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Updated on: 18 Dec 2023, 23:38
Originally posted by Zareensoobuh on 18 Dec 2023, 15:22.
Last edited by Bunuel on 18 Dec 2023, 23:38, edited 1 time in total.
Last edited by Bunuel on 18 Dec 2023, 23:38, edited 1 time in total.
Renamed the topic.
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Difficulty:
55%
(hard)
Question Stats:
56% (02:23) correct
44%
(01:32)
wrong
based on 9
sessions
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From a container 6 litres milk was drawn out and was replaced by water. Again 6 litres of mixture was drawn out and replaced by water. Thus the quantity of milk in the container after these two operations is 9: 16. Calculate the quantity of the milk initially in the container:
A. 12
B. 15
C. 9
D. 6
E. 10
A. 12
B. 15
C. 9
D. 6
E. 10
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23 Dec 2023, 22:18
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Zareensoobuh
Faulty question.
If you are comparing quantity of milk after each operation, the quantity would go down further after second time. So, it cannot be 9:16
It should be water:milk or milk:water after second operation is 9:16.
Take milk:water = 9:16
Let x be the initial quantity.
I: Milk = \(x-6\) and water = \(6\)
II: Milk = \((x-6)-\frac{x-6}{x}*6\) and water = \(6+6-\frac{6}{x}*6\)
Ratio = \(\frac{x^2-6x-6x-36}{x}:\frac{12x-36}{x} = 9:16\)
\((x^2-12x-36)16 = 9(12x-36)=108(x-3).......4(x^2-12x-36)=27(x-3)......4x^2-75x+225=0\)
\(4x^2-60x-15x+225=0........4x(x-15)-15(x-15)=0.......(4x-15)(x-15)=0\)
Thus x is 15/4 or 15. But x cannot be less than 6, so x=15
B
24 Dec 2023, 11:39
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Final milk : Total mixture = 9 : (9+16) = 9 : 25
Here, Initial milk/mix. = Final mixture
So, Initial milk : Final milk = 25 : 9
Two times replacement, so taking square root ⇒ √25 : √9 = 5 : 3
That means, each time (5-3)/5 i.e. 2/5 part is replaced.
Now, 2 part ≡ 6 L
∴ 5 part ≡ 15 L [Ans.]
Here, Initial milk/mix. = Final mixture
So, Initial milk : Final milk = 25 : 9
Two times replacement, so taking square root ⇒ √25 : √9 = 5 : 3
That means, each time (5-3)/5 i.e. 2/5 part is replaced.
Now, 2 part ≡ 6 L
∴ 5 part ≡ 15 L [Ans.]
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
