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

Question

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?

zhanbo · Accepted Answer

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

Sr1994 · Answer

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

yoursimran · Answer

(100 - x) = y - (2y/5)
100 - x = 3y/5

Y = 75
X = 55

GMATBusters · Answer

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 )

Advait01 · Answer

I think
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

egmat · Answer

Let's break this down step by step.

Understanding the Setup: 
- Two beakers of equal size. Beaker X starts  x%  full, Beaker Y starts  y%  full.
- Water is poured from Y into X until X is completely full ( 100% ).
- After pouring, Y is  2/5  =  40%  full.
-  Key constraint:  y > x (Y initially had MORE water than X).

Step 1: How much water was poured? 
X went from  x%  to  100% , so the amount poured INTO X is  ( 100  - x)% .

Step 2: Set up the equation for Y. 
Y started at  y%  and lost  (100 - x)% , ending at  40% .

y - (100 - x) = 40 
y -  100  + x =  40 
  x + y = 140

So the two unknowns must add up to  140 .

Step 3: Check which pair from the choices sums to  140 . 
-  55  +  85  →  85  is not a choice
-  60  +  80  →  80  is not a choice
-  65  +  75  → Both are choices! Sum =  140 . Check:  75  >  65 , so y > x is satisfied.
-  70  +  70  → Sum =  140 , but this means y = x, which  violates y > x .
-  50  +  90  →  50  is not a choice

The only valid pair is x =  65  and y =  75 .

Quick verification:  X starts at  65% , needs  35%  more to be full. Y starts at  75% , gives away  35% , and is left with  75  -  35  =  40%  =  2/5 . Y initially had more water ( 75%  >  65% ). Everything checks out.

Answer: x = 65 (Row 3, Column A) and y = 75 (Row 5, Column B)

Adit_ · Answer

Simplest approach:
Take 100ml.
2/5*100 = 40
What is the final result after the pouring?
X = 100, Y=40
Total = 140
Find x+y=140.
Answer:  65,75

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

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