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:
198 comments Page 8 of 20.
Fazle Rasool said:
8 years ago
Given : w:s=3:5,
After removal of x litter the ratio becomes w:s=1:1,
Quantity of water before and after is,
3/8 - 3x/8 +x = 1/2.
5x/8 = 1/8.
x = 1/5 this is the answer and it is a simple method.
After removal of x litter the ratio becomes w:s=1:1,
Quantity of water before and after is,
3/8 - 3x/8 +x = 1/2.
5x/8 = 1/8.
x = 1/5 this is the answer and it is a simple method.
Radheshyam said:
1 decade ago
Let volume is 8.
Water = 3/8 syrup = 5/8.
According to question,
Suppose y mixture(water,syrup)are drawn.
Water quantity is 3y/8 and syrup quantity is 5y/8.
3/8-3y/8+y = 5/8-5y/8.
y = 1/5.
Water = 3/8 syrup = 5/8.
According to question,
Suppose y mixture(water,syrup)are drawn.
Water quantity is 3y/8 and syrup quantity is 5y/8.
3/8-3y/8+y = 5/8-5y/8.
y = 1/5.
(1)
Rajnish jha said:
9 years ago
New quantity of syrup = 5x - 5y.
New quantity of water = 3x - 3y + 8y = 3x + 5y.
They are equal to each other.
=> 5x - 5y = 3x + 5y.
=> 2x = 10y.
=> y/x = 2/10 = 1/5.
=> 8y/8x = 1/5.
New quantity of water = 3x - 3y + 8y = 3x + 5y.
They are equal to each other.
=> 5x - 5y = 3x + 5y.
=> 2x = 10y.
=> y/x = 2/10 = 1/5.
=> 8y/8x = 1/5.
Sybil said:
1 decade ago
A solution of sugar syrup has 15% sugar. Another solution has 5% sugar. How many litres of the second solution must be added to 20% litres of the first solution to make a solution of 10% sugar?
(1)
Haniffa said:
9 years ago
WATER : SYRUP
3 : 5 = 8.
2 : 0 = 2 (TO BOTH BECOME EQUAL ADD 2 UNITS TO WATER).
5 : 5 = 10 (RATIO IS EQUAL NOW).
SO, ADDED AMOUNT OF WATER IS 2/10 OF FULL AMOUNT. IF YOU SIMPLIFIED IT'S 1/5.
3 : 5 = 8.
2 : 0 = 2 (TO BOTH BECOME EQUAL ADD 2 UNITS TO WATER).
5 : 5 = 10 (RATIO IS EQUAL NOW).
SO, ADDED AMOUNT OF WATER IS 2/10 OF FULL AMOUNT. IF YOU SIMPLIFIED IT'S 1/5.
Krishjlk@gmail.com said:
1 decade ago
Your great surender. Thanks a lot. Really that is very good explanation.
Instead of remembering the formulas you came with the theory.
And concept. Good excellent. Thank you for your post.
Instead of remembering the formulas you came with the theory.
And concept. Good excellent. Thank you for your post.
Shivangi said:
5 years ago
Final(syrup)=initial(syrup)*(1-(qt_replaced/total_mix).
X is a mixture replaced with water.
p is the total quantity of mixture.
4/8=5/8(1-x/p).
We need to find x/p which is 1/5 i.e answer.
X is a mixture replaced with water.
p is the total quantity of mixture.
4/8=5/8(1-x/p).
We need to find x/p which is 1/5 i.e answer.
ABHIJIT said:
9 years ago
w:s
3:5
Consider in tens
30 : 50
3 : 5
1:1
40:40
To get 40 syrup you need to remove 10.
as the ratio is 3:5.
So, to remove 10 you need to remove 8x2=16 volume of liquid = 16/80 = 1/5.
3:5
Consider in tens
30 : 50
3 : 5
1:1
40:40
To get 40 syrup you need to remove 10.
as the ratio is 3:5.
So, to remove 10 you need to remove 8x2=16 volume of liquid = 16/80 = 1/5.
Priyanka said:
8 years ago
Initially W : S=3:5.
After adding water W:S=1:1
Since only water is adding Qty of Syrup will change.
W:S=3:5.
W:S=5:5.
Difference in water =2 part.
Total =(5+5)=10 part.
2/10=1/5 Ans.
After adding water W:S=1:1
Since only water is adding Qty of Syrup will change.
W:S=3:5.
W:S=5:5.
Difference in water =2 part.
Total =(5+5)=10 part.
2/10=1/5 Ans.
Rahul Wiley said:
1 decade ago
Hey Keep it simple.
Fraction of syrup in final Mixture = (Fraction of syrup in initial mixture)*(1 - x/v).
Where,
x is the Quantity of mixture removed.
V is the volume of the Mixture.
Fraction of syrup in final Mixture = (Fraction of syrup in initial mixture)*(1 - x/v).
Where,
x is the Quantity of mixture removed.
V is the volume of the Mixture.
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