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11 Jul 2025, 19:52
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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.
estimate range of storage : 1250 - 1350 (1300 +- off up to 50 liters)
Estimate of time 3.5 - 4.5 hours (4 hours +- 0.5 hours)
Maximum: highest amount / least of hours : 1350/3.5
Minimum : lowest amount of storage / highest amount of hours = 1250/4.5
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.
estimate range of storage : 1250 - 1350 (1300 +- off up to 50 liters)
Estimate of time 3.5 - 4.5 hours (4 hours +- 0.5 hours)
Maximum: highest amount / least of hours : 1350/3.5
Minimum : lowest amount of storage / highest amount of hours = 1250/4.5
Rahul_Sharma23
GMAT Focus 1: 695 Q87 V83 DI83 
Posts: 145
11 Jul 2025, 20:01
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estimated volume of water pumped = 1300 ltr with margin of error of 50 ltr
estimated time taken = 4 hours with margin of error of 0.5 hours
rate of water pumping = volume of water pumped in ltrs/ time taken in hours
we will get the lowest possible actual pumping rate by making numerator smaller and denominator larger
volume of water pumped in ltrs = 1300-50 = 1250 ltr
time taken for the same = 4 + 0.5 = 4.5 hrs
minimum rate = 1250/4.5
we will get the highest possible actual pumping rate by making numerator bigger and denominator smaller
volume of water pumped in ltrs = 1300+50 = 1350 ltr
time taken for the same = 4 - 0.5 = 3.5 hrs
maximum rate = 1350/3.5
estimated time taken = 4 hours with margin of error of 0.5 hours
rate of water pumping = volume of water pumped in ltrs/ time taken in hours
we will get the lowest possible actual pumping rate by making numerator smaller and denominator larger
volume of water pumped in ltrs = 1300-50 = 1250 ltr
time taken for the same = 4 + 0.5 = 4.5 hrs
minimum rate = 1250/4.5
we will get the highest possible actual pumping rate by making numerator bigger and denominator smaller
volume of water pumped in ltrs = 1300+50 = 1350 ltr
time taken for the same = 4 - 0.5 = 3.5 hrs
maximum rate = 1350/3.5
Bunuel
11 Jul 2025, 21:50
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Bunuel
1300 L off by up to 50 L => Range: 1250 L to 1350 L
4 h off by up to 0.5h => Range: 3.5 h to 4.5 h
Pumping rate: Litres of water pumped/ Time taken
Now, we know that for the fraction to have the minimum value, the numerator should be as small as possible and the denominator should be as large as possible.
=> 1250/4.5
Now, we know that for the fraction to have the maximum value, the numerator should be as large as possible and the denominator should be as small as possible.
=> 1350/3.5
Minimum: 1250/4.5
Maximum: 1350/3.5
