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
The 20kg mixture comprised of water and sand is mixed evenly
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 157
- Joined: Sat Nov 19, 2016 5:34 am
- Thanked: 2 times
- Followed by:4 members
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Original mixture: W/S = 8/2ziyuenlau 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
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- [email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7243
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We can let the number of kilograms of water = x and the number of kilograms of sand = y. Thus, x + y = 20.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 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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
GMAT/MBA Expert
- Jay@ManhattanReview
- GMAT Instructor
- Posts: 3008
- Joined: Mon Aug 22, 2016 6:19 am
- Location: Grand Central / New York
- Thanked: 470 times
- Followed by:34 members
Hi ziyuenlau,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
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.
So let's try option B = 5 kg.
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.
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!
Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide
Jay
_________________
Manhattan Review GMAT Prep
Locations: Almaty | Minsk | Aarhus | Vilnius | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.