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10 Apr 2023, 08:51
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A
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Difficulty:
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25% (02:09) correct
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If x and y are non-zero integers, is \(\frac{2x}{5} - \frac{3y}{4}= 1\) ?
(1) 4x – 5y = 15
(2) 2x - 4y = 6
(1) 4x – 5y = 15
(2) 2x - 4y = 6
10 Apr 2023, 12:25
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ChandlerBong
Pretty straightforward question!
Let's solve the premise, by taking the LCM of the left-hand side of the equation -
\(\frac{2x}{5} - \frac{3y}{4}= 1\)
\(8x - 15y = 20\)
Statement 1
(1) 4x – 5y = 15
Given \(4x – 5y = 15\), let's try to find out if it is possible to conclude \(8x - 15y = 20\). If this were the case, we will have only one pair value for (x,y) which satisfies both equations.
Multiplying 2 on both sides of the equation \(4x – 5y = 15\) we get \(8x - 10y = 30\)
Let's solve for the value of x and y using both the equations -
\(8x - 15y = 20\) --- (1)
\(8x - 10y = 30\) --- (2)
Subtracting equation (2) from equation (1) we get,
\(-5y = -10\)
\(y = 2\)
Value of x, if y = 2
\(8x - 20 = 30\)
\(x = \frac{50}{3}\)
The question premise states that x and y are integers, however, we end up getting a fractional value of x.
Hence we can conclude that for no integer value of x and y \(8x - 15y = 20\) given equation \(4x – 5y = 15\). As we have a definite answer, the statement is sufficient. We can eliminate B, C, and E.
Statement 2
(2) 2x - 4y = 6
Similar to solving statement 1, let's find out if it is possible to conclude that \(8x - 15y = 20\), given \(2x - 4y = 6\)
Multiplying 4 on both sides of the equation \(2x - 4y = 6\) we get \(8x - 16y = 24\)
Solve for the value of x and y using both the equations -
\(8x - 16y = 24\) --- (1)
\(8x - 15y = 20\) --- (2)
Subtracting equation (2) from equation (1) we get,
y = -4
Value of x, when y = -4
x = -5
So, for x = -5 & y = -4, we can conclude that \(8x - 15y = 20\).
However, the statement \(2x - 4y = 6\) holds true for other values of x and y as well. For such pairs the value of \(8x - 15y \neq 20\). Hence, we don't have a definite answer to this statement.
Eliminate D.
Option A
29 Jun 2024, 08:23
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