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 * 50 grams

vscid wrote: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 * 50 grams In the first coin since the volume of silver and aluminium are same and silver is twice as heavy as aluminum we can say that the weight ratio of aluminium to that of silver is 1:2 so in coin 1 weight of aluminium is 30x(1/3)=10gm so in coin 1 weight of silver is 30x(2/3)=20gm if we replace silver with aluminium in first coin i.e the first is of pure aluminium the weight would have been 10+10=20gm second coin has twice of the volume of the first (diameter is double and thickness is half) hence weight of second coin is 20x2=40 gm ans is b 40gm

vscid wrote: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 * 50 grams weight is proportional to thickness * (dia)^2. Therefore, 2*(15)^2 / 1 * (30)^2 = 30/x => X(weight of coin in the second case) = 60 ---> had the coin be of the same alloy. However, we now have only aluminium which is half heavy as silver. Therefore, the weight will be 2/3 * 60 =40.