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The average of 4 consecutive odd numbers is half that of the

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The average of 4 consecutive odd numbers is half that of the  [#permalink]

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New post 18 Apr 2010, 23:59
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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
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Re: average problem  [#permalink]

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New post 17 Apr 2012, 01:35
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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.

Answer: D.

Hope it's clear.
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Re: average problem  [#permalink]

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New post 19 Apr 2010, 00:01
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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
OA d
please explain

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
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Re: average problem  [#permalink]

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New post 19 Apr 2010, 00:22
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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)

16 - 3 = 3.

QED.
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Re: average problem  [#permalink]

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New post 19 Apr 2010, 00:31
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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.
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Re: average problem  [#permalink]

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New post 16 Apr 2012, 21:32
Is there a different approach to this problem? I find the explanations above tough!! :-(
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Re: The average of 4 consecutive odd numbers is half that of the  [#permalink]

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New post 19 Sep 2016, 03:00
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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


Start with what you have been given so that you don't need to take variables. One average is half of the other and the sum of both is 18.
So a + 2a = 18
a = 6

Avg of 4 consecutive odd numbers is 6. The consecutive odd numbers will be 3, 5, 7 and 9. (avg lies in between the middle two numbers)
Avg of 5 consecutive even numbers is 12. The consecutive even numbers will be 8, 10, 12, 14, 16 (avg is the middle number).

Largest - smallest number = 16 - 3 = 13

Answer (D)
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The average of 4 consecutive odd numbers is half that of the  [#permalink]

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New post 19 Sep 2016, 15:19
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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

odd average=(4x+12)/4=x+3
even average=(5y+20)/5=y+4
y+4=2(x+3)➡2x-y=-2
(x+3)+(y+4)=18➡x+y=11
adding, 3x=9
x=3
y=11-3=8
8+4*2=16
16-3=13
D.
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Re: The average of 4 consecutive odd numbers is half that of the  [#permalink]

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New post 08 Aug 2019, 10:56
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Clearly there are two equations possible.
And by the averages' property -
1.x=y/2 and x+y=18 Solving- x=6x=6and y=12.
Now the highest and lowest - 16 and 3.
So 16-3=13
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Re: The average of 4 consecutive odd numbers is half that of the   [#permalink] 08 Aug 2019, 10:56
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