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

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

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New post Updated on: 02 May 2020, 04:51
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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?

A. 36 grams
B. 40 grams
C. 42 grams
D. 48 grams
E. 50 grams

Originally posted by eyunni on 27 Nov 2007, 19:14.
Last edited by Bunuel on 02 May 2020, 04:51, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 13 Sep 2011, 04:25
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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


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.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 03 Aug 2010, 00:20
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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


Coin is in the shape of a cylinder.

Coin 1

Volume = pi* \(7.5^2\)*2 = 112.5pi cubic mm

Total coin weight(including both Aluminium and Silver)= 30g but Silver is twice as heavy as aluminium.
So weight of silver = 20g and weight of aluminium =10g

Volume of aluminium = Volume of silver
volume of aluminium = volume of silver = 112.5pi/2 = 56.25pi
So 56.25pi cubic mm of aluminium weighs 10g

Coin 2

Volume = pi* \(15^2\)*1 = 225pi cubic mm
Coin is made of pure aluminium . Also 56.25pi cubic mm of aluminium weighs 10g from above
So total weight of the second coin = (10/56.25pi)*225pi = 40g
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 09 May 2009, 13:29
Let's move from one coin to another:

1) m1 = 30 g.
2) decrease in thick from 2mm to 1mm: m = 30 * 1/2
3) increase in diameter from 15mm to 30mm: m= 30 * 1/2 * 2^2
4) substitute 1/2 of silver to 1/2 of aluminum: it reduces weight by 3/2 times: m = 30 * 1/2 * 4 * 2/3 = 40 g.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 09 May 2009, 15:12
walker wrote:
Let's move from one coin to another:

1) m1 = 30 g.
2) decrease in thick from 2mm to 1mm: m = 30 * 1/2
3) increase in diameter from 15mm to 30mm: m= 30 * 1/2 * 2^2
4) substitute 1/2 of silver to 1/2 of aluminum: it reduces weight by 3/2 times: m = 30 * 1/2 * 4 * 2/3 = 40 g.


You make it look so simple but I am afraid I did not get the tail or head of what you tried to explain. How are you relating weight thickness and diameter?

In 2 Are you saying that as thickness decreases weight also decreases and hence divide by 2??

Diameter increased by 2 but why are you multiplying by 2^2 rather than 2? Is it because to account for 2 sides of coin?

What really perplexed me is this Sentence

The weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver.

Can you explain why the weight reduces by 3/2 times?


It is talking about weight of the coin and then says volume of AL and Si are same. Its not like they are 15 gms each.

Here is the OE and OA



Denote S the weight of silver and A the weight of aluminum in the first coin. We can compose an equation: S + A = 30 or 2A + A = 30. Thus, the aluminum half of the coin weighs 10 grams. If the coin were made of pure aluminum, the second half of the coin would also weigh 10 grams and the whole coin would weigh 20 grams.

If we look at the proportions of the second coin, we will see that it is twice as large as the first coin (volume2 = \(1(\frac{30}{2})^2 \pi\) ; volume1 = \(2(\frac{15}{2})^2 \pi)\) . The second coin were made of pure aluminum would be twice as heavy as the first coin made of pure aluminum. The answer to the question is therefore 20*2 = 40 grams.

The correct answer is B.

While the OE makes a lot of sense in the first part, I dont understand what kind of an object are they considering the coin to be? Sphere?
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 10 May 2009, 06:00
There is a direct way to solve it, as you've posted here: to write a formula of weight for first coin and second coin. After that, express weight of first coin through weight of second coin.

I used indirect way. I actually transform our first coin to second coin by means of a few steps as a blacksmith. It seems obvious for me that if we change the thickness of a coin from 2mm to 1mm we have 1/2 m (or 2 coins instead). The same with radius, changing the radius of a coin from 15mm to 30mm we increase area of the coin by 4 and volume (weight) by the same factor. In the case of [Al][Ag] coin, it has weight that equals to [Al][Al][Al] coin, or coin with volume that 3/2 times more. So, we need use 2/3. Anyway, it was the fastest way for me, but if it is not so obvious or fast for you, it would be better to use the direct way.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 03 Aug 2010, 02:07
Volume of aluminium = Volume of silver
Would you please explain this step??
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 03 Aug 2010, 18:07
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bibha wrote:
Volume of aluminium = Volume of silver
Would you please explain this step??


See the 2nd statement in the question "volume of aluminum in the alloy equals that of silver". Coin 1 is a mixture of aluminium and silver and in that mixture volume of aluminium is equal to the volume of the silver. So if the total volume of the coin is 112.5pi cubic mm , volume of aluminium in coin1 is 112.5pi/2 cubic mm
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 12 Sep 2011, 08:26
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 10gm of aluminum
Volume of pi (15)^2*1 will contain = 10/ pi(7.5)^ * pi (15)^2 * 1= 40gm

B.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 12 Sep 2011, 08:41
Total volume 225*2pi/4
volume of aluminum in the alloy equals that of silver
so volume of only aluminum is half of 225*2pi/4
which is 225*pi/4

Total weight 30 gm
Silver is twice as heavy as aluminium. So
Silver weight in the coin 20 gm and aluminium weight 10 gm

so weight of 225*pi/4 cubic mm of aluminum is 10 gms

Now volume of 1 x 30 mm coin is 225 pi which is 10*4 =40 gms
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 28 Nov 2015, 17:55
ratio of aluminum weight to volume in smaller coin=10 grams/56.25⫪ mm^3
ratio of aluminum weight to volume in larger coin=w grams/225⫪ mm^3
10/56.25=w/225
w=40 grams
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 20 Dec 2015, 17:02
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it's a tough one...
was confused at the beginning with volume/gr...but
since first one has diameter 15, then r=7.5 and h=2.
volume is pi*r^2*h = 7.5^2 * 2 * pi = 102.50
now, volume of silver is equal to volume of aluminium, or each 56.25pi.
we are told that silver is twice as heavy as aluminium. which means that S+A=30, and 2A+A=30, or A=10gr.
now we know that aluminium consists of 56.25pi volume and has 10gr in the mixture.

we need to get a new coin with volume 15^2*1*h, or 225pi.
now, 225pi/56,25pi = 4.
so we have 4 "volumes" of aluminium. 1 volume of aluminium is 10 gr, thus 4 "volumes" is equal to 4*10=40.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 20 Dec 2015, 20:20
The trick is not to confuse weight and volume. They play separate roles here.
V/2 measures 10 grams.
So,

2V measures 40 grams.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 03 Jul 2018, 00:56
VeritasPrepKarishma wrote:
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


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).



Hi Experts: VeritasPrepKarishma, Bunuel

Just for understanding purposes of this really good sum, what's the ans for "(Think what would happen if it was given that volume of aluminum was twice the volume of silver)" ?
Would it be 30?
Va/ Vs = 2/1
so now weight 10 corresponds to 66.66% of volume and thus 100% of volume is 15 and twice that is 30.

Please confirm.

Thanks.
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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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New post 03 Jul 2018, 04:00
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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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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is  [#permalink]

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Re: A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is   [#permalink] 02 May 2020, 04:18

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