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# The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2

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The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2  [#permalink]

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05 Jun 2018, 02:24
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35% (medium)

Question Stats:

77% (02:08) correct 23% (02:31) wrong based on 38 sessions

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The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2 lightest crates weigh between 200 and 205 pounds each, inclusive, and the 2 heaviest crates weigh between 300 and 310 pounds each, inclusive. If the weight of the fifth crate is x pounds, then x is expressed by which of the following?

A) 220 ≤ x ≤ 250

B) 230 ≤ x ≤ 260

C) 240 ≤ x ≤ 270

D) 250 ≤ x ≤ 270

E) 260 ≤ x ≤ 280

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Re: The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2  [#permalink]

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05 Jun 2018, 03:14
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Bunuel wrote:
The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2 lightest crates weigh between 200 and 205 pounds each, inclusive, and the 2 heaviest crates weigh between 300 and 310 pounds each, inclusive. If the weight of the fifth crate is x pounds, then x is expressed by which of the following?

A) 220 ≤ x ≤ 250

B) 230 ≤ x ≤ 260

C) 240 ≤ x ≤ 270

D) 250 ≤ x ≤ 270

E) 260 ≤ x ≤ 280

Minimum $$X=5*250-2*205-2*310=220$$

Maximum $$X=5*250-2*200-2*300=250$$

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Re: The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2  [#permalink]

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05 Jun 2018, 05:13

Solution

Given:
• Average weight of 5 crates in 250 pounds
• Lightest 2 crates weigh between 200 and 205 pounds each, inclusive
• Heaviest 2 crates weigh between 300 and 310 pounds each, inclusive
• Weight of the fifth crate is x pound

To find:
• From the given options, the range that best describes x

Approach and Working:
If we want to find the minimum value of x, we need to maximise the weight of other crates
• Minimum value of x = (250 * 5) – [(205 * 2) + (310 * 2)] = 1250 – (410 + 620) = 1250 – 1030 = 220

Similarly, if we want to find the maximum value of x, we need to minimise the weight of other crates
• Maximum value of x = (250 * 5) – [(200 * 2) + (300 * 2)] = 1250 – (400 + 600) = 1250 – 1000 = 250

• Therefore, we can say: 220 ≤ x ≤ 250

Hence, the correct answer is option A.

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Re: The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2  [#permalink]

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20 Sep 2018, 18:58
Bunuel wrote:
The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2 lightest crates weigh between 200 and 205 pounds each, inclusive, and the 2 heaviest crates weigh between 300 and 310 pounds each, inclusive. If the weight of the fifth crate is x pounds, then x is expressed by which of the following?

A) 220 ≤ x ≤ 250

B) 230 ≤ x ≤ 260

C) 240 ≤ x ≤ 270

D) 250 ≤ x ≤ 270

E) 260 ≤ x ≤ 280

$$?\,\,\,:\,\,\,x \in \left[ {{x_{\min }}\,\,,\,\,\,{x_{\max }}} \right]$$

By the homogeneity nature of the average, we know that:

$$\sum\nolimits_5 = \,\,\,5 \cdot 250 = 1250$$

$$\sum\nolimits_{2\,{\text{light}}}^{\min } = \,\,\,2 \cdot 200 = 400\,\,\,\,\,\,;\,\,\,\,\,\,\sum\nolimits_{2\,{\text{light}}}^{\max } = \,\,\,2 \cdot 205 = 410$$

$$\sum\nolimits_{2\,{\text{heavy}}}^{\min } = \,\,\,2 \cdot 300 = 600\,\,\,\,\,\,;\,\,\,\,\,\,\sum\nolimits_{2\,{\text{heavy}}}^{\max } = \,\,\,2 \cdot 310 = 620$$

$${x_{\min }} = \sum\nolimits_5 { - \left( {\,\sum\nolimits_{2\,{\text{light}}}^{\max } { + \sum\nolimits_{2\,{\text{light}}}^{\max } {} } } \right)} = 1250 - \left( {410 + 620} \right) = 220$$

$${x_{\max }} = \sum\nolimits_5 { - \left( {\,\sum\nolimits_{2\,{\text{heavy}}}^{\min } { + \sum\nolimits_{2\,{\text{heavy}}}^{\min } {} } } \right)} = 1250 - \left( {400 + 600} \right) = 250$$

$$?\,\,\,:\,\,\,x \in \left[ {220,250} \right]\,\,\,\,\, \Leftrightarrow \,\,\,\,\,220 \leqslant x \leqslant 250$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2 &nbs [#permalink] 20 Sep 2018, 18:58
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