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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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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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 Hence 1/5 of the mix must have been replaced by sand.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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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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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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let x=fraction of mixture to be substituted
3/8-3x/8+x=1/2
x=1/5

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.

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.

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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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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