Zeke's average score on his five Pre-Algebra tests was an 88%. If he scored a 78% on his first test, an 83% on the second test, a 91% on the third test, and a 93% on the fourth test, what was his score on the fifth test?
Please answer the whole problem, this is not homework, or anything, i just wanna how good at math people really are.. Answer nicely too..!
2006-12-12 08:12:32 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'm pretty good, but I'm not sure I feel like being tested right now. :)
Okay, I give in...
Let n be the score on the fifth test.
(78 + 83 + 91 + 93 + n) / 5 = 88
Add the numbers:
(345 + n) / 5 = 88
Multiply both sides by 5:
345 + n = 440
Subtract 345 from both sides:
n = 95
The score on the last test was 95%
2006-12-12 08:15:17 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
If the average score on 5 tests is 88%, then the total sum of the scores is 88*5 = 440
The sum total of the first 4 tests is 78+83+91+93 = 345
So the mark on the 5th test is 440 - 345 = 95%
2006-12-12 16:17:00 · answer #2 · answered by claudeaf 3 · 0⤊ 0⤋
total of the five tests=5*88=440
total of the first four tests=78+83+91+93=345
score in the fifth test=440-345=95%
2006-12-12 16:16:27 · answer #3 · answered by raj 7 · 0⤊ 0⤋
ok so 88 = (78+83+91+93+x)/5
so 5*88= 78+83+91+93+x
440 -78-83-91-93 = x
95 = x
2006-12-12 16:18:57 · answer #4 · answered by JimE 2 · 0⤊ 0⤋
(78 + 83 + 91 + 93 + x) / 5 = 88
x = 95
2006-12-12 16:16:49 · answer #5 · answered by teacher16 1 · 0⤊ 0⤋
[78+83+91+93+ x]/5 = 88
Solve for x. Nice enough, for ya?
2006-12-12 16:18:05 · answer #6 · answered by S. B. 6 · 0⤊ 0⤋
95 on the 5th test
2006-12-12 16:15:42 · answer #7 · answered by hanumistee 7 · 0⤊ 0⤋