Problema Solution
A student walks and jogs to college each day. She averages 5km/h walking and 9km/hjogging. The distance from home to college is 8km, and she makes the trip in 1 hr. How far does the student jog?
Answer provided by our tutors
let W = rate of walking = 5 km/h
let J = rate of jogging = 9 km/h
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D = 8 km (given)
T = 1 hour (given)
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rate * time = distance.
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let x = amount of time walking.
let y = amount of time jogging.
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x * W = distance walking.
y * J = distance jogging.
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since total time = 1 hour, then x + y = 1
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since total distance = 8 km, then x*W + y*J = 8
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you have 2 equations:
x + y = 1
x * W + y * J = 8
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since W = 5 km/h and J = 9 km/h, then second equation becomes:
x * 5 + y * 9 = 8
which is the same as
5x + 9y = 8
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two equations become:
x + y = 1
5x + 9y = 8
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if x + y = 1, then y = 1 - x
second equation of 5x + 9y = 8 becomes:
5x + 9*(1-x) = 8
remove parentheses:
5x + 9 - 9x = 8
combine like terms:
-4x + 9 = 8
subtract 9 from both sides:
-4x = -1
divide both sides by 4:
x = 1/4
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if x = 1/4 and x + y = 1, then y = 3/4
you have:
x = 1/4 of an hour.
y = 3/4 of an hour.
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you can now solve for distance since you have the rate and the time.
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equation to use is:
5x + 9y = 8
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5*x = 5 * 1/4 = 5/4 miles.
9*y = 9 * 3/4 = 27/4 miles.
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she walked 5/4 miles.
she jogged 27/4 miles.
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how does this work out?
in the formula 5x + 9y = 8, we can substitute 1/4 for x since she walked 1 quarter of an hour, and we can susbtitute 3/4 for y since she jogged 3/4 of an hour.
1/4 * 5 + 3/4 * 9 must equal 8
5/4 + 27/4 = 32/4 = 8.
since 8 = 8 the equation is true.
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student jogged 27/4 miles = 6.75 miles.
she walked 5/4 miles = 1.25 miles.
she jogged for 3/4 of an hour.
she walked for 1/4 of an hour.