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22 Nov 2021, 04:01
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D
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38% (02:36) correct
63%
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If \(x\) and \(y\) are positive integers and \((5\frac{3}{x})*(y\frac{1}{2})=19\), where \(5\frac{3}{x}\) and \(y\frac{1}{2}\) are mixed fractions, then what is the values of \(x + y\)?
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
22 Nov 2021, 04:01
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Official Solution:
If \(x\) and \(y\) are positive integers and \((5\frac{3}{x})*(y\frac{1}{2})=19\), where \(5\frac{3}{x}\) and \(y\frac{1}{2}\) are mixed fractions, then what is the values of \(x + y\)?
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
\((5\frac{3}{x})*(y\frac{1}{2})=19\)
\((\frac{5x+3}{x})*(\frac{2y+1}{2})=19\)
\((5x+3)(2y+1)=38x\)
From, \((5\frac{3}{x})*(y\frac{1}{2})=19\) it follows that \(y\) must be less than 4 (\(5.something*4.something=20.something\)). So, \(y\) can be 3, 2, or 1.
If \(y=3\), then from \((5x+3)(2y+1)=38x\), \(x=7\), and \(x+y=10\).
If \(y=2\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{15}{13}\neq integer\). Not possible since \(x\) must be integer.
If \(y=1\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{9}{23}\neq integer\). Not possible since \(x\) must be integer.
Answer: D
If \(x\) and \(y\) are positive integers and \((5\frac{3}{x})*(y\frac{1}{2})=19\), where \(5\frac{3}{x}\) and \(y\frac{1}{2}\) are mixed fractions, then what is the values of \(x + y\)?
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
\((5\frac{3}{x})*(y\frac{1}{2})=19\)
\((\frac{5x+3}{x})*(\frac{2y+1}{2})=19\)
\((5x+3)(2y+1)=38x\)
From, \((5\frac{3}{x})*(y\frac{1}{2})=19\) it follows that \(y\) must be less than 4 (\(5.something*4.something=20.something\)). So, \(y\) can be 3, 2, or 1.
If \(y=3\), then from \((5x+3)(2y+1)=38x\), \(x=7\), and \(x+y=10\).
If \(y=2\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{15}{13}\neq integer\). Not possible since \(x\) must be integer.
If \(y=1\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{9}{23}\neq integer\). Not possible since \(x\) must be integer.
Answer: D
18 Oct 2024, 16:36
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Bunuel, I assumed that to get 19.something then it could only be (5.something)*(3.something) and directly went to solve it for Y=3
It is the correct approach or the mixed fraction could be 5 3/1 for example and being greater than 5?
Thanks in advance
It is the correct approach or the mixed fraction could be 5 3/1 for example and being greater than 5?
Thanks in advance
Bunuel
Official Solution:
If \(x\) and \(y\) are positive integers and \((5\frac{3}{x})*(y\frac{1}{2})=19\), where \(5\frac{3}{x}\) and \(y\frac{1}{2}\) are mixed fractions, then what is the values of \(x + y\)?
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
\((5\frac{3}{x})*(y\frac{1}{2})=19\)
\((\frac{5x+3}{x})*(\frac{2y+1}{2})=19\)
\((5x+3)(2y+1)=38x\)
From, \((5\frac{3}{x})*(y\frac{1}{2})=19\) it follows that \(y\) must be less than 4 (\(5.something*4.something=20.something\)). So, \(y\) can be 3, 2, or 1.
If \(y=3\), then from \((5x+3)(2y+1)=38x\), \(x=7\), and \(x+y=10\).
If \(y=2\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{15}{13}\neq integer\). Not possible since \(x\) must be integer.
If \(y=1\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{9}{23}\neq integer\). Not possible since \(x\) must be integer.
Answer: D
If \(x\) and \(y\) are positive integers and \((5\frac{3}{x})*(y\frac{1}{2})=19\), where \(5\frac{3}{x}\) and \(y\frac{1}{2}\) are mixed fractions, then what is the values of \(x + y\)?
A. \(3\)
B. \(5\)
C. \(7\)
D. \(10\)
E. \(11\)
\((5\frac{3}{x})*(y\frac{1}{2})=19\)
\((\frac{5x+3}{x})*(\frac{2y+1}{2})=19\)
\((5x+3)(2y+1)=38x\)
From, \((5\frac{3}{x})*(y\frac{1}{2})=19\) it follows that \(y\) must be less than 4 (\(5.something*4.something=20.something\)). So, \(y\) can be 3, 2, or 1.
If \(y=3\), then from \((5x+3)(2y+1)=38x\), \(x=7\), and \(x+y=10\).
If \(y=2\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{15}{13}\neq integer\). Not possible since \(x\) must be integer.
If \(y=1\), then from \((5x+3)(2y+1)=38x\), \(x=\frac{9}{23}\neq integer\). Not possible since \(x\) must be integer.
Answer: D
