Math, asked by amritarautela1209, 8 months ago

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​

Answers

Answered by shahanafarook8
3

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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Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
10

\huge\sf\pink{Answer}

☞ The value of x is 20 km/h

━━━━━━━━━━━━━

\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

━━━━━━━━━━━━━

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

◈ The value of x?

━━━━━━━━━━━━━

\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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