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27 May 2018, 05:31
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (03:06) correct
77%
(03:06)
wrong
based on 496
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A three digit number A with hundreds digit a, tens digit b and units digit c is written in reverse order to form another three digit number B. If B > A , is B + A > 1200?
(1) B - A is divisible by 8
(2) A is a multiple of 5
(1) B - A is divisible by 8
(2) A is a multiple of 5
06 Jun 2018, 22:13
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GMATSkilled
A is 100a+10b+c
B is 100c+10b+a
Let's see the statements
1) B-A is div by 8
So (100c+10b+a)-(100a+10b+c)=99(c-a) is div by 8...
Since C and a are digits, ways the equation is satisfied
a) c is 9 and a is 1 , number becomes 9b1 and 1b9
Even you take b as max possible 9, sum becomes 991+199=1190, <1200
b) c is 8 and a is 0, but then A becomes two digit number so not possible..
So number can be only 9b1 and 1b9, and answer will be NO irrespective of value of B
Sufficient
2) A is a multiple of 5..
So C can be either 5 or 0 but 0 is not possible as the other number B becomes two digit number.
Hence ab5 and 5ba where 5ba>ab5 but the larger number has to be > 1200/2 or 600.. hence Ans is NO
Sufficient
D
27 May 2018, 06:50
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Statement I alone is not sufficient to answer the question, since it gives us multiple possibilities.
Let's take statement II:
Since A is a multiple of 5, it should end with 5 or 0, since the reverse order number should also be a three digit number, the number cannot end with 0. Hence, A is of the form ab5. Then B should be 5ba, then B+A=5ba + ab5. Since B>A, a can take values (5,4,3,2,1), in any case A+B will always be <1200. Hence statement II alone is sufficient to answer.
IMO, option B.
Thanks and regards,
Sharan Salem
Let's take statement II:
Since A is a multiple of 5, it should end with 5 or 0, since the reverse order number should also be a three digit number, the number cannot end with 0. Hence, A is of the form ab5. Then B should be 5ba, then B+A=5ba + ab5. Since B>A, a can take values (5,4,3,2,1), in any case A+B will always be <1200. Hence statement II alone is sufficient to answer.
IMO, option B.
Thanks and regards,
Sharan Salem
