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 6 of 21.
Umesh said:
8 years ago
1/5 of 8 is 1.6 ------->remaining=6.4.
We have to remove 1/5 of liquid and add 1/5 of water(1.6 of water).
1/5 of 3=.6 -----> remaining=2.4.
1/5 of 5=1 -----> remaining=4.
6.4 of liquid =2.4 of water + 4 of syrup.
Which means add 1.6 of water to make to make composition equal and 1.6 is 1/5 of 8.
We have to remove 1/5 of liquid and add 1/5 of water(1.6 of water).
1/5 of 3=.6 -----> remaining=2.4.
1/5 of 5=1 -----> remaining=4.
6.4 of liquid =2.4 of water + 4 of syrup.
Which means add 1.6 of water to make to make composition equal and 1.6 is 1/5 of 8.
Shriyash sharma said:
9 years ago
Formula.
Final volume or the ratio of solute A=initial volume or ratio solution A(1-solute to be replaced/total volume)^n.
let A=syrup, B=water.
Putting in formula:
assuming volume is 1litre.
A=5/8, B=3/8.
applying formula on solute A because it will be reduced after an operation.
so,
4/8=5/8(1-x/1),
x=1/5.
Final volume or the ratio of solute A=initial volume or ratio solution A(1-solute to be replaced/total volume)^n.
let A=syrup, B=water.
Putting in formula:
assuming volume is 1litre.
A=5/8, B=3/8.
applying formula on solute A because it will be reduced after an operation.
so,
4/8=5/8(1-x/1),
x=1/5.
Karthik said:
8 years ago
Let take 5 parts of syrup as 50% of the liquid and 3 parts of water as 30% of the liquid. In order to make both the liquids equal, we have to reduce syrup by 10% and raise water by 10%. So, the total quantity of water we have to add or quantity of liquid we have to draw is 20% which is 1/5 of the liquid.
Anand Shaw said:
1 decade ago
Water in one mixture : 3/8.
Water in another mixture : 1.
Water in new mixture : 1/2.
According allegation formula:
Ratio of quantity of two mixtures mixed : (3/8-1/2) : (1-1/2) = 4:1.
So 1/5 parts of another mixture is mixed with one.
Another mixture is pure and is same as mixture replaced.
Water in another mixture : 1.
Water in new mixture : 1/2.
According allegation formula:
Ratio of quantity of two mixtures mixed : (3/8-1/2) : (1-1/2) = 4:1.
So 1/5 parts of another mixture is mixed with one.
Another mixture is pure and is same as mixture replaced.
Abhishek Kumar JhaJi said:
3 years ago
Let 100 ltr mixture.
So, 5/8x 100 = 62.5 ltr syrup.
3/8x100 = 37.5 ltr water.
We need to remove syrup and add that much quantity of water so make it 1:1,
62.5- x/37.5+x = 1/1,
62.5- x = 37.5 +x.
2x = 25.
x = 12.5 (need to remove from syrup and this quantity into water),
12.5/62.5(syrup) = 1/5.
So, 5/8x 100 = 62.5 ltr syrup.
3/8x100 = 37.5 ltr water.
We need to remove syrup and add that much quantity of water so make it 1:1,
62.5- x/37.5+x = 1/1,
62.5- x = 37.5 +x.
2x = 25.
x = 12.5 (need to remove from syrup and this quantity into water),
12.5/62.5(syrup) = 1/5.
(57)
Anil said:
10 years ago
Quantity of syrup is 5. Quantity of water is 3. You have to add and subtract 1 from water quant and syrup quant respectively. Keep that one in numerator and the amount of which has to be reduced (syrup 5) should come in the denominator. 1/5. I have tried with other problems also it worked.
Kamaksha Puri said:
5 years ago
Let 100 ltr mixture.
So 5/8x 100=62.5 ltr syrup.
3/8x100= 37.5 ltr water.
We need to remove syrup and add that much quantity of water so make it 1:1
62.5- x/37.5+x=1/1,
62.5- x =37.5 +x.
2x =25.
x = 12.5 (need to remove from syrup and this quantity into water),
12.5/62.5(syrup) = 1/5.
So 5/8x 100=62.5 ltr syrup.
3/8x100= 37.5 ltr water.
We need to remove syrup and add that much quantity of water so make it 1:1
62.5- x/37.5+x=1/1,
62.5- x =37.5 +x.
2x =25.
x = 12.5 (need to remove from syrup and this quantity into water),
12.5/62.5(syrup) = 1/5.
(23)
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)
Manish said:
1 decade ago
Let there be 1kg of mixture with water concentration 3/8 i.e. 3/8 kg water present.
X kg mixture is removed so water removed = 3X/8.
Now X kg pure water is added.
In the required mixture water = 1/2 kg. So balancing over water.
3/8 " 3X/8 + X = 1/2 solving we get X = 1/5.
X kg mixture is removed so water removed = 3X/8.
Now X kg pure water is added.
In the required mixture water = 1/2 kg. So balancing over water.
3/8 " 3X/8 + X = 1/2 solving we get X = 1/5.
Mir said:
7 years ago
When a mixture is drawn, both syrup and water is drawn equally.
So, if 1 part from water and 1 part of syrup(total 2 parts) is drawn, the ratio of 3:5 becomes 2:4. When replaced wit 2 parts of water, gives the ratio 4:4 i..e 1:1.
Therefore mixture drawn is 2/8 or 1/4.
So, if 1 part from water and 1 part of syrup(total 2 parts) is drawn, the ratio of 3:5 becomes 2:4. When replaced wit 2 parts of water, gives the ratio 4:4 i..e 1:1.
Therefore mixture drawn is 2/8 or 1/4.
(1)
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