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A chemist starts with two equally sized beakers, labeled X and Y, that

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Sajjad1994
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A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   Updated on: 30 Sep 2023, 13:26
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Question Stats:

73% (02:46) correct 27% (02:50) wrong based on 73 sessions
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A chemist starts with two equally sized beakers, labeled X and Y, that each contains a different amount of water. X is x% full and Y is y% full. The chemist then pours water from beaker Y until beaker X is completely full and Y is now 2/5 full. If beaker Y initially contained more water than did beaker X, which of the following could be the values of x and y?
x y
55
60
65
70
75
90
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Originally posted by Sajjad1994 on 03 Mar 2022, 08:00.
Last edited by BottomJee on 30 Sep 2023, 13:26, edited 1 time in total.
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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   04 Mar 2022, 04:46
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My answer is:
X 65
Y 75

Let N be the percentage of water transferred from Y to X. We have:
X+N=100
Y-N=40
X<Y

So: X+Y=140
X<Y

The only combination that satisfies is
X=65
Y=75
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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   03 Mar 2022, 11:07
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Let's take volume of X = Y = 1

X = \(\frac{x}{100}\) full

Y = \(\frac{y}{100}\) full

\(\frac{x}{100}\) > \(\frac{y}{100}\)

Let \(\frac{a}{100}\) = amount of water transferred from Y to X

\(\frac{x}{100}\) + \(\frac{a}{100}\) = 1

\(\frac{(x+a)}{100 }\)= 1

x+a = 100 ------- (1)

\(\frac{y}{100}\) - \(\frac{a}{100}\) = \(\frac{2}{5}\)

\(\frac{(y-a)}{100}\) = \(\frac{2}{5}\)

y-a = \(\frac{2}{5}\) * 100

y-a = 40 ------- (2)

(1) + (2) => (x+a) + (y-a) = 140
x+y = 140 --------- (3)

It is already mentioned in the question that Y initially contained more water than did beaker X.
So, going through the options, Y = 75 & X = 65 should satisfy the equation 3.

Answer X = 65 & Y = 75
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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   10 Mar 2022, 13:26
(100 - x) = y - (2y/5)
100 - x = 3y/5

Y = 75
X = 55
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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   11 Mar 2022, 09:55
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Competition mode is off now.

OA is:

x=65
y=75
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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink] New post   23 Mar 2024, 07:33
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Let us take the capacity of each beaker as 100 ml

water in beaker in X and Y = x ml and y ml respectively ( x% amd y%)

After transfer X is full = 100, Y is 2/5 full = 40ml

Hence total = 100+40= 140ml

or x+y = 140

Only (65, 75) satisfies the above relation. 


(PS : Follow this checking approach:  x cannot be 55 , coz corresponding y should be 85 )
Sajjad1994 wrote:
A chemist starts with two equally sized beakers, labeled X and Y, that each contains a different amount of water. X is x% full and Y is y% full. The chemist then pours water from beaker Y until beaker X is completely full and Y is now 2/5 full. If beaker Y initially contained more water than did beaker X, which of the following could be the values of x and y?

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Re: A chemist starts with two equally sized beakers, labeled X and Y, that [#permalink]
23 Mar 2024, 07:33
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