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Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Is the two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 61% (01:46) correct 39% (01:33) wrong based on 76 sessions

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Is the two-digit positive integer, RM, a multiple of 7?

(1) R + M = 13
(2) R is divisible by 3

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Re: Is the two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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(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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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Bunuel wrote:
Is the two-digit 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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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 C

Did I miss anything?

eswarchethu135

Archit3110 wrote:
Bunuel wrote:
Is the two-digit 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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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 E

Did I miss anything?

eswarchethu135

Archit3110 wrote:
Bunuel wrote:
Is the two-digit 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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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 two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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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GMAT 1: 760 Q51 V42 Re: Is the two-digit positive integer, RM, a multiple of 7? (1) R + M = 13  [#permalink]

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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 2-digit number and not the product of R and M
hope it is clear
_________________ Re: Is the two-digit positive integer, RM, a multiple of 7? (1) R + M = 13   [#permalink] 18 Dec 2018, 18:37
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