Consider three mixtures - the first having water and liquid A in the r

Question

Consider three mixtures - the first having water and liquid A in the ratio 1 : 2, the second having water and liquid B in the ratio 1 : 3, and the third having water and liquid C in the ratio 1 : 4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4 : 3 : 2. Then the resulting mixture has

A. The same amount of water and liquid B
B. The same amount of liquids B and C
C. More water than liquid B
D. More water than liquid A
E. Less water than liquid B

A. The same amount of water and liquid B
B. The same amount of liquids B and C
C. More water than liquid B
D. More water than liquid A
E. Less water than liquid B

Are You Up For the Challenge: 700 Level Questions

CareerGeek · Accepted Answer

Mixtures A, B, & C are mixed in the ratio 4 : 3 : 2
--> Let the volumes = 4k, 3k, 2k respectively

Mixtures ---- Ratio ---- Water ------ Liquid
Mixture A --- 1 : 2 ---- 4k(1/3) ---- 4k(2/3)
Mixture B --- 1 : 3 ---- 3k(1/4) ---- 3k(3/4)
Mixture C --- 1 : 4 ---- 2k(1/5) ---- 2k(4/5)

Assume k = 60  
Volume of Liquid A = 8k/3 = 160
Volume of Liquid B = 9k/4 = 135
Volume of Liquid C = 8k/5 = 96
Volume of water = 4k/3 + 3k/4 + 2k/5 = 80 + 45 + 24 = 149

--> There is more water than Liquid B

Option C

axeller · Answer

Took longer than the allocated 2 mins so I hope someone could provide a more elegant way of answering.

Water in Mixture A - 1/3
Water in Mixture B - 1/4 (25% water)
Water in Mixture C - 1/5

ratio of mixed in the proportion 4 : 3 : 2 
Water in Liquid A - (1/3)(4/9) = 4/27
Water in Liquid B - (1/4)(3/9) = 3/36
Water in Liquid C - (1/5)(2/9) = 2/45

Sum of all 3 -> (4/27+3/36+2/45) * 100% = approx 28%

which is > water in Mixture B

Phoenix19 · Answer

Aim here is to arrive at some common parameters. 
 Step 1  W:A = 1:2 , W:B = 1:3, W:C = 1:4

Step 2 Take LCM of 3, 4, & 5 = 60 ....which becomes our qty of soln

Step 3 Now convert ratio into absolute terms ie (Soln A : W 20, A 40) , (Soln B :W 15, B 45) , (Soln C: W 12, C 48)

Step 4 Multiply the absolute amount with desired ratio ie 4:3:2 So Water W (80+45+24), Soln B (45*3 = 135) and hence option C

I hope this is helpful IMO!!

youcouldsailaway · Answer

Since water is a common element amongst all the three old mixtures and is also the an important element in the new mixtures, let's pick water to make estimations.

Mixture .......  Water component .......  Mixture share in new proportion ....... Final Share of water, per mixture
 A...............\frac{1}{3}............................\frac{4}{9}....................\frac{1}{3} * \frac{4}{9} = \frac{4}{27} 
 B...............\frac{1}{4}............................\frac{3}{9}....................\frac{1}{4} * \frac{3}{9} = \frac{3}{36} 
 C...............\frac{1}{5}............................\frac{2}{9}....................\frac{1}{5} * \frac{2}{9} = \frac{2}{45}

Now, let's equalize the denominators by taking LCM. 
LCM (27, 36, 45) = 540

Thus, final share of water, per mixture is as follows:

Mixture.........Final Share of water, per mixture
A..................\frac{80}{540}
B..................\frac{45}{540}
C..................\frac{24}{540}

Total Water =  (\frac{80}{540}+\frac{45}{540}+\frac{24}{540}) * 540 = 149

Now, we need to find the share of LIQUID in each of the mixtures, as each mixture is comprised of a liquid and water.

Thus, 
Share of liquid from Mixture A =  \frac{4}{9} * 540 * \frac{2}{3} = 160 
Share of liquid from Mixture B =  \frac{3}{9} * 540 * \frac{3}{4} = 135 
Share of liquid from Mixture C =  \frac{2}{9} * 540 * \frac{4}{5} = 96

Thus, answer C is correct.

deletus · Answer

I took a bizarre route, so let me know if there are errors in my way.

W:A = 1:2
W:B = 1:3
W:C = 1:4

to get to A:B:C = 4:3:2, we need 1 W:C, 2 W:B, and 4 W:A to get 8:6:4 = 4:3:2 
this also means that we have W:A:B:C = 7:8:6:4 -> more water than B or C

Option C: W > B

pappal · Answer

mixture 
 water : liquid  
 liquid type 
 mixed in volume(let) 
 
  I 
 1:2 
 A 
 4x 
 
  II 
 1:3 
 B 
 3x 
 
  III 
 1:4 
 C 
 2x

volume of water in final mixture={1/3*4x+ 1/4*3x+1/5*2x} / 9x
 =(80+45+24)/60*9=149/540 which is more than 1/4(volume of water in liquid B) and less than 1/3(volume of water in liquid A)
hence C

Aboyhasnoname · Answer

Mixture 
 Water  
 Liquid A/B/C

A 
 1 
 2

B 
 1 
 3

C 
 1 
 4

Now Taking the mixtures in the ratio 4:3:2 ..Le

Mixture 
 Water  
 Liquid A/B/C 
 Water  
 Liquid A/B/C 
 
  A 
 1 X 4  
 2 X 4  
 4  
 8 
 
  B 
 1 X 3  
 3 X 3  
 3 
 9 
 
  C 
 1 X 2 
 4 X 2 
 2 
 8 
 
  TOTAL

9

So My Water: Liquid A: Liquid B: Liquid C Becomes 9:8:9:8

Where is it that I am wrong? Please Help

Bunuel

salonijain684 · Answer

Mannisha  could you please help me out with this problem?

Consider three mixtures - the first having water and liquid A in the r

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mixture	water : liquid	liquid type	mixed in volume(let)
I	1:2	A	4x
II	1:3	B	3x
III	1:4	C	2x

Mixture	Water	Liquid A/B/C	Water	Liquid A/B/C
A	1 X 4	2 X 4	4	8
B	1 X 3	3 X 3	3	9
C	1 X 2	4 X 2	2	8
TOTAL			9

GMAT Problem Solving (PS) Questions

Consider three mixtures - the first having water and liquid A in the r

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