Question

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

Answer 1

Answer:

Step-by-step explanation:

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{</strong>x-x+120<strong>}{</strong>x(x-120)<strong>} = \frac{</strong>x+180-x<strong>}{</strong>(x+180)x<strong>}

⇒  \frac{</strong>120<strong>}{</strong>x(x-120)<strong>} = \frac{</strong>180<strong>}{</strong>(x+180)x<strong>}

⇒ 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