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

The height of the second coin (1) is half the height of the first coin (2).
The radius of the second coin (15) is twice the radius of the first coin (7.5).
We can plug in ANY VALUES for the two heights and two radii such that the relationships above are satisfied.

In the first coin, let h=2 and r=1, with the result that the volume of the first coin = \(πr^2h = π(1^2)(2) = 2π\).
In the second coin, let h=1 and r=2, with the result that the volume of the second coin = \(πr^2h = π(2^2)(1) = 4π\).
The results above indicate that the volume of the second coin is double the volume of the first coin.

Implication:
If the second coin had the same composition as the first coin -- 1/2 silver, 1/2 aluminum -- the weight would double from 30 grams to 60 grams.
Since silver is twice as heavy as aluminum, this 60-gram coin would be composed of 40 grams of silver and 20 grams of aluminum.
Since the second coin is actually 100% aluminum, the 40-gram silver portion above is replaced by another 20 grams of aluminum, reducing the total weight to 40 grams.

Hi KarishmaB
Could you please explain the relation between volume and weight in this question? How does it impact the solution?
Also, it'll be great if you could answer your own question around what would happen if it was given that volume of aluminum was twice the volume of silver?

We know that weight = density * volume for a material.
Since the density of the material at a certain temperature does not change, weight is directly proportional to volume.
So more the volume more the weight, less the volume less the weight.

When comparing two materials, if we are given that both have same volume, then their weights depend on their densities. We are told that silver is twice as heavy as aluminium. This means that the density of silver is twice as much as aluminium. So given same volumes of silver and aluminium, silver will have twice the weight of aluminium.

Since in the 2*15 mm coin, the volume of both silver and aluminium is same (say it is V each), we can say that the weight of that amount of silver will be twice the weight of that amount of aluminium. Since overall weight of the coin is 30 gms, silver must be 20 gms and aluminium must be 10 gms to give total 30 gms. The total volume of this coin is 2V. Volume of silver is V and volume of aluminium is V.

The volume of 1*30 gms coin is twice of the first coin so its volume is 4V. Now note that this is completely made of aluminium so we have 4V volume of aluminium. Since the volume of aluminium is 4 times the previous volume of aluminium, the weight of this aluminium will be 4 times the previous weight i.e. it will be 10 * 4 = 40 gms

Official Solution:

A right circular cylinder-shaped coin made of an aluminum-silver alloy weighs 30 grams and has a diameter of 15 mm and a height of 2 mm. The coin contains equal volumes of aluminum and silver, with silver being twice as heavy as aluminum. What is the weight of a pure aluminum circular cylinder-shaped coin with a diameter of 30 mm and a height of 1 mm?

A. 36 grams
B. 40 grams
C. 42 grams
D. 48 grams
E. 80 grams


Let \(s\) be the weight of silver and \(a\) be the weight of aluminum in the first coin. Since \(s + a = 30\), and silver is twice as heavy as aluminum, we can write \(2a + a = 30\), which simplifies to \(a=10\) grams. Therefore, the weight of the aluminum in the coin is 10 grams. If the coin were made of pure aluminum, then its weight would be 20 grams since it contains equal volumes of aluminum and silver.

The volume of the first coin is \(\pi*r^2*h = \pi*(\frac{15}{2})^2*2 =\pi*\frac{225}{2}\).

The volume of the second coin is \(\pi r^2 h = \pi*(\frac{30}{2})^2*1 =\pi*225\).

Since the first coin, if it were made of pure aluminum, would weigh 20 grams. we can conclude that the second coin, which is twice as large as the first, weighs 40 grams.


Answer: B

Can't believe this was so easy Q

Just do a quick calculation of Volume of first cylinder coin and volume of second cylinder coin = This is pure formula (Pie* radius ^2 * h)
You will realise Volume of second cylinder coin is 2 times volume of first cylinder coin

Now its just 20 more seconds from here.

Now two conditions are given

Statement 1: Coin 1 has equal volume of aluminium and silver.
So if V is total volume of coin 1 THEN Volume of ALuminium = Volume of Silver = V/2..................(1)

Statement 2: Coin 1's weight is 30 g but silver weight is 2x that of aluminium
So weight of aluminium is 10g....................(2)

Using 1 and 2
v/2 volume weight is 10 g
v volume weight will be 20 g
2v volume (Remember volume of 2 coin which is pure aluminium is twice that of first coin) weight will be 40 g

Option b