I am getting 12.5 so would go with C)

Whats the OA?

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.

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.

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

Guys, the OA - E. I think there is a typo with answer choice E. The answer is 25.

Thanks

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?

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

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.

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

I am also getting = 25

I dont understand this one...I think it should be 18?

What do you say guys?

Please share your analysis