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by tanyajoseph » Mon Aug 06, 2007 10:32 am
Can someone explain these 2 problems? I am missing somethin very trivial !
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by givemeanid » Mon Aug 06, 2007 11:17 am
Let the 3 weights be x, y and z.
Median = middle value = y = 9

The three weights are x, 9 and z
Mean = 7
x + 9 + z = 21
x + z = 12

For x to be max, z needs to be minimized. The min value of z has to be 9 otherwise the median changes. That means x = 3 is the max possible weight of the lightest box.
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by cveluswamy » Tue Aug 07, 2007 6:48 am
600 * 500/10^6 to identify the cost per milligram.

Easiest way is 600 * 500/10^6 = 300000/10^6 Since 1 kilogram equals to 10^6. put the decimal points 6 digits from left which is 0.3

What is the OA?
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by tanyajoseph » Tue Aug 07, 2007 12:35 pm
Its 0.30. You were right:) Thanks
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by maolivie » Wed Aug 08, 2007 7:23 am
for the first question:

x + z = 12. You stated that for the x to be minimum, z has to be 9 otherwise the median changes. I'm confused by this because if z was 11, x would be 1. Then we have 1, 9, 11. 9 is still the median. Whats up with that?
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by givemeanid » Wed Aug 08, 2007 8:00 am
maolivie wrote:for the first question:

x + z = 12. You stated that for the x to be minimum, z has to be 9 otherwise the median changes. I'm confused by this because if z was 11, x would be 1. Then we have 1, 9, 11. 9 is still the median. Whats up with that?

The question is asking for MAXIMUM possible value of LIGHTEST box.
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