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gmatquant25
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But what makes E correct, and why not C?
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gmatquant25
Is J + K > L + M ?

(1) J > L + M

(2) K > L + M

we can clearly see 1 and 2 alone are insufficient.

(1)+(2)

assume all terms are +ve
add the inequality
(j+k) > 2(l+m)

assume all terms are -ve
add the inequality
(j+k) < 2(l+m)

so E. :)
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But how will we do this type of que in real exam? If J>L+M and K>L+M then we can say that both J and K are greater than (L+M) so on right side we have two quantities J and K which both are individually greater than (L+M). then their sum(j+k) should be logically greater than L+M?
Bunuel


Is J + K > L + M ?

(1) J > L + M. If J = K = 1 and L + M = 0, then J + K > L + M but if If J = 1, K = -1 and L + M = 0, then J + K = L + M. Not sufficient.

(2) K > L + M. If J = K = 1 and L + M = 0, then J + K > L + M but if If J = -1, K = 1 and L + M = 0, then J + K = L + M. Not sufficient.

(1)+(2):

If J = K = 1 and L + M = 0, then J + K > L + M
If J = K = -1 and L + M = -2, then J + K = L + M.

Not sufficient.

Answer: E.
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mkeshri185
But how will we do this type of que in real exam? If J>L+M and K>L+M then we can say that both J and K are greater than (L+M) so on right side we have two quantities J and K which both are individually greater than (L+M). then their sum(j+k) should be logically greater than L+M?

Even with J > L + M and K > L + M, the sum is not forced to be greater.

Example from the solution: J = K = -1, L + M = -2.
Then J > L + M and K > L + M, but J + K = -2, so not greater.
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