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Mizan42
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Container A, B and C contain water, milk and acid in equal quantities 09 Dec 2023, 19:26
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95% (hard)

Question Stats:

55% (03:18) correct 45% (03:42) wrong based on 85 sessions

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Container A, B and C contain water, milk and acid in equal quantities. 25% of content A is taken out and poured into B. Then, same percentage from B is transferred to C. Finally, 25% from C is transferred to A. The ratio of acid in contain A is:

a) 13/1125
b)55/2357
c)16/69
d)18/77
e)25/76
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C
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gmatophobia
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Container A, B and C contain water, milk and acid in equal quantities 09 Dec 2023, 23:24
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Mizan42
Container A, B and C contain water, milk and acid in equal quantities. 25% of content A is taken out and poured into B. Then, same percentage from B is transferred to C. Finally, 25% from C is transferred to A. The ratio of acid in contain A is:

a) 13/1125
b)55/2357
c)16/69
d)18/77
e)25/76
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Container A, B and C contain water, milk and acid in equal quantities.

Let's assume each container has 256 units of liquid as shown below

Attachment:
1.png
1.png [ 37.89 KiB | Viewed 5182 times ]

25% of content A is taken out and poured into B.

25% of 256 is 64. Hence, 64 units of water are transferred from A to B. The amount of liquid in each of the containers becomes as shown below -

Attachment:
2.png
2.png [ 39.98 KiB | Viewed 5147 times ]

Then, same percentage from B is transferred to C

Using the same logic in the previous step we can determine the volume of acid, milk and water in container C after the transfer.

Attachment:
3.png
3.png [ 44.15 KiB | Viewed 5076 times ]

Finally, 25% from C is transferred to A.

After the last transfer, the volume of milk, acid and water in A is shown below

Attachment:
4.png
4.png [ 48.89 KiB | Viewed 4990 times ]

After the process -

The Volume of acid in A = 64
The Volume of water in A = 256 - 64 + 4 = 196
The Volume of milk in A = 16

Required Ratio = \(\frac{64}{(64+196+16)} = \frac{64}{276} = \frac{16}{69}\)

Option C
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Container A, B and C contain water, milk and acid in equal quantities 11 Dec 2023, 06:27
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gmatophobia
Mizan42
Container A, B and C contain water, milk and acid in equal quantities. 25% of content A is taken out and poured into B. Then, same percentage from B is transferred to C. Finally, 25% from C is transferred to A. The ratio of acid in contain A is:

a) 13/1125
b)55/2357
c)16/69
d)18/77
e)25/76

Container A, B and C contain water, milk and acid in equal quantities.

Let's assume each container has 256 units of liquid as shown below

Attachment:
1.png

25% of content A is taken out and poured into B.

25% of 256 is 64. Hence, 64 units of water are transferred from A to B. The amount of liquid in each of the containers becomes as shown below -

Attachment:
2.png

Then, same percentage from B is transferred to C

Using the same logic in the previous step we can determine the volume of acid, milk and water in container C after the transfer.

Attachment:
3.png

Finally, 25% from C is transferred to A.

After the last transfer, the volume of milk, acid and water in A is shown below

Attachment:
4.png

After the process -

The Volume of acid in A = 64
The Volume of water in A = 256 - 64 + 4 = 196
The Volume of milk in A = 16

Required Ratio = \(\frac{64}{(64+196+16)} = \frac{64}{276} = \frac{16}{69}\)

Option C
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How do we know what number to assume in these type of sums?
Is it based on the number of steps?
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Ankur17000
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Container A, B and C contain water, milk and acid in equal quantities 16 Dec 2023, 03:40
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Guntabulla
gmatophobia
Mizan42
Container A, B and C contain water, milk and acid in equal quantities. 25% of content A is taken out and poured into B. Then, same percentage from B is transferred to C. Finally, 25% from C is transferred to A. The ratio of acid in contain A is:

a) 13/1125
b)55/2357
c)16/69
d)18/77
e)25/76

Container A, B and C contain water, milk and acid in equal quantities.

Let's assume each container has 256 units of liquid as shown below

Attachment:
1.png

25% of content A is taken out and poured into B.

25% of 256 is 64. Hence, 64 units of water are transferred from A to B. The amount of liquid in each of the containers becomes as shown below -

Attachment:
2.png

Then, same percentage from B is transferred to C

Using the same logic in the previous step we can determine the volume of acid, milk and water in container C after the transfer.

Attachment:
3.png

Finally, 25% from C is transferred to A.

After the last transfer, the volume of milk, acid and water in A is shown below

Attachment:
4.png

After the process -

The Volume of acid in A = 64
The Volume of water in A = 256 - 64 + 4 = 196
The Volume of milk in A = 16

Required Ratio = \(\frac{64}{(64+196+16)} = \frac{64}{276} = \frac{16}{69}\)

Option C

How do we know what number to assume in these type of sums?
Is it based on the number of steps?
Show more


For this one, you need to find a number that can be divided by 4 --> 3 times
Due to the 25% or 1/4, and the replacement happening a total number of 3 times. A-->B--->C--->A (no. of arrows)
That number has to be 4X4X4 = 64

Any further multiple of 4 can be used in this succession, 64*4 = 256 and so on.
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Container A, B and C contain water, milk and acid in equal quantities 15 Jun 2024, 22:59
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25% of each quantitity is being poured into the other almost three time.

So, let's take each quantity to be 256 litres for ease of calculation.


Initial Stage:

Mixture A ; Mixture B ; Mixture C
Water - 256 litres ;  Milk - 256 litres ; Acid - 256.  litres

Round 1 -

25% of Mixture A poured to Mixture B.
25% of 256 litres = 64 litres.

Mixture A ; Mixture B ; Mixture C
Water - 192 litres ;  Water - 64 litres + Milk 256 litres; Acid - 256 litres.

Round 2  -

25% of Mixture B poured to Acid.
25% of (256+64) litres = 80 litres of Mixture B.

80litres of Mixture B has 25% of 64 litres of water and 25% of 256litres of Milk. (80 litres of Mixture B = 16litres of water + 64litres of milk)

Mixture A ; Mixture B ; Mixture C
Water - 192 litres ;  Water - 48 litres + Milk 192 litres; Acid - 256 litres + 16litres of water + 64litres of milk.

Round 3 -

25% of Mixture C poured to Water.
25% of (256+16+64) litres = 84 litres of Mixture C.

84 litres of Mixture C has 25% of 256 litres of acid 25% of 64 litres of milk and 25% of 16 litres of water. (84 litres of Mixture C = 64litres of acid + 16litres of milk + 4litres of water)

Mixture A ; Mixture B ; Mixture C
Water - 192 litres + 4 litres, Milk - 16litrs, Acid - 64litres ;  Mixture B - Unnecessary ; Mixture C - Unnecessary


Ratio of Acid in Container/ Mixture A =  64/ 196+16+64 = 64/276 = 16/69
 
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Container A, B and C contain water, milk and acid in equal quantities 01 Nov 2025, 21:45
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