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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.

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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)
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petu
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?


(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
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petu
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?
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?

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 on 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?

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.

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