Question

(i) If the speed of an aeroplane is decreased by 120 km/h, it takes 18 minutes more
to travel some distance. If the speed is increased by 180 km/h, it takes 18 minutes
less to travel the same distance. Find the average speed of the aeroplane and the
distance.​

Answer 1

If the average speed decreases by 120 km/hr & the time increase by 18 min and if the speed is increased by 180 km/hr & the time reduces by 18 mins then the average speed of the car is  720 km/hr and the distance travelled is 1080 km.

Step-by-step explanation:

Required formula:

Average Speed = $\frac{Total Distance}{Total Time}$

Let the average speed of the aeroplane be denoted by “x” km/hr and the distance travelled by it be denoted by “y” km/hr.

Step 1:

According to the first condition given in the question and based on the above formula, we can write the eq. as,

$\frac{y}{x-120}  - \frac{y}{x}  = \frac{18}{60}$

⇒ $y [\frac{1}{x-120}  - \frac{1}{x}]  = \frac{18}{60}$ …… (i)

According to the second condition given in the question and based on the above formula, we can write the eq. as,

$\frac{y}{x}  - \frac{y}{x+180}  = \frac{18}{60}$

⇒ $y [\frac{1}{x}  - \frac{1}{x+180}]  = \frac{18}{60}$ …… (ii)

Step 2:

On dividing the eq. (i) by (ii), we get

$\frac{1}{x-120} - \frac{1}{x} = \frac{1}{x}  - \frac{1}{x+180}$

⇒ $\frac{x-x+120}{x(x-120)} = \frac{x+180-x}{}$

⇒ $\frac{120}{x(x-120)} = \frac{180}{(x+180)x}$

⇒ 120 (x+180) = 180 (x-120)

⇒ 2x + 360 = 3x - 360  

⇒ x = 720 km/hr

Step 3:

Substituting the value of x = 720 km/hr in eq. (i), we get

$y [\frac{1}{720-120}  - \frac{1}{720}]  = \frac{18}{60}$

⇒  $y [\frac{1}{600}  - \frac{1}{720}]  = \frac{3}{10}$

⇒ $y[\frac{720-600}{720*600}] = \frac{3}{10}$

⇒ y = $\frac{3}{10} * 720 * 600 * \frac{1}{120}$

⇒ y = 1080 km

Thus, the average speed of the car is 720 km/hr and the distance travelled by it is 1080 km.

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