Question

Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised. The remaining apples were perfect. How many apples were too small, too green, bruised, or perfect? What if n apples were harvested?
@chefducky

Answer 1

What kind of question is this?! Ohmigosh, math and it's math problems. (-.-)

I'm not sure if I'm going to be right, but I THINK what you can do is divde them by the numbers. So every third apple was too small, so...

- represents the apples that weren't too small, and ~ represents the ones that were small.

--~
--~
--~
etc, etc...

So I think that you would have to divide 6000 by 3 and then get 2000. So I am assuming 2000 would be apples that came out too small.

***
Then, I guess you would do this with the other mathematics problems.
6000 / 4 = 1500 too green apples
6000 / 10 = 600 bruised apples.

Now what I don't know how much it would be in total. It cannot be the sum of these numbers because, well, some might be too green AND too small...

Answer 2

Hope I'm right. :)

Answer 3

im kind of confused by this honestly

Answer 4

Think of a Venn diagram (making assumptions that you line up 6000 apples, and count down the line for every 3rd, every 4th, and every 10th)

6000 / 3 = 2000 apples that were too small
6000 / 4 = 1500 apples that were too green
6000 / 10 = 600 apples that were bruised

however: every 12th apple was too small and too green ==> 500 were counted twice (in the intersection of too small and too green)

every 30th apple were too small and bruised ==> 6000 / 30 = 200 too small and too bruised (in the intersection of too small and bruised)

every 40th apple was too green and bruised ==> 6000 / 40 = 150 too green and too bruised

every 60th apple was the trifecta: too green, too small, and bruised
6000 / 60 = 100 in the intersection of all 3

In the Venn Diagram:
2000 too small
--> 1400 too small only
--> 400 too small and too green
--> 100 too small and bruised

1500 too green
--> 950 too green only
--> 50 too green and bruised

600 bruised
--> 350 bruised only

100 bruised, too green, and too small

add these to give a total of 3350 defective apples

thus, 2650 were perfect

(again, this uses the assumption that the "every x apple" was effectively done in a lineup of the apples, thus multiples of 3 , 4 , or 10 were counted with multiple defects)

Answer 5

@ameenbriscoe13 explained it better and correctly. xD

Answer 6

hmmm... 10, 4 and 3 are not multiples of each other, I don't see any intersections..
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Unless I'm missing something.
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
oh wait.. 20 and 4
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
yea you're right, hang on we'll have to draw a venn diagram.
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
The innermost circle has:
6000/120 = 50 apples
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
x has: (6000/40) - 50 = 150-50 = 100 apples
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Reply Using Drawing

Now we have: 6000/30=200 apples that are both bruised and too small, out of those 200, 50 are in the innermost circle, so 150 are in the other intersection.
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Now, small and green:
6000/12 = 500, but 50 are in the innermost, so you have 450 left:
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Every tenth apple is bruised, so:
6000/10 = 600 apples are in B, but we already put 150, 50 and 100 apples in B, so the rest of B has:
600-150-50-100=600-300=300
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Every fourth apple is too green, so: 6000/4=1500, and the rest of green is:
1500 - 100 - 50 - 450 = 1500 - 600 = 900
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
Finally, every third apple is too small, so:
6000/3 = 2000, but we already included 150, 50, and 450 in S, so:
2000 - 150 - 50 - 450 = 2000 - 650 = 1350 left for S:
Reply Using Drawing

2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
So how many perfect apples were perfect? lol:
6000 - 1350 - 150 - 50 - 450 - 300 - 100 - 900
2 years agoReport AbuseQuote
99
bahrom7893 Honorary Professor of Mathematics Best Response Medals 1
6000 - 1350 - 150 - 50 - 450 - 300 - 100 - 900 = 2700 of perfect apples were perfect :P

Answer 7

@mebengel 6000 - 6000/3 - 6000/4 - 6000/10

Answer 8

A girl in my class had said the answer is 2800 perfect but I am not sure how she got it

Answer 9

@ameenbriscoe13 I am not sure how the girl in my class got 2800 as an answer but the teacher said she knows the right answer.

Answer 10

oh kool