A car ferry can hold up to 50 tons of cargo. What is the greatest

Question

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?

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made,
Rich

AjChakravarthy · Answer

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.

EMPOWERgmatRichC · Answer

Hi AjChakravarthy,

By taking an Algebraic approach, you would need to be more specific with the details. The end result would actually be an inequality, NOT an equation.....

0.75C + 5C < 50

and C = INTEGER

Thus, you'd end up with....

5.75C < 50

The maximum value of C would be 8, so there would be 8+8=16 vehicles.

GMAT assassins aren't born, they're made,
Rich

OptimusPrepJanielle · Answer

Let x represent the number of vehicles. 
.75x + 5x = 50
5.75 x = 50
5.75 = 5 3/4 or 23/4
(23/4) x = 50
23x=200
23 is almost 25, so x is approximately 8.
That means 8 cars + 8 trucks = 16 vehicles (Choice D)

Bowtye · Answer

(1/2)*(3/4)*x+(1/2)*(5)*x = 50

(47/8)x = 50 -> x = 8.5....... so the max is x = 8. So 8 cars + 8 trucks = 16

umg · Answer

This is a Wrong Question. Please Read CAREFULLY the Question Stem.

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?

Question: What Exactly do you want me to do?? 
Answer: Find the  greatest number of vehicles

But Question has already given the constraint (Half of the vehicles are Cars and half of them are Trucks) because of which Maximum Number of Vehicles cannot be found.

Question should be changed to

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?

Otherwise, Answer 16 is the wrong one.

Here is the reason.

If there are 16 Vehicles, half of which are Cars and the rest Trucks, then
Total weight of 16 Vehicles = 8*0.75 + 8*5 = 6 + 40 = 46.

To load up the ferry to its  Maximum Capacity , 4 More Cars can be loaded on it, making the Total Weight = 46 + 4*0.75 = 46 + 3 = 49 Tons

Therefore, Total Number of Vehicles = 16 + 4 = 20 => 12 Cars and 8 Trucks => Breaks the condition that half the vehicles are Cars and Rest are Trucks.

A car ferry can hold up to 50 tons of cargo. What is the greatest

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A car ferry can hold up to 50 tons of cargo. What is the greatest

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