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In the correctly-worked addition operation above, A, B, C, and D each

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In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 05:53
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (01:24) correct 29% (01:34) wrong based on 126 sessions

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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 06:14
A. D = 2B

B. B > 5

C. C = 2A

D. A < 5

E. D > B


We know that D= 2B from the question itself so hold option A

B) B can also be less than 5 (can take 0, 1, 2, 3, 4) rejected

C) C=2A what if their is a carryover rejected

D) A can take 1,2,3,4 but cannot take 0 rejected

E) D can be equal to B when B=0 rejected

Hence A
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 06:21
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mynamegoeson

what if A = 3 and B = 9
C = 7 and D = 8

D = 8 is not equal to 2* B = 18
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 06:28
Leo8 wrote:
mynamegoeson

what if A = 3 and B = 9
C = 7 and D = 8

D = 8 is not equal to 2* B = 18


Yes ohh i missed it the trick here was "A, B, C, and D each represent unique digits"

so B=D=0 is not possible
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 06:38
Leo8 wrote:
mynamegoeson

what if A = 3 and B = 9
C = 7 and D = 8

D = 8 is not equal to 2* B = 18


Wondering what would be the answer
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 06:58
IMO ans must be D

a< 5
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 07:10
mynamegoeson

Attachment:
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FullSizeRender (8).jpg [ 63.39 KiB | Viewed 1568 times ]

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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 12:09
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Answer is D.

A. D = 2B => this cannot be true when value of B is equal to or greater than 5.

B. B > 5 => B can be less than 5 also, since unique digit can range from 0 to 9 without any conditions.

C. C = 2A => this cannot be true, when value of B is equal to or greater than 5. Let's say if B=5, A =4, then C=4+4+1=9, which conflicts the statement.

D. A < 5 => This condition must be true because if A is equal to or greater than 5, then it will have value equal to or above 10, but the given statement is -> C is a unique digit, so value of C can be maximum 9.

E. D > B => This statement cannot be true when value of B is equal to or greater than 5. eg. If B=5, then D=0, which conflicts the given statement.
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 12:17
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Yes, you're doubling a two-digit number AB here (adding it to itself), and getting a two-digit number as a result. So AB has to be less than 50, or else you'd get a three-digit result. And if AB < 50, that means A < 5.
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 19:07
If A>5, then the answer would be a three digit number, so A must be less than 5. Ans - D.
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 08 Jul 2017, 22:30
It should be D as anything greater then 5 (for A position ) will give the end result in 3 digits


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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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New post 14 Jul 2017, 22:33
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Re: In the correctly-worked addition operation above, A, B, C, and D each  [#permalink]

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Re: In the correctly-worked addition operation above, A, B, C, and D each   [#permalink] 03 Mar 2019, 06:25
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