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Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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10 Dec 2018, 03:11
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62% (01:48) correct 38% (01:33) wrong based on 78 sessions
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Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13 (2) R is divisible by 3
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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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10 Dec 2018, 03:41
(1) R + M = 13 R = 4 and M = 9 RM is divisible by 7. R = 5 and M = 8 RM is not divisible by 7. INSUFFICIENT (2) R is divisible by 3 R = 6 or 9 M = 7 or 4 RM = 67 or 94. Both are not multiples of 7. SUFFICIENT. OPTION: B
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Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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Updated on: 10 Dec 2018, 09:23
Bunuel wrote: Is the twodigit positive integer, RM, a multiple of 7?
(1) R + M = 13 (2) R is divisible by 3 from 1: R,M can be ( 49,94,58,85,67,76) so 49 is divisible by 7 so in sufficient from 2: r divisible by 3 so RM can be ( 67 or 94) both are not divisible by 7 or 63 divisible by 7 insufficient Combining 1 & 2 we get 94,67 both are in sufficient , so RM is not multiple of 7 IMO C
Originally posted by Archit3110 on 10 Dec 2018, 05:45.
Last edited by Archit3110 on 10 Dec 2018, 09:23, edited 1 time in total.



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Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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Updated on: 10 Dec 2018, 09:34
Archit3110 I am thinking C is the answer. Here is how: Is RM multiple of 7? St 1: RM could be 67 (insufficient), 85 (sufficient), or 94 (insufficient). So St 1 is insufficient St 2: RM could be 63 (sufficient) or 67 (insufficient). So St 2 is insufficient Combining Sts 1 and 2 (R + M = 13 and R is divisible by 3), RM = 67 (insufficient) or 94 (insufficient). So answer is CDid I miss anything? eswarchethu135Archit3110 wrote: Bunuel wrote: Is the twodigit positive integer, RM, a multiple of 7?
(1) R + M = 13 (2) R is divisible by 3 from 1: R,M can be ( 49,94,58,85,67,76) so 49 is divisible by 7 so in sufficient from 2: r divisible by 3 so RM can be ( 67 or 94) both are not divisible by 7 IMO B
Originally posted by bebs on 10 Dec 2018, 08:11.
Last edited by bebs on 10 Dec 2018, 09:34, edited 1 time in total.



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Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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10 Dec 2018, 09:19
funsogu wrote: Archit3110 I am thinking E is the answer. Here is how: Is RM multiple of 7? St 1: RM could be 67 (insufficient), 85 (sufficient), or 94 (insufficient). So St 1 is insufficient St 2: RM could be 63 (sufficient) or 67 (insufficient). So St 2 is insufficient Combining Sts 1 and 2 (R + M = 13 and R is divisible by 3), RM = 67 (insufficient) or 94 (insufficient). So answer is EDid I miss anything? eswarchethu135Archit3110 wrote: Bunuel wrote: Is the twodigit positive integer, RM, a multiple of 7?
(1) R + M = 13 (2) R is divisible by 3 from 1: R,M can be ( 49,94,58,85,67,76) so 49 is divisible by 7 so in sufficient from 2: r divisible by 3 so RM can be ( 67 or 94) both are not divisible by 7 IMO B funsogu : you seem to correct on stmnt 2 , it would be insufficient.. Combining 1 & 2 we get 94,67 ,so RM is not multiple of 7 IMO C



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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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10 Dec 2018, 09:36
Archit3110 You are correct. I meant to say C. This how I lose easy points on DS questions.



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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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10 Dec 2018, 09:41
funsogu wrote: Archit3110 You are correct. I meant to say C. This how I lose easy points on DS questions. funsogu: no worries mate , its part of learning .. thanks to you for highlighting the error in my solution.. even i initially got the question wrong..



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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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18 Dec 2018, 10:30
Can someone please explain where I went wrong: 1.) R+M = 13 E.g. 1+12,2+11,3+10,4+11...... 12+1 2.) R is divisible by 3 E.g. 3,6,9,12 1&2 combined could be: R = 3,6,9,12 M = 10,7,4,1 30/7 does not work 56/7 would work Please help me out
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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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18 Dec 2018, 18:37
Arro44 wrote: Can someone please explain where I went wrong: 1.) R+M = 13 E.g. 1+12,2+11,3+10,4+11...... 12+1 2.) R is divisible by 3 E.g. 3,6,9,12 1&2 combined could be: R = 3,6,9,12 M = 10,7,4,1 30/7 does not work 56/7 would work Please help me out R and M are digits. They can only take values from 0 to 9 RM is a 2digit number and not the product of R and M hope it is clear
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Re: Is the twodigit positive integer, RM, a multiple of 7? (1) R + M = 13
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