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10 May 2015, 02:25
Kudos
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (02:24) correct
38%
(03:21)
wrong
based on 82
sessions
History
Date
Time
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Not Attempted Yet
A car ferry can hold up to 50 tons of cargo. What is the greatest number of vehicles that the ferry can carry if half the vehicles are cars with an average (arithmetic mean) weight of 0.75 tons and half of the vehicles are trucks with an average (arithmetic mean) weight of 5 tons?
A. 10
B. 14
C. 15
D. 16
E. 17
Could someone walk me through please?
A. 10
B. 14
C. 15
D. 16
E. 17
Could someone walk me through please?
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10 May 2015, 16:21
Kudos
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Hi petu,
This question is written in a very specific way - if you can catch the subtle clues in it, then you can answer this question relatively quickly.
We're told a few facts about a ferry:
1) It can hold UP TO 50 tons of cargo
2) HALF the vehicles on the ferry will be trucks while HALF the vehicles will be cars
3) Cars weigh an average of .75 tons each and trucks weigh an average of 5 tons each
We're asked for the GREATEST number of vehicles that the ferry can hold.
Since half the vehicles are trucks and half are cars, there will be an EQUAL number of each. By extension, that means there MUST be an EVEN number of vehicles (since 1+1=2, 2+2=4, 3+3=6, etc.).
Adding the weight of 1 car and 1 truck gets us .75 + 5 = 5.75 tons, so we really just have to figure out the MAXIMUM number of 5.75 "pairs" that go into 50 tons. You can do this math either by division OR by using the answers that are given, since one of them MUST be the answer.
Let's start with 16....that would be 8 "pairs" of vehicles.....8(5.75) is less than 8(6) = 48. So 16 vehicles COULD end up on the ferry. Since there is no other bigger answer that's EVEN, 16 must be the answer.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This question is written in a very specific way - if you can catch the subtle clues in it, then you can answer this question relatively quickly.
We're told a few facts about a ferry:
1) It can hold UP TO 50 tons of cargo
2) HALF the vehicles on the ferry will be trucks while HALF the vehicles will be cars
3) Cars weigh an average of .75 tons each and trucks weigh an average of 5 tons each
We're asked for the GREATEST number of vehicles that the ferry can hold.
Since half the vehicles are trucks and half are cars, there will be an EQUAL number of each. By extension, that means there MUST be an EVEN number of vehicles (since 1+1=2, 2+2=4, 3+3=6, etc.).
Adding the weight of 1 car and 1 truck gets us .75 + 5 = 5.75 tons, so we really just have to figure out the MAXIMUM number of 5.75 "pairs" that go into 50 tons. You can do this math either by division OR by using the answers that are given, since one of them MUST be the answer.
Let's start with 16....that would be 8 "pairs" of vehicles.....8(5.75) is less than 8(6) = 48. So 16 vehicles COULD end up on the ferry. Since there is no other bigger answer that's EVEN, 16 must be the answer.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
12 May 2015, 11:08
Kudos
Bookmarks
[quote="EMPOWERgmatRichC"
Since half the vehicles are trucks and half are cars, there will be an EQUAL number of each. By extension, that means there MUST be an EVEN number of vehicles (since 1+1=2, 2+2=4, 3+3=6, etc.).
/quote]
This is a very important conclusion. I got it wrong because I didn't realize this.
I solved it this way: So the number of cars and trucks is equal. So , 0.75C + 5C = 50
So soling for C gives 8.6(something). This gives C, our answer is total number hence 2C. So 8.6 X 2, I estimated this as 17 (something greater than 16)
We need to round down not round up the answer because of the even number constraint.
Since half the vehicles are trucks and half are cars, there will be an EQUAL number of each. By extension, that means there MUST be an EVEN number of vehicles (since 1+1=2, 2+2=4, 3+3=6, etc.).
/quote]
This is a very important conclusion. I got it wrong because I didn't realize this.
I solved it this way: So the number of cars and trucks is equal. So , 0.75C + 5C = 50
So soling for C gives 8.6(something). This gives C, our answer is total number hence 2C. So 8.6 X 2, I estimated this as 17 (something greater than 16)
We need to round down not round up the answer because of the even number constraint.
