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# A shipment of watermelons weighs 899 pounds. If each watermelon weighs

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A shipment of watermelons weighs 899 pounds. If each watermelon weighs  [#permalink]

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23 Oct 2018, 21:23
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15% (low)

Question Stats:

85% (01:08) correct 15% (01:26) wrong based on 33 sessions

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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

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Re: A shipment of watermelons weighs 899 pounds. If each watermelon weighs  [#permalink]

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23 Oct 2018, 21:27
Bunuel 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

Total Weight = 899

One watermelon ≥ 15

Maximum Number of watermelons = ?

Number of watermelons will be maximum when the weight of each watermelon is minimum

i.e. Maximum watermelons = 899 / 15 = 59.93

i.e. 59 Watermelon

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Re: A shipment of watermelons weighs 899 pounds. If each watermelon weighs  [#permalink]

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24 Oct 2018, 00:04
Bunuel 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

$$899 ~ 900$$

$$\frac{900}{15} = 60$$

Since we have approximated the value to next integer, the answer must be anything less than 60, the correct Answer must be (C) 59
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Re: A shipment of watermelons weighs 899 pounds. If each watermelon weighs  [#permalink]

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28 Oct 2018, 18:23
Bunuel 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

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.

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Re: A shipment of watermelons weighs 899 pounds. If each watermelon weighs  [#permalink]

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20 May 2019, 20:40
Bunuel 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

First Concept- To maximize the number of watermelons we have to assume the least possible weight. Let each watermelon is of weight 15 pounds.
Now 899/15 gives
Quotient 59
Remainder 14

So 899 = 59 watermelons * 15 pounds + 1 watermelon 14 pounds
But 14 pounds watermelon is not possible, so
899= 58 watermelons * 15 pounds + 1 watermelon (14+15) pounds
Total watermelons 58+1=59
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Re: A shipment of watermelons weighs 899 pounds. If each watermelon weighs   [#permalink] 20 May 2019, 20:40
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