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The average (arithmetic mean) of six interfere is 32. If the number are all different, and if none is led then 10, what is the greatest possible value of any of these intergers? A) 127 B) 132 C) 137 D) 142 E) 147

Guest Apr 19, 2015

Best Answer 

 #1
avatar+17653 
+13

If the average of the six numbers is 32, then the total of these numbers is 6 x 32  =  192.

If five of these numbers are all at the minimum (10), these five numbers have a total of 5 x 10 = 50. (If any number is larger than 10, then the largest won't be the largest possible.)

Subtracting 50 from 192 gives 142, the maximum any number can be for this problem.

geno3141  Apr 19, 2015
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2+0 Answers

 #1
avatar+17653 
+13
Best Answer

If the average of the six numbers is 32, then the total of these numbers is 6 x 32  =  192.

If five of these numbers are all at the minimum (10), these five numbers have a total of 5 x 10 = 50. (If any number is larger than 10, then the largest won't be the largest possible.)

Subtracting 50 from 192 gives 142, the maximum any number can be for this problem.

geno3141  Apr 19, 2015
 #2
avatar+248 
+10

Just in regards to the answer above, thats the minimum if all the numbers dont have to be different. Since the question says they all have to be different, the lowest five values are 10,11,12,13,14 which add up to 60. Subtract this from 192 gives you 132 as the largest value for your 6'th number.

Brodudedoodebrodude  Apr 20, 2015

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