A coin made of alloy of Aluminium and silver measures 2 x 15 mm (it

Question

A coin made of alloy of Aluminium and silver measures 2 x 15 mm ( it is 2 mm thick and diameter is 15 mm).If the weight of the coin is 30gms and volume of Aluminium in the alloy equals that of silver,what will be the weight of the a coin measuring 1 x 30 mm made of pure aluminium if silver is twice as heavy as aluminium?

a: 36gm
b: 40gm
c: 42 gm
d: 48 gm
4: 50 gm

a: 36gm
b: 40gm
c: 42 gm
d: 48 gm
4: 50 gm

beyondgmatscore · Answer

We are given that first coin has similar volume of Aluminium and Silver. Also, weight of silver is twice that of Aluminium for same volume. So, Weight of Aluminium and Weight of Silver is in the ratio 1:2 in original coin and hence the weights are 10gms and 20gms respectively.

We know, volume of 1st coin is  pi*r^2*h = pi*(7.5)^2*2 .
Volume of Aluminium in first coin is  1/2*pi*(7.5)^2*2 = pi*(7.5)^2

So, we know that  pi*(7.5)^2  of Al weighs 10 gms

Now, for the pure Aluminium coin, volume is  pi*(15)^2*1

or  pi*(7.5*2)^2*1

or  4*pi*(7.5)^2

We know that  pi*(7.5)^2 of Al weighs 10 gms

So, this pure Aluminium coin would weigh 4*(10) = 40 gms

Answer B

Another approach:

We can see that second cylinder would be twice the volume of first cylinder as radius is double (so 4 times) and height is halved (so 0.5 times).

Volume of Aluminium in second cylinder is 4 times the volume in first one.

We also know that weight of Aluminium and Silver in first one is in the ratio 1:2 and hence weight of Aluminium in first one is 10 gms, so weight in second one is 4 times, so 40 gms

Answer B

gmat1220 · Answer

Weight of Sl : Weight of Al = 2:1
Hence in 30 gm alloy coin 
Sl = 20 gm
Al = 10 gm
Volume of Sl = Volume of Al. r = 7.5 and h = 2
If the coin were pure Al, then its weight would be 10 + 10 = 20 gm.
 
When radius is doubled and height is halfed  , volume goes up by twice.

If the coin is pure Al, its weight = 2 * 20 gm = 40 gm

subhashghosh · Answer

W = k * V, where K is a constant.

V1 = Vol of Alumunium, W1 = Wt of Alumunium -> W1 = K1 * V1

V2 = Vol of Silver, W2 = Wt of Silver -> W2 = K2 * V2 Now K2 = 2K1

So W2 = 2 * K1 * V1

V1 + V2 = pi * (15/2)^2 * 2

So V1 = pi * (15/2)^2

So W1 + W2 = 3W1 = 3 * K1 * pi * (15/2)^2

30/w = {3 * K1 * pi * (15/2)^2}/{k1 * pi * (15)^2 * 1}

=> w = 10 * 2 * 2 = 40

So the answer is B.

A coin made of alloy of Aluminium and silver measures 2 x 15 mm (it

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