bibha 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

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.

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