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As per an estimate, the depth D(t), in centimeters, of the water in a 09 Apr 2021, 07:30
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As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12. At what time does the depth of the water in the tank becomes the maximum?

(A) 5:30 a.m.
(B) 7:00 a.m.
(C) 7:30 a.m.
(D) 8:00 a.m.
(E) 9:00 a.m.
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B
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As per an estimate, the depth D(t), in centimeters, of the water in a 09 Apr 2021, 08:31
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Given: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 15 Apr 2021, 23:46
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Bunuel
As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12. At what time does the depth of the water in the tank becomes the maximum?
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Quadratic equations have the maximum (or minimum) at the vertex.
This equation is already presented in vertex form. Vertex form is y=(x-a)+b with the vertex at (-a, b)
\(D(t) = −10(t − 7)^2 + 100\)
(-(-7), 100); (7,100)
After 7 seconds the depth of the water will be 100 centimetres
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As per an estimate, the depth D(t), in centimeters, of the water in a 02 Jun 2025, 17:08
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Can anyone break this down further? I'm still not seeing how to get to 7.
Kinshook
Given: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 02 Jun 2025, 19:16
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Same, I dont understand whhy did we choose to equate to 0

JLambert
Can anyone break this down further? I'm still not seeing how to get to 7.
Kinshook
Given: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 03 Jun 2025, 01:01
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Another way of thinking about it is to look at the signs of what goes into \(D(t) = −10(t − 7)^2 + 100\)

\((t − 7)^2\) is always positive or 0 since it's the square of a number

\(−10(t − 7)^2\) is therefore always negative or 0 since we're just multiplying the previous line by a negative answer

\(D(t)\) is the previous line plus a positive 100 (which never changes, no matter what t is)

So the maximum value of D will be when the negative or zero part will be 0 (we want to subtract the least possible from 100). That happens when the square is 0, e.g. when t = 7

Hope this helps!
ʕ•ᴥ•ʔ

JLambert
Can anyone break this down further? I'm still not seeing how to get to 7.
Kinshook
Given: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 04 Jun 2025, 20:28
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Thank you! That makes perfect sense and is more along the lines of what I'm trying to work on: 'logic-ing' my way out of things if I'm uncertain. :D
MBABackstage
Another way of thinking about it is to look at the signs of what goes into \(D(t) = −10(t − 7)^2 + 100\)

\((t − 7)^2\) is always positive or 0 since it's the square of a number

\(−10(t − 7)^2\) is therefore always negative or 0 since we're just multiplying the previous line by a negative answer

\(D(t)\) is the previous line plus a positive 100 (which never changes, no matter what t is)

So the maximum value of D will be when the negative or zero part will be 0 (we want to subtract the least possible from 100). That happens when the square is 0, e.g. when t = 7

Hope this helps!
ʕ•ᴥ•ʔ

JLambert
Can anyone break this down further? I'm still not seeing how to get to 7.
Kinshook
Given: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 05 Jun 2025, 02:25
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No worries! Your approach is the correct way to do it. The GMAT is more about pattern recognition and learning these little tricks rather than a maths test. People who understand the concept of a derivative in maths can use it here, but in most cases it's an overkill for the GMAT

Good luck
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JLambert
Thank you! That makes perfect sense and is more along the lines of what I'm trying to work on: 'logic-ing' my way out of things if I'm uncertain. :D
MBABackstage
Another way of thinking about it is to look at the signs of what goes into \(D(t) = −10(t − 7)^2 + 100\)

\((t − 7)^2\) is always positive or 0 since it's the square of a number

\(−10(t − 7)^2\) is therefore always negative or 0 since we're just multiplying the previous line by a negative answer

\(D(t)\) is the previous line plus a positive 100 (which never changes, no matter what t is)

So the maximum value of D will be when the negative or zero part will be 0 (we want to subtract the least possible from 100). That happens when the square is 0, e.g. when t = 7

Hope this helps!
ʕ•ᴥ•ʔ

JLambert
Can anyone break this down further? I'm still not seeing how to get to 7.
KinshookGiven: As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00a.m. is given by \(D(t) = −10(t − 7)^2 + 100\), for 0 ≤ t ≤ 12.
Asked: At what time does the depth of the water in the tank becomes the maximum?

For maximum depth: -
d D(t) /dt = -20(t-7) = 0
t = 7

IMO B
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As per an estimate, the depth D(t), in centimeters, of the water in a 13 Sep 2026, 09:21
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