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A three digit number A with hundreds digit a, tens digit b and units

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A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 27 May 2018, 05:31
1
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  95% (hard)

Question Stats:

18% (02:59) correct 82% (03:05) wrong based on 82 sessions

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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
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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 06 Jun 2018, 22:13
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GMATSkilled wrote:
A three digit number A with hundreds digit a, tens digit b and units digit c is written in the reverse order to form another three digit number B. If B > A, is B + A > 1200?

A. B - A is divisible by 8

B. A is a multiple of 5



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

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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 27 May 2018, 06:50
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1
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
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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 27 May 2018, 07:11
Sorry mate, i missed some information.
From statement I, we get:
B-A = (100c+10b+a) - (100a+10b+c) = 99c-99a. This should be a multiple of 8, this is possible only when c is 9 and a is 1, then in such case A+B is always <1200.
Hence, both the statements individually are sufficient to answer.
Imo, option D.

Thanks and regards,
Sharan Salem
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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 06 Jun 2018, 17:39
A three digit number A with hundreds digit a, tens digit b and units digit c is written in the reverse order to form another three digit number B. If B > A, is B + A > 1200?

A. B - A is divisible by 8

B. A is a multiple of 5
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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 08 Jun 2018, 02:29
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A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 10 Jun 2018, 12:50
Hi


Is it possible for digit c to have different values in different Statements.

As per St.1, c=9
As per St.2, c=5
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Re: A three digit number A with hundreds digit a, tens digit b and units  [#permalink]

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New post 22 Sep 2019, 00:55
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Re: A three digit number A with hundreds digit a, tens digit b and units   [#permalink] 22 Sep 2019, 00:55
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