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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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philipssonicare
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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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JLambert
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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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GraemeGmatPanda
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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
ʕ•ᴥ•ʔ
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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