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A shipment of watermelons weighs 899 pounds. If each waterme

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by Brent@GMATPrepNow » Fri Aug 16, 2019 4:44 am
BTGmoderatorDC wrote:A shipment of watermelons weighs 899 pounds. If each watermelon weighs at least 15 pounds, what is the greatest number of watermelons that could be in the shipment?

A. 51
B. 52
C. 59
D. 60
E. 61

OA C

Source: Magoosh
To MAXIMIZE the number of watermelons in the shipment, we must MINIMIZE the weight per watermelon.
The minimum weight per melon is 15 pounds.

To determine the maximum number of water melons, we'll determine how many times 15 pounds divides into 899 pounds.
Calculating 899/15 is a pain, but if we recognize that 900/15 = 60, then we know that 899/15 = 59.something
In other words, the shipment can hold 59.something watermelons.
Since we need WHOLE watermelons, the maximum number is 59

Answer: C

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by Scott@TargetTestPrep » Fri Aug 23, 2019 10:42 am
BTGmoderatorDC wrote:A shipment of watermelons weighs 899 pounds. If each watermelon weighs at least 15 pounds, what is the greatest number of watermelons that could be in the shipment?

A. 51
B. 52
C. 59
D. 60
E. 61

OA C

Source: Magoosh
Let's say 58 watermelons each weighs 15 pounds, so they weigh a total of 58 x 15 = 870 pounds.

Thus, one more watermelon must weigh 29 pounds.

So the maximum number of watermelons is 59.

Alternate Solution:

Since we want the greatest number of watermelons, we want as many watermelons as possible to weigh the minimum amount, which is 15 pounds. Dividing the entire shipment weight of 899 pounds by 15, we obtain 59.93. This means that we cannot have 60 watermelons in the shipment; instead we will have 59 watermelons. We see that 58 watermelons will weigh a total of 58 x 15 = 870 pounds, and the 59th watermelon will weigh 899 - 870 = 29 pounds.

Answer: C

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