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A mixture of sand and cement contains, 3 parts of sand and 5
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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. 1/3 B. 1/4 C. 1/5 D. 1/7 E. 1/8 Source: Indian CAT
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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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01 Jun 2014, 15:05




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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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04 Jun 2014, 02:08
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 \(= \frac{\frac{1}{8} * 1}{\frac{5}{8}}\) \(= \frac{1}{5}\) Answer = C
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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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02 Jun 2014, 13:32
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: \((\frac{3}{8})*(1x) + (1)*(x)=\frac{1}{2}\) Thus, \(x=\frac{1}{5}\)



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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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02 Jun 2014, 21:15
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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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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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"?



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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, 21:01
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 Adding "x" quantity of sand \(\frac{3}{8}  \frac{3x}{8} + x\) .......... \(\frac{5}{8}  \frac{5x}{8}\) .............. 1x+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\(\frac{5}{8}  \frac{5x}{8} = \frac{1}{2}\) \(x = \frac{1}{5}\) Answer = C
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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, 21:09
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"? 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
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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.
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A mixture of sand and cement contains, 3 parts of sand and 5
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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.



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A mixture of sand and cement contains, 3 parts of sand and 5
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Updated on: 11 Nov 2015, 20:39
let x=fraction of mixture to be substituted 3/83x/8+x=1/2 x=1/5
Originally posted by gracie on 06 Sep 2015, 15:53.
Last edited by gracie on 11 Nov 2015, 20:39, edited 1 time in total.



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Re: A mixture of sand and cement contains, 3 parts of sand and 5
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07 Sep 2015, 13:36
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
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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.



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A mixture of sand and cement contains, 3 parts of sand and 5
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26 Dec 2016, 05:14
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...................Sandonly mix 3/8......................1/1 .......\.................../ .........\............../ ..............4/8 1/2......................1/8 Therefore the resulting mix will have a ratio Old Mix to Sandonly 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 sandonly 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
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15 Jan 2017, 19:47
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.



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A mixture of sand and cement contains, 3 parts of sand and 5
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10 Jun 2017, 20:10
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. ((xy) *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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