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Math Expert V
Joined: 02 Sep 2009
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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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Difficulty:   75% (hard)

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

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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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Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1684
Location: India
Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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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

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Intern  Joined: 25 Feb 2016
Posts: 4
Schools: Alberta "22
Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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Is there a shorter way for finding sum of all the 2 digit factors of 8?
Intern  Joined: 25 Feb 2016
Posts: 4
Schools: Alberta "22
Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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Is there a shorter way for finding sum of all the 2 digit factors of 8?
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
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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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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.
Intern  Joined: 23 Oct 2015
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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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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

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 Marshall & McDonough Moderator D
Joined: 13 Apr 2015
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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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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

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.
Intern  Joined: 23 Oct 2015
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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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Right, got it. Also read Veritas Prep's approach on this as well and it made it clearer, thanks!
Manager  Joined: 01 Mar 2014
Posts: 108
Schools: Tepper '18
Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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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

Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96???
Marshall & McDonough Moderator D
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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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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

Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96???

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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Joined: 12 Aug 2015
Posts: 2567
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GRE 1: Q169 V154 Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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Nice One...
The remainders must be => 16,24,,,,,96=> using the AP sum sequence => 616
hence B
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Joined: 03 Apr 2017
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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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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?
Intern  B
Joined: 19 Feb 2017
Posts: 42
Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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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.
Manager  B
Joined: 24 Jul 2019
Posts: 71
X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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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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Go kick in the door. 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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