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Han and tyler both started running toward each other from opposite ends of a 10-mile path along the river. Han runs at a pace of 12 minutes per mile. Tyler runs at a pace of 15 minutes per mile. How long does it take until Han and Tyler meet?

Respuesta :

Answer:

The time taken by both Han and Tyler to meet at point on 10 miles path is 0.37 minutes

Step-by-step explanation:

Given as :

The distance of path on which Han and Tyler run from opposite end = d = 10 miles

The speed of Han = [tex]s_1[/tex] = 12 miles per min

The speed of Tyler = [tex]s_2[/tex] = 15 miles per min

Let The time at which both meet at a points = t minutes

Let The distance cover by Han = [tex]d_1[/tex] = d miles

And The distance cover by Tyler = [tex]d_2[/tex] = (10-d) miles

Now, According to question

Time = [tex]\dfrac{\textrm Distance}{\textrm Speed}[/tex]

So, For Han

T = [tex]\dfrac{d_1}{s_1}[/tex]

i.e T = [tex]\dfrac{\textrm d miles}{\textrm 12 miles per min}[/tex]             ....1

Again For Tyler

T = [tex]\dfrac{d_2}{s_2}[/tex]

i.e T = [tex]\dfrac{\textrm (10-d) miles}{\textrm 15 miles per min}[/tex]        .....2

Now, Equating both equations

[tex]\dfrac{\textrm d miles}{\textrm 12 miles per min}[/tex] =  [tex]\dfrac{\textrm (10-d) miles}{\textrm 15 miles per min}[/tex]  

cross multiplying

15 × d = 12 × (10 - d)

Or, 15 d = 120 - 12 d

Or, 15 d + 12 d = 120

or, 27 d = 120

∴ d = [tex]\dfrac{120}{27}[/tex] = [tex]\dfrac{40}{9}[/tex]

I.e d = 4.44 miles

Now, Put the value of d in eq 1

So, Time taken = T = [tex]\dfrac{\textrm 4.44 miles}{\textrm 12 miles per min}[/tex]

I.e T = 0.37 min

So, The time taken to meet at a point = T = 0.37 min

Hence, The time taken by both Han and Tyler to meet at point on 10 miles path is 0.37 minutes  . Answer

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