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Anonymous asked in Science & MathematicsMathematics · 2 months ago

# math problem help?

A jogger ran 3 ​miles, decreased her speed by 1 mile per​ hour, and then ran another 4 miles. If her total jogging time was 2 and 1/12 hours, find her speed for each part of her run.

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• LET THE SPEED BE X MPH FOR THE FIRST 3 MILES

SPEED FOR THE NEXT 4 MILES = X-1   MPH

TIME TAKEN =  3 /X + 4 / (X-1) = 25/12

36( X-1) + 48X = 25X ( X-1)

25X^2 - 109X  +36 = 0

SOLVE  X = 9/25  OR 4

X CANNOT BE 9/25

X = 4

ANSWER SPEEDS = 4MPH AND 3 MPH

CHECK

3/4 +4/3 = 9+16 /12 = 25/12 = 2 1/12 HOURS

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• A jogger ran 3 ​miles, decreased her speed by 1 mile per​ hour,

and then ran another 4 miles.

If her total jogging time was 2 and 1/12 hours,

find her speed for each part of her run.

let x = speed for 1st part of her run

x - 1 = speed for 2nd part of her run

travel time = distance/speed

3/x + 4/(x - 1) = 25/12

36/x + 48/(x - 1) = 25

lcd = x(x - 1)

36(x - 1) + 48x = 25x^2 - 25x

36x - 36 + 48x = 25x^2 - 25x

25x^2 - 109x + 36 = 0

(x - 4) (25 x - 9) = 0

speed for 1st part of her run was 4 mph

speed for 2nd part of her run  was 3 mph

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• let s be speed in first leg

then s–1 is speed in second leg

t = d/s = 3/s + 4/(s–1) = 2 1/12 = 25/12

3/s + 4/(s–1) = 25/12

multiply by 12s(s–1)

36(s–1) + 48s = 25s(s–1)

36s – 36 + 48s = 25(s²–s)

84s – 36 = 25s² – 25s

25s² – 25s – 84s + 36 = 0

25s² – 109s + 36 = 0

using a solver,

s = 4 or 9/25 (second solution results in s–1 negative, rejectted_

s–1 = 3

check

lap1, t = d/s = 3/4 = 0.75 hr

lap2, t = 4/3 = 1.333 hr

sum = 2.083 hr correct

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• 3/s + 4/(s - 1)= 25/12

(3s - 3 + 4s)/(s² - s)= 25/12

36s - 36 + 48s= 25s² - 25s

25s² - 109s + 36= 0

Solve by using the quadratic equation.

s= 0.36, 4.  Use s= 4

first speed= 4 mi/h, second speed= 3 mi/h

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• Let s be the speed for the 3 mile portion.

Let s -1 be the speed for the rest of the run.

Total Time = 25/12 = Distance / Speed = (3 / s) + [4 / (s - 1)]

25/12 = (3 / s) + [4 / (s - 1)]

25 = (36 / s) + (48 / (s - 1))

25(s)(s - 1) = 36(s - 1) + 48s

25s^2 - 25s = 36s - 36 + 48s

25s^2 - 25s = 84s - 36

25s^2 - 109s + 36 = 0

s = 4 mph, 9/25 mph.

9/25 is an extraneous solution.

So, the jogger ran 4 mph for the first part of the run and 3 mph for the second.

• thank you!

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