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When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y? I. 15.15 II.18.16 III. 17.17

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

I don't have a clue how this will be solved. Can someone please help? Unfortunately, I don't have an OA either.

"x is divided by the integer y, the remainder is 60" --> \(x=qy+60\) --> or \(\frac{x}{y}=q+\frac{60}{y}\), so basically \(\frac{60}{y}\) is a decimal portion of \(\frac{x}{y}\).

For example: \(\frac{x}{y}=15.15\) means that \(\frac{x}{y}=15+0.15=15+\frac{60}{y}\)

Let's look at the options:

I. If \(\frac{x}{y}=15.15\) then \(\frac{60}{y}=0.15\) and \(y=400=integer\), which is possible;

II. If \(\frac{x}{y}=18.16\) then \(\frac{60}{y}=0.16\) and \(y=375=integer\), which is possible;

III. If \(\frac{x}{y}=17.17\) then \(\frac{60}{y}=0.17\) and \(y=\frac{6000}{17}\neq{integer}\), which is not possible as \(y\) must be an integer.

Re: When the integer x is divided by the integer y [#permalink]

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01 Aug 2013, 07:24

Bunuel,

I am confused by the terminology of the question. When x is divided by y the remainder 60. is x=yk+60. "y" is the quotient and the question is asking for possible quotients, ie, possible y values.

But in your solution you have x/y=quotient. Very confused.

I am confused by the terminology of the question. When x is divided by y the remainder 60. is x=yk+60. "y" is the quotient and the question is asking for possible quotients, ie, possible y values.

But in your solution you have x/y=quotient. Very confused.

In x=yk+60, y = divisor and k = quotient.

On dividing both sides of the equation by Y --------> \(\frac{x}{y} = \frac{(yk)}{y} + \frac{60}{y}\) --------> \(\frac{x}{y} = k + \frac{60}{y}\)

So k = quotient and \(\frac{60}{y}\)is Remainder. -------- Quotient can not be fraction.

So while considering 15.15 we should recognize that 15 must be quotient and 0.15 must be remainder, but the remainder is \(\frac{60}{y}\)---------> Hence \(0.15 = \frac{60}{y}\) -------> \(Y = \frac{60}{0.15}\) -------> Y = 400

Note that we are told that Y is integer so basically we are just looking for whether given expression could give the integer value of Y

Consider second option 18.16 ------> Y = 60/0.16 -------> 6000/16 ------> some integer

In 17.17 ---------> 60/0.17 -------> 6000/17 ------> Not an Integer

Re: When the integer x is divided by the integer y [#permalink]

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02 Aug 2013, 10:46

Hello!

Would it be possible to solve the task as follows:

I. We know that the reminder is 15. As 60 is divisible by 15 we can assume that 15.15 could be one solution. II. We know that the reminder is 16 which is 2*8. As 60 can be divided by 2 we can assume that 18.16 could be one solution. III. We know that the reminder is 17. As 60 is not divisible by 17 or vice versa, we can assume that 17.17 cannot be one of the solutions.

Would it be possible to solve the task as follows:

I. We know that the reminder is 15. As 60 is divisible by 15 we can assume that 15.15 could be one solution. II. We know that the reminder is 16 which is 2*8. As 60 can be divided by 2 we can assume that 18.16 could be one solution. III. We know that the reminder is 17. As 60 is not divisible by 17 or vice versa, we can assume that 17.17 cannot be one of the solutions.

Also, please explain if my approach has a flaw.

Thank you in advance!

I think, it will not work every time. I will show you how.

Let's change the third option. Lets assume Remainder is 21 and quotient is 5.10

We know that the remainder is 0.10 As 21 is not divisible by 10 or vice versa, we can assume that 5.10 cannot be one of the solutions. Can this be true?? 21/0.10 ------> 2100 / 10 ---------> well divisible by 10.

in the division 21/0.10 when you move the decimal sign of denominator to the right by two places, you add two zeros in the numerator to create the fraction 2100/10 When you are awarding two zeros to the numerator (i.e. to 21) remember you are actually giving 2 twos and 2 fives to the numerator. Why is so. That is because One zero can be obtained by multiplying 5 with 2 Now since the denominator (i.e. 10) contains 1 five and 1 two and since you just have given 2 fives and 2 twos, you can surely divide numerator by the denominator.
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Re: When the integer x is divided by the integer y [#permalink]

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05 Feb 2014, 06:41

The question is simply asking IF the decimal part of the values given in I, II & III yield 60 when multiplied by y, while keeping y as integer.

Let me show how,

I. 15.15 ===> .15*y=60 is y integer? yes, y= 6000/15 = 400 hence 15.15 can be quotient. II 18.16 ===> .16*y=60 is y integer? yes, y= 6000/16 = 375 hence 18.16 can be quotient. III 17.17 ===> .17*y=60 is y integer? NO, y=6000/17 hence 17.17 can not be quotient.

Re: When the integer x is divided by the integer y [#permalink]

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30 Apr 2014, 09:37

When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?

I was totally confused...Although the source is MGMAT.. I don't agree with the terminology.. The question demands that we find the value of the quotient of X/Y...i.e the value of Green colored portion in X/Y= Q + 60/Y But the question implies us to find out the value of X/Y i.e Q+60/Y Consider the case: 7=1.2*5 + 1...Although this is unreasonable..but this is what I took from the Question
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Re: When the integer x is divided by the integer y [#permalink]

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01 May 2014, 00:31

[quote="enigma123"]When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?

I. 15.15 II.18.16 III. 17.17

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

I don't have a clue how this will be solved. Can someone please help? Unfortunately, I don't have an OA either.[/quo

It is better to evaluate possible quotients as integer + fraction manner (decimal part), exploiting idea that decimals come after dividing remainder to divisor, so:

I. 15.15. is 15+15/100 where 15 is factor of 60, so it can be quotient

II. 18.16 is 18+16/100 or 4/25 where 4 is a factor of 60, so it can be quotient

III. 17.17 is 17+17/100 where 17 is not a factor of 60, so it cannot be quotient

Re: When the integer x is divided by the integer y [#permalink]

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28 May 2014, 09:34

Many seemingly difficult remainder problems can be simplified using the following concept: If you take the decimal portion of the resulting number when you divide ''x' by "y", and multiply it to "y", you will get the remainder.

Let's look at the options:

I. 15.15 ===> .15*y=60 is y integer? yes, y= 6000/15 = 400 hence 15.15 can be quotient. II 18.16 ===> .16*y=60 is y integer? yes, y= 6000/16 = 375 hence 18.16 can be quotient. III 17.17 ===> .17*y=60 is y integer? NO, y=6000/17 hence 17.17 can not be quotient.

w.k.t x = yn (i.e Q) + yk=60 (i.e remainder) therefore : 1) y*0.15=60 -- we can get y an integer. 2) y*0.16=60 -- we can get y an integer. 3) y*0.17=60 -- we can't get y an integer.

Therefore 1 and 2 are possible answers. Ans D.
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Re: When the integer x is divided by the integer y [#permalink]

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14 Oct 2015, 08:07

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Re: When the integer x is divided by the integer y [#permalink]

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14 Mar 2016, 17:40

enigma123 wrote:

When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?

I. 15.15 II.18.16 III. 17.17

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

I don't have a clue how this will be solved. Can someone please help? Unfortunately, I don't have an OA either.

first of all..rebuild the formula: x/y=quotient+remainder x=q*y + r r=x-q*y

1. 15.15 q=15. r = 60/y = 15/100 simplify 60 with 15 = 4/y=1/100 y=400 so it's possible. x=15*400+60 note that 60/400 = 0.15

i'll start with 2, since the numbers are even and are easier to work with.. q=18, and r=60/y=16/100 simplify 60 with 4, and 16 with 4, then 4 and 100 with 4. 15/y=1/25 y=375.

again, it is possible. 60/375 = 0.16

so far, we could have eliminated all but D. but let's test for III.

q=17, r=60/y = 17/100 17y=6000 6000 is not divisible by 17. but we are told that y is an integer.

Re: When the integer x is divided by the integer y [#permalink]

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17 Mar 2016, 04:17

Excellent Question.. here is what we need to know prior to solving this one=> When a positive integer X is Divided by a Positive integer Y => the decimal part in the quotient => Remainder Hence In all the cases the quotient is viable only if the remainder which is 60 can be divided by the decimal part. now as i one and 2 .15 and .16 both divide 60 but .17 does not Hence D is the answer.
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