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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]  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]  01 Jun 2014, 14:05
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Expert's post
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]  02 Jun 2014, 12:32
1
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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]  02 Jun 2014, 20:15
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Expert's post
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]  04 Jun 2014, 01:08
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

$$= \frac{\frac{1}{8} * 1}{\frac{5}{8}}$$

$$= \frac{1}{5}$$

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Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]  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]  25 Jun 2014, 20:01
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

$$\frac{3}{8} - \frac{3x}{8} + x$$ .......... $$\frac{5}{8} - \frac{5x}{8}$$ .............. 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]  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]  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] 25 Jun 2014, 20:19
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