Question

a man covers a distance of 200km traveling with a uniform speed of X km per hour the distance could have been covered in 2 hours less had the speed been X+5 km/hr. calculate the value of x​

Answer 1

Answer:

Step-by-step explanation:

Speed of the car =x km /hr

Time taken =t hrs

xt=200

∴t=  

x

200

​  

 

Now, (x+5)(t−2)=200

(x+5)(  

x

200

​  

−2)=200

(x+5)(200−2x)=200x

200x−2x  

2

+1000−10x=200x

x  

2

+5x−500=0

x  

2

+25x−20x−500=0

(x+25)(x−20)=0

x=−25,x=20

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

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Answer 2

$\huge\sf\pink{Answer}$

☞ The value of x is 20 km/h

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$\huge\sf\blue{Given}$

✭ Distance = 200 km

✭ Time = x hour

✭ If the speed was x+5 then time would have been reduced by 2 hours

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$\huge\sf\gray{To \:Find}$

◈ The value of x?

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$\huge\sf\purple{Steps}$

☯ $\underline{\sf As \ Per \ the \ Question}$

➠ Speed of man = x

➠ Distance = 200 km

So we know that,

$\underline{\boxed{\sf Speed = \dfrac{Distance}{Time}}}$

So then,

➝ $\sf x = \dfrac{200}{T}$

➝ $\sf T_1 = \dfrac{200}{x} \:\:\: -eq(1)$

Similarly

➠ Speed = x+5

➠ Distance = 200 km

Which means,

➝ $\sf x+5 = \dfrac{200}{T}$

➝ $\sf T_2 = \dfrac{200}{x+5} \:\:\: -eq(2)$

On Subtracting eq(1) & eq(2)

➳ $\sf\dfrac{200}{x} - \dfrac{200}{x+5} = 2$

$\sf Taking \ 200 \ common$

➳ $\sf200\bigg\lgroup \dfrac{1}{x} - \dfrac{1}{x+5} \bigg\rgroup = 2$

➳ $\sf200\bigg\lgroup \dfrac{\cancel{x}+5-x}{(x)(x+5)} \bigg\rgroup = 2$

➳ $\sf\dfrac{1000}{x^2+5} = 2$

➳ $\sf1000 = 2(x^2+5)$

➳ $\sf1000 = 2x^2+10$

➳ $\sf2x^2+10-1000 = 0$

➳ $\sf x^2+5-500 = 0$

➳ $\sf x^2 + 25x-20 x-500 = 0$

➳ $\sf x(x+25) - 20(x+25) = 0$

➳ $\sf (x-20)(x+25)$

➳ $\sf \red{x=20 \ or \ -25}$

But speed cannot be Negative

➳ $\sf\orange{Speed \ (x) = 20 \ km/h}$

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