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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, 10:28
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58% (02:05) correct 42% (02:24) wrong based on 152 sessions
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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, 11:33
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, 12:37
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, 12:37
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, 16:55
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, 03:40
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, 03:59
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, 05:46
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, 04:31
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, 04:45
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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16 Mar 2016, 23:17



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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, 10:16
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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