Question

A Car travels a distance of 300km at a Unifrom speed. If the speed had been 10km/h less, then it would have taken 1 hour more
to to cover the sene distance represent the situation in the foo form of a quadratic equation.​

Answer 1

Step-by-step explanation:

A car travels first 300 kms at an average rate of 80 km/hr and further travels the same distance at an average of 60km/hr.

Answer 2

The given situation can be represented in the form of a quadratic equation as x² - 10x - 300 = 0.

• Let the initial speed of the car be taken as x km / hr.

• Given,

Distance travelled by the car at x km / hr = 300 km

• Now, time taken by the car to cover 300 km (T1) = Distance / Speed

= (300 km) / (x km / hr)

= (300 / x) hr

• Now, if the speed of the car decreases by 10 km / hr, then the speed of the car becomes ( x - 10 ) km / hr

• Time taken by the car to cover 300 km at the decreased speed (T2) = Distance / New Speed

= 300 km / ( x - 10 ) km / hr

= [ 300 / ( x - 10 ) ] hr

• It has been said that the car takes 1 hour more than the usual while travelling at a speed of ( x - 10 ) km / hr.

This means, the difference between T2 and T1 is 1 hr.

• Therefore, according to the question,

$=> [ \frac{300}{x \ - \ 10} ] \ hr - [ \frac{300}{x} ]\ hr = 1 \ hr\\\\Taking \ out\ 300 \ hr \ as \ the\ common\ factor,\\=> 300 \ hr [ \frac{1}{x \ - \ 10} - \frac{1}{x} ] = 1 \ hr\\\\=> 300 [ \frac{1}{x \ - \ 10} - \frac{1}{x} ] = \frac{1 \ hr}{hr} \\\\=> 300\ [ \frac{x \ - \ (x \ - \ 10)}{x ( x\ - \ 10\ )}= 1\\\\=> 300 \ ( x - x + 10 ) = x \ ( x - 10 )\\\\=> 300 \ . \ 10 = x^{2} - 10 x\\\}=> x^{2} - 10x - 300 = 0$

∴   The required quadratic equation is x² - 10x - 300 = 0.