The mean of four numbers is 71.5. If three of the numbers are 58, 76, and 88, what is the value of the fourth number?

(222+x) / 4 = 71.5

4( 222+ x)/4 = 71.5 x 4

222+x = 286
222 + x - 222 = 286-222
x=64
 
JeffM said:
The whole practical point of algebra is to reduce the burden on your imagination.

Had you said to yourself

What do I need to do?

Find a number. OK, name it.

To find a single number, I need one equation.

What is it?
Click to expand...
piece of cake for you guys, math is second nature to you.
 
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blamocur said:
No, 4 is fine without parentheses. You need parentheses around the some because otherwise it would mean "58 + 76 + 88 + (n / 4) = 71.5" since division has higher precedence than addition.
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yes, i see it now. the importance of the grouping symbols!. wow!
 
I'm in doubt about another road that a high school teacher took to solve the problem posted here.
He said:
Let's replace the pair of opposite outliers with their mean.
58+76+88+x=4 ×71.5
73 +76+73+x=4×71.5
Now take 70 as the reference value.
3+6+3+y=6
y= -6
And so
x=64.

Except that I didn't understand at all what he did here.
Can you elucidate on this method?
 
eddy2017 said:
I'm in doubt about another road that a high school teacher took to solve the problem posted here.
He said:
Let's replace the pair of opposite outliers with their mean.
58+76+88+x=4 ×71.5
73 +76+73+x=4×71.5
Now take 70 as the reference value.
3+6+3+y=6
y= -6
And so
x=64.

Except that I didn't understand at all what he did here.
Can you elucidate on this method?
Click to expand...
I can try elucidating, but this method does not seem natural to me. The teacher replaced 58 and 88 by their average of 73. The sum, and thus the mean, has not changed, and the teacher seems to like having the values closer to each other. Now 70 can be subtracted from every term, thus reducing both sides by 280, and the resulting equation has smaller numbers in it. Once the equation is solved the subtracted value of 70 has to be added back.
Definitely not my favorite way to solve this problem.
 
blamocur said:
I can try elucidating, but this method does not seem natural to me. The teacher replaced 58 and 88 by their average of 73. The sum, and thus the mean, has not changed, and the teacher seems to like having the values closer to each other. Now 70 can be subtracted from every term, thus reducing both sides by 280, and the resulting equation has smaller numbers in it. Once the equation is solved the subtracted value of 70 has to be added back.
Definitely not my favorite way to solve this problem.
Click to expand...
Wow. Thank you
blamocur said:
I can try elucidating, but this method does not seem natural to me. The teacher replaced 58 and 88 by their average of 73. The sum, and thus the mean, has not changed, and the teacher seems to like having the values closer to each other. Now 70 can be subtracted from every term, thus reducing both sides by 280, and the resulting equation has smaller numbers in it. Once the equation is solved the subtracted value of 70 has to be added back.
Definitely not my favorite way to solve this problem.
Click to expand...
Does that method have to do with this:
Outlier treatment is the process of removing or replacing conversions, visits, or visitors with a “normal” data point. Removal involves eliminating the data point from the sample. Replacement involves swapping the data point for the mean or median of the sample.
 
blamocur said:
I can try elucidating, but this method does not seem natural to me. The teacher replaced 58 and 88 by their average of 73. The sum, and thus the mean, has not changed, and the teacher seems to like having the values closer to each other. Now 70 can be subtracted from every term, thus reducing both sides by 280, and the resulting equation has smaller numbers in it. Once the equation is solved the subtracted value of 70 has to be added back.
Definitely not my favorite way to solve this problem.
Click to expand...
I understood now. Thank you. Way too involved for my taste. But I was curious about how he did that.
 
blamocur said:
I can try elucidating, but this method does not seem natural to me. The teacher replaced 58 and 88 by their average of 73. The sum, and thus the mean, has not changed, and the teacher seems to like having the values closer to each other. Now 70 can be subtracted from every term, thus reducing both sides by 280, and the resulting equation has smaller numbers in it. Once the equation is solved the subtracted value of 70 has to be added back.
Definitely not my favorite way to solve this problem.
Click to expand...
You said: thus reducing both sides by 280. I don't see how. At least not the left side. Can you post it
 
The numerator of a fraction ALWAYS equals the denominator times the other side.


10/2 = 5, so 2*5 =10
8/4=2, so 4*2=8
a/b=c, so b*c=a

(a+b+c+d+e)/11 =22, then (a+b+c+d+e) = 11*22
 
Jomo said:
The numerator of a fraction ALWAYS equals the denominator times the other side.


10/2 = 5, so 2*5 =10
8/4=2, so 4*2=8
a/b=c, so b*c=a

(a+b+c+d+e)/11 =22, then (a+b+c+d+e) = 11*22
Click to expand...
Thanks a lot Jomo ?
 
Actually, I think the teachers method makes perfect sense. If the mean of p disparate numbers is q, then the sum of those numbers ALWAYS equals the sum of p instances of q.

[math]4 * 71.5 = 58 + 76 + 88 + x \implies 286 = 222 + x \implies x = 286 - 222 = 64.[/math]
Proof

[math]\text {By definition, the arithmetic mean } q \text { of } p \text { numbers } \{x_ 1, \ … x_p \} \text { is}[/math]
[math]\dfrac{\displaystyle \sum_{i=1}^p x_i}{p} = q \implies pq = \sum_{i=1}^p x_i \implies \sum_{i=1}^p q = \sum_{i=1}^p x_i.[/math]
 
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Jomo said:
The numerator of a fraction ALWAYS equals the denominator times the other side.


10/2 = 5, so 2*5 =10
8/4=2, so 4*2=8
a/b=c, so b*c=a

(a+b+c+d+e)/11 =22, then (a+b+c+d+e) = 11*22
Click to expand...
eddy2017 said:
Not an easy method but interesting nonetheless.
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Eddy,
This is how you check division! You claim 816/2 = 408. I disagree with you. How would you convince me that you're correct? Easy, you multiply 2 by 408 and get 816. That would surely convince me that you're correct.
 
JeffM said:
Actually, I think the teachers method makes perfect sense. If the mean of p disparate numbers is q, then the sum of those numbers ALWAYS equals the sum of p instances of q.

[math]4 * 71.5 = 58 + 76 + 88 + x \implies 286 = 222 + x \implies x = 286 - 222 = 64.[/math]
Proof

[math]\text {By definition, the arithmetic mean } q \text { of } p \text { numbers } \{x_ 1, \ … x_p \} \text { is}[/math]
[math]\dfrac{\displaystyle \sum_{i=1}^p x_i}{p} = q \implies pq = \sum_{i=1}^p x_i \implies \sum_{i=1}^p q = \sum_{i=1}^p x_i.[/math]
Click to expand...
Thanks, I was studying this yesterday. What the summation notation meant aka sigma. This comes in very handy.
 
JeffM said:
Actually, I think the teachers method makes perfect sense. If the mean of p disparate numbers is q, then the sum of those numbers ALWAYS equals the sum of p instances of q.

[math]4 * 71.5 = 58 + 76 + 88 + x \implies 286 = 222 + x \implies x = 286 - 222 = 64.[/math]
Proof

[math]\text {By definition, the arithmetic mean } q \text { of } p \text { numbers } \{x_ 1, \ … x_p \} \text { is}[/math]
[math]\dfrac{\displaystyle \sum_{i=1}^p x_i}{p} = q \implies pq = \sum_{i=1}^p x_i \implies \sum_{i=1}^p q = \sum_{i=1}^p x_i.[/math]
Click to expand...
Can this problem be also solved the basic summation notation formula?.
 
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