bibha
A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?
36 grams
40 grams
42 grams
48 grams
You need to do very few calculations if you understand the relation between the variables.
Let's consider the first coin:
Volume of Silver = Volume of Aluminum
Weight of Silver = 2* Weight of Aluminum
Total weight is 30 gms. Since they are in equal volume but silver is twice as heavy, out of 30 gms, silver must be 20 gms and aluminum must be 10 gms.
(Think what would happen if it was given that volume of aluminum was twice the volume of silver)
Now we need to figure out the relation between the volumes of the two coins.
Volume of first coin \(= pi * (15/2)^2 * 2 = (225/2)*pi\)
Volume of second coin \(= pi * (15)^2 * 1 = (225)*pi\)
Volume of second coin is twice the volume of the first coin. If volume of aluminum in the first coin is V, the volume of the first coin is 2V. The volume of the second coin is 4V. Since it is all Aluminum, volume of Aluminum in the second coin is 4V. Since weight of Aluminum of weight V was 10 gms, weight of Aluminum of volume 4V will be 40 gms.
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?