SOLUTION: Here is the word problem I am trying to write an equation for: " Mike received 73%, 71%, 79% and 75% on his first four math tests. What grade on his fifth test would allow him

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Here is the word problem I am trying to write an equation for: " Mike received 73%, 71%, 79% and 75% on his first four math tests. What grade on his fifth test would allow him       Log On


   



Question 572864: Here is the word problem I am trying to write an equation for:
" Mike received 73%, 71%, 79% and 75% on his first four math tests. What grade on his fifth test would allow him to make an average of at least 75%? Assume the test scores are whole numbers."
Here is what I have tried so far:
Let X represent fifth test score
(73 +71+79+75 + X)/5 = 75
(298 + X)/5 = 75
59.6 + 1/5X = 75
X = (75 - 59.6)/5
X = 3.08

I know this is not correct but I am not sure what is wrong with my equation. Any help showing me how to set up the right equation would really help me. thanks

Found 3 solutions by stanbon, mananth, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
" Mike received 73%, 71%, 79% and 75% on his first four math tests. What grade on his fifth test would allow him to make an average of at least 75%? Assume the test scores are whole numbers."
Here is what I have tried so far:
Let X represent fifth test score
(73 +71+79+75 + X)/5 = 75
----
(298+x) = 5*75
298+x = 375
x = 77
=============
Cheers,
Stan H.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Here is what I have tried so far:
Let X represent fifth test score
(73 +71+79+75 + X)/5 = 75
(298 + X)/5 = 75
59.6 + (1/5)X = 75
X = (75 - 59.6)*5
X = 77

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the scores of the 5 tests has to be 5 times the desired average. 5 times 75 is 375. The sum of the first 4 scores is 298. Subtract that from 375 to get 77.

You were actually on the right track:



is correct. At this point, I would have multiplied both sides by 5 to get rid of that pesky denominator, but there is really nothing wrong with your next step though.



is also correct, as is:



and



But right here is where you made your error. You were already dividing by 5 in the LHS, but then you chose to divide by 5 on the right (???).



what you wanted to do was multiply both sides by 5 to get:



John

My calculator said it, I believe it, that settles it
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