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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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Difficulty:

55% (hard)

Question Stats:

38% (02:00) correct
62% (01:46) wrong based on 65 sessions

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?

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
01 Jun 2014, 14:05

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.

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
02 Jun 2014, 12:32

1

This post received KUDOS

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...

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:

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
02 Jun 2014, 20:15

1

This post received KUDOS

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.

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.

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"?

Re: A mixture of sand and cement contains, 3 parts of sand and 5 [#permalink]
25 Jun 2014, 20:09

1

This post received KUDOS

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"?

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. _________________