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X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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07 Mar 2016, 11:28
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X and Y are both integers. If X/Y = 59.32, then what is the sum of all the possible two digit remainders of X/Y? A. 560 B. 616 C. 672 D. 900 E. 1024
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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07 Mar 2016, 12:33
Remainder = 0.32 > 32/100 > Can be written as (32/4) / (100/4) = 8/25
So remainders can be 8, 16, 24, 32, ..... 96.
We need the sum of only 2 digit remainders > 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616
Answer: B




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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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08 Mar 2016, 13:37
Is there a shorter way for finding sum of all the 2 digit factors of 8?



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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08 Mar 2016, 13:37
Is there a shorter way for finding sum of all the 2 digit factors of 8?



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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08 Mar 2016, 17:55
mosurok wrote: Is there a shorter way for finding sum of all the 2 digit factors of 8? Yes. Once you realize that the numbers are multiple of 8 and they are in AP, you can apply the formula \((first term + last term)/2\) * Number of terms. > 56 * 11 = 616.



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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09 Mar 2016, 04:40
Vyshak wrote: Remainder = 0.32 > 32/100 > Can be written as (32/4) / (100/4) = 8/25
So remainders can be 8, 16, 24, 32, ..... 96.
We need the sum of only 2 digit remainders > 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616
Answer: B Can you please explain the first step? I did not understand how doing that step could decide what the remainders could be, escpecially the 8/25. Thank you



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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09 Mar 2016, 04:59
nilleskold wrote: Vyshak wrote: Remainder = 0.32 > 32/100 > Can be written as (32/4) / (100/4) = 8/25
So remainders can be 8, 16, 24, 32, ..... 96.
We need the sum of only 2 digit remainders > 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616
Answer: B Can you please explain the first step? I did not understand how doing that step could decide what the remainders could be, escpecially the 8/25. Thank you We have to reduce 32/100 to its lowest ratio by dividing both numerator and denominator with a common factor. In other words the original ratio (0.32) gets unaffected by dividing both numerator and denominator with a common factor. We cannot reduce it beyond 8/25 as there are no common factors beyond this point.



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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09 Mar 2016, 06:46
Right, got it. Also read Veritas Prep's approach on this as well and it made it clearer, thanks!



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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16 Mar 2016, 05:31
Vyshak wrote: Remainder = 0.32 > 32/100 > Can be written as (32/4) / (100/4) = 8/25
So remainders can be 8, 16, 24, 32, ..... 96.
We need the sum of only 2 digit remainders > 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616
Answer: B Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96??? Please explain??



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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16 Mar 2016, 05:45
MeghaP wrote: Vyshak wrote: Remainder = 0.32 > 32/100 > Can be written as (32/4) / (100/4) = 8/25
So remainders can be 8, 16, 24, 32, ..... 96.
We need the sum of only 2 digit remainders > 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616
Answer: B Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96??? Please explain?? Once you find out 8/25, you just have to multiply the numerator and denominator by same values. (8*2)/(25*2) > Fractional part = 0.32 (8*3)/(25*3) > Fractional part = 0.32 . . . (8*12)/(25*12) > Fractional part remains unchanged, but the remainders vary. Since we are asked only 2 digit remainders we take 16, 24, .... upto 96.



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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17 Mar 2016, 00:17
Nice One... The remainders must be => 16,24,,,,,96=> using the AP sum sequence => 616 hence B
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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16 Aug 2018, 11:16
Bunuel wrote: X and Y are both integers. If X/Y = 59.32, then what is the sum of all the possible two digit remainders of X/Y?
A. 560 B. 616 C. 672 D. 900 E. 1024 So what if the question were If X/Y = 59.325, then what is the sum of all the possible two digit and 3 digit remainders of X/Y? Would it have been 325/1000 = 13/40 Hence 13, 26, 39,..., (13*76) = 988. Am I getting this correctly? I guess the crux if my query is, why are we considering 100 as the denominator, because of the '2 digit remainder' that's mentioned?



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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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31 Aug 2019, 07:01
Bunuel wrote: X and Y are both integers. If X/Y = 59.32, then what is the sum of all the possible two digit remainders of X/Y?
A. 560 B. 616 C. 672 D. 900 E. 1024 Hi Bunuel, Could you please explain the concept needed to solve this problem? This would be of great help. Thanks.



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X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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13 Sep 2019, 04:29
aviejay wrote: Bunuel wrote: X and Y are both integers. If X/Y = 59.32, then what is the sum of all the possible two digit remainders of X/Y?
A. 560 B. 616 C. 672 D. 900 E. 1024 Hi Bunuel, Could you please explain the concept needed to solve this problem? This would be of great help. Thanks. Well I am not him but just went through that problem. There is a rule that you can apply with decimals and remainder question that the digits behind the decimal (in this case 0.32) equivalent "Remainder / Divisor" In this case this would mean "32/100" The tricky thing about this rule is that you can't determine with a hundert percent certainty what the actual combination of remainder and divisor is. Think about it > 8 / 25 = 0.32 as well which would be the same as a remainder of 32 and a divisor of 100. In this question they only ask about the sum of the two digit remainders so we go from 8 / 25 up to all the way to 100 in steps of 8. which means 16 , 24 , 32 .... 96 \((96+16)/2 =56\) \((56*11)=616\)
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X and Y are both integers. If X / Y = 59.32, then what is the sum of a
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