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14 May 2021, 09:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
33% (02:01) correct
67%
(01:28)
wrong
based on 9
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In a factory X out of every Y items produced were found to be defective. If 5% of the items produced during a day were found to be defective, what was the total number of items produced that day?
(A) x/.05
(B) 95x
(C) 20x/y
(D) 20y
(E) None of these
Posted from my mobile device
(A) x/.05
(B) 95x
(C) 20x/y
(D) 20y
(E) None of these
Posted from my mobile device
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14 May 2021, 09:46
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Plugging in x=5 and y=100. In this case, as 5% were defective 100 items were made in total.
(A) x/.05 = 5/(5/100) = 5*(100/5) = 100 correct
(B) 95x = 95*5 = 475
(C) 20x/y = 20*5/100 = 1
(D) 20y = 20(100) = 2000
Answer A
(A) x/.05 = 5/(5/100) = 5*(100/5) = 100 correct
(B) 95x = 95*5 = 475
(C) 20x/y = 20*5/100 = 1
(D) 20y = 20(100) = 2000
Answer A
Emdad
Every y item defective is x
So, every 100 item defective is \(\frac{100x}{y}\)
\(\frac{100x}{y}=5\)% of the production
So 100% of production is \(5×20\)%=\(\frac{20×100x}{y}\)
=\(\frac{2000x}{y}\)
So, answer is E
