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What is the smallest possible number? HELP

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by gmatpup » Fri Dec 23, 2011 6:35 pm
A certain meter records voltage between 0 volts and 10 volts, inclusive. If the average value of 3 recordings was 3 volts, what was the smallest possible recording, in volts?

A. 2
B. 3
C. 4
D. 5
E. 6

Answer is E


How do I solve this? I am confused! Thanks so much :)
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Source: — Problem Solving |

by Anurag@Gurome » Fri Dec 23, 2011 7:43 pm
gmatpup wrote:A certain meter records voltage between 0 volts and 10 volts, inclusive. If the average value of 3 recordings was 3 volts, what was the smallest possible recording, in volts?

A. 2
B. 3
C. 4
D. 5
E. 6

Answer is E


How do I solve this? I am confused! Thanks so much :)
Can you check the question again?, i think the average value of 3 recordings has to be 8 volts.
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by gmatpup » Fri Dec 23, 2011 7:51 pm
Oh my goodness I am so sorry, you are absolutely correct. I meant to type the average is 8, not 3! Thanks for catching that:)
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by Anurag@Gurome » Fri Dec 23, 2011 7:56 pm
gmatpup wrote:Oh my goodness I am so sorry, you are absolutely correct. I meant to type the average is 8, not 3! Thanks for catching that:)
Thanks for confirming!

The average value of 3 recordings on the meter is 8 volts, means that the sum of three recording should be 24 as 24/3 = 8.
3 recording can be from 0-10, inclusive.
For one of the recording to be the smallest, the other two must take the maximum values and the maximum value can be 10.
So, if the smallest value is x, then (x + 10 + 10) = 24
Hence, x = 24 - 20 = 4

The correct answer is C.
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GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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by arpan20690 » Mon Dec 26, 2011 2:45 am
There are 3 recordings. We have to find the one which gives minimum.
Also we know sum of 3 recordings gives (3*8)= 24. now to find the minimum,
take the other 2 recordings as maximum i.e 10. so, two maximum recordings give (2*10)= 20 as total.
So the minimum one give (24-20)= 4
4 is the right answer.
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