On a ferry, there are 50 cars and 10 trucks. The cars have an average

50(1200)+10(3000) / 60
= 1500

Answer: B

Answer is B
(50*1200+10*3000)/60 = 1500 .. this is normal method for calculation.

but if i were running out of time and needed to mark this question in very less time then also i would mark B.
See, Number of cars are 5 times trucks. So mean of entire set should be closer to Car's mean. 1200 is not logical.
So next closer option 1500 i would mark as answer . ( this depends on options provided , but if you need to mark it in very less time then this could be an educated guess)

Answer = B = 1500

\(\frac{50*1200}{60} + \frac{10*3000}{60} = 1000 + 500 = 1500\)

Weighted average question :
\(\frac{(0*50 + 1800*10)}{60} = 300\) so weighted average = 300+1200 = 1500 Ans B.

that's risky . what if trucks were 16 , in that case w.avg is 1800 .
i suggest that one apply the very basic steps which are as follows
as cars are more in number than truck w.avg will be close to cars' weight .
C:R=5:1
so we know w.avg will be 1unit away from cars'weight and 5 unit away from trucks' weight . total unit of gap will be there between
cars' weight and trucks' weight

6X=3000- 1200 = 1800
X=300

w.avg , which is 1unit away from cars' weight ,will be 1200 + X = 1500 .
alternately , i have also provided algebric approach which is also little faster than conventional brute force method.

Reduce the lower weight to 0 , and other weight as well by same magnitude . we will add it later.
Weighted average question :
\(\frac{(0*50 + 1800*10)}{60} = 300\)
so weighted average = 300+1200 = 1500 Ans B.

Hi All,

There's a great Weighted Average 'shortcut' to this question. Since we know that there are 5 times as many cars as trucks, the actual number of cars and trucks doesn't matter AS LONG as the ratio remains.

This means that instead of calculating the average of 50 cars and 10 trucks, we can calculate the average of 5 cars and 1 truck (again, the resulting average will be the SAME)...

[5(1200) + 1(3000)]/6 =

(6,000 + 3,000)/6 =

9,000/6 =

1500

Final Answer:

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

Ans- B

\(total average= \frac{no. of cars*cars average weight+ no. of trucks *trucks average weight}{total no. of vehicles}\)
= \(\frac{50*1200+10*3000}{(50+10)}\)
= 1500

MAGOOSH OFFICIAL SOLUTION:

Method I: using sums

We will divide the two masses by 1000, 1.2 and 3 respectively, to simplify calculations. Note the use of the Doubling and Halving trick in the first multiplication.

sum for cars = 50*1.2 = 100*0.6 = 60

sum for trucks = 10*3 = 30

total sum = 60 + 30 = 90

To find the total average, we need to divide this total sum by the total number of vehicles, 60.

total average = 90/60 = 1.5.

Since we divided masses by 1000 earlier, we need to multiply by 1000 to get the answer. Total average = 1500 kg.

Answer = (B)

Method II: proportional placement of the total average

Cars to trucks is 5:1, so if the distance between the car’s average and truck’s average were divided into 6 parts, the car’s average is 1 part away from the total average, and the truck’s average is 5 parts away.

Well, the difference in the two group averages is 3000 – 1200 = 1800 kg. Divide that by six: each “part” is 300 kg. Well, the total average must be 300 kg bigger than 1200 kg, or 5*300 kg smaller than 3000 kg. Either way, that’s 1500 kg.

Answer = (B)
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We calculate the weighted average to determine the average mass of all 60 vehicles on the ferry:

(50 x 1200 + 10 x 3000) / 60 = 50 x 20 + 10 x 50 = 1000 + 500 = 1500 kg

Answer: B
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