amirhakimi wrote:

For age ranging from 12 (A=12) to 30, the average mass is within which of the following?

A)1000 to 2000
B)2000 to 3000
C)3000 to 4000
D)4000 to 5000
E)5000 to 6000

Answer is D

Ages 12-20:
For these 9 ages, the graph seems symmetrical about point B, which corresponds to a mass of 3000.
Implication:
The average mass for these 9 ages is about 3000.
Approximate SUM of these 9 masses = (number)(average) = 9*3000 = 27,000.

Ages 21-30:
For these 10 ages, the corresponding masses are all around 5500, as indicated by the points to the right of point C on the graph.
Approximate SUM of these 10 masses = (number)(average) = 10*5500 = 55,000.

Resulting average:
Approximate AVERAGE of all 19 masses = sum/number = (27000+55000)/19 = 82000/19 ≈ 4300.

The correct answer is D.

Resulting average:
Approximate AVERAGE of all 19 masses = sum/number = (27000+55000)/19 ≈ 84000/19 ≈ 4400.

I agree with this approach. However, just to take note, there's an error up there; the 84000 should be 82000.

Cheers

Arnav

kackerarnav wrote:the 84000 should be 82000

Thanks for pointing out the typo. I've edited the post accordingly.

Does anyone think that a good approximation for the curve is: y =5500/2(-cos(6x) + 1)
which could be integrated between x = 12 and x = 30 to get the area.

Divide this by the range of 18 years and then find the closest match in the answer list?

Just an idea.

Brent@GMATPrepNow wrote:

Mathsbuddy wrote:Does anyone think that a good approximation for the curve is: y =5500/2(-cos(6x) + 1)
which could be integrated between x = 12 and x = 30 to get the area.

Divide this by the range of 18 years and then find the closest match in the answer list?

Just an idea.

I like the "outside the box" thinking, but it'd be hard to find a cosine (or sine) curve that approximates this very well. Also, the equation y = 5500/2(-cos(6x) + 1) has a period of (pi)/3, which is too short. I think we'd need the period to be much bigger.

Plus, of course, integration is wayyyy beyond the scope of the GMAT.

Cheers,
Brent

Thanks Brent,

Of course the simplest way would be to notice that the shape can be APPROXIMATED to a trapezium:
Area = (20 + 18)*4.5k /2 = 85.5k (where k = 1000)
Average mass = Area/age range = 85.5k/18 = 4750

ANSWER D