y}=1$

Since y = x - 10

Therefore, $\frac{200}{x}-\frac{200}{(x-10)}=1$

$200[\frac{x-10-x}{x(x-10)}]=1$

x(x - 100) = 2000

x² - 100x - 2000 = 0

By quadratic expression,

x = $\frac{100\pm \sqrt{(100)^{2}+4\times 2000} }{2}$

x = 50 ± 30√5

x = 58.54 km per hours

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