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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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Amateur wrote:
Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....



Let me try

X/Y = 13+ .8
X/Y = 13 + 4/5

X and Y are integers...
Possibilities of Remainder/Divisor = 4/5, 8/10, 12/15
if 4/5
X = 69 Y = 5
If 8/10
X=138 Y =10
If 12/15
X= 207 Y =15


B says we should have the total digits of X and Y less than 5
only 4/5 suffice.
Hence OA.
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
BangOn wrote:
Amateur wrote:
Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....



Let me try

X/Y = 13+ .8
X/Y = 13 + 4/5

X and Y are integers...
Possibilities of Remainder/Divisor = 4/5, 8/10, 12/15
if 4/5
X = 69 Y = 5
If 8/10
X=138 Y =10
If 12/15
X= 207 Y =15


B says we should have the total digits of X and Y less than 5
only 4/5 suffice.
Hence OA.



B says we should have the total digits of X and Y less than 5
only 4/5 suffice.


Didnt this mean total no of digits less than 5?
Can u explain ur soln Bangon
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
shanmugamgsn wrote:
BangOn wrote:
Amateur wrote:
Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....



Let me try

X/Y = 13+ .8
X/Y = 13 + 4/5

X and Y are integers...
Possibilities of Remainder/Divisor = 4/5, 8/10, 12/15
if 4/5
X = 69 Y = 5
If 8/10
X=138 Y =10
If 12/15
X= 207 Y =15


B says we should have the total digits of X and Y less than 5
only 4/5 suffice.
Hence OA.



B says we should have the total digits of X and Y less than 5
only 4/5 suffice.


Didnt this mean total no of digits less than 5?
Can u explain ur soln Bangon



Yeah Sure. Total Digits in both X and Y should be less than 5.
if 4/5
X = 69 Y = 5
No. Of digits in X and Y is 3
If 8/10
X=138 Y =10
No. Of digits in X and Y is 5
If 12/15
X= 207 Y =15
No. Of digits in X and Y is 5

So from C, we have only one possible solution X = 69 Y = 5
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
BangOn wrote:
Amateur wrote:
Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....



Let me try

X/Y = 13+ .8
X/Y = 13 + 4/5

X and Y are integers...
Possibilities of Remainder/Divisor = 4/5, 8/10, 12/15
if 4/5
X = 69 Y = 5
If 8/10
X=138 Y =10
If 12/15
X= 207 Y =15


B says we should have the total digits of X and Y less than 5
only 4/5 suffice.
Hence OA.


thank you for the explanation, statement B makes sense now. one thing- how did you get the possibilities of remainder/divisor?
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
[/quote] thank you for the explanation, statement B makes sense now. one thing- how did you get the possibilities of remainder/divisor?[/quote]

Possibilities by just multiplying the fraction .8 = 4/5
1) 4/5*1/1 = 4/5
2) 4/5 * 2/2 = 8/10
.
.
similarly
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
JJ2014 wrote:
For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8
(2) The numbers x and y have a combined total of less than 5 digits

BangOn wrote:
Amateur wrote:
Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....

B says we should have the total digitsof X and Y less than 5
only 4/5 suffice.
Hence OA.

BangOn wrote:
thank you for the explanation, statement B makes sense now. one thing- how did you get the possibilities of remainder/divisor?[/quote]

Possibilities by just multiplying the fraction .8 = 4/5
1) 4/5*1/1 = 4/5
2) 4/5 * 2/2 = 8/10
.
.
similarly[/quote]
estimating reminders is fine.... but calculating x and y values when you have remainders, I think it is time consuming process for a 600-700 problem.....
Also, B didnot say we should have total digits less than 5, b said x and y have a combined total of less than 5 digits.... if it did say the number of digits of x and y have a combined total of less than 5 you must have been correct.... but since it said less than 5 digits.... the equation will be x+y<10^4
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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Statement -1 :
X= 13 Y + .8 Y
As X & Y are Non negative integers ,hence .8Y integer
Possible values of Y for which R =.8 Y is an integers : 5,10,15,...
As there are multiple values of Y for which R is an integers- Not sufficient


Statement 2 : Insufficient

Together :
As ,Total digits ( X & Y ) less than 5 : this implies Y cant be in double digit ,so Y=5
Hence C is sufficient
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
Not a good question from veritas .

As I saw the explanations and the question stem, it revealed two flaws.

(1).Non - negative integer x,y '0' is also non-negative integer.

(2). Combined total of digits of x,y mean the sum total of x and y or as stated in the explanation

Rgds,
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
Could somebody help with this tough question:-

If zt < -3, is z < 4? E
a. z < 9
b. t < -4

thanks
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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semwal wrote:
Could somebody help with this tough question:-

If zt < -3, is z < 4? E
a. z < 9
b. t < -4

thanks



From your post I perceive that you are a new bee in this forum.You have to post your question at the right place and in a right way where in you provide OA/source/difficulty level.

PS: Rules of posting

rules-for-posting-please-read-this-before-posting-133935.html
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
what does combined total mean? That is a redundant usage...question not well formed

Source please
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
I was having a lot of trouble with this one.

I could cross of the first statement as both 138/10 and 1380/10 would give the same answer, but the remainder would've been different.

For the second one I got that we have a max of 4 numbers. I could not really solve it, but I figured it was possible given the information so I picked C. (However, it was impossible with just the information from B, I sometimes just jump ahead of time-.-)
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8
(2) The numbers x and y have a combined total of less than 5 digits

There are 2 variables (x,y) and 2 equations, so there is high chance that (C) will be our answer.
Looking at the conditions together, we will substitute in values directly, as it is the fastest way to solve remainder problems. From x=yQ+r(Q=integer, 0<=r<y), x/y=(yQ+r)/y=Q+(r/y)=13+0.8
r/y=0.8에서 r=0.8y
r is an integer, so y=5,10,15... . Then we can see that (x,y,r)=(69,5,4),(138,10,8),(207,15,12)...
But the only set of values that suits conditions 2 is (x,y,r)=(69,5,4). The conditions give a unique value, so they are sufficient and the answer becomes (C).

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
x,y >0, integers
x=yQ+r
r=?

(1) x/y =13.8
138/10 = 13.8 --> R=8
138/10 can be reduced to 69/5 --> R=4

Two different remainders, thus INSUFFICIENT

(2) We don't have enough information about x,y to go off of

(1)+(2)
We know that digits are less than 5 and based on (1), that leaves 69/5 as the correct fraction and thus the correct values for x and y (because 69 & 5 have a combined total of 3 digits)
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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JJ2014 wrote:
For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8
(2) The numbers x and y have a combined total of less than 5 digits


This is a poorly written question in my honest opinion..
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]
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JJ2014 wrote:
For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8
(2) The numbers x and y have a combined total of less than 5 digits


We need to determine the remainder when x is divided by y.

Statement One Alone:

x/y = 13.8

We can rewrite statement one as:

x/y = 13 + 8/10

There are infinite possible values for the remainder when x is divided by y. That is because the remainder is the numerator of any equivalent fraction of 8/10. For example, the remainder can be 8 if x = 138 and y = 10, or it can be 16 if x = 276 and y = 20 (note: 276/20 = 13 + 16/20 = 13.8). Thus, statement one is not sufficient to answer the question.

Statement Two Alone:

The numbers x and y have a combined total of less than 5 digits.

Statement two does not provide enough information to answer the question. For example, x can be 3 digits and y can be 1 digit, or x can be 2 digits and y can be 2 digits. Without knowing the exact values of x and y, statement two is not sufficient to answer the question.

Statements One and Two Together:

From statement one, we know that x/y = 13.8 or x = 13.8y. Since x is at least 10 times as much as y, x has either one or two more digits than y. For example, if y = 10, then x = 138 (so x has one more digit than y); if y = 80, then x = 1,104 (so x has two more digits than y). Combining statement one with statement two, we see that either x has 3 digits and y has 1 digit or x has 2 digits and y has 1 digit.

Recall that x = 13.8y, and if y has 1 digit and x has 3 digits, a 1-digit number times 13.8 cannot yield a 3-digit whole number. Therefore, x must have 2 digits and y must be 1 digit. In that case, the only way to satisfy x = 13.8y is if y = 5 and x = 13.8(5) = 69. Hence, the remainder when x is divided y is 4 (69/5 = 13 + 4/5).

Answer: C
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