A can contains a mixture of two liquids A and B in the ratio 7:5 when

Question

A can contains a mixture of two liquids A and B in 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?

A. 45 litres

B. 36 litres

C. 28 litres

D. 25 litres

E. 21 litres

A. 45 litres

B. 36 litres

C. 28 litres

D. 25 litres

E. 21 litres

A can contains a mixture of 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 liter of liquid A was contained by the can initially?

A. 10
B. 20
C. 21
D. 25
E. 27

A. 10
B. 20
C. 21
D. 25
E. 27

KarishmaB · Accepted Answer

First of all, the wording of the question is a bit unclear. It needs to specify that 9 litres were removed and 9 litres was added back. "can is filled" is not sufficient because we do not know whether the can was full initially. Say initial volume of mix is V1. After you remove 9 litres, the concentration of liquid A remains 7/12 and volume of the mix is (V1 - 9). After you add back liquid B, the concentration of A becomes 7/16 and volume comes back to V1. Ci*Vi = Cf*Vf (7/12)*(V1 - 9) = (7/16)*V1 (V1 - 9)/V1 = 3/4 9 accounts for the difference of 1 on ratio scale so Initial volume = V1 = 4*9 = 36 litres. 7/12 of the initial mixture was liquid A so liquid A was (7/12)*36 = 21 litres. Check this post for more on replacement: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... -mixtures/

LMP · Answer

Suppose the can initially contain 7x and 5x litres of mixtures A and B respectively. The quantity of A in the mixture left =   litres =   litres.

The quantity of B in mixture left =   litres =   litres.

Therefore we can write: the ratio of the two quantities as  /   = 7/9

In other words, we may say that 252x – 189 = 140x + 147

Hence, x = 3. Therefore the can had 7(3) = 21 litres of the quantity A. Thus the correct option is C.

Or 
WAY 2 : 
let A = 7x & B = 5x
total mixture = 7x+5x=12x

after 9 litre of mixture are drawn off remaining misture = 12x-9

after filling can with B:
A = (12x-9) * 7/12= 28x-21
B = {(12x-9)* 5/12}+9= 20x+21

So, (28x-21)/(20x+21)= 7/9
by solving above equation, we get
x=3

So liquid A was contained by the can initially = 7x = 7*3 = 21

ENGRTOMBA2018 · Answer

Using the values here might be the most straightforward way for this question.

As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total

If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4

Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4

Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.

Hence E is the correct answer.

FYI, the algebraic equations will become:

A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.

Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) = 7*3 = 21.

Hope this helps.

gracie · Answer

let x=total liters in can
5x/12-(9)(5/12)+9=9x/16
x=36
(7/12)(36)=21 liters of A in can initially

sanabirm · Answer

7x/12 - 9*7/12 = x*7/16
Solving, x=36
A = 7x/12 = 7*36/12 = 21

Method 2 ROA
5/12 - - - - 1
----9/17----
 3 : 1
1 =9
4=36
A = 36*7/12 = 21

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CrackverbalGMAT · Answer

To solve removal and replacement questions like the one above, we can use the following equation.

\frac{Final}{Initial} = (1 - \frac{b}{a})^n

Final = The Final Quantity of that component who's   concentration is decreasing  .

Initial = The Initial Quantity of that component who's   concentration is decreasing  .

b = Amount of liquid replaced

a = Final Volume in the container   after the replacement   of quantity b.

n = number of times the operation is done

Let the initial volume = x

b = 9

Since, we are removing and replacing the same amounts i.e 9 liters, therefore a = x

Initial quantity = 7x/12

Final Quantity = 7x/16

n = 1

Substituting in the equation above, we get  \frac{\frac{7x}{16}}{\frac{7x}{12}} = (1 - \frac{9}{x})^1 
 
 \frac{12}{16} = 1 - \frac{9}{x}

\frac{9}{x} = 1 - \frac{3}{4} = \frac{1}{4}

x = 36

Therefore Initial amount = 7x/12 = (7 * 36) / 12 = 21 L

Option E

Arun Kumar

SangamJena · Answer

Using the formula will help us in arriving the answer C1* V1 = C2 * V2

Initial Volume = V1
9lts removed Volume now = V1-9
Concentration of A is 7/12

Now, B is added Volume is V1 again.

Hence 
 7/12 (V1-9) = 7/16 *V1

V1=36
Liquid A = 7/12*36 = 21

ScottTargetTestPrep · Answer

Solution:

We can let 7k and 5k be the number of liters of liquids A and B, respectively. When 9 liters of the mixture is drawn, 7/12 x 9 = 7/4 x 3 = 21/4 liters are liquid A and 5/12 x 9 = 5/4 x 3 = 15/4 liters are liquid B. We can create the following equation for the ratio of the two liquids after the removal:

(7k - 21/4) / (5k - 15/4 + 9) = 7/9

(28k - 21) / (20k - 15 + 36) = 7/9

7(20k + 21) = 9(28k - 21)

140k + 147 = 252k - 189

336 = 112k

3 = k

Initially, there were 7(3) = 21 liters of liquid A.

Answer: E

Shadyshades · Answer

Conclusion= Solution(A&B) & B are getting mixed.

Let x be the total volume of the solution 
A in solution = 7/12 x ----------Refer 1 
9 L of mixture is drawn off so in 9 L how much is A ? 
A= 7/12 * 9 = 63/12

till now the equation is 7/12 x - 63/12

now the same quantity is filled up with B in the solution. What is the quantity of A in pure B ? It's 0.
So now the equation becomes 7/12 x - 63/12 + 0
How much is A in the resultant mixture ? 
7/16 x ( remember the total quantity is x)

Therefore, 7/12 x - 63/12 + 0 = 7/16 x
x= 36 
How much is A in 36 ? 
7/12 x -----Refer (1)
7/12 * 36
= 21 L
 Option E

hadimadi · Answer

Hi,

let x be volume of the mixture. We know that A=(7/12)x and B=(5/12)x. Now according to the text, what happens is:

((7/12)x-(7/12)*9) / ((5/12)x-(5/12)*9+9)=A/B

Solve for x to get A=21L.

bv8562 · Answer

Initial \ Ratio: \frac{A}{B} = \frac{7}{5}

Initial \ Quantity \ of \ A = \frac{7}{12}

Final \ Ratio: \frac{A}{B} = \frac{7}{9}

Final \ Quantity \ of \ A = \frac{7}{16}

Formula: \ F = I*(1-\frac{y}{t})^n

F = Final Quantity
I = Initial Quantity
y = Amount being replaced
t = Total Quantity
n = No. of times it is repeated

F =  \frac{7}{16}  
I =  \frac{7}{12} 
y = 9 litres
t = 12x
n = 1

\frac{7}{16} = \frac{7}{12}*(1-\frac{9}{12x})^1

\frac{3}{4} = \frac{12x-9}{12x}

12x - 9 = 9x 
 3x = 9 
 x = 3

Thus, \ initial \ quantity \ of \ liquid \ A \ was = 7x = 7*3 = 21 (E)

YuktiMaheshwari · Answer

Let 7x and 5x be number of litres of liquid of A and B respectively:

A removed 7/12*9 = 21/4
B removed 5/12*9 = 154
B added - 9

(7x - 21/4) / (5x - 15/4 + 9) = 7/9

(28x - 21) / (20x - 15 + 36) = 7/9

7(20x + 21) = 9(28x - 21)

140x + 147 = 252x - 189

336 = 112x

3 = x

So is 7*3 = 21 (E)

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Dereno · Answer

Let the can contain mixtures in ratio A:B = 7: 5

When 9 litres are removed once. The ratio A:B = 7: 9.

Initial liters of A ???

Final concentration = Initial concentration ( 1- (fraction removed/ total capacity)) ^n

let’s take for liquid A, final conc : total = 7:16

initial conc of liquid A , final conc : total = 7/12

n = number of times the fraction is removed. Or the process is carried out . Here n =1

7/16 = 7/12 * (1-(9/capacity) )

12/16 = (1-(9/capacity) )

3/4 = (1-(9/capacity) )

1-3/4 = 9/ capacity

capacity = 36

This implies 7x:5x = 12x units =36. X =3

Capacity of A = 7*3 = 21 litres

A can contains a mixture of two liquids A and B in the ratio 7:5 when

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A can contains a mixture of two liquids A and B in the ratio 7:5 when

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