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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
Hello from the GMAT Club BumpBot!

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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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