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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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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  [#permalink]

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New post 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.

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

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New post 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

\(= \frac{\frac{1}{8} * 1}{\frac{5}{8}}\)

\(= \frac{1}{5}\)

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

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

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

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New post 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

Adding "x" quantity of sand

\(\frac{3}{8} - \frac{3x}{8} + x\) .......... \(\frac{5}{8} - \frac{5x}{8}\) .............. 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

\(\frac{5}{8} - \frac{5x}{8} = \frac{1}{2}\)

\(x = \frac{1}{5}\)

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

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New post 25 Jun 2014, 20:09
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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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New post Updated on: 11 Nov 2015, 19:39
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let x=fraction of mixture to be substituted
3/8-3x/8+x=1/2
x=1/5

Originally posted by gracie on 06 Sep 2015, 14:53.
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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New post 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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New post 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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New post 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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New post 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.
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A mixture of sand and cement contains, 3 parts of sand and 5  [#permalink]

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

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New post 13 Dec 2018, 23:24
Approach 1

5/8 - x*5/8 = 1/2
Solving, x = 1/5

Approach 2
3 : 5 = 6 : 10
1 : 1 = 8 : 8

(10-8)/10 = 1/5

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

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New post 17 Dec 2018, 23:18
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.
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Re: A mixture of sand and cement contains, 3 parts of sand and 5  [#permalink]

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

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New post 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.

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Re: A mixture of sand and cement contains, 3 parts of sand and 5 &nbs [#permalink] 19 Dec 2018, 08:14
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