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12 Jul 2017, 01:55
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:21) correct
21%
(02:13)
wrong
based on 149
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Not Attempted Yet
On the line above, if AB < BC < CD < DE, which of the following must be true?
(A) AC < CD
(B) AC < CE
(C) AD < CE
(D) AD < DE
(E) BD < DE
Attachment:
2017-07-12_1253.png [ 3.09 KiB | Viewed 4584 times ]
12 Jul 2017, 03:10
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I tried to solve this question by plugging in nos. but really didn't get anything out of it.
Just my best guess:
Is it B?
AC<CE
That's because: AB<BC and BC<CD and DE
So AB is also less than CD and DE.
So the sum of the smaller two nos. would be less than sum of the bigger two nos.
(I am not sure if this correct).
Waiting for your method Bunuel.
Thanks
Just my best guess:
Is it B?
AC<CE
That's because: AB<BC and BC<CD and DE
So AB is also less than CD and DE.
So the sum of the smaller two nos. would be less than sum of the bigger two nos.
(I am not sure if this correct).
Waiting for your method Bunuel.
Thanks
Luckisnoexcuse
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12 Jul 2017, 04:43
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Bunuel
Given AB<BC<CD<DE
Please note that all the values which AB, BC, CD or DE take will be positive. Lets assume
AB = a
BC=a+x
CD=a+x+y
DE=a+x+y+z
All a,x,y,z are positive
(A) AC < CD => AB+BC<CD => a+a+x<a+x+y (a could be greater than y) Nope
(B) AC < CE => AB+BC<CD+DE => a+a+x<a+x+y+a+x+y+z => 0<y+x+y+z and since x,y,z is positive..this is our answer
(C) AD < CE => AB+BC+CD<CD+DE => AB+BC<DE => a+a+x<a+x+y+z (a could be greater than y+z) Nope
(D) AD < DE => AB+BC+CD < DE => a+a+x+a+x+y<a+x+y+z (2a+x could be larger than z) Nope
(E) BD < DE => BC+CD<DE => a+x+a+x+y<a+x+y+z (a+x could be larger than z) Nope
Hence B
We can also use smart numbers like a=6, x=2, y=3, z=4
