Question

(ALGEBRA!) Joe's average score on 3 tests was 79. His average for his next 4 tests were 78.5. There are 2 tests left in the semester and Joe needs to raise his average to at least 80. What must he average on the next 2 tests?

Answer 1

HELP PLEASE!

Answer 2

@nabizag help?

Answer 3

@lgbasallote

Answer 4

let x be the total for the first three
y = total of the next 4 tests
z = total of the next 2 tests..

okay so far?

Answer 5

yes

Answer 6

@nincompoop help?

Answer 7

sorry i got lagged up. anyway the final average would be (x+y+z)/9 >= 80 correct?

Answer 8

80%?

Answer 9

uhh...no...im just interpreting your problem into algebra...it's not the answer yet

Answer 10

your question said the final average is 80% so that's what i wrote

Answer 11

yes

Answer 12

anyway

(x+ y + z)/9 \(\ge\) 80

now cross multiply..what do you get?

Answer 13

do i add 78.5, 80, and 79?

Answer 14

then divide by 9?

Answer 15

no..

Answer 16

78.5 and 79 are averages...x y and z are totals

Answer 17

do you see what im planning to do?

basically what you're aiming to do is:

score for first quiz + score of second quiz + score of third quiz + ....+ score of 9th quiz then divide it all by 9 and the result should be greater than or equal to 80

make sense?

Answer 18

yes

Answer 19

92.61 I THINK... I shall now try and figure out how the hell I came to this figure......

Answer 20

good.

so you have \[\frac {x + y + z}9 \ge 80\]
cross multiply..what do you get?

Answer 21

26.3 less than or equal to 80 then subtract the 26.3?

Answer 22

Actually no, Im pretty sure im wrong...

Answer 23

why do you have 26.3?

Answer 24

I divided the answer i got of all 3 numbers by 9

Answer 25

just focus on the equation im just showing you right now

\[\frac{x+y+z}9 \ge 80\]
cross multiply

don't put in any additional information

Answer 26

okay I got 237.7 greater than or equal to 80

Answer 27

How do i cross multiple that, Im not sure how too.

Answer 28

why are you getting numbers? i wrote letters and they come out numbers?

Answer 29

you just multiply both sides by 9. nothing more

Answer 30

okay
so i got 9x+9y+18>=720

Answer 31

if you multiply (x+y+x)/9 by 9..the 9 just cancels out..so it should be \[x + y + z \ge 720\]
got it?

Answer 32

okay yes

Answer 33

good so remember that equation..we'll come back to that later. for now..we gve x and y some values

Answer 34

it says the average of the first 3 tests is 79..so you can equate this as \[\frac x3 = 79\]
right? since x is the total of the first 3 tests

Answer 35

The answer is definitely 84.5

Answer 36

yes

Answer 37

so now solve for x by multiplying both sides by 3

Answer 38

x=237

Answer 39

right

Answer 40

now it says the average of the next 4 quizzes is 78.5 so \[\frac y4 = 78.5\]
solve for y

Answer 41

y=314

Answer 42

right. so you have x = 237 and y = 314

now substitute that into x +y + z > = 720 then solve for z

Answer 43

ok

Answer 44

what do you get?

Answer 45

z>169

Answer 46

close..

z \(\ge\) 169

since z is the total of the next two tests, to get the average, divide 169 by 2

Answer 47

okay

Answer 48

84.5

Answer 49

right. so he needs at least 84.5 average in the next two tests

Answer 50

thank you!

Answer 51

indeed it was

Answer 52

I concur! ;)