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what's your question?

Good Question @chihiroasleaf

I have worked over this problem

Faizan :o

AM=4
x+3+2+0+8+y+3+4/8=4
x+y+20=32
x+y=12

going good so far

Looks like the question is wrong

why x AND y ??
any one of them can be 4

No but as its the mode so it should be the most recurring.....

I mean to say that 3 occurs 2 times and it is not the mode.So 4 should occur 3 times to be the mode

And if 3 is 2 times 4 is 2 times and then 8 would be 2 times,We have 3 modes here

yes so what? we can select mode as any 1 of the 3, and we select here 4

Okay got it,What should my step be now?

What should i do after getting an eqn?

take any one of 'x' or 'y' as 4

Okay got it thanks a lot,but what should i state

Why am i taking x or y as 4?

I need to do some reasoning?

because mode =4
if none of x or y is 4, then mode cannot be 4

Got it bro thanks :D

welcome :)