Question 280523: The front wheel of a wagon makes 110 more revolutions than the rear wheel in going a mile . if the circumference of the front wheel is increased by 25% and that of the rear wheel decreased 25%, the rear wheel then revolves 88 more times than the front wheel in going a mile . the original circumference of the front wheel in feet is
A 8 b 10 c 12 d 14 e 16
Found 2 solutions by richwmiller, Theo:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! (x+110)*a=5280
x*b=5280
(y)*1.25*a=5280
(y+88)*.75b=5280
(x+110)*a=5280 and x*b=5280 and (y)*1.25*a=5280 and (y+88)*.75b=5280
a = 12, b = 16, x = 330, y = 352
the original front wheel circumference is 12
check
original
330+110=440
440 front wheel revolutions
12 is the circumference
440*12=5280
rear wheel revolutions 330
440-330=110 more in front wheel
330*16=5280
rear wheel circumference 16
rear wheel revolutions 330
second arrangement
352+88=440
front wheel circumference 15
front wheel revolutions 352
15*352=5280
rear wheel circumference 12
rear wheel revolutions 440
rear wheel revolutions 88 more than front
12*440=5280
ok
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! Let x = the number of revolutions that the front wheel makes in a mile.
Let a = the circumference of the smaller wheel.
Let b = the circumference of the bigger wheel.
If the smaller wheel makes 110 more revolutions than the bigger wheel in a mile, this means that the bigger wheel makes 110 less revolutions than the smaller wheel in a mile.
Since the number of revolutions that the smaller wheel makes is x, then the number of revolutions that the bigger wheel makes is (x-110).
Since we are looking for the circumference of the wheel in feet, we will convert 1 mile to 5280 feet so all the measurements will be the same as what we are looking for.
We get the following formulas:
x*a = 5280 (first equation)
(x-110)*b = 5280 (second equation)
We increase the circumference of the front wheel by 25% to get 1.25*a
We decrease the circumference of the rear wheel by 25% to get .75*b
Now the rear wheel is revolving 88 more times than the front wheel when going a mile.
Our formulas becomes:
y*1.25*a = 5280 (third equation)
(y+88)*.75*b = 5280 (fourth equation)
Since all equations equal to 5280, we can pick and choose which equations we want to make equal to each other.
We have x*a = 5280 and y*1.25*a = 5280.
Since both these equations equal 5280, we set them equal to each other to get:
x*a = y*1.25*a which becomes:
x = 1.25 * y (equation between fourth and fifth equation)
We substitute for x in the second equation to get:
(x-110)*b = 5280 becomes (1.25*y - 110)*b = 5280 (fifth equation)
Since the fifth equation and the fourth equation both equal 5280, we now set the fifth equation equal to the fourth equation to get:
(1.25*y - 110)*b = (y+88)*.75*b (sixth equation).
We simplify this equation to get:
1.25*y*b - 110*b = .75*y*b + .75*88*b
we simplify this equation further to get:
1.25*y*b - 110*b = .75*y*b + 66*b
We subtract .75*y*b from both sides of this equation and we add 110*b to both sides of this equation to get:
1.25*y*b - .75*y*b = 66*b + 110*b
We simplify further to get:
.5*y*b = 176*b
We divide both sides of this equation by b to get:
.5*y = 176
We divide both sides of this equation by .5 to get:
y = 352 (seventh equation)
Now that we have y, we can go back and solve for a, and then b.
We go back to the first equation which is:
x*a = 5280
We know that x = 1.25*y (from the equation between fourth and fifth equation), so this equation becomes:
1.25*y*a = 5280
We know that y = 352 (from the seventh equation), so we substitute to get:
1.25*352*a = 5280 which becomes:
440*a = 5280 which becomes:
a = 5280 / 440 = 12
Since a is the circumference of our front wheel and the problem wanted us to solve for the circumference of the front wheel, we are done.
Our answer is that the circumference of the front wheel is originally 12 feet.
We could stop here, but we'll go on to confirm our results are good meaning that all equations in the problem check out.
Since we have a = 12, we now look to solve for b.
Our second equation states that:
(x-110)*b = 5280
We know that x = 1.25 * y (from the equation between the fourth and fifth equation).
We substitute 1.25 * y for x to get:
(1.25*y - 110)*b = 5280
We know that y = 352 (from the seventh equation), so we substitute for y in this equation to get:
(1.25*352 - 110)*b = 5280
We simplify to get:
(440 - 110)*b = 5280
We simplify further to get:
330*b = 5280
We divide both sides of this equation by 330 to get:
b = 5280/330 = 16.
We now know that the original size of the front wheel was 12 and the original size of the rear wheel was 16.
We now need to get the value of x.
We know that x = 1.25 * y and we know that y = 352, so x = 1.25 * 352 = 440.
We now have:
x = 440
y = 352
a = 12
b = 16
a is the original circumference of the front wheel.
b is the original circumference of the rear wheel.
Our first equation is x*a = 5280 which becomes 440*12 = 5280 which becomes 5280 = 5280 which is true.
Our second equation is (x-110)*b = 5280 which becomes 330*16 = 5280 which becomes 5280 = 5280 which is true.
Our third equation is y*1.25*a = 5280 which becomes 352*1.25*12 = 5280 which becomes 352*15 = 5280 which becomes 5280 = 5280 which is true.
In this case, the 1.25 factor made the circumference of the front wheel equal to 15 which is 25% larger than 12 which satisfies the requirements of the problem.
Our fourth equation is (y+88)*.75*b = 5280 which becomes (352+88)*.75*16 = 5280 which becomes 440 * 12 = 5280 which becomes 5280 = 5280 which is true.
In this case, the .75 factor made the circumference of the rear wheel equal to 12 which is 25% smaller than 16 which satisfies the requirements of the problem.
It appears that we started off with the front wheel at 12 and the rear wheel at 16 and ended up with the front wheel at 15 and the rear wheel at 12.
In the second equation, the front wheel was revolving 110 more times than the rear wheel.
This means that the 12 inch wheel was revolving 110 more times than the 16 inch wheel.
Since x was equal to 440, this meant that 440*12 = 330*16 = 5280.
In the fourth equation, the rear wheel was revolving 88 more times than the front wheel.
Since y was equal to 352, this meant that 440*12 = 352*15 = 5280
All the numbers check out so the answer is good.
The answer is that the original circumference of the front wheel was 12 feet.
That would be selection C.
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