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08 Jul 2020, 07:53
Kudos
Bookmarks
I have used allegations here.
30% 100%(as we are replacing with pure sand)
50%
50% 20%
Which means 50% of the mixture has to be replaced with 20% of pure sand.
Posted from my mobile device
30% 100%(as we are replacing with pure sand)
50%
50% 20%
Which means 50% of the mixture has to be replaced with 20% of pure sand.
Posted from my mobile device
tejal777
So we have a mixture with 30% sand to which we need to add 100% sand to get 50% sand mixture.
w1/w2 = (100 - 50)/(50 - 30) = 5/2
So for every 5 parts initial mixture, we should add 2 parts 100% sand. This means that of the 7 parts of total initial mixture, we should remove 2 parts and put 2 parts of 100% sand in its place.
(2/7)*10 kg = 20/7 = 2.85 kg
Check these posts:
https://anaprep.com/arithmetic-mixtures/
https://anaprep.com/arithmetic-replacement-in-mixtures/
13 Sep 2020, 03:47
Kudos
Bookmarks
I solved this using weighted average VeritasKarishma.
Drawing the scale: Let's say x kg of sand needs to be added. This means 10-x of original mixture needs to be removed
10-x x
-----------------------------------
30% 50% 100%
Writing the equation:
(50-30) / (100-50) = x / (10 - x)
20 (10 -x) = 50x
200 - 20x = 50x
200 = 70x
x = 20/7
20/7 kgs of mixture needs to be replaced by pure sand.
Drawing the scale: Let's say x kg of sand needs to be added. This means 10-x of original mixture needs to be removed
10-x x
-----------------------------------
30% 50% 100%
Writing the equation:
(50-30) / (100-50) = x / (10 - x)
20 (10 -x) = 50x
200 - 20x = 50x
200 = 70x
x = 20/7
20/7 kgs of mixture needs to be replaced by pure sand.
