Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 1)
1.
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Answer: Option
Explanation:
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
| Quantity of water in new mixture = | ![]() |
3 - | 3x | + x | ![]() |
litres |
| 8 |
| Quantity of syrup in new mixture = | ![]() |
5 - | 5x | ![]() |
litres |
| 8 |
|
![]() |
3 - | 3x | + x | ![]() |
= | ![]() |
5 - | 5x | ![]() |
| 8 | 8 |
5x + 24 = 40 - 5x
10x = 16
x = |
8 | . |
| 5 |
| So, part of the mixture replaced = | ![]() |
8 | x | 1 | ![]() |
= | 1 | . |
| 5 | 8 | 5 |
Discussion:
203 comments Page 19 of 21.
Chacha chodhary said:
2 decades ago
Here is simple approach
3/5-1/2=1:1
3/5-1/2=1:1
Victor said:
2 decades ago
Simple solution for this;
Given: Vessel contains...water -- 3 parts. Syrup -- 5 parts.
Required: Both should be half-- half
Sol: take one part of syrup and add it to water i.e then both will have 4 parts.
So divide syrup into 5 parts.then one part is 1/5, that is the solution.
Given: Vessel contains...water -- 3 parts. Syrup -- 5 parts.
Required: Both should be half-- half
Sol: take one part of syrup and add it to water i.e then both will have 4 parts.
So divide syrup into 5 parts.then one part is 1/5, that is the solution.
(3)
Manasa said:
2 decades ago
Uttam, surender and ameer has cleared out my confusion regarding this problem. Thanks a lot.
Ramakhanna said:
2 decades ago
Thanks ameer for your detailed explanation...
Ameer said:
2 decades ago
Thank you Surender and Utham...i hav tried to explain the answer in detail
Let the volume of vessel be x
quantity of water = 3/8 x
quantity of syrup = 5/8 x
Let the amount of mixture removed be y
The removed mixture contains 3/8 y of water and 5/8 y of syrup
So quantity of water left in the mixture=3/8[x-y]
quantity of syrup left in the mixture 5/8 [x-y]
According to the question,the amount of mixture removed is replaced later by water,so that the syrup an water would be exactly half each of the mixture...
so when more water is added to the existing water in the mixture..
3/8[x-y] + y = 5/8[x-y]
3x/8 - 3y/8 +y = 5x/8 - 5y/8
y = 2x/8 - 2y/8
8y=2[x-y]
4y=x-y
x=5y
y=1/5 x
Let the volume of vessel be x
quantity of water = 3/8 x
quantity of syrup = 5/8 x
Let the amount of mixture removed be y
The removed mixture contains 3/8 y of water and 5/8 y of syrup
So quantity of water left in the mixture=3/8[x-y]
quantity of syrup left in the mixture 5/8 [x-y]
According to the question,the amount of mixture removed is replaced later by water,so that the syrup an water would be exactly half each of the mixture...
so when more water is added to the existing water in the mixture..
3/8[x-y] + y = 5/8[x-y]
3x/8 - 3y/8 +y = 5x/8 - 5y/8
y = 2x/8 - 2y/8
8y=2[x-y]
4y=x-y
x=5y
y=1/5 x
(1)
Ravi said:
2 decades ago
Uttam and surender. Your answer is perfect. Kiran is wrong.
Megha said:
2 decades ago
Thanks it is really a good way to solve it.
Uttam said:
2 decades ago
Let X=volume of vessel
Y=volume of mix taken out.
Initially,
volume of water in vessel=3X/8
volume of syrup in vessel=5X/8
Y=volume of mix taken out=volume of water added.
Now,
drawn off mix may contain,
volume of water=3Y/8.
volume of syrup=5Y/8.
after mix drawn off,
volume of water in the vessel=(3X/8)-(3Y/8)=(3/8)(X-Y).
volume of syrup in the vessel=(5X/8)-(5Y/8)=(5/8)(X-Y).
After water added, volume of water in the vessel
=[(3X/8)-(3Y/8)]+Y
=[(3/8)(X-Y)]+Y.
hence from the Question,in the vessel, volume of water
=volume of syrup
[(3/8)(X-Y)]+Y=(5/8)(X-Y)
10Y=2X
Y=(1/5)*X
So volume of water should be added is (1/5)th of volume of vessel.
Y=volume of mix taken out.
Initially,
volume of water in vessel=3X/8
volume of syrup in vessel=5X/8
Y=volume of mix taken out=volume of water added.
Now,
drawn off mix may contain,
volume of water=3Y/8.
volume of syrup=5Y/8.
after mix drawn off,
volume of water in the vessel=(3X/8)-(3Y/8)=(3/8)(X-Y).
volume of syrup in the vessel=(5X/8)-(5Y/8)=(5/8)(X-Y).
After water added, volume of water in the vessel
=[(3X/8)-(3Y/8)]+Y
=[(3/8)(X-Y)]+Y.
hence from the Question,in the vessel, volume of water
=volume of syrup
[(3/8)(X-Y)]+Y=(5/8)(X-Y)
10Y=2X
Y=(1/5)*X
So volume of water should be added is (1/5)th of volume of vessel.
Kalyan said:
2 decades ago
@Mahi and Kiran
Since it is 3 parts and 5 parts...totally 8.
50% of 8 is 4 and 40% of 8 is 3.
So 60% of 8 is 5 parts.
Hope you understand.
Since it is 3 parts and 5 parts...totally 8.
50% of 8 is 4 and 40% of 8 is 3.
So 60% of 8 is 5 parts.
Hope you understand.
Vikas said:
2 decades ago
Yes mahi, something wrong with Kiran's answer.
3/8= 37.5%
5/8= 62.5%
And its not 40% and 60% as told by kiran
3/8= 37.5%
5/8= 62.5%
And its not 40% and 60% as told by kiran
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