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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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01 Jun 2014, 12:14

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A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

Source: Indian CAT

We have total of 8 parts: 3 parts of sand and 5 parts of cement.

In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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02 Jun 2014, 13:32

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ConnectTheDots wrote:

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

This is just a weighted average question, so we can apply the formula for that: with \(C\) as the concentrations and \(V\) as the volumes...

If sand:cement=3:5, then the concentration of sand in the initial mixture is \(\frac{3}{8}\). Since we are asked for the proportion of the mixture that should be replaced, we can assume a total volume of \(1\) and let \(x\) be the "amount" of the mixture to be replaced by pure sand (i.e., concentration of 1). We can then write the following equation:

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

Source: Indian CAT

You can use the scale method here too.

A mix has 3/8 of sand. Another is all sand so fraction of sand is 1. You have to mix them to get 1/2 sand.

w1/w2 = (1 - 1/2)/(1/2 - 3/8) = 4/1

So the mix should be 4 parts and only sand should be 1 part. Hence 1/5 of the mix must have been replaced by sand.

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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25 Jun 2014, 16:42

Bunuel wrote:

ConnectTheDots wrote:

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

Source: Indian CAT

We have total of 8 parts: 3 parts of sand and 5 parts of cement.

In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.

Answer: C.

Bunuel: I do not quite understand the highlighted section above. 1 part of the mixture will contain 3/8 sand and 5/8 cement. How did you come up with "1 part of cement contains 3/5 parts of sand"?

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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25 Jun 2014, 21:09

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Game wrote:

Bunuel wrote:

ConnectTheDots wrote:

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

Source: Indian CAT

We have total of 8 parts: 3 parts of sand and 5 parts of cement.

In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.

Answer: C.

Bunuel: I do not quite understand the highlighted section above. 1 part of the mixture will contain 3/8 sand and 5/8 cement. How did you come up with "1 part of cement contains 3/5 parts of sand"?

A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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25 Jun 2014, 21:19

S1 - 3 parts element 1 , 5 parts element 2 S2 - x parts of mixture removed i.e, -(3/8 * x) parts element 1, -(5/8 * x) parts element 2 S3 - x parts of element 1 added

We thus have 3 - (3*x)/8 +x / 5 - (5*x)/8 = 1/1 x=1.6 . i.e, 1.6/8 = 1/5 of the mixture removed.
_________________

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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18 Aug 2015, 23:18

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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18 Aug 2015, 23:53

ConnectTheDots wrote:

A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?

A. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8

Source: Indian CAT

It is easier to answer this type of question using allegation rule. 3/8 parts are sand in the original mixture. we are adding with only sand to make the new mixture 1:1 therefore, 1/1 part is sand that we add. 3/8 1/1

1/2

1/2 : 1/8 so, the ratio is 1/2 : 1/8 or 4:1 That is, if you remove 4 parts from the original mixture and add 1 part sand, the resultant mixture is 1:1 To elaborate further, if the original mixture is 8 kg (3 kg sand and 5 kg cement), you remove 4 kg of the mixture which contains 1.5 kg sand and 2.5 kg of cement. Now, add 1 kg of sand. The new mixture becomes 2.5 kg of sand and 2.5 kg of cement.

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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13 Nov 2016, 15:21

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]

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24 Dec 2016, 13:40

Lets say we have 3 Kg of Sand and 5 Kg of Cement in the total 8 Kg mixture. We want to make it 4 Kg Sand and 4 Kg Cement. To do so, we need to remove 1 Kg of Cement. Each 800 grams of mixture will have 500 grams Cement and 300 grams of Sand. So we need to remove 1.6 Kg ( i.e. 1/5th) of mixture to remove 1 Kg of cement. In the process we have also removed 600 grams of Sand (i.e. remaining sand is 2.4 Kg) Now we replace the mixture with 1.6 Kg of Sand making it 4 Kg Cement and 4 Kg Sand. Hence the answer is replace 1/5th of the mixture with Sand.

Old Mix...................Sand-only mix 3/8......................1/1 .......\.................../ .........\............../ ..............4/8 1/2......................1/8

Therefore the resulting mix will have a ratio Old Mix to Sand-only mix of (1/2)/(1/8) = 4 to 1. From here we know that every 4 part of the old mix we need to have one part of sand only mix. So in the final mix the old mix will be 4/5 and the sand-only mix will be 1/5. Answer is C we need to substitute 1/5 of the old mix and replace with san.

Anyone could suggest if I have applied the method correctly? (Bunuel )

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