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Updated on: 19 Apr 2023, 23:55
Originally posted by ChandlerBong on 19 Apr 2023, 20:53.
Last edited by Bunuel on 19 Apr 2023, 23:55, edited 1 time in total.
Last edited by Bunuel on 19 Apr 2023, 23:55, edited 1 time in total.
Renamed the topic and edited the question.
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91%
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If x and y are non-zero integers, is \(\frac{x}{20}\)+ \(\frac{y}{40} = 50\)?
(1) x – 2y = 1000
(2) 2x – y = 2300
(1) x – 2y = 1000
(2) 2x – y = 2300
19 Apr 2023, 23:15
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ChandlerBong
Let's simplify the question first.
\(\frac{x}{20}\)+ \(\frac{y}{40}\) = 50
Taking 40 as the LCM on the RHS of the equation we get
\(\frac{2x + y}{40} = 50\)
Multiplying by 40 on both sides of the equation, we get
\(2x + y = 2000\)
Target question: Is 2x + y = 2000? (Given: x and y are non-zero integers)
Statement 1
(A) x – 2y = 1000
Let's find the value of x, y considering x – 2y = 1000 & 2x + y = 2000 both hold true.
2x - 4y = 2000
2x + y = 2000
Subtracting both equations we get -
-5y = 0
y = 0
Both equations hold true if y = 0. However, from the constraints given, we are told that \(y \neq 0\). Hence, we can conclude that given x – 2y = 1000, \(2x + y \neq 2000\).
The statement is sufficient to conclude, and we can eliminate options B, C, and E.
Statement 2
(B) 2x – y = 2300
Following a similar approach as that of Step 1, let's find out if we value(s) of x,y for which both the equations satisfy
2x + y = 2000
2x – y = 2300
Adding both the equations
4x = 4300
x = 4300 / 4 = 1075
Value of y (at x = 1075) = 150
Hence, for (x,y) = (1075, 150) both equations hold true.
However, we can have more than one solution for 2x – y = 2300, for example, y = 2 and x = 1149. For such values of (x,y), the equation \(2x + y \neq 2000\), hence this statement alone is not sufficient to answer.
Eliminate D.
Option A
