A certain expressway has Exits J, K, L, and M, in that order. What is the road distance from Exit K to Exit L ?
(1) The road distance from Exit J to Exit L is 21 kilometers.
(2) The road distance from Exit K to Exit M is 26 kilometers.
I thought the answer would be D because, we know the distance from J to L and K to M.
Can someone further explain why E is the correct choice?
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We want to find distance for K to LMalcolmW wrote:A certain expressway has Exits J, K, L, and M, in that order. What is the road distance from Exit K to Exit L ?
(1) The road distance from Exit J to Exit L is 21 kilometers.
(2) The road distance from Exit K to Exit M is 26 kilometers.
I thought the answer would be D because, we know the distance from J to L and K to M.
Can someone further explain why E is the correct choice?
1) J to L = 21; nothing about K so statement is insufficent
2) K to M = 26; nothing about L so statement is insufficent
3) Combining statements, lets use a number line where J is at point 0
possibility: J=0; L=21; K=1; M=27 -> K to L = 20
possibility: J=0; L=21; K=5; M=31 -> K to L = 16
Because we can have diferent values, statement is not sufficent
ans = e
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There is nothing that anchors these points to each other. K could be very close to J, or it could be very close to L. These different position would create different distances from K to L.
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Thank you. I understand your explanation. The problem is when I think this way I am usually wrong and when I do not its the opposite.Jim@StratusPrep wrote:There is nothing that anchors these points to each other. K could be very close to J, or it could be very close to L. These different position would create different distances from K to L.
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Test EXTREMES.MalcolmW wrote:A certain expressway has Exits J, K, L, and M, in that order. What is the road distance from Exit K to Exit L ?
(1) The road distance from Exit J to Exit L is 21 kilometers.
(2) The road distance from Exit K to Exit M is 26 kilometers.
When the statements are combined, the following cases are possible:
Case 1:
J <--20--> K <--1--> L <--25--> M.
In this case, KL = 1.
Case 2:
J <--1--> K <--20--> L <--6--> M.
In this case, KL = 20.
Since KL can be different values, the two statements combined are INSUFFICIENT.
The correct answer is E.
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For these kind of questions use number line.
We need to find distance between K and L
--------------J-------K--------L-------M-------
J=0
Statement 1: distance between J to L is given.
-------------(J=0)--------K(not given)-----------(L=21)---------M
so we know distance between J and L BUT we do not know what is value of K.
so statement is not sufficient to find KL.
Statement 2)
J-------------(K=0)--------L(not given)-----------(M=26)------
Here also we don't know what's value of K.
So statement 2 is not sufficient.
Now take statement 1 and 2 together
J to l = 21 and K to M = 26
-------------(J=0)--------K(not given)-----------(L=21)---------------M
J---------(K=0)--------------------L(not given)------(M=26)------
We don't know what is value of both K and L.
therefore, OA is E.
We need to find distance between K and L
--------------J-------K--------L-------M-------
J=0
Statement 1: distance between J to L is given.
-------------(J=0)--------K(not given)-----------(L=21)---------M
so we know distance between J and L BUT we do not know what is value of K.
so statement is not sufficient to find KL.
Statement 2)
J-------------(K=0)--------L(not given)-----------(M=26)------
Here also we don't know what's value of K.
So statement 2 is not sufficient.
Now take statement 1 and 2 together
J to l = 21 and K to M = 26
-------------(J=0)--------K(not given)-----------(L=21)---------------M
J---------(K=0)--------------------L(not given)------(M=26)------
We don't know what is value of both K and L.
therefore, OA is E.
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That does lead me to one of my favorite GMAT principles: if you're down to two answers on the GMAT on test day -- not at home working out of a book -- and one of the answers seems easy and obvious, you typically want to pick the other answer!MalcolmW wrote:Thank you. I understand your explanation. The problem is when I think this way I am usually wrong and when I do not its the opposite.
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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.
A certain expressway has Exits J, K, L, and M, in that order. What is the road distance from Exit K to Exit L ?
(1) The road distance from Exit J to Exit L is 21 kilometers.
(2) The road distance from Exit K to Exit M is 26 kilometers.
In the original condition, if we let the distance between J&K=a, K&L=b, L&M=c, the question is asking about the value of b, so there are 3 variables and 2 equations are given from the conditions, so there is high chance (E) will be our answer.
Looking at them together, a+b=21, b+c=26, we cannot get a unique value for b, so this is insufficient, and the answer is (E).
For cases where we need 3 more equation, 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.
A certain expressway has Exits J, K, L, and M, in that order. What is the road distance from Exit K to Exit L ?
(1) The road distance from Exit J to Exit L is 21 kilometers.
(2) The road distance from Exit K to Exit M is 26 kilometers.
In the original condition, if we let the distance between J&K=a, K&L=b, L&M=c, the question is asking about the value of b, so there are 3 variables and 2 equations are given from the conditions, so there is high chance (E) will be our answer.
Looking at them together, a+b=21, b+c=26, we cannot get a unique value for b, so this is insufficient, and the answer is (E).
For cases where we need 3 more equation, 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.
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