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T is some point on the hypotenuse D F of the right-angled triangle
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11 Aug 2026, 02:50
11 Aug 2026, 02:50
Expert Reply
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Question Stats:
0% (00:00) correct
100% (02:54) wrong based on 4 sessions
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\(T\) is some point on the hypotenuse \(D F\) of the right-angled triangle \(D E F\) , right-angled at \(E\) , where \(D E=9\) and \(E F=12\) .
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Probability that the length of \(E T\) is less than 9 |
0.7 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
T is some point on the hypotenuse D F of the right-angled triangle
[#permalink]
15 Aug 2026, 02:50
15 Aug 2026, 02:50
Expert Reply
Set coordinates:
E=(0,0)
D=(9,0)
F=(0,12)
Then \(D F=\sqrt{9^2+12^2}=15\) .
Let T be a point on D F . Parametrize:
$$
T=(9-9 t, 12 t), \quad 0 \leq t \leq 1
$$
Then
$$
E T^2=(9-9 t)^2+(12 t)^2=225 t^2-162 t+81
$$
We need \(E T<9 \), so:
$$
\begin{gathered}
225 t^2-162 t+81<81 \\
225 t^2-162 t<0 \\
9 t(25 t-18)<0
\end{gathered}
$$
Since \(t \geq 0\) , this gives:
$$
0<t<\frac{18}{25}
$$
So the portion of the hypotenuse where $E T<9$ has length:
$$
\frac{18}{25} \cdot 15=10.8
$$
Total hypotenuse length is 15, so:
$$
\text { Probability }=\frac{10.8}{15}=0.72
$$
Quantity A is 0.72, which is greater than Quantity B 0.7.
A
E=(0,0)
D=(9,0)
F=(0,12)
Then \(D F=\sqrt{9^2+12^2}=15\) .
Let T be a point on D F . Parametrize:
$$
T=(9-9 t, 12 t), \quad 0 \leq t \leq 1
$$
Then
$$
E T^2=(9-9 t)^2+(12 t)^2=225 t^2-162 t+81
$$
We need \(E T<9 \), so:
$$
\begin{gathered}
225 t^2-162 t+81<81 \\
225 t^2-162 t<0 \\
9 t(25 t-18)<0
\end{gathered}
$$
Since \(t \geq 0\) , this gives:
$$
0<t<\frac{18}{25}
$$
So the portion of the hypotenuse where $E T<9$ has length:
$$
\frac{18}{25} \cdot 15=10.8
$$
Total hypotenuse length is 15, so:
$$
\text { Probability }=\frac{10.8}{15}=0.72
$$
Quantity A is 0.72, which is greater than Quantity B 0.7.
A
gmatclubot
T is some point on the hypotenuse D F of the right-angled triangle [#permalink]
15 Aug 2026, 02:50
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