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 2 of 10.
Varsha said:
1 decade ago
Thanks silu.
Naveen said:
1 decade ago
How should I know in this case I should take one variable.
Shravan said:
1 decade ago
@Silu good explaintion.
Sus said:
1 decade ago
@Silu your explanation is good.
Amisha said:
1 decade ago
Suppose intial quantity of mixture is x
withdraw 9 lit so,qut. remain is x-9
(it can't change ratio A:B=7:5)
so, B is 5/12
alligation
5/12 1 9/16-5/12=7/48
1 - 9/16= 7/16
9/16(A:B=7:9 so, B=9/16)
7/16 7/48
qt of mixture (after withdraw of 9 lit)= 9lit
qt of pure liquid B=9 lit.
so, x-9/9= (7/16)/(7/48)
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
intial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/12 = 21 lit
withdraw 9 lit so,qut. remain is x-9
(it can't change ratio A:B=7:5)
so, B is 5/12
alligation
5/12 1 9/16-5/12=7/48
1 - 9/16= 7/16
9/16(A:B=7:9 so, B=9/16)
7/16 7/48
qt of mixture (after withdraw of 9 lit)= 9lit
qt of pure liquid B=9 lit.
so, x-9/9= (7/16)/(7/48)
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
intial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/12 = 21 lit
Karthik RS said:
1 decade ago
The Value of A is 7*x ... the only answer which is divisible by 7 is 21 .. so there you go :)
Amisha patel said:
1 decade ago
Suppose intial qt. of mixture is 9
withdraw 9 lit. so qt. remaining is x-9,but it can't change ratio so A:B = 7:5 and B = 5/12
apply alligation where,
c = 5/12
d = 1 (pure B is added)
mean = 9/16 (we want to make A:B= 7:9,so B= 9/16)
qt of mixture/qt of pure liquid B= (1-mean)/(mean-c)
x-9/9 = (1-9/16)/(9/16-5/12)
x-9/9 = (7/16)/ (7/48)
x-9/9 = 48/16
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
initial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/(7+5) = 21 lit
withdraw 9 lit. so qt. remaining is x-9,but it can't change ratio so A:B = 7:5 and B = 5/12
apply alligation where,
c = 5/12
d = 1 (pure B is added)
mean = 9/16 (we want to make A:B= 7:9,so B= 9/16)
qt of mixture/qt of pure liquid B= (1-mean)/(mean-c)
x-9/9 = (1-9/16)/(9/16-5/12)
x-9/9 = (7/16)/ (7/48)
x-9/9 = 48/16
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
initial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/(7+5) = 21 lit
ANIL KUMAR JAT said:
1 decade ago
A B
7 : 5
7 : 9
9-5=4unit=9
1unit=9/4
16unit=36
A=36*7/12=21.
B=36*5/12=15.
7 : 5
7 : 9
9-5=4unit=9
1unit=9/4
16unit=36
A=36*7/12=21.
B=36*5/12=15.
Srividhya said:
1 decade ago
Simple Method: Forget B.
Let the total mixture be x litres.
So A will be (7x/12) lts.
After removing 9 lts, (7x9)/12 litres of A is gone.
So A will become (7x/12)-((7x9)/12).
Which is 7x/16 litres of the new mixture.
=>(7x/12)-((7x9)/12) = 7x/16.
=> 7(x-9)/12 = 7x/16.
=> (x-9)/3 = x/4.
=> 4x-36 = 3x.
=> x = 36 litres.
A in the original mixture = (7x36)/12 = 21 litres.
Let the total mixture be x litres.
So A will be (7x/12) lts.
After removing 9 lts, (7x9)/12 litres of A is gone.
So A will become (7x/12)-((7x9)/12).
Which is 7x/16 litres of the new mixture.
=>(7x/12)-((7x9)/12) = 7x/16.
=> 7(x-9)/12 = 7x/16.
=> (x-9)/3 = x/4.
=> 4x-36 = 3x.
=> x = 36 litres.
A in the original mixture = (7x36)/12 = 21 litres.
(1)
Unknown said:
1 decade ago
Can't we find it by forming two linear equations?
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