Question

If the speed of an aeroplane is decreased by 120km/hr it take 18 minutes more to travel some distance .if the speed increased by 180 km/hr it takes 18 minutes less to travel same distance .find the average speed of the aeroplane and the distance .

Answer 1

Answer:

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Answer 2

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.

Required formula:

Required formula:Average Speed = Total distance/ total time

step 1:

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.

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

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

$y \times ( \frac{1}{x - 120} - \frac{1}{x} ) = \frac{18}{60} ......(1)$

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

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

$y \times ( \frac{1}{x} - \frac{1}{x + 180} ) = \frac{18}{60} .......(2)$

Step 2:

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

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

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

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

$120(x + 180) = 180(x - 120)$

$2x + 360 = 3x - 360$

$x =720$

x = 720 km/hr

Step 3:

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

$y \times ( \frac{1}{720 - 120} - \frac{1}{720} ) = \frac{18}{6 0 }$

$y \times ( \frac{1}{600} - \frac{1}{720} ) = \frac{3}{10}$

$y \times ( \frac{720 - 600}{720 \times 600} ) = \frac{3}{10}$

$y = \frac{3}{10} \times \frac{720 \times 600}{120}$

$y = 1080$

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