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25 Aug 2021, 08:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (02:53) correct
49%
(03:03)
wrong
based on 320
sessions
History
Date
Time
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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
M37-23
A. 3
B. 5
C. 7
D. 10
E. 11
|
M37-23
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22 Nov 2021, 04:02
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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
General Discussion
25 Aug 2021, 08:18
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IMO D
Since the final product is 19, y must be 3
[As 5*3 < 19 < 5*4]
Thus we are left with solving for x:
[(5x+3)/3]*7/2=19
Solving for x, we get x=7
Therefore x+y=7+3=10 ANS
Since the final product is 19, y must be 3
[As 5*3 < 19 < 5*4]
Thus we are left with solving for x:
[(5x+3)/3]*7/2=19
Solving for x, we get x=7
Therefore x+y=7+3=10 ANS
