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Bunuel
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While I agree with the answer, I do not think that the approach used here is correct.
The stem says x is a two-digit number, I am not getting why are you guys considering it to be a multiple of 5?
Stmt 1 says, X when multiplied with 3 will yield a 3 digit number. So X could be anything above 33.
Because 3 times 33 will give 99 and any 2 digit number above this will yield more than 2 digits when multiplied by 3.
So, from stmt 1, X has a lot of options. Hence insufficient.

Coming to stmt 2, it says a < 7 , meaning, the last digit of 3X is less than 7. Insuff.

1 + 2. take number 34, 3*34 = 102.
take number 35, 3*35 = 105.
take number 36, 3*36 = 108, last digit exceeds 7.
So, we are left out with 2 results 102 and 105 whose last digits are below 7. i.e., we have 2 options for X, 34 & 35.
Multiple answers, So both stmts are insuff.

Hence E.

Still, I would like to know how you guys considered X to be a multiple of 5. Please let me know if there is any logic that I might have missed while analyzing the stem.
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reachskishore
While I agree with the answer, I do not think that the approach used here is correct.
The stem says x is a two-digit number, I am not getting why are you guys considering it to be a multiple of 5?
Stmt 1 says, X when multiplied with 3 will yield a 3 digit number. So X could be anything above 33.
Because 3 times 33 will give 99 and any 2 digit number above this will yield more than 2 digits when multiplied by 3.
So, from stmt 1, X has a lot of options. Hence insufficient.

Coming to stmt 2, it says a < 7 , meaning, the last digit of 3X is less than 7. Insuff.

1 + 2. take number 34, 3*34 = 102.
take number 35, 3*35 = 105.
take number 36, 3*36 = 108, last digit exceeds 7.
So, we are left out with 2 results 102 and 105 whose last digits are below 7. i.e., we have 2 options for X, 34 & 35.
Multiple answers, So both stmts are insuff.

Hence E.

Still, I would like to know how you guys considered X to be a multiple of 5. Please let me know if there is any logic that I might have missed while analyzing the stem.

Since x=ba which when multiplied by 3 gives 3 digit number with last digit a, the only possibilities are a=0 or 5 . 0 and 5 are the only digits which when multiplied by 3 gives last digit 0 or 5.
Hope it is clear.
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souvonik2k
reachskishore
While I agree with the answer, I do not think that the approach used here is correct.
The stem says x is a two-digit number, I am not getting why are you guys considering it to be a multiple of 5?
Stmt 1 says, X when multiplied with 3 will yield a 3 digit number. So X could be anything above 33.
Because 3 times 33 will give 99 and any 2 digit number above this will yield more than 2 digits when multiplied by 3.
So, from stmt 1, X has a lot of options. Hence insufficient.

Coming to stmt 2, it says a < 7 , meaning, the last digit of 3X is less than 7. Insuff.

1 + 2. take number 34, 3*34 = 102.
take number 35, 3*35 = 105.
take number 36, 3*36 = 108, last digit exceeds 7.
So, we are left out with 2 results 102 and 105 whose last digits are below 7. i.e., we have 2 options for X, 34 & 35.
Multiple answers, So both stmts are insuff.

Hence E.

Still, I would like to know how you guys considered X to be a multiple of 5. Please let me know if there is any logic that I might have missed while analyzing the stem.

Since x=ba which when multiplied by 3 gives 3 digit number with last digit a, the only possibilities are a=0 or 5 . 0 and 5 are the only digits which when multiplied by 3 gives last digit 0 or 5.
Hope it is clear.

Oh yes !!! Had missed that detail. Sorry about that.
Thanks for clearing out my doubt.

+1Kudos to you
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