Question

If the average speed of car is increased by 10 km/hr. the time required to
travel a certain distance is reduced by 36 min. If the average speed is
decreased by 15 km/hr. then 1 hr 12 mins more are required to travel the
same distance. Find the average speed of the car and the distance travelled​

Answer 1

If the average speed increases by 10 km/hr & the time reduce by 36 min and if the speed is decreased by 15 km/hr & the time increases by 1 hr 12 mins then the average speed of the car is 90 km/hr and the distance travelled is 540 km.

Step-by-step explanation:

Required formula:

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

Let the average speed of the car be denoted by “x” km/hr and the distance travelled by car 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,

$\frac{y}{x}$ – $\frac{y}{x+10}$ = $\frac{36}{60}$

⇒ y [$\frac{y}{x}$ – $\frac{1}{x+10}$] = $\frac{3}{5}$ …… (i)

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

$\frac{y}{x-15}$ – $\frac{y}{x}$ = [1+$\frac{12}{60}$]

⇒ y [$\frac{1}{x-15}$ – $\frac{1}{x}$] = $\frac{6}{5}$ …… (ii)

Step 2:

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

[$\frac{y}{x}$ – $\frac{1}{x+10}$] / [$\frac{1}{x-15}$ – $\frac{1}{x}$] = $\frac{3}{5}$ * $\frac{6}{5}$

⇒ $\frac{</strong>x+10-x<strong>}{</strong>x(x+10)<strong>}$ / $\frac{</strong>x-x+15<strong>}{</strong>(x-15)x<strong>}$ = ½  

⇒ $\frac{</strong>10<strong>}{</strong>x(x+10)<strong>}$ / $\frac{</strong>15<strong>}{</strong>(x-15)x<strong>}$ = ½  

⇒ 20 (x-15) = 15 (x+10)

⇒ 20x – 300 = 15x + 150

⇒ 5x = 450

⇒ x = 90 km/hr

Step 3:

Substituting the value of x = 90 km/hr in eq. (i), we get

y [$\frac{1}{90} - \frac{1}{10 + 90}$] = $\frac{3}{5}$

⇒ y [$\frac{1}{90} - \frac{1}{100}$] = $\frac{3}{5}$

⇒ y [$\frac{</strong>100-90<strong>}{</strong>9000<strong>}$] = $\frac{3}{5}$

⇒ y [$\frac{</strong>10<strong>}{</strong>9000<strong>}$] = $\frac{3}{5}$

⇒ y = $\frac{3*900}{5}$

⇒ y = 540 km

Thus, the average speed of the car is 90 km/hr and the distance travelled by it is 540 km.

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