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You can use alligation to solve this.
Ratio X=0.75-0.5=0.25

Ratio Y=0.8-0.75=0.05
Now the ratio y/x is:
\(\frac{Y}{X}=\frac{0.25}{0.05}\) or \(\frac{5}{1}\)

So for 6 products there were 5 Y and 1 X. 5/6 is the answer
If you wanna know more about this method check my signature : mixture!
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navigator123
Each of the products produced yesterday was checked by x or y. 0.5% of the products checked by x are defective and 0.8% of the products checked by y are defective. If the total defective rate of all the products checked by x and y is 0.75%, what fraction of the products was checked by y?

A. 3/5
B. 5/6
C. 7/11
D. 4/5
E. 5/8


This is a allegation problem.

We can use a shortcut.

See the attachment for the shortcut.Its a simple one


So X/Y = 0.05/0.25 = 1/5

Therefore the fraction of products checked by Y = 5/6


Hi, I am not aware of allegation. can you please explain about this or point me to some useful but short source.
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Differential approach

---50------------75------80----->

25x=5y

x/y=5/25=1/5, so y/x+y=5/6

B
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navigator123
Each of the products produced yesterday was checked by x or y. 0.5% of the products checked by x are defective and 0.8% of the products checked by y are defective. If the total defective rate of all the products checked by x and y is 0.75%, what fraction of the products was checked by y?

A. 3/5
B. 5/6
C. 7/11
D. 4/5
E. 5/8

0.5X + 0.8Y = 0.75 ( X + Y )
=> 0.05 Y = 0.25 X
=> Y / X = 5 / 1

Fraction = 5/6

Answer : B
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