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Word Prob

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by beater » Fri Sep 26, 2008 8:26 pm
Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24
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by cramya » Fri Sep 26, 2008 9:19 pm
I am getting 12.5 so would go with C)

Whats the OA?
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by cramya » Fri Sep 26, 2008 9:24 pm
Box W

Let x - > no of red sticks
Let k be its length

Let y be the number of blue sticks
Let m be the length

Box V

Let a - > no of red sticks
Let b be its length

Let c be the number of blue sticks
Let d be the length

I did the weighted average

kx+my/k+m

ab+cd/a+b

Since the red sticks in both the boxes are of same length and based on the given info(19 less than and 6 greater than)


(kx+my/k+m) - 19 = (ab+cd/a+b) + 6

(kx+my/k+m) - (ab+cd/a+b) = 25 which is what we need.

My bad I meant to say 25 not 12.5

I would go with 24.
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by stop@800 » Sat Sep 27, 2008 3:11 am
As per given information
length = red = avgW - 19 = avgV + 6

avgW - 19 = avgV + 6
avgW - avgV = 19 + 6 = 25

I think there is something wrong with question or my understanding.

As 25 is not part of answer choices.

I will not go with 24 since there is no word approximately in the question.
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by cramya » Sat Sep 27, 2008 7:28 am
I did see that. All I am saying if the real exam had a question like this (doubt if it would be since there will not be errros like this with the answer choices) I wil choose 24 since you have to choose some answer before you could move on to the next.

I am also saying it should be 25 and I just want to make that clear :lol:
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by beater » Sat Sep 27, 2008 7:54 am
Guys, the OA - E. I think there is a typo with answer choice E. The answer is 25.

Thanks
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by priyankamishra11 » Sat Sep 27, 2008 2:37 pm
There is an error in question. In actual question it is Average of W- 18 and not 19. With 18 we are getting 24 as an answer .

Here is the revised question:
Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 18 inches less than the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
Regards,
Priyanka
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by vishubn » Sat Sep 27, 2008 11:04 pm
I go with stop@800.... its kinda straight forward ! and the answer has to be 25


question states:
Red(stick)=avg(box[w])-19------(1)
also
Red(stick)=avg(box[v]+6------(2)

but we know length of each red stick is equal

(1)=(2)
avg(box[w])-19=avg(box[v]+6

avg(box[w])-avg(box[v]=6+19
which is 25

Oa????

Vishu
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by [email protected] » Wed Jul 08, 2009 3:08 am
The question reads-
Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 18 inches less than the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
A. 3
B. 6
C. 12
D. 18
E. 24

The OA is : E

If you could pls. explain in detail.

Regards,
Niharika.
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Response:

by [email protected] » Wed Jul 08, 2009 3:19 am
Let 'a' be th average sticks in Box W.
Let 'b' be th average sticks in Box V.


Therefore the equation reads-
a - 18 = b +6
a-b(Diff in Box W & Box V) = 18+6 = 24 .

Ans. is E
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by imhimanshu » Wed Jul 08, 2009 8:10 am
I am also getting = 25
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by ssgmatter » Sun Jun 13, 2010 5:23 am
I dont understand this one...I think it should be 18?

What do you say guys?

Please share your analysis
Best-
Amit
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