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A mixture of sand and cement contains, 3 parts of sand and 5
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01 Jun 2014, 11:14
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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?
Re: A mixture of sand and cement contains, 3 parts of sand and 5
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01 Jun 2014, 14:05
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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
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.
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02 Jun 2014, 12:32
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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:
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02 Jun 2014, 20:15
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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
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.
Answer (C)
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Karishma Veritas Prep GMAT Instructor
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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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25 Jun 2014, 15:42
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
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25 Jun 2014, 20:09
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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
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25 Jun 2014, 20:19
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.
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A mixture of sand and cement contains, 3 parts of sand and 5
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18 Aug 2015, 22:53
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
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24 Dec 2016, 12:40
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 )
A mixture of sand and cement contains, 3 parts of sand and 5
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10 Jun 2017, 19:10
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
1.(Initial quantity - quantity removed) *( strength of the mixture) + (Quantity of sand added *strength of sand)/ (initial quantity) =1/2 2. Let initial quantity of the mixture be x and let the quantity removed be y. 3. ((x-y) *5/8) +y*0) / x =1/2, y/x=1/5
Note: By strength, I mean the proportion of cement.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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19 Dec 2018, 05:46
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MWithrock wrote:
So, I'm having a Hard time working through this with the weighted average formula which I thought I understood.
3/8ths of the Mix is sand. the volume added is 100%sand. the resulting mix we have is 1/2 sand
3/8x + 1y = 1/2 (x + y)
4 parts mix and 1 parts pure sand. Cool.
But how do I relate this back? The difference in the first compound and the second? I have no idea.
4 parts mix and 1 part pure sand. This means that your new mix should have 4 parts mix and 1 part pure sand. So out of total 5 parts of new mix, 4 parts should be mix and 1 part pure sand. So take 5 parts of the old mix, remove 1 part of it and replace that with pure sand. Now you have 4 parts of old mix and 1 part of pure sand. Hence 1/5 of the old mix was replaced with pure sand.
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Karishma Veritas Prep GMAT Instructor
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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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19 Dec 2018, 08:14
VeritasKarishma wrote:
MWithrock wrote:
So, I'm having a Hard time working through this with the weighted average formula which I thought I understood.
3/8ths of the Mix is sand. the volume added is 100%sand. the resulting mix we have is 1/2 sand
3/8x + 1y = 1/2 (x + y)
4 parts mix and 1 parts pure sand. Cool.
But how do I relate this back? The difference in the first compound and the second? I have no idea.
4 parts mix and 1 part pure sand. This means that your new mix should have 4 parts mix and 1 part pure sand. So out of total 5 parts of new mix, 4 parts should be mix and 1 part pure sand. So take 5 parts of the old mix, remove 1 part of it and replace that with pure sand. Now you have 4 parts of old mix and 1 part of pure sand. Hence 1/5 of the old mix was replaced with pure sand.
Holy Heck. ok. I got it.
THANK YOU SO MUCH! Kudos
gmatclubot
Re: A mixture of sand and cement contains, 3 parts of sand and 5 &nbs
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19 Dec 2018, 08:14
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39% (02:00) correct 
