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10kg of a mixture contains 30% sand and 70% clay. In order t 08 Jul 2020, 07:53
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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
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10kg of a mixture contains 30% sand and 70% clay. In order t Updated on: 01 Apr 2025, 22:17
Originally posted by KarishmaB on 21 Jul 2020, 05:41.
Last edited by KarishmaB on 01 Apr 2025, 22:17, edited 1 time in total.
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tejal777
10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand how much of the mixture is to be removed and replaced with pure sand?

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20/7

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/
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10kg of a mixture contains 30% sand and 70% clay. In order t 13 Sep 2020, 03:47
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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.
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10kg of a mixture contains 30% sand and 70% clay. In order t 04 Oct 2021, 10:25
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Given: 10kg of a mixture contains 30% sand and 70% clay.
Asked: In order to make the mixture contain equal quantities of clay and sand how much of the mixture is to be removed and replaced with pure sand?

10 kg mixture: -
Sand = 30%*10 = 3 kg
Clay = 70%*10 = 7 kg

In replace mixture: -
Sand = 3 - .3x + x = 3 + .7x
Clay = 7 - .7x = 7 - .7x
Since Sand = Clay
3 + .7x = 7 - .7x
4 = 1.4x
x = 4/1.4 = 40/14 = 20/7

IMO C
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10kg of a mixture contains 30% sand and 70% clay. In order t 25 Jul 2024, 16:46
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Fastest way is to use alligations, like others have mentioned.

30% 100%
50%

(100-50) (50-30)
50 + 20 =70 parts if the total

Thus proportion that needs to be added is 20/70 = 2/7

To know how many kg is that, we need to multiply by the total 10kg value
2/7 x 10 = 20/7 Answer
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10kg of a mixture contains 30% sand and 70% clay. In order t 05 Aug 2024, 11:37
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now there are 3 kg of sand out of 10kg and you want them to be 5kg: so 5-3 equal to 2kg
to do that you need to remove x amount with 30% concentration of sand and add x amount with 100% concentration of sand

2kg = -30%x+100%x
2kg = 70%x
x=20/7

you can also do it on clay side:
now there are 7 kg of clay out of 10kg and you want them to be 5kg: so 5-7 equal to -2kg
-2kg=-70%x
x=20/7
in words: how much of something that is 70% I need to remove to subtract 2kg?­
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10kg of a mixture contains 30% sand and 70% clay. In order t 20 Dec 2024, 00:03
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Easiest Method:
When Sand and Clay would be equal it would be 5 Kg each. Apply formula for clay:
Final Quantity = Initial quantity(1 - x/C) x-> Replacing amount; C is total Capacity

5 = 7(1 - x/10).
x= 20/7
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10kg of a mixture contains 30% sand and 70% clay. In order t 28 Mar 2025, 23:20
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KarishmaB ma'am can you please tell me where am I going wrong using allegations ?
A = mixture quantity
B = pure sand quantity
30 -- 50 -- 100%
A- ---- B
so 20(A) = 50(B)
so A/B= 5/2 now A = 10kg given so B = 4 so 4 kgs of pure sand was added hence 4kgs must also have been removed initially ??
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10kg of a mixture contains 30% sand and 70% clay. In order t 01 Apr 2025, 22:22
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djangobackend
KarishmaB ma'am can you please tell me where am I going wrong using allegations ?
A = mixture quantity
B = pure sand quantity
30 -- 50 -- 100%
A- ---- B
so 20(A) = 50(B)
so A/B= 5/2 now A = 10kg given so B = 4 so 4 kgs of pure sand was added hence 4kgs must also have been removed initially ??

If we were to mix pure sand with 10 kg of mixture, then your answer would be correct. But that is not what we are supposed to do. We must remove some mixture and then mix pure sand.
All we know is that they need to be in the ratio 5:2 finally.
Hence total mixture would be 7 parts which is actually equal to 10 kg. Out of this 2/7th would be pure sand i.e. 20/7 kg would be pure sand.

Check the first part of this blog post: https://anaprep.com/arithmetic-replacement-in-mixtures/
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