Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 3)
3.
A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
Answer: Option
Explanation:
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
| Quantity of A in mixture left = |  |
7x - | 7 | x 9 |  |
litres = | |
7x - | 21 | litres. |
| 12 | 4 |
| Quantity of B in mixture left = | |
5x - | 5 | x 9 | |
litres = | |
5x - | 15 | litres. |
| 12 | 4 |
 |
|
= | 7 | |||||
|
9 |
 |
28x - 21 | = | 7 |
| 20x + 21 | 9 |
252x - 189 = 140x + 147
112x = 336
x = 3.
So, the can contained 21 litres of A.
Discussion:
100 comments Page 1 of 10.
Anand said:
2 years ago
Ist part = 7 : 5=(7+5) = 12.
2nd part = 12-9 = 3.
3rd part = 7 * 3 = 21.
2nd part = 12-9 = 3.
3rd part = 7 * 3 = 21.
(89)
Jamshaid said:
3 years ago
The Correct solution is;
Change in old ratio to new ratio;
7X / (5X+9) = 7/9,
X = 9/4.
Old A = 7X + 9litre * 7/12 (A's part in 9l drew).
= 7*(9/4)+9*7/12 = 28.7/20L A's share in old mixture.
Change in old ratio to new ratio;
7X / (5X+9) = 7/9,
X = 9/4.
Old A = 7X + 9litre * 7/12 (A's part in 9l drew).
= 7*(9/4)+9*7/12 = 28.7/20L A's share in old mixture.
(16)
Anonymous said:
4 years ago
@All.
Here, after performing n operations on a total of x litres replacing y liters, the final amount present is x(1+y/x)^n.
Substituting values, 12(1+9/12) = 12+9 = 21.
Here, after performing n operations on a total of x litres replacing y liters, the final amount present is x(1+y/x)^n.
Substituting values, 12(1+9/12) = 12+9 = 21.
(29)
Supriya Balaji said:
4 years ago
We can also solve if easily by using a formula, final QT/initial QT=(1-(x/c) ^t.
Where, initial QT = quantity.
x=how much QT is removed,
C=total capacity,
T=how many times the process is repeated, that is one time as per the quetion,
Final Qt = 12(1-(9/12)) ^1,
=12(12-9/12),
=3.
Where, initial QT = quantity.
x=how much QT is removed,
C=total capacity,
T=how many times the process is repeated, that is one time as per the quetion,
Final Qt = 12(1-(9/12)) ^1,
=12(12-9/12),
=3.
(12)
Tejashwini said:
5 years ago
@All.
Simply, the solution is;
Initially, 7/12 amount of liquidA is present in the mixture.
Finally, 7/16 amount of liquidA is present in the mixture.
Let us assume there is x litres of mixture.
9litres of the mixture is drawn out of x litres.
Final proportion = initial proportion(after removing/final volume).
You will get x = 36.
Therefore 7/12(x) = 21.
Simply, the solution is;
Initially, 7/12 amount of liquidA is present in the mixture.
Finally, 7/16 amount of liquidA is present in the mixture.
Let us assume there is x litres of mixture.
9litres of the mixture is drawn out of x litres.
Final proportion = initial proportion(after removing/final volume).
You will get x = 36.
Therefore 7/12(x) = 21.
(8)
Kalai said:
6 years ago
@Yamini.
Just simple initially there is 7:5 ratio of A & B now they need initial liters of A so in the answer which is a multiple of 7 is 21.
Simple one.
Just simple initially there is 7:5 ratio of A & B now they need initial liters of A so in the answer which is a multiple of 7 is 21.
Simple one.
(33)
Yamini@usha said:
6 years ago
It's very difficult to understand. Anyone please help me to get it.
(5)
Haider said:
6 years ago
9 litres of the withdrawn mixture will also contain A:B = 7:5.
So (7+5)n = 12*n.
12*n = 9(given)
n=3/4.
So the quantity of A in withdrawn mixture = 7*3/4=21/4 litres.
So in original mixtures A will be remained = 7*x-21/4.
Which will be equal to the quantity of A in 2nd mixture.
So (7*x-21/4)=7*y................equation 1.
Also, sum of A & B in both the mixtures will be same,
(7+5)*x=(7+9)*y....................equation 2.
Solving both we will get x=3.
So the quantity of A =7*3=21 litres.
So (7+5)n = 12*n.
12*n = 9(given)
n=3/4.
So the quantity of A in withdrawn mixture = 7*3/4=21/4 litres.
So in original mixtures A will be remained = 7*x-21/4.
Which will be equal to the quantity of A in 2nd mixture.
So (7*x-21/4)=7*y................equation 1.
Also, sum of A & B in both the mixtures will be same,
(7+5)*x=(7+9)*y....................equation 2.
Solving both we will get x=3.
So the quantity of A =7*3=21 litres.
(3)
Ankush Singh said:
6 years ago
7:5.
7:9.
Increase is part b 4.
4=9.
1=9/4.
2nd of total mix=16.
16= 9/4*16=36.
1st mix A is. 7/12*36 =21.
7:9.
Increase is part b 4.
4=9.
1=9/4.
2nd of total mix=16.
16= 9/4*16=36.
1st mix A is. 7/12*36 =21.
(36)
Veeresh nj said:
6 years ago
7:5 = 7 * 5 = 35.
7:9 = 7 * 9 = 63.
Now subtract both values;
63 - 35 = 28.
Now u see common value in 7:5 and 7:9.
Here 7 is common to subtract this value with 28.
A = 28 - 7 = 21.
A = 21.
7:9 = 7 * 9 = 63.
Now subtract both values;
63 - 35 = 28.
Now u see common value in 7:5 and 7:9.
Here 7 is common to subtract this value with 28.
A = 28 - 7 = 21.
A = 21.
(27)
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