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If x and y are both integers
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12 Aug 2025, 01:26
12 Aug 2025, 01:26
Expert Reply
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Question Stats:
50% (00:38) correct
50% (02:20) wrong based on 2 sessions
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\($$
\begin{gathered}
4<x<7 \\
9<y<12
\end{gathered}
$$\)
If $\(x\)$ and $\(y\)$ are both integers that satisfy the inequalities, what is the lowest value of $\(\frac{x+y}{(-y)^2}\)$
Express your answer as a fraction.
\begin{gathered}
4<x<7 \\
9<y<12
\end{gathered}
$$\)
If $\(x\)$ and $\(y\)$ are both integers that satisfy the inequalities, what is the lowest value of $\(\frac{x+y}{(-y)^2}\)$
Express your answer as a fraction.
Show: ::
16/121
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Re: If x and y are both integers
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30 Aug 2025, 08:15
30 Aug 2025, 08:15
Expert Reply
Given:
- $\(4<x<7\)$ where $x$ is an integer, so $x=5$ or 6 ,
- $\(9<y<12\)$ where $y$ is an integer, so $y=10$ or 11 .
We want to find the lowest value of:
$$
\(\frac{x+y}{(-y)^2}=\frac{x+y}{y^2}\)
$$
(As $\((-y)^2=y^2\)$ )
Calculate possible values for all combinations of $x$ and $y$ :
1. $\(x=5, y=10\)$ :
$$
\(\frac{5+10}{10^2}=\frac{15}{100}=0.15\)
$$
2. $x=5, y=11$ :
$$
\(\frac{5+11}{11^2}=\frac{16}{121} \approx 0.132\)
$$
3. $x=6, y=10$ :
$$
\(\frac{6+10}{10^2}=\frac{16}{100}=0.16\)
$$
4. $x=6, y=11$ :
$$
\(\frac{6+11}{11^2}=\frac{17}{121} \approx 0.1405\)
$$
The minimum among these values is:
$$
\(\frac{16}{121} \approx 0.132\)
$$
which corresponds to $x=5, y=11$.
Hence, the lowest value of $\(\frac{x+y}{(-y)^2}\)$ is:
$$
\(\frac{16}{121}\)
$$
- $\(4<x<7\)$ where $x$ is an integer, so $x=5$ or 6 ,
- $\(9<y<12\)$ where $y$ is an integer, so $y=10$ or 11 .
We want to find the lowest value of:
$$
\(\frac{x+y}{(-y)^2}=\frac{x+y}{y^2}\)
$$
(As $\((-y)^2=y^2\)$ )
Calculate possible values for all combinations of $x$ and $y$ :
1. $\(x=5, y=10\)$ :
$$
\(\frac{5+10}{10^2}=\frac{15}{100}=0.15\)
$$
2. $x=5, y=11$ :
$$
\(\frac{5+11}{11^2}=\frac{16}{121} \approx 0.132\)
$$
3. $x=6, y=10$ :
$$
\(\frac{6+10}{10^2}=\frac{16}{100}=0.16\)
$$
4. $x=6, y=11$ :
$$
\(\frac{6+11}{11^2}=\frac{17}{121} \approx 0.1405\)
$$
The minimum among these values is:
$$
\(\frac{16}{121} \approx 0.132\)
$$
which corresponds to $x=5, y=11$.
Hence, the lowest value of $\(\frac{x+y}{(-y)^2}\)$ is:
$$
\(\frac{16}{121}\)
$$
gmatclubot
Re: If x and y are both integers [#permalink]
30 Aug 2025, 08:15
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