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Updated on: 06 Dec 2015, 12:49
Originally posted by EMPOWERgmatRichC on 20 Nov 2015, 15:18.
Last edited by EMPOWERgmatRichC on 06 Dec 2015, 12:49, edited 1 time in total.
Last edited by EMPOWERgmatRichC on 06 Dec 2015, 12:49, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (02:01) correct
55%
(02:06)
wrong
based on 183
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QUANT 4-PACK SERIES Data Sufficiency Pack 4 Question 1 Each of the canisters in...
Each of the canisters in a warehouse is classified as a small canister or a large canister. What is the ratio of small canisters to large canisters in the warehouse?
1) An average large canister holds 6 ounces more than the average of all the canisters in the warehouse.
2) An average small canister holds 4 ounces less than the average of all the canisters in the warehouse.
48 Hour Window Answer & Explanation Window
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.
This question is part of the Quant 4-Pack series
Scroll Down For Official Explanation
Each of the canisters in a warehouse is classified as a small canister or a large canister. What is the ratio of small canisters to large canisters in the warehouse?
1) An average large canister holds 6 ounces more than the average of all the canisters in the warehouse.
2) An average small canister holds 4 ounces less than the average of all the canisters in the warehouse.
48 Hour Window Answer & Explanation Window
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.
This question is part of the Quant 4-Pack series
Scroll Down For Official Explanation
camlan1990
Joined: 11 Sep 2013
Last visit: 19 Sep 2016
Posts: 96
Given Kudos: 26
Schools: Kenan-Flagler '17 SMU '17 McCombs '19
20 Nov 2015, 22:13
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EMPOWERgmatRichC
An average large container = A with the quantity: a
An average small container = B with the quantity: b
a/b =?
1) An average large container holds 6 ounces more than the average of all the containers in the warehouse
=> A - (Aa+Bb)/(a+b)=6
=>b(A-B)/((a+b)=6
Insufficient
2) An average small container holds 4 ounces less than the average of all the containers in the warehouse.
=> a(A-B)/((a+b)=4
Insufficient
Combine 1 and 2 => a/b = 4/6=2/3
Sufficient
Ans: C
22 Nov 2015, 01:54
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Each of the canisters in a warehouse is classified as a small canister or a large canister. What is the ratio of small canisters to large canisters in the warehouse?
1) An average large container holds 6 ounces more than the average of all the containers in the warehouse.
2) An average small container holds 4 ounces less than the average of all the containers in the warehouse.
This is a "2by2" question, one of the most common type of questions in GMAT math
GCDS EMPOWERgmatRichC Data Sufficiency Pack 4, Question 1) Each (20151121).jpg [ 26.55 KiB | Viewed 3970 times ]
There are 4 variables (L,S,a,b), but only 2 equations are given by the 2 conditions, so there is high chance (E) will be the answer.
Looking at the conditions together,
a=[aL+bS/L+S]+6, b=[aL+bS/L+S]-4.
If both sides of b=[aL+bS/L+S]-4 are multiplied by 3/2,
3b/2=3/2[aL+bS/L+S]-6, and if this is added to the equation of a,
a+3b/2=(3/2)[aL+bS/L+S]+[aL+bS/L+S]=(5/2)[aL+bS/L+S], and if 2(L+S) is multiplied to both sides,
(2a+3b)(L+S)=5(aL+bS), 2aL+2aS+3bL+3bS=5aL+5bS,
2aS+3bL=3aL+2bS, 2S(a-b)=3L(a-b).
If both sides are divided by a-b, we get 2S=3L, so the ratio of S:L becomes 3:2 and we can obtain the answer.
The answer therefore becomes (C).
For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
Each of the canisters in a warehouse is classified as a small canister or a large canister. What is the ratio of small canisters to large canisters in the warehouse?
1) An average large container holds 6 ounces more than the average of all the containers in the warehouse.
2) An average small container holds 4 ounces less than the average of all the containers in the warehouse.
This is a "2by2" question, one of the most common type of questions in GMAT math
Attachment:
GCDS EMPOWERgmatRichC Data Sufficiency Pack 4, Question 1) Each (20151121).jpg [ 26.55 KiB | Viewed 3970 times ]
There are 4 variables (L,S,a,b), but only 2 equations are given by the 2 conditions, so there is high chance (E) will be the answer.
Looking at the conditions together,
a=[aL+bS/L+S]+6, b=[aL+bS/L+S]-4.
If both sides of b=[aL+bS/L+S]-4 are multiplied by 3/2,
3b/2=3/2[aL+bS/L+S]-6, and if this is added to the equation of a,
a+3b/2=(3/2)[aL+bS/L+S]+[aL+bS/L+S]=(5/2)[aL+bS/L+S], and if 2(L+S) is multiplied to both sides,
(2a+3b)(L+S)=5(aL+bS), 2aL+2aS+3bL+3bS=5aL+5bS,
2aS+3bL=3aL+2bS, 2S(a-b)=3L(a-b).
If both sides are divided by a-b, we get 2S=3L, so the ratio of S:L becomes 3:2 and we can obtain the answer.
The answer therefore becomes (C).
For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
