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The average of 4 consecutive odd numbers is half that of the [#permalink]
18 Apr 2010, 22:59
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65% (hard)
Question Stats:
64% (04:13) correct
36% (02:53) wrong based on 157 sessions
The average of 4 consecutive odd numbers is half that of the average of 5 consecutive even numbers. If the sum of these two average is 18, then the difference between the largest and smallest of these numbers is
Re: average problem [#permalink]
18 Apr 2010, 23:01
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gmat2012 wrote:
The average of 4 consecutive odd numbers is half that of the average of 5 consecutive even numbers. If the sum of these two average is 18, then the difference between the largest and smallest of these numbers is a.10 b.21 c.7 d.13 e.5
Let odd numbers be 2n-3, 2n-1, 2n + 1, 2n + 3. Average = 2n. Let even numbers be 2m- 4, 2m - 2, 2m, 2m + 2, 2m + 4. Average = 2m it is given that 2m = 4n Also 2n + 2m = 18 => 2n + 4n = 18. 6n = 18, 2n = 6 & 2m = 12. Largest = 16, smallest = 3. Difference = 16 - 3 = 13. hope this will help
Re: average problem [#permalink]
18 Apr 2010, 23:22
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There's a simple solution to this.
To find the average for a set of consecutive numbers, you add the first and last terms and divide by 2. In other words, the average is essentially center/pivot point of the series, whether or not it is a number in the series. (e.g. 1, 3, 5, 7 - the average is 4)
Now we look at the other information given. the average of the odd series is half the average of the even series and they sum up to 18. So let e be the average of the even series. We get 1.5e = 18 => e = 12
12 will be the middle term of the series and since there are 5, we now know the series look like this: (8, 10, 12, 14, 16) 12/2 = 6, the pivot point of the odd series, since there are 4, we know the series look like this: (3, 5, 7, 9)
Re: average problem [#permalink]
18 Apr 2010, 23:31
thanatoz wrote:
There's a simple solution to this.
To find the average for a set of consecutive numbers, you add the first and last terms and divide by 2. In other words, the average is essentially center/pivot point of the series, whether or not it is a number in the series. (e.g. 1, 3, 5, 7 - the average is 4)
Now we look at the other information given. the average of the odd series is half the average of the even series and they sum up to 18. So let e be the average of the even series. We get 1.5e = 18 => e = 12
12 will be the middle term of the series and since there are 5, we now know the series look like this: (8, 10, 12, 14, 16) 12/2 = 6, the pivot point of the odd series, since there are 4, we know the series look like this: (3, 5, 7, 9)
16 - 3 = 3.
QED.
good thought, i essentially solved using conventional method like assuming even and odd series numbers.. thanks for giving different prospective to the solution.
Re: average problem [#permalink]
17 Apr 2012, 00:35
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ENAFEX wrote:
Is there a different approach to this problem? I find the explanations above tough!!
The average of 4 consecutive odd numbers is half that of the average of 5 consecutive even numbers. If the sum of these two average is 18, then the difference between the largest and smallest of these numbers is A. 10 B. 21 C. 7 D. 13 E. 5
Some notes: The average of evenly spaced set with even number of terms (4 in our case) is the average of two middle terms. The average of evenly spaced set with odd number of terms (5 i our case) is the middle term.
Say the average of 4 consecutive odd numbers is \(x\) and the average of 5 consecutive even numbers is \(y\).
Given: \(x=\frac{y}{2}\) and \(x+y=18\) --> solve for \(x\) and \(y\): \(x=6\)and \(y=12\).
So, we have that the average of 4 consecutive odd numbers is 6, which means that those numbers are: {3, 5, 7, 9} (6 is the average of two middle terms);
Similarly we have that the average of 5 consecutive even numbers is 12, which means that those numbers are: {8, 10, 12, 14, 16} (12 is the middle term);
The difference between the largest and smallest of these numbers is 16-3=13.
Re: The average of 4 consecutive odd numbers is half that of the [#permalink]
23 Feb 2014, 02:45
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Re: The average of 4 consecutive odd numbers is half that of the [#permalink]
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