Math, asked by narsikodur22, 3 months ago

In a race, x beats y by 50 seconds. In the same race y beats z by 70 seconds. Y's speed is the average of the speeds of x and y. The time (in minutes) taken by Y to run the race is

Answers

Answered by amitnrw

Given : In a race, x beats y by 50 seconds.

In the same race y beats z by 70 seconds.

Y's speed is the average of the speeds of x and z.

To Find : The time (in minutes) taken by Y to run the race is

Solution:

x beats y by 50 seconds

y beats z by 70 seconds.

Let say time taken by y to run the race in minutes is t/60 mins

then time in secs = t secs

x beats y by 50 seconds

=> x take t - 50 second

y beats z by 70 seconds.

=> z take t + 70 secs

speed of x = D/(t - 50)

speed of z = D / (t + 70)

Speed of y = D/t = ( D/(t - 50) + D/(t + 70)) / 2

=> 2/t = 1/(t - 50) + 1/(t + 70)

=> 2/t = (t + 70 + t - 50)/(t - 50)(t + 70)

=> 2 (t²+ 20t - 3500) = 2t² + 20t

=> t²+ 20t -3500 = t² + 10t

=> 10t = 3500

=> t = 350

350 secs

Time taken by y to run the race in minutes is t/60 mins

= 350/60 mins

= 35/6 mins

Time (in minutes) taken by Y to run the race is 35/6 mins

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Answered by RvChaudharY50

Given :-

In a race, x beats y by 50 seconds.
In the same race y beats z by 70 seconds.
Y's speed is the average of the speeds of x and z.

To Find :-

The time (in minutes) taken by Y to run the race is ?

Solution :-

Let us assume that speed of y to run the race is x seconds and length of race is D m .

since it is given that, x beats y by 50 seconds.

so,

→ Time taken by x to complete the race = 50 seconds less then time taken by y to complete the race = (x - 50) seconds.

also,

we have given that , in the same race y beats z by 70 seconds.

so,

→ Time taken by z to complete the race = 70 seconds more then time taken by y to complete the race = (x + 70) seconds.

as we have assumed that, length of race is D m .

we know that,

Speed = Distance / Time taken .

then,

→ Speed of x = D / T = D/(x - 50) m / s .

→ Speed of y = D / T = (D / x) m /s .

→ Speed of z = D / T = D/(x + 70) m /s .

now, we have given that, Y's speed is the average of the speeds of x and z. Since average of 2 numbers is sum of them divided by 2,

therefore,

→ Speed of y = [Speed of x + speed of z] / 2

→ 2 * Speed of y = [Speed of x + speed of z]

putting values of speed, we get,

→ 2 * (D/x) = [ {D/(x - 50)} + {D/(x + 70)} ]

→ 2D/x = D[ 1/(x - 50) + 1/(x + 70) ]

→ 2/x = 1/(x - 50) + 1/(x + 70)

→ 2/x = (x + 70 + x - 50)/(x - 50)(x + 70)

→ 2(x - 50)(x + 70) = x(2x + 20)

→ 2(x² + 70x - 50x - 3500) = 2x² + 20x

→ 2x² + 40x - 7000 = 2x² + 20x

2x² will be cancel from both sides,

→ 40x - 20x = 7000

→ 20x = 7000

→ x = (7000/20)

→ x = 350 s .

since we know that,

60 seconds = 1 minute .

hence,

→ The time (in minutes) taken by Y to run the race is = x seconds = (x/60) = (350/60) = (35/6) = 5(5/6) minutes . (Ans.)