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?
Correct answer: B
Explanation
Coin is basically a cylinder.
So volume of coin T= pi r^2 h = pi (7.5)^2 * 2
Coin=Silver+Aluminum
Now total volume of coin(T) = volume of silver + volume of aluminum
Also, volume of silver(Vs)= volume of aluminum(Va)
T= Va+Vb
T=2Va
Va=T/2= pi (7.5)^2 * 2 /2 = pi (7.5)^2
Silver is twice as heavy as aluminum.
Let the weight of aluminum in coin be x
Weight of Silver = 2x
Total weight of coin = 30
x+2x=30
x=10
Weight of Aluminum in coin is 10gm
Wright of Silver in coin is 20gm.
Weight of Aluminum in coin is 10gm and volume is pi (7.5)^2
Now new Aluminum coin is made with dimension 1x30mm.
Volume of this new coin = pi (15)^2*1.
Volume of pi (7.5)^2 contains weight of 10 gm of aluminum
Volume of pi (15)^2*1 will contain = 10/ pi(7.5)^ * pi (15)^2 * 1= 40gm
B
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AQuA-RAT items are crowdsourced algebra problems, not ACT questions. They carry five answer choices in the pre-2025 ACT Mathematics style and no ACT reporting category. Useful for drilling Preparing for Higher Math content; not a substitute for a real form.