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Updated on: 30 Sep 2023, 13:26
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.
Last edited by BottomJee on 30 Sep 2023, 13:26, edited 1 time in total.
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x: 65
y: 75
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:49) correct
31%
(02:58)
wrong
based on 724
sessions
History
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Not Attempted Yet
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 |
ShowHide Answer
Official Answer
x: 65
y: 75
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
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
General Discussion
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
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
