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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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X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 07 Mar 2016, 11:28
1
32
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A
B
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E

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  75% (hard)

Question Stats:

57% (02:18) correct 43% (02:25) wrong based on 177 sessions

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Marshall & McDonough Moderator
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 07 Mar 2016, 12:33
4
2
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  [#permalink]

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New post 08 Mar 2016, 13:37
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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  [#permalink]

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New post 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  [#permalink]

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New post 08 Mar 2016, 17:55
3
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 17 Mar 2016, 00:17
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 16 Aug 2018, 11:16
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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  [#permalink]

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New post 31 Aug 2019, 07:01
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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  [#permalink]

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New post 13 Sep 2019, 04:29
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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   [#permalink] 13 Sep 2019, 04:29
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