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x, y, z, and t are different nonzero digits.
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20 Dec 2021, 01:31
20 Dec 2021, 01:31
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33% (01:38) correct
66% (01:28) wrong based on 18 sessions
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x, y, z, and t are different nonzero digits.
x
y
+ z
-------
2t
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
x
y
+ z
-------
2t
Quantity A |
Quantity B |
4 |
The maximum possible value for t |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: x, y, z, and t are different nonzero digits.
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20 Dec 2021, 08:02
20 Dec 2021, 08:02
1
\(x, y, z,\) and \( t\) are different nonzero digits.
\(x + y + z = 2\)t
We need the max value of \(t\) so we need the max sum, which can be obtained with \(x, y\) & \(z\) being the highest possible digits.
Take \(x = 9, y = 8, z = 7\)
\(x+y+z = 24\)
Max value of \(t = 4\)
Qt A = Qt B
Answer \(C\)
\(x + y + z = 2\)t
We need the max value of \(t\) so we need the max sum, which can be obtained with \(x, y\) & \(z\) being the highest possible digits.
Take \(x = 9, y = 8, z = 7\)
\(x+y+z = 24\)
Max value of \(t = 4\)
Qt A = Qt B
Answer \(C\)
Re: x, y, z, and t are different nonzero digits.
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22 Aug 2022, 01:33
22 Aug 2022, 01:33
rx10 wrote:
\(x, y, z,\) and \( t\) are different nonzero digits.
\(x + y + z = 2\)t
We need the max value of \(t\) so we need the max sum, which can be obtained with \(x, y\) & \(z\) being the highest possible digits.
Take \(x = 9, y = 8, z = 7\)
\(x+y+z = 24\)
Max value of \(t = 4\)
Qt A = Qt B
Answer \(C\)
\(x + y + z = 2\)t
We need the max value of \(t\) so we need the max sum, which can be obtained with \(x, y\) & \(z\) being the highest possible digits.
Take \(x = 9, y = 8, z = 7\)
\(x+y+z = 24\)
Max value of \(t = 4\)
Qt A = Qt B
Answer \(C\)
I'm not following you here.
we need to find the maximum possible value of t
and by doing your example we get 24 as sum of x,y,z
24 by logic should equal to 2t= t=12!!!
