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20 Mar 2024, 08:29
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:37) correct
32%
(01:45)
wrong
based on 136
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In an experiment, the rate at which a certain bacterium grows is G. If G is a function of time t, such that \(G = xt^3 + yt^2 + z\), where x, y, and z are constants and t > 0. Is there a value of t for which G is negative?
(1) x > y > z
(2) z > 0
(1) x > y > z
(2) z > 0
21 Mar 2024, 01:59
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Gangadhar111990
(1) x > y > z
If \(x\) \(y\) and \(z\) are \(-ve\), we can answer YES
If \(x\) \(y\) and \(z\) are \(+ve\), can we can answer NO.
INSUFF.
(2) z > 0
We have no idea about \(x\) and \(y \)
INSUFF.
1+2
Now we know that \(x,y\) and \(z\) are all \(+ve\), Therefore we can answer NO to the question.
SUFF.
Ans C
Hope it helped.
23 Mar 2024, 08:25
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Gangadhar111990
G(t) = \( xt^3 \) +\( yt^2 \) + z
Statement 1
(1) x > y > z
Inference: x lies to the right of y which lies to the right of z on a number line.
Attachment:
Screenshot 2024-03-23 204850.png [ 5.24 KiB | Viewed 3185 times ]
We do not know the position of zero.
Case 1: zero lies to the left of \(z\)
In this case, G(t) is always positive.
Is there a value of t for which G is negative? ⇒ No
Case 2: zero lies to the right of \(x\)
In this case, G(t) is always negative.
Is there a value of \(t\) for which G is negative? ⇒ Yes
Attachment:
Screenshot 2024-03-23 205007.png [ 17.02 KiB | Viewed 3093 times ]
As we are getting two answers, the statement alone is not sufficient to answer the question.
Statement 2
(2) z > 0
In this case, zero lies to the left of z. Hence, x, y, and z are always positive.
Therefore the value of G(t) is always positive.
Is there a value of t for which G is negative? ⇒ No
Attachment:
Screenshot 2024-03-23 205350.png [ 6.59 KiB | Viewed 3006 times ]
Option B
