The 20kg mixture comprised of water and sand is mixed evenly

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hazelnut01: Master | Next Rank: 500 Posts; Posts: 157; Joined: Sat Nov 19, 2016 5:34 amFollowed by: 4 members; Thanked: 2 times

by hazelnut01 » Sun Apr 02, 2017 2:57 pm

The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0Â Â Â Â
B. 5.0Â Â Â Â
C. 6.8Â Â Â Â
D. 7.6Â Â Â Â
E. 8.4

Source : Math Revolution
Official Answer : B

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Apr 02, 2017 4:44 pm

ziyuenlau wrote:The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0Â Â Â Â
B. 5.0Â Â Â Â
C. 6.8Â Â Â Â
D. 7.6Â Â Â Â
E. 8.4

Original mixture: W/S = 8/2
Thus, of every 10kg, 8kg is water and 2kg is sand, implying that 20% of the original mixture is sand.

Added pure sand = 100% sand.

Final mixture: W/S = 6/4
Thus, of every 10kg, 6kg is water and 4kg is sand, implying that 40% of the final mixture is sand.

Let O = the original mixture and S = the added pure sand.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for O and S on the ends and the percentage for the final mixture in the middle.
O 20%-----------40%-----------100% S

Step 2: Calculate the distances between the percentages.
O 20%----20-----40%----60-----100% S

Step 3: Determine the ratio in the final mixture.
The ratio of O to S is equal to the RECIPROCAL of the distances in red.
O:S = 60:20 = 3:1 = 15:5.

Thus:
The 20kg of final mixture is composed of 15kg original mixture and 5kg pure sand.

The correct answer is B.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

Last edited by GMATGuruNY on Tue Apr 04, 2017 7:52 am, edited 1 time in total.

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by [email protected] » Sun Apr 02, 2017 6:41 pm

Hi ziyuenlau,

This question can be solved by TESTing THE ANSWERS.

We're told that a 20kg mixture is made up of water and sand (in a ratio of 8:2). This means that 16 kg of the mixture is water and 4 kg is sand. We're asked to determine what amount of the mixture must be removed - and replaced with pure sand - so that the new ratio of water to sand is 6:4 (meaning that there will be 12 kg of water and 8 kg of sand).

Let's TEST Answer B: 5 kg

If we remove 5 kg from the original 20 kg mixture, then we'll end up with a 15 kg mixture that is 12 kg water and 3 kg sand. If we add 5 kg of pure sand to these totals, then we'll end up with a new 20 kg mix that is 12 kg water and 8 kg sand. This new ratio of water to sand is 12:8 = 6:4. This is an exact match for what we are after, so this MUST be the answer.

Final Answer: B

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Rich

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by Scott@TargetTestPrep » Tue Apr 11, 2017 11:23 am

ziyuenlau wrote:The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0Â Â Â Â
B. 5.0Â Â Â Â
C. 6.8Â Â Â Â
D. 7.6Â Â Â Â
E. 8.4

We can let the number of kilograms of water = x and the number of kilograms of sand = y. Thus, x + y = 20.

We also see that:

x/y = 8/2

x/y = 4/1

x = 4y

Thus:

4y + y = 20

5y = 20

y = 4, so x = 16

We are given that part of the mixture is replaced by sand to make the ratio of water to sand 6 to 4. We can let n represent the number of kilograms of mixture that is replaced; thus, the number of kilograms of sand that is added to the mixture is also n.

When we replace n kilograms of mixture, (1/5)n kilograms will be sand and (4/5)n kilograms will be water, since the ratio of water to sand is 4 : 1. So, we are removing (4/5)n kilograms of water from the existing 16 kilograms of water, and we are removing (1/5)n kilograms of sand from the existing 4 kilograms of sand.

However, we are also adding back n kilograms of sand. Thus:

water/sand = 6/4

(16 - (4/5)n)/(4 - (1/5)n + n) = 6/4

(16 - (4/5)n)/(4 + (4/5)n) = 3/2

2(16 - (4/5)n) = 3(4 + (4/5)n)

32 - (8/5)n = 12 + (12/5)n

Multiplying the whole equation by 5, we have:

160 - 8n = 60 + 12n

100 = 20n

n = 5

Alternate solution:

The original water:sand ratio of 8:2 indicates that sand comprises 20% of the original 20 kg mixture. We will remove x kilograms of the original mixture which was 20% sand, and we will replace what was removed with pure sand (100% sand). The result will be 20 kilograms of a mixture which is now 40% sand. We can express this algebraically as:

20(0.2) - x(0.2) + x(1.0) = 20(0.4)

4 - 0.2x + x = 8

0.8 x = 4

x = 5

Thus we replace 5 kg of the original mixture with pure sand to create the new mixture.

Answer: B

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by Jay@ManhattanReview » Tue Apr 11, 2017 9:28 pm

ziyuenlau wrote:The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0Â Â Â Â
B. 5.0Â Â Â Â
C. 6.8Â Â Â Â
D. 7.6Â Â Â Â
E. 8.4

Source : Math Revolution
Official Answer : B

Hi ziyuenlau,

Get another approach for this question.

We know that Water = 16 kg and Sand = 4 kg.

Let's go by testing option values. Since the options are arranged in an ascending order, we can plug-in at the maximum two option values and reach the conclusion.

Since the resultant ratio of Water : Sand after removing some part of the mixture and replacing it with the same amount of Sand = 6 : 4 = 3 : 2, we must check which option renders the ratio = 3 : 2.

If the option value results in a ratio > 3 : 2, it means that we replaced lesser quantity of mixture with Sand, we must try with relatively larger value; however, if the option value results in a ratio < 3 : 2, it means that we replaced more quantity of mixture with Sand, we must try with relatively smaller value.

Always try option B (if the options are arranged in an ascending/descending order)

1a. If option B value results in a ratio = 3 : 2, option B is the correct answer.

1b. If option B value results in a ratio < 3 : 2, option A is the correct answer.

1c. If option B value results in a ratio > 3 : 2, try option D and not C.

2a. If option D value results in a ratio = 3 : 2, option D is the correct answer.

2b. If option D value results in a ratio < 3 : 2, option C is the correct answer.

2c. If option D value results in a ratio > 3 : 2, option E is the correct answer.

So let's try option B = 5 kg.

Water coming out of 5 kg. mixture = (8/10)*5 = 4 kg, thus Water in the final mixture = 16 - 4 = 12 kg

Sand coming out of 5 kg. mixture = (2/10)*5 = 1 kg, thus Sand in the final mixture = 4 - 1 + 5 = 8 kg

The final ratio = Water : Sand : : 12 : 8 = 3 : 2. Option B is the correct answer.

Hope this helps!

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