text: A mixture of sand and cement contains, 3 parts of sand and 5 : GMAT Problem Solving (PS)
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  # A mixture of sand and cement contains, 3 parts of sand and 5
  
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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, 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?
  
  A. 1/3
  B. 1/4
  C. 1/5
  D. 1/7
  E. 1/8
  
  Source: Indian CAT
  [Reveal] Spoiler: OA
  
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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...
  
  $$C_1*\frac{V_1}{V_1 + V_2} + C_2*\frac{V_2}{V_1 + V_2} = C_{final}$$
  
  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:
  
  $$(\frac{3}{8})*(1-x) + (1)*(x)=\frac{1}{2}$$
  
  Thus, $$x=\frac{1}{5}$$
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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.
  
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  04 Jun 2014, 01:08
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  Initial
  
  Sand ............... Cement ............... Total
  
  $$\frac{3}{8}$$ ..................... $$\frac{5}{8}$$ ...................... 1
  
  Requirement
  
  $$\frac{4}{8} ..................... \frac{4}{8} ..................... 1$$
  
  Just focus on cement:
  
  To have $$\frac{4}{8}$$ cement, we require to remove
  
  $$\frac{5}{8} - \frac{4}{8} = \frac{1}{8}$$cement
  
  In Mixture, the portion of cement is $$\frac{5}{8};$$so to remove $$\frac{1}{8}$$ cement, mixture required to be removed
  
  $$= \frac{\frac{1}{8} * 1}{\frac{5}{8}}$$
  
  $$= \frac{1}{5}$$
  
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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.
  
  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"?
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  25 Jun 2014, 20:01
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  One more method:
  
  Sand ............. Cement .............. Total
  
  $$\frac{3}{8}$$ .................. $$\frac{5}{8}$$ ........................ 1
  
  Let "x" quantity of mixture be removed;
  
  $$\frac{3}{8} - \frac{3x}{8}$$ ........... $$\frac{5}{8} - \frac{5x}{8}$$ ............... 1 - x
  
  $$\frac{3}{8} - \frac{3x}{8} + x$$ .......... $$\frac{5}{8} - \frac{5x}{8}$$ .............. 1-x+x
  
  Resultant should be half sand & half cement
  
  Two options to set up the equation
  
  Option I
  
  $$\frac{3}{8} - \frac{3x}{8} + x = \frac{1}{2}$$
  
  $$x = \frac{1}{5}$$
  
  Option II
  
  $$\frac{5}{8} - \frac{5x}{8} = \frac{1}{2}$$
  
  $$x = \frac{1}{5}$$
  
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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.
  
  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"?
  
  Not Bunuel, but seems I can explain it
  
  Mixture .................. Sand ............... Cement
  
  1 ............................ $$\frac{3}{8}$$ ................... $$\frac{5}{8}$$
  
  Multiply by $$\frac{8}{5}$$ to all above
  
  $$\frac{8}{5}$$ ........................ $$\frac{3}{8} * \frac{8}{5}$$ ............. $$\frac{5}{8} * \frac{8}{5}$$
  
  $$\frac{8}{5}$$ ........................ $$\frac{3}{5}$$ ....................... 1
  
  From the above, we can say that 1 part of cement comes with $$\frac{3}{5}$$ parts of sand in $$\frac{8}{5}$$ quantity of mixture.
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  A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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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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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  18 Aug 2015, 22:18
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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, 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.
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  A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  06 Sep 2015, 14:53
  let x=fraction of mixture to be substituted
  3/8-3x/8+x=1/2
  x=1/5
  
  Last edited by gracie on 11 Nov 2015, 19:39, edited 1 time in total.
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  07 Sep 2015, 12:36
  How can we use the alligation method/shortcut to solve this?
  
  My attempt:
  
  Sand...................Cement
  3/8......................5/8
  .......\.................../
  .........\............../
  ..............4/8
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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  13 Nov 2016, 14:21
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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, 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.
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  A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  26 Dec 2016, 04:14
  miva0601 wrote:
  How can we use the alligation method/shortcut to solve this?
  
  My attempt:
  
  Sand...................Cement
  3/8......................5/8
  .......\.................../
  .........\............../
  ..............4/8
  
  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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  Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
  
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  15 Jan 2017, 18:47
  Think of "x" as the original amount of mixture & "y" as the amount we are replacing of the original mixture
  
  (3/8)x - (3/8)y + y = (1/2)x
  5y = x --> therefore, y=(1/5)x
  
  C.
  Re: A mixture of sand and cement contains, 3 parts of sand and 5   [#permalink] 15 Jan 2017, 18:47
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