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Re: Circumference of the front wheel of a cart is 40ft long and that of [#permalink]
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Let x = distance travelled by the vehicle.

Rotations by front wheel = x/40, Rotations by rear wheel = x/48.

Then, x/40 = x/48 + 5
Solving for x = 1200 feet.

Qa = 1200 feet = 400 yards
Qb = 500 yards

Hence answer is B.
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Re: Circumference of the front wheel of a cart is 40ft long and that of [#permalink]
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I had difficulty conceiving of the equation here, so I just tried to get the pattern:

- STEP 1: If the cart goes 48ft
--The back wheel has 1 revolution
--The front wheel has 48/40 = 1.2 revolutions
--This is 0.2 more for the front than the back

- STEP 2: If the cart goes 96 ft
--The back wheel has 2 revolutions
--The front wheel has 2.4 revolutions
--This is 0.4 more for the front and than the back

- Based on the above, I observed that for every 48 ft (let's call that "1 STEP"), the front wheel goes 0.2 more revolutions than the back.
- So to go 5 revolutions more, if would have to go 5/0.2 = 25 "STEPS"
- If a STEP = 48ft, that's 25*48 = 1200 ft
- 1200 ft = 400 yards (from 1200/3), so Quantity B (500 yards) is bigger.

(I observed that on the real GRE, which I took one time recently, whenever there are unit conversions, the question tells you what they are. For example, on the real GRE, I think the above question would include the sentence, "3 ft = 1 yard," which would clue you in about the need to convert between feet and yards.)
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Re: Circumference of the front wheel of a cart is 40ft long and that of [#permalink]
The LCM of 48 and 40 is 240.

240/48 =5
240/40 =6

For every 240, 1 extra revolution front wheel is made. hence 240 *5 = 1200
the conversion is 1 ft = 0.33333 yd
so 1200 ft = 400 ft

hence A< B
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Re: Circumference of the front wheel of a cart is 40ft long and that of [#permalink]
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