Question

47. A train travels 180 km at a uniform speed. If the speed had been
9 km/hr more, it would have taken 1 hour less for the same journey
Find the speed of the train.​

Answer 1

$\huge\star{\red{\underline{\mathfrak{Answer :-}}}}$

36 km/hr.

$\huge\star{\red{\underline{\mathfrak{Explanation :-}}}}$

Let the speed of the train be x km/hr.

We know that,

$Speed =\dfrac {Distance}{Time}$

Distance given is 180 km.

Therefore, time when the speed is x.

$T1 = \dfrac{180}{x}$

And time when the speed if increased is is 9 km per hour more than the previous speed i.e. x+9.

$T2 = \dfrac{180}{x+9}$

It given that the difference between the two speeds ( how much less time the train will take to reach the destination if the speed is increased by 9 ) is 1 hour less.

So,

T2 - T1 = 1

$\leadsto {\dfrac{180}{x}-\dfrac {180}{x+9}=1}$

Taking LCM,

$\leadsto { \dfrac{180x + 1620 - 180x}{9x + {x}^{2} } = 1}$

$\leadsto {\dfrac{ 1620}{9x + {x}^{2} } = 1}$

By cross multiplication,

$\leadsto{ 1620 = 9x + {x}^{2}}$

$\leadsto { {x}^{2} + 9x - 1620 = 0}$

Now, solving the equation by splitting the middle term.

$\leadsto {{x}^{2} + 45x - 36x - 1620 = 0}$

$\leadsto {(x + 45) - 36(x + 45) = 0}$

$\leadsto {(x + 45)(x - 36) = 0}$

$\leadsto {x = (- 45) \: and \: 36}$

As speed cannot be in negative. So, the value (-45) is to be neglected.

Therefore, the speed of the train = x = 36 km/hr.

__________________________

Answer 2

$\bf{\huge{\underline{\boxed{\sf{\green{ANSWER\:}}}}}}$

Given:

A train travels 180km at a uniform speed. If the speed had been 9km/hr more,it would have taken 1 hour less for the same journey.

To find:

<The speed of the train.>

Explanation:

Let the speed of the train be R km/hrs.

We have,

• Distance of train travels= 180km.
• Speed of the train= R km/hrs,

We know that time= $\frac{Distance}{Speed}$

Therefore,

Time taken by train to cover 180km with speed= $(\frac{180}{x} )\:hours$

&

Assume the more speed of the train= (R+9)km/hrs.

Time taken by train to cover 180km with this speed= $(\frac{180}{R+9})\:hours$

Give that speed be (R+9)km/hrs, then the train will take 1 hous less the speed to travel same distance.

A/q,

→ $\frac{180}{R+9}\:-\frac{180}{R} =1$

→ $\frac{180R-180(R+9)}{R(R+9)} =1$

→ 180R - 180(R+9)= R(R+9)

→ 180R -180R - 1620= R² + 9R

→ 0 - 1620= R² +9R

→ R² +9R -1620= 0

[factorise]

→ R² -36R + 45R -1620= 0

→ R(R -36) + 45(R-36)= 0

→ (R-36)(R+45)= 0

→ R- 36= 0   or   R + 45= 0

→ R= 36    or   R= -45

Here, we obtained two value but we know that negative value is not acceptable the speed.

So,

The speed of the train is 36km/hrs.