M15-06

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation

Hi,
second coin is 1mm thick and 30mm diameter, and first coin is 2mm thick and 15mm diameter. I didn't understand why second coin is double weight.

First case
Weight of aluminium=w, weight of silver 2w, ;it is given w+2w=30; so w=10
now volume of aluminium =v (assume) = volume of of silver
total volume = 2v

Second case
(volume of new coin ):volume of old coin =(π*((30²)/4)))*1/(π*((15²)/4))*2= (π*((15²*2²)/4)))*1/(π*((15²)/4))*2)=2:1

so volume of new coin = 2 total volume of 1st coin = 4V
Now weight of aluminium =weight of volume v in coin 1 = 10 gm
so weight of aluminium in coin 2 = 40 gm

Hope this helps

I think this is a high-quality question and I agree with explanation.

Since weight = volume x density, we need to be careful when equating weight anywhere as the two metals have different densities.
It is given that silver is twice as heavy as aluminium. So, for the same volume, silver is denser.

Now, in the first case, the coin is made up of half silver and half aluminium (volume is same) --> Weight of aluminium in coin = 10 and silver = 20 (because silver : aluminium in weight = 2:1)

For half coin --> 0.5*pi*\((15/2)^2\)*2 (volume) -> wt of aluminium = 10 gms
For 1 unit of volume = 10/[0.5*pi*\((15/2)^2\)*2]
For new coin -> pi*\((30/2)^2\)*1 -> multiple above by this new volume to get --> 40

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I don't understand about the aluminum weight being 20 instead of 10 grams? Please explain

Posted from my mobile device

The first coin is half aluminum and half silver by volume. The aluminum part weighs 10 grams. The the solution says, IF the coin were made of pure aluminum, then its weight would be 20 grams.

Does this make sense ?

I think this is a high-quality question and I agree with explanation.

I have done this problem using mixture table. Please can any expert (@bunuel) weigh in and let me know if this approach is accurate?

From the first coin - ALUMINIUM is 10g/30g = 33% (For easier calculation I would approximate it to 30%)
From the second coin- ALUMINIUM is 100% as the second coin is pure aluminum.
We also know the total weight of the first coin = 30g.
I would assume for a moment that I am combining the two coins to get a "mean value" and will then end up using that mean value to find the weight of the coin.

1st coin..........2nd coin

30%..............100%
.........70%(mean value)...........
30.................(70-30=40)

All the values in pink are the information provided by the question. The mean value in blue can be found by subtracting 30 from 100. The value in green is the answer we have found out. Hence, the weight of the 2nd coin will be 40%.

I think this is a high-quality question and I agree with explanation. Please elaborate

­Bunuel - such questions would be classified as geomtery right? I could find them in the sectional test for GMAT Focus Edition on the Club Practice Tests - request your team to check and remove these Qs from the selection if my understanding is correct to avoid confusion. 

 ­

Updated the question by adding the formula and removed from free tests.

I think this is a high-quality question and I agree with explanation.

I think that if the formula is provided, the question could be perfectly valid for the new edition of the GMAT.