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Manager  G
Joined: 13 Aug 2018
Posts: 50
A bucket contains a mixture of two liquids A & B  [#permalink]

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15 00:00

Difficulty:   95% (hard)

Question Stats: 49% (02:35) correct 51% (02:39) wrong based on 152 sessions

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A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device
Math Expert V
Joined: 02 Aug 2009
Posts: 8199
A bucket contains a mixture of two liquids A & B  [#permalink]

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jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

Responding to a PM

I would think of the problem in two ways..

(I) logical..
Now 16 litres of MIXTURE is removed, so A removed is $$16*\frac{5}{8}=10$$...
This removal of 10 litres leads to 2 units of A going down, so 2 units =10
1 unit =10/2=5...
B is 3 units ( from 5:3) , so 3*5=15

(II) Algebraic
Let the initial ratio be 5x:3x....
Removal of 16 in same ratio MEANS 10:6..
So A becomes = 5x-10 and B becomes =3x-6
When we add 16 litres to B ...
$$\frac{5x-10}{3x-6+16}=\frac{3}{5}$$..
$$25x-50=9x+30.......16x=80......x=5$$
So 3x=3*5=15

B
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Intern  B
Joined: 19 Sep 2017
Posts: 2
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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I didn't find any clearer explanation, thats why I am just sharing my approach. Hope this helps.

Initially the ratio is $$\frac{5x}{3x}$$. We know that 16 litres is removed from TOTAL MIXTURE, so the removal is in the same ratio. After removal of the total 16 litres, the removal will be in the ratio of 5:3, which will result is 10 ltrs from A and 6 ltrs from B.

So, $$\frac{(5x-10)}{(3x-6)}$$. Further, 16 litres is added to B.

$$\frac{(5x-10)}{(3x-6+16)} = \frac{(5x-10)}{(3x+10)}$$ this ratio now becomes $$\frac{3}{5}$$. hence solving for $$\frac{(5x-10)}{(3x+10)}$$= $$\frac{3}{5}$$, we get x= 5.

now using $$x=5$$, we know the initial Value of $$\frac{A}{B}$$= $$\frac{5x}{3x}$$ is $$\frac{25}{15}$$.
Hence B is 15.

Please Give kudos if this helped.
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Manager  P
Joined: 01 Mar 2015
Posts: 68
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

Here the fraction A:B changes from 5:3 to 3:5 when the 16 litres of A is replaced by 16 litres of B

If we take solution to be 5x + 3x = 8x
Initially
A = 5x
B = 3x

As total solution remains same, 16 litres is removed and 16 litres is added. Therefore final in solution

A' = 3x
B' = 5x

A' = A - 16
=> 5x - 16 = 3x
=> 2x = 16
=> x = 8 litres

B = 3x = 3*8 = 24 litres

So, initially 24 litres of liquid B was there in the bucket.

Manager  G
Joined: 13 Aug 2018
Posts: 50
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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v12345 wrote:
jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

Here the fraction A:B changes from 5:3 to 3:5 when the 16 litres of A is replaced by 16 litres of B

If we take solution to be 5x + 3x = 8x
Initially
A = 5x
B = 3x

As total solution remains same, 16 litres is removed and 16 litres is added. Therefore final in solution

A' = 3x
B' = 5x

A' = A - 16
=> 5x - 16 = 3x
=> 2x = 16
=> x = 8 litres

B = 3x = 3*8 = 24 litres

So, initially 24 litres of liquid B was there in the bucket.

if initially B is 24 A must be 40.
after removal B should be 18 and A 30 .
after adding 16 litre of B , B should be 32 and A 32 which isn't a 3:5 ratio .
VP  P
Joined: 07 Dec 2014
Posts: 1221
A bucket contains a mixture of two liquids A & B  [#permalink]

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jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

let x=total original mixture
3/8*(x-16)+16=5x/8
x=40
3/8*40=15
B
Senior Manager  D
Joined: 18 Jun 2018
Posts: 257
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

OA:B

Initial Amount of A: $$5x$$
Initial Amount of B: $$3x$$
Total mixture amount: A+B $$= 5x+3x = 8x$$

$$16$$ Litre of this mixture is taken out

Amount of A left :$$\frac{5}{8}*(8x-16)=5x-10$$

Amount of B left :$$\frac{3}{8}*(8x-16)=3x-6$$

Final Amount of A :$$\frac{5}{8}*(8x-16)=5x-10$$

Final Amount of B :$$\frac{3}{8}*(8x-16)+16=3x+10$$

According to the question,

$$\frac{5x-10}{3x+10} =\frac{3}{5}$$

$$25x-50=9x+30$$

$$16x=80$$

$$x=\frac{80}{16}=5$$

Initial Amount of B $$= 3x = 3*5 = 15$$ Litres
Founder  V
Joined: 04 Dec 2002
Posts: 18387
Location: United States (WA)
GMAT 1: 750 Q49 V42 GPA: 3.5
A bucket contains a mixture of two liquids A & B  [#permalink]

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chetan2u - what do you think of this question?

(testing notifications - pardon for the interruption)
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 8199
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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bb wrote:
chetan2u - what do you think of this question?

(testing notifications - pardon for the interruption)

BB,

no notifications in the mails. I just checked the notifications on site and saw this...
_________________
SVP  V
Joined: 26 Mar 2013
Posts: 2344
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

Posted from my mobile device

Dear GMATGuruNY

Can you share your thoughts in this problem?
Senior Manager  G
Joined: 04 Aug 2010
Posts: 493
Schools: Dartmouth College
A bucket contains a mixture of two liquids A & B  [#permalink]

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1
jackfr2 wrote:
A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

(A) 12 litres
(B) 15 litres
(C) 18 litres
(D) 6 litres
(E) 24 litres

We can use ALLIGATION.
Let:
S = the original solution
B = the 16 liters of pure B
M = the final mixture
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS with the same denominator.
S --> Since A:B = 5:3, $$\frac{B}{total} = \frac{3}{8}$$
B --> $$\frac{B}{total} = \frac{{16}}{{16}} = \frac{8}{8}$$
M --> Since A:B = 3:5, $$\frac{B}{total} = \frac{5}{8}$$

Step 2: Plot the 3 numerators on a number line, with the numerators for S and B on the ends and the numerator for the mixture in the middle.
S 3------------5-----------8 B

Step 3: Calculate the distances between the numerators.
S 3-----2-----5-----3-----8 B

Step 4: Determine the ratio in the mixture.
The ratio of S to B is equal to the RECIPROCAL of the distances in red.
S:B = 3:2 = 24:16.

The ratio in blue indicates that the mixture is composed of 24 liters of original solution and 16 liters of pure B, implying that the total volume in the bucket = 40 liters.
Since B constitutes $$\frac{3}{8}$$ of the original 40 liters in the bucket, we get:
$$\frac{3}{8} * 40 = 15$$ liters

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Intern  S
Joined: 05 Jan 2017
Posts: 34
Re: A bucket contains a mixture of two liquids A & B  [#permalink]

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by weighted avg approach:
B initial- 3/8
B after - 5/8

applying in the formula- (5-8)/(8-3)
=> 3/5

hence ratio a:b = 3:2
16% of sol was replaced. hence a:b can be 24:16
total sol = 40
after replace B= 3/8 * 40 = 15 (B)
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Kudos if my post helps! Re: A bucket contains a mixture of two liquids A & B   [#permalink] 22 Oct 2018, 04:29
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