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If d1 and d2 are the numbers of defective items in production runs yie

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If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 26 Aug 2019, 23:18
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A
B
C
D
E

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  25% (medium)

Question Stats:

83% (01:03) correct 17% (01:07) wrong based on 41 sessions

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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 26 Aug 2019, 23:35
1
We are to determine which production run has a greater ratio of defects per run. d1 is defects in production 1 and d2 is defects in production 2. r1 is the runs in production 1 and r2 is the runs in production 2.

1 is insufficient because merely knowing d1>d2 is not enough to determine if d1/r1 is greater or otherwise than d2/r2.

2. Is also insufficient because it only tells us r2>r1. We need information on d1 and d2 in order to make a judgement.

1+2 clearly is sufficient.
Numerator d1>d2 and denominator r1<r2, hence d1/r1>d2/r2. We can therefore conclude that the defects per runs in production 1 is higher than that for production 2.

Answer is C.

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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 26 Aug 2019, 23:41
1
We have to find the ratio of defects per run of the two production runs i.e d1/r1 and d2/r2

From (1) d1 > d2 say 6>5

without knowing any info regarding r1 and r2 we cannot conclude anything, so clearly insufficient

From (2) r2 > r1 say 5>1

similarly, without knowing any info regarding d1 and d2 we cannot conclude anything, so clearly insufficient

From (1) , (2)
we can say that 6/1 > 5/3 so ans : (C)
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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 27 Aug 2019, 01:08
1
To find which ratio of defect per runs is greater it is to be identified whether d1 > or < d2 and r1 > or < r2 individually. Of course d1 < r1 and d2 < r2

(1) d1 > d2
Clearly, INSUFFICIENT.

(2) r2 > r1
Clearly, INSUFFICIENT.

Together Statement 1) and 2).

Refer table snapshot.

Thus, for sufficiency of \(\frac{d1}{r1} > \frac{d2}{r2}\), d1 > d2 and r1 < r2 – Case 1.

And for sufficiency of \(\frac{d1}{r1} < \frac{d2}{r2}\), d1 < d2 and r1 > r2 – Case 4.

I.e. for a ratio to be greater than other numerator has to be greater than the other and denominator should be lesser than the other.

Answer (C).
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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 27 Aug 2019, 01:35
1
Given: d1 & d2 are defects in production runs yielding r1 & r2 total items

Asked: Number of defects are greater for which of the two production runs?

(1) d1 >d2
Since relationship between r1 & r2 is unknown
NOT SUFFICIENT

(2) r2 > r1
Since relationship between d1 & d2 is unknown
NOT SUFFICIENT

Combining (1) & (2)
(1) d1>d2
(2) r2>r1
d1/r1 > d2/r2
Number of defects per run are greater for production run 1 than production run 2.
SUFFICIENT

IMO C

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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 27 Aug 2019, 02:50
If d1 and d2 are the numbers of defective items in production runs yielding r1 and r2 total items, respectively, the ratio of defects per run is greater for which of the two production runs?

(1) d1 > d2
(2) r2 > r1
need to determine ratio r1/d1 and r2/d2

#1 d1>d2 but runs not given insufficient
#2
r2>r1 but defects not given insufficient
from 1&2
we can say that d1/r1> d2/r2
IMO C
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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 27 Aug 2019, 06:03
E is the answer. Because even on combining we will not get a single result.

Because if (d1/r1) and (d2/r2) are (4/6) & (2/3) then both are equal.
and if (d1/r1) and (d2/r2) are (4/6) & (1/3) then 1st is greater.
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Re: If d1 and d2 are the numbers of defective items in production runs yie  [#permalink]

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New post 27 Aug 2019, 12:22
IMO C
Statement 1 : InSufficient: Tells us about no of defect is greater in d 1>d2. But doesn't tell us total items, It may be possible that d1 > d2, but if r2>r1, ratio can reverse.
e.g : d1 : 10, d2: 6 : r1 : 50 ; r2 : 240
r1/d1 : 5 ::1 , but r2/d2 : 40::1 ,
2nd case: d1 : 10, d2: 6 : r1 : 50 ; r2 : 18
r1/d1 : 5 ::1 , but r2/d2 : 3::1 ,

Insufficient

Statement 2: r2>r1: doesn't tell about defects: Insufficient

Combined: We can eliminate 1 condition from statement 1 : and we have only 1 solution:
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Re: If d1 and d2 are the numbers of defective items in production runs yie   [#permalink] 27 Aug 2019, 12:22
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