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A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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Updated on: 07 Jul 2017, 00:32
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A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long? (1 mile = 5,280 feet and 1 foot = 12 inches) A. \(\frac{5,280 mt}{12 p}\) B. \(\frac{5,280 pt}{12m}\) C. \(\frac{5,280pmt}{12}\) D. \(\frac{5,280*12m}{pt}\) E. \(\frac{5,280*12p}{mt}\)
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Originally posted by TheBigCheese on 12 Jul 2008, 09:04.
Last edited by Bunuel on 07 Jul 2017, 00:32, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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12 Jul 2008, 11:50
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Quote: A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long? (1mile = 5,280 feet and 1 foot = 12 inches) Let’s translate all measure units to feet and calculate the area of the stripe: Area = (m*5280)*(t/12) square feet. Now, all we need is to divide this by p – this will give us the number of gallons: N of gallons = 5280*mt/12p. This is A. (I wonder why they didn’t ask us to divide 5280 by 12… The result is 440, an integer)
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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12 Jul 2008, 12:04
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POE
unit of m = distance unit of t = distance unit of p = area = distance X distance
we are looking for an answer in gallon ... so all units of distance should cancel each other ... eliminate B, C, D, .. between A and E
p sq feet ------------- 1 gallon xyz sq feet ----------- xyx/p gallon ---> p should come in denominator
Answer A,
30 seconds ... max
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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12 Jul 2008, 12:12
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one thing i've noticed about algebra and problem solving on the GMAT is that when you have a problem that appears to require a lot of multiplication of large numbers, it's best to just keep those number un-multiplied and then set up the equation. It also helps to take a look at the answer choices and get a feel for what format the answers are in. If you did the multiplication and did 5280 * 12, and then divided by 144, or whatever, you'd have the correct answer, but you would have a hard time identifying it in the answer choices. You'd come up with \(\frac{440tm}{p}\) Like here t = how many feet wide the stripe is. 5280 = feet long in 1 mile, m = length of the yellow stripe in 1 mile so 5280 * 12 = number of inches in a mile. 5280 * 12 * t = number of square inches per mile of yellow stripe. 5280 * 12 * t * m = total number of squre inches we need to cover...but the coverage is in square feet p so 144 * p = number of inches per square foot the paint will cover. \(\frac{5280 * 12 * t * m}{144p}\) becomes \(\frac{5280 * t * m}{12p}\)
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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19 Dec 2010, 17:02
Can you please describe how to attack this problem? I just stuck to the p's, m's and t's and did a quick calculation on where they end up, but the p square feet is throwing me off. How does this compare to the inch wide yellow line? thanks.
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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20 Dec 2010, 01:55
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gettinit wrote: Can you please describe how to attack this problem? I just stuck to the p's, m's and t's and did a quick calculation on where they end up, but the p square feet is throwing me off. How does this compare to the inch wide yellow line? thanks. A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long? (1mile = 5,280 feet and 1 foot = 12 inches)A. (5,280 mt) / 12 p B. (5,280 pt) / 12m C. (5,280 pmt) /12 D. (5,280)(12m) / pt E. (5,280)(12p) / mt Given that: 1 gallon of paint covers an area of p square feet. Question: how many gallons of paint will be needed ... In any case you will have: (total area in square feet)/(gallons per feet)=(total area in square feet)/p, so p must be in the denominator: eliminate all but A and D. Now, lets see where should be t: (area in square feet)=(width in feet)*(length in feet) --> width=t inches as 1 feet=12 inches then t inches=t/12 feet, so (area in square feet)=(t/12) * (length in feet), so t must be in the nominator: only A is left. Answer: A. Direct way: (total are in square feet)/(gallons per feet)=(total are in square feet)/p=? (area in square feet)=(width in feet)*(length in feet)=(t/12)*(5,280m) --> (total are in square feet)/p=(5,280mt)/(12p). Answer: A. Hope it's clear.
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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23 Dec 2010, 19:19
Thanks Bunuel very clear!
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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12 Feb 2015, 13:49
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Hi All, I'm a big fan of TESTing VALUES in these types of questions. This question is written in such a way though that you can solve it with just a few math concepts... There are two "logic shortcuts" to this question that can help you to zero-in on the correct answer. First, we're dealing with "square feet" of a stripe, which implies the area formula for a rectangle (L x W)…. In this case, we have a length of M MILES, so L = M(5280 feet) We have a width of T INCHES; since we're dealing in SQUARE FEET though, we have to convert inches to feet: W = T/12 So, if we just focus on the area, we'll have 5280(M)(T) / 12. Eliminate B, D and E. Second, we're told that 1 gallon of paint covers P SQUARE FEET. This is "ratio data", so to calculate the number of gallons…..whichever side of the fraction has the area….the other side has the P. This means that the P must be in the denominator. Eliminate C. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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22 May 2016, 11:26
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TheBigCheese wrote: A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long? (1 mile = 5,280 feet and 1 foot = 12 inches)
A. (5,280 mt) / 12 p B. (5,280 pt) / 12m C. (5,280 pmt) /12 D. (5,280)(12m) / pt E. (5,280)(12p) / mt area of the strip = m miles *t inches 5280m * t/12 square feet 1 gallon is required to paint p square feet, then 5280 mt/12p is required to paint the whole stretch. A should be the answer.
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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03 Jun 2016, 10:10
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Attached is a visual that should help.
Attachments

Screen Shot 2016-06-03 at 10.09.50 AM.png [ 185.09 KiB | Viewed 10973 times ]
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Re: A solid yellow stripe is to be painted in the middle of a certain high [#permalink]
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03 Jun 2016, 11:14
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TheBigCheese wrote: A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet of highway, how many gallons of paint will be needed to paint a stripe of t inches wide on a stretch of highway m miles long? (1 mile = 5,280 feet and 1 foot = 12 inches)
A. (5,280 mt) / 12 p B. (5,280 pt) / 12m C. (5,280 pmt) /12 D. (5,280)(12m) / pt E. (5,280)(12p) / mt Visualize the problem - Attachment:
Road.png [ 1.66 KiB | Viewed 10947 times ]
Notice you have three denominations -(A) 1 gallon of paint covers an area of p square feet of highway (B) A stripe of t inches wide(C) The stretch of highway m miles longGiven 1 Mile = 5,280 feet ; So, Length of m miles = 5280m feet And , Breadth of t inches = t/12 feet Area = 5280 mt/12 sq Feet Quote: 1 gallon of paint covers an area of p square feet of highway So, quantity of paint required = 5280 mt/12p Hence answer will be (A)
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Re: A solid yellow stripe is to be painted in the middle of a certain high
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