Question

A man covers a distance of 200 km travelling with a uniform speed of x km/hr.The distance could have been covered in 2hrs less,had the speed been (x+5)km/hr.Calculate the value of x.
quadratic equation

Answer 1

Speed of man = x km/hr
Distance = 200 km
Therefore, time = 200/x hrs
In the second case,
Speed = (x + 5) km/hr
Therefore, time = 200/(x + 5) hrs
According to the condition,
200/x - 200/(x + 5) = 2
=> 200[1/x - 1/(x + 5)] = 2
=> 200[(x + 5 - x) / x(x + 5)] = 2
=> (200 × 5) / (x² + 5x) = 2
=> 1000 = 2x² + 10x
=> 2x² + 10x - 1000 = 0
=> x² + 5x - 500 = 0  (Dividing by 2)
=> x² + 25x - 20x - 500 = 0
=> x(x + 25) -20(x + 25) = 0
=> (x + 25)(x-20) = 0
Either (x + 25) = 0, then x = -60, but it is not possible as speed can't be negative
or (x - 20) = 0, then x = 20 Ans.


 

Answer 2

The value of x is, x=20 km/hr.

What is speed distance and time?

• Speed tells us how fast something or someone is travelling.
• You can find the average speed of an object if you know the distance travelled and the time it took.
• The formula for speed is, speed = distance ÷ time.
• To work out what the units are for speed, you need to know the units for distance and time.

What are quadratic equation with example?

• In a quadratic equation, the variable x is an unknown value, for which we need to find the solution.
• Examples of quadratic equations are: 6x² + 11x – 35 = 0, 2x² – 4x – 2 = 0, 2x² – 64 = 0, x² – 16 = 0, x² – 7x = 0, 2x² + 8x = 0 etc.

According to the question:

Speed of the car =x km /hr.

Time taken = t hrs.

xt =200.

$t=\frac{200}{x}$

Now, (x+5) (t−2)=200.

$(x+5)\left(\frac{200}{x}-2\right)=200$

(x+5)(200−2x)=200x.

$200 x-2 x^{2}+1000-10 x=200 x\\ x^{2}+5 x-500=0\\ x^{2}+25 x-20 x-500=0\\ (x+25)(x-20)=0\\ x=-25, x=20$

Speed cannot be negative value, x=20 km/hr.

Hence, The value of x are  x=20 km/hr.

Learn more about quadratic equation here,

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