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GMAT Club Olympics 2025 (Day 10): A maintenance team estimated that

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bart08241192

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We aim to maximize and minimize the "rate = actual volume ÷ actual time," both of which are subject to estimation errors:

The actual volume ranges within [1300 – 50, 1300 + 50] = [1250, 1350] liters
The actual time ranges within [4 – 0.5, 4 + 0.5] = [3.5, 4.5] hours
— Minimum rate: achieved by the smallest volume and largest time ⇒ 1250 ÷ 4.5
— Maximum rate: achieved by the largest volume and smallest time ⇒ 1350 ÷ 3.5

Corresponding options:
Minimum (Minimum) selects 1250/4.5
Maximum (Maximum) selects 1350/3.5

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Actual volume range = 1250 to 1350 litres.
Time range = 3.5 to 4.5 hours

Minimum pumping rate= min volume/ max time
=1250/4.5= 278 liters/ hour.

Maximum pumping rate = max volume/ min time
= 1350/ 3.5 = 386 liters/ hour.

Therefore ,
Minimum the lowest possible actual pumping rate, can take the value of 1250/ 4.5 liters/ hour.
Maximum the highest possible actual pumping rate of 1350/3.5 liters/ hour.

LucasH20

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To find the minimum pumping rate, we divide the smallest possible volume (1300 - 50 = 1250 liters) by the longest possible time (4 + 0.5 = 4.5 hours), resulting in 1250/4.5 liters per hour. For the maximum pumping rate, we divide the largest possible volume (1300 + 50 = 1350 liters) by the shortest possible time (4 - 0.5 = 3.5 hours), resulting in 1350/3.5 liters per hour.

Regards,
Lucas

Bunuel

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A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours.

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges. Make only two selections, one in each column.

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Bunuel

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Given: 1300 Litres of water in 4 hours.

Error margin for litres of volume = 50 Litres and Error margin for hours = 0.5 hours

Hence the flow rate can be maximum only when highest amount of water flows in lost amount of time which is 1300 +50 Litres of water in 4- 0.5 hours Hnece the seer is 1350/.5 Option 2

Similarly, the minimum flow rate can be won lowest amount of water flows through the pipe in most amount of time which is 1300 - 50 Litres and 4+0.5 hours
Hence the answer is 1250/4.5 which is option 5

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Work = 1300 with error of 50, so we can say,

W(min) = 1250
W(max) = 1350

Time = 4 with error of 0.5, so we can say,

T(min) = 3.5
T(max) = 4.5

$Rate = \frac{Work}{Time}$

$Rate(max) = \frac{Work(max)}{Time(min)}$ = $\frac{1350}{3.5}$

$Rate(min) = \frac{Work(min)}{Time(max)}$ = $\frac{1250}{4.5}$

Answer:
Minimum = $\frac{1250}{4.5}$
Maximum = $\frac{1350}{3.5}$

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The volume estimate may be off by up to and including 50 liters: OFF BY UTMOST OR EQUAL TO 50
The time estimate may be off by up to and including 0.5 hours: OFF BY UTMOST OR EQUAL TO 0.5

Maximim-

Maximum numerator and minimum denominator: 1300/3.5

Minimum-

Minimum numerator and maximum denominator: 1250/4.5

Bunuel

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1300L in 4 hrs i.e., 325L/Hr

We have an off amt of 50L and 0.5 Hrs, which could be 1350 or 1250L and 3.5 or 4.5 hrs

In order to minimize:
1250/4.5 would be correct

To Maximise
1350/35 would be correct

Bunuel

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Rate = Volume / Time

Estimated Volume = 1300 liters
Volume error = +/- 50 liters

Minimum Actual Volume = 1300−50=> 1250 liters
Maximum Actual Volume = 1300+50=> 1350 liters

Estimated Time = 4 hours
Time error = +/- 0.5 hours

Minimum Actual Time = 4−0.5=> 3.5 hours
Maximum Actual Time = 4+0.5=> 4.5 hours

To get the minimum rate, we need the smallest possible volume and the largest possible time.
$Minimum Rate = \frac{ Minimum Volume }{ Maximum Time }$
Minimum Rate = 1250/4.5

To get the maximum rate, we need the largest possible volume and the smallest possible time.
$Maximum Rate = \frac{ Maximum Volume }{ Minimum Time }$
Maximum Rate = 1350/3.5

The final answer is Minimum:1250/4.5,Maximum:1350/3.5

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Bunuel

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Volume given 1300 litres

Time taken = 4 hours.

The volume might be either + or - 50 litres. Hence, the volume is (1300+50) or ( 1300-50)

1250 <= Volume <= 1350

Similarly, the time taken is (4+0.5) or (4-0.5) . So the time is 4.5 or 3.5 hours.

3.5 < = time < = 4.5

Minimum lowest pumping rate = ?

Pumping rate = volume / time.

To get the lowest rate, the volume should be low as possible and given time should be high as possible.

Minimum pumping rate = 1250/4.5

Maximum pumping rate :

To get the maximum rate, the volume must be high and the time should be low as possible.

Maximum Pumping rate = 1350/3.5

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To make it minimum decrease num, increase denom- 1250/4.5

To make it max, increase denom, reduce denom- 1350/3.5

Ans E,B

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Given data,
Estimated Water Volume: 1,300 liters
Estimated Time: 4 hours

Given conditions,
Volume Error = Up to 50 liters (meaning the real amount could be 50 liters less or 50 liters more than 1,300).
Time Error = Up to 0.5 hours (meaning the real time could be 0.5 hours less or 0.5 hours more than 4 hours).

we know that, Pumping rate( Rate) = (Amount of Water) / (Time taken)

Now lets find the lowest possible pumping rate:
Smallest Actual Water Volume = The smallest amount of water that could have been pumped - The longest time it could have taken to pump that water
= 1,300 liters - 50 liters = 1,250 liters
longest Actual Time = 4 hours + 0.5 hours = 4.5 hours
Minimum Rate = 1,250 liters / 4.5 hours

Finding the Highest Possible Pumping Rate:
Largest Actual Water Volume = The largest amount of water that could have been pumped - The shortest time it could have taken to pump that water
=1,300 liters + 50 liters = 1,350 liters
Shortest Actual Time = 4 hours - 0.5 hours = 3.5 hours
Maximum Rate = 1,350 liters / 3.5 hours

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Volume = 1300 lts
Time = 4 hours

Considering the error tolerance:
Volume = 1300 +- 50lts
Time = 4 +- 0.5 hours

For minimum rate minimize numerator and maximize denominator.
For maximum rate, maximize numerator and minimize denominator.

Answer:
Minimum: 1,250/4.5
Maximum: 1,350/3.5

Bunuel

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1300 in 4h
1250 in 4.5h
1350 in 3.5h

Lowest rate:
1250/4.5

Highest rate:
1350/3.5

Bunuel

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Max Pumping rate = Max volume/ Min Time = 1300 + 50 / 4 - 0.5 = 1350/3.5

Min Pumping rate = Min volume / Max Time = 1300 - 50 / 4 + 0.5 = 1250/4.5

Bunuel

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The correct answer is Minimum: 1250/4.5 and Maximum: 1350/3.5

To calculate the rates, first write out the possibilities with the smallest and largest amounts of water with the shortest and fastest speeds. The combination of the largest amount with the smallest amount of time will be our fastest rate and the smallest amount with the longest time will be our slowest rate.

Amounts: 1300 is our initial amount but with a possible largest value of 1350 and smallest value of 1250 based on the error of up to 50 liters.

Times: 4 hours is our initial time but with a possible largest value of 4.5 hours and smallest value of 3.5 hours based on the error of up to 0.5 hours.

Maximum Rate: 1350/3.5
Minimum Rate: 1250/4.5

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"A maintenance team estimated that a certain pipe pumped 1,300 liters of water into a storage tank in 4 hours. The volume estimate may be off by up to and including 50 liters, and the time estimate may be off by up to and including 0.5 hours."

Select for Minimum the lowest possible actual pumping rate (in liters per hour), and
Rate = Work / Time

To minimize Rate we need to minimize Work and maximize Time.

Therefore:

Minimum = (1300 - 50) / (4 + 0.5)
Minimum = 1250 / 4.5

select for Maximum the highest possible actual pumping rate (in liters per hour), consistent with these estimation ranges.

To maximize Rate we need to maximize Work and minimize Time.

Therefore:

Minimum = (1300 + 50) / (4 - 0.5)
Minimum = 1350 / 3.5

Answer:
Minimum: 1,250/4.5
Maximum: 1,350/3.5

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with estimation
given volume of water pumped=1300+50 ltrs or 1300-50 ltrs
time taken to pump=4+0.5 hrs. or 4-0.5 hrs.
maximum rate=maximum volume/minimum time=1350/3.5
minimum rate=minimum volume/maximum time=1250/4.5

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Bunuel

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The minimum is reached by the following modifications:

Very low output and high input.
So we go with 1300-50 = 1250 liters
and 4+0.5 = 4.5 liters

Vice versa for the maximum output, we have:

Very high output and low input.
So we go with 1300+50 = 1350 liters
and 4-0.5 = 3.5 liters

Therefore:

Low: 1250/4.5
High: 1350/3.5

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The estimated rate is $\frac{1300}{4}$ Liters per hour.
The water volume could be 50 liters more or less, and the time could be 0.5 hours more or less.
The minimum possible rate is when the water volume (numerator) is minimized and the time (denominator) is maximized.
So, the lowest possible actual pumping rate is $\frac{1250}{4.5}$
The maximum possible rate is when the water volume (numerator) is maximized and the time (denominator) is minimized.
So, the highest possible actual pumping rate is $\frac{1350}{3.5 }$

Bunuel

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Estimated volume=1300 litres
Estimated time=4 hrs
Estimated Pumping rate= 1300/4

Give that Range of volume = 1300-50 to 1300+50
=1250 to 1350
Max volume =1350
Min vol= 1250

Range of time = 4-0.5 to 4+0.5= 3.5 to 4.5
Max time =4.5
Min time =3.5

So
Maximum pumping rate= 1350/3.5
Minimum pumping rate=1250/4.5