( a+b+c+d+e)/5 = 82
a+b+c+d+e = 410
but 90*4 = 360
so he got at least 50
Tom took five math tests and got an integer score on each of them. He never scored higher than 90, and his lowest score was the fourth test. If his average score, rounded to the nearest integer, was 82, what is the lowest possible score he could have gotten on the fourth test?
12 answers
that was not quite right
try again please
Here's an idea, Mr. Al.
Why don't you try something else?
Or, maybe you can find a flaw in Damon's logic, and get a better answer.
Why don't you try something else?
Or, maybe you can find a flaw in Damon's logic, and get a better answer.
I do not know very well. Help me~
Why did you say that Damon was wrong?
Cause I tried entering the problem, it was wrong :(
sometimes answer keys have errors. As far as I can see, Damon is right.
4 tests of 90 score a total of 360 points.
The other test must be at least 50 to achieve the 410 points of 5 tests averaging 82 each.
If any of the other high scores is less than 90, then the 4th test must be even greater than 50.
4 tests of 90 score a total of 360 points.
The other test must be at least 50 to achieve the 410 points of 5 tests averaging 82 each.
If any of the other high scores is less than 90, then the 4th test must be even greater than 50.
Well, try 49 :)
49 +4*90 = 409
409/5 = 81.8 rounds to 82 so ok
try 48
48 +4*90 = 408
408/5 = 81.6 rounds to 82 so ok
try 47
47 + 4*90 = 407
407/5 = 81.4 no good
49 +4*90 = 409
409/5 = 81.8 rounds to 82 so ok
try 48
48 +4*90 = 408
408/5 = 81.6 rounds to 82 so ok
try 47
47 + 4*90 = 407
407/5 = 81.4 no good
The key is "rounded to 82" :)
LOL - thanks
The answer is obviously 50.