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# Data Sufficiency Pack 4, Question 1) Each of the canisters in....

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Re: Data Sufficiency Pack 4, Question 1) Each of the canisters in.... [#permalink]
Cannisters ---> Containers? Also, I was thinking that the question asked for the ratio of containers.
Is there a typo in the question? I did not understand it at all.
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Re: Data Sufficiency Pack 4, Question 1) Each of the canisters in.... [#permalink]
To solve this problem, a formula for weighted averages is very helpful. Why should we care about weighted averages? This is because all the statements talk is concerned with difference among overall average ounce per can, average ounce per large can and average ounce per small can.

let L be the number of large cans , S be the number of small cans, oa be overall weighted average, al be average ounce per large can and as be overall average per small can
then
L/S = oa-as/al-oa
statement #1 tells that al-oa is 6
L/S = oa-as/6.. This still needs numerator to be determined
combining we know that L/S = 4/6.
So why do we expect large cans to be less in number? This is because the average of the entire set is closer to small can average
cheers!
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Re: Data Sufficiency Pack 4, Question 1) Each of the canisters in.... [#permalink]
Good question - Test of weighted average concept.

S ------------- WA --------------- L

Statement 1) An average large container holds 6 ounces more than the average of all the containers in the warehouse.
S-------------- WA ----------------L = WA+6 => no idea about S => insuff

Statement 2) An average small container holds 4 ounces less than the average of all the containers in the warehouse.
S = WA-4 ----------WA ---------------S => no idea about L => insuff

1 + 2
WA-4 ------------WA ---------------WA + 6

so weights are in ratio (6/4), i.e. weightage of small canister(number of small canister)/weight of Large canister (number of large canister) = 3/2 => (C)
Re: Data Sufficiency Pack 4, Question 1) Each of the canisters in.... [#permalink]
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