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:
101 comments Page 3 of 11.
Deepu said:
11 months ago
Let's assume B is reduced by x.
Then 7/12 X "x"=7/16 {A in intial 7/12 and left 7/16}
x=3/4, so reduce by 1/4.
If reducing by 1/4 requires the removal 9L.
Then full amount removal required 36L---Capacity of vessel{1/4part=9L}.
36l x 7/12 = 21L of A.
Then 7/12 X "x"=7/16 {A in intial 7/12 and left 7/16}
x=3/4, so reduce by 1/4.
If reducing by 1/4 requires the removal 9L.
Then full amount removal required 36L---Capacity of vessel{1/4part=9L}.
36l x 7/12 = 21L of A.
(1)
Teja said:
2 decades ago
Can You explain in other way ?
Amara said:
2 decades ago
How to calculate mixture left?
Debjyoti said:
2 decades ago
Total mixer is x.
A=7x/12 B=5x/12
after drawn 9 lit of mixer........
mixer remain(x-9)
So the new ratio of A:B is ...
A=7(x-9)/12 and B=5(x-9)/12
now drawn 9 lit filled with B
so B now contain B=5(x-9)+9/12...........
now the new ratio of mixer is A:B=7:9
we know the new value of A and B...........put that in this equation
7(x-9)/12:(5(x-9)+9)=7:9-------if u calculate further u ll get the X value as 36
Put that x value in A=7x/12, so here you can get answer of A contain initially.
A=7x/12 B=5x/12
after drawn 9 lit of mixer........
mixer remain(x-9)
So the new ratio of A:B is ...
A=7(x-9)/12 and B=5(x-9)/12
now drawn 9 lit filled with B
so B now contain B=5(x-9)+9/12...........
now the new ratio of mixer is A:B=7:9
we know the new value of A and B...........put that in this equation
7(x-9)/12:(5(x-9)+9)=7:9-------if u calculate further u ll get the X value as 36
Put that x value in A=7x/12, so here you can get answer of A contain initially.
Sree said:
2 decades ago
Hi Debjyoti,
Can you give the explination how you got x=36 using below equation?
7(x-9)/12:(5(x-9)+9) = 7:9
Can you give the explination how you got x=36 using below equation?
7(x-9)/12:(5(x-9)+9) = 7:9
Karthikeyini said:
2 decades ago
(7x-(21/4))/((5x-(15/4))+9)
In this term how 9 occured at the end ? Can anyone explain this ?
In this term how 9 occured at the end ? Can anyone explain this ?
Prithvi chauhan said:
2 decades ago
Please can you explain in any other simple way?
Gowtham said:
1 decade ago
Hi guys,
Find the pure liqued of A in initial
Assume pure liqued x,
Quantity of a measured left formula=(pure A liquid - mix liquid)
Pure liquid of A=7x
Mix liquid of A=(7/12)*9
here 9 is also mixture
So, quantity of a measured left A=(7x-(7/12)*9)=7x-21/4
the same rule
quantity of a measured left B=(5x-(5/12)*9)=5x-15/4
(7x-(21/4)/5x-(15/4)+9)=7/9
Given here that 9 is add with b
ans x=3
ratio of 7
So=7*3=21 that all
Find the pure liqued of A in initial
Assume pure liqued x,
Quantity of a measured left formula=(pure A liquid - mix liquid)
Pure liquid of A=7x
Mix liquid of A=(7/12)*9
here 9 is also mixture
So, quantity of a measured left A=(7x-(7/12)*9)=7x-21/4
the same rule
quantity of a measured left B=(5x-(5/12)*9)=5x-15/4
(7x-(21/4)/5x-(15/4)+9)=7/9
Given here that 9 is add with b
ans x=3
ratio of 7
So=7*3=21 that all
San said:
1 decade ago
Is there any easy way to do it ?
Preethi said:
1 decade ago
Any simple method please.
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