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In an integer division operation, the divisor is x, the quotient is y,

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In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 22 Apr 2015, 03:08
3
10
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

53% (02:14) correct 47% (02:06) wrong based on 229 sessions

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In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 22 Apr 2015, 03:30
3
4
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


We can write this task in such way:
\(z = x*y + r\)

1) we know only \(z\) and this is insufficient.
2) we know only \(r\) and this is insufficient

1+2) \(151 = x*y + 11\)
This is looks like insufficient, but we know that \(x\) and \(y\) should be integer and remainder should be less than divisor: \(x>11\).
And from this information we can infer that \(x*y = 151-11 = 140\)
and as \(x\) should be more than 11, the least possible number is 14 and \(y\) should be equal to 10
So \(x > y\) and answer is C
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Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 22 Apr 2015, 20:30
1
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


z=xy+r

(1) z = 151
Insufficient.
(2) r = 11
Insufficient.

x*y= 140
x>11
so minimum 'x' is 14 and maximum 'y' is '10' .

Answer C .
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Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 24 Apr 2015, 07:31
1
Hi,
IMO it should be C. that is after combining both statements.
St 1) just be talking about z-151 nothing comes out.
151/2 or 151/150. so remainder being same still x and y cant be quantified.

st 2) r = 11.
still nothing comes out

combining both:
find the number which when divides 151 leaves remainder of 11
x should be 14 and y be 10.

Though while solving I assigned x as 140 / 28 too.. which leaves remainder of 11 when divided by 151
hence C
Regards
Celestial


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Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.
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Posts: 58446
Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 27 Apr 2015, 03:10
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.

Recognizing the algebra of division can be quite helpful here, too. Since Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.

Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:

z = xy + 11

And clearly not enough information to solve.

But taken together, the information is sufficient. Why, if it still leaves two variables?

151 = xy + 11, so:

140 = xy

The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.
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Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 26 Apr 2016, 07:25
Bunuel wrote:
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.

Recognizing the algebra of division can be quite helpful here, too. Since Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.

Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:

z = xy + 11

And clearly not enough information to solve.

But taken together, the information is sufficient. Why, if it still leaves two variables?

151 = xy + 11, so:

140 = xy

The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.



Typo:
Should be
Dividend=Divisor*Quotient + Remainder
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In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 21 Dec 2018, 05:32
Bunuel wrote:
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.

Recognizing the algebra of division can be quite helpful here, too. Since Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.

Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:

z = xy + 11

And clearly not enough information to solve.

But taken together, the information is sufficient. Why, if it still leaves two variables?

151 = xy + 11, so:

140 = xy

The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.


Bunuel , can you help me out with my confusion here, i read that Y is the quotient, and X is the divisor..
Why x has to be greater than y ? as We can also divide 140 by 7 i.e X and in that case quotient i.e Y would be 20.
Please correct me if i am missing out on some information, but it's causing me a lot of confusion here.
How's X > Y here ?
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Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 21 Dec 2018, 05:49
1
Shrey9 wrote:
Bunuel wrote:
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.

Recognizing the algebra of division can be quite helpful here, too. Since Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.

Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:

z = xy + 11

And clearly not enough information to solve.

But taken together, the information is sufficient. Why, if it still leaves two variables?

151 = xy + 11, so:

140 = xy

The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.


Bunuel , can you help me out with my confusion here, i read that Y is the quotient, and X is the divisor..
Why x has to be greater than y ? as We can also divide 140 by 7 i.e X and in that case quotient i.e Y would be 20.
Please correct me if i am missing out on some information, but it's causing me a lot of confusion here.
How's X > Y here ?


If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Notice that \(0\leq{r}<x\) means that remainder is a non-negative integer and always less than divisor.

It's easier to understand if you consider a simple example: when you divide by 3, the remainder could only be 0, 1, or 2.

For more on this please check the links below:

6. Remainders



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Re: In an integer division operation, the divisor is x, the quotient is y,  [#permalink]

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New post 30 May 2019, 21:26
Bunuel wrote:
Bunuel wrote:
In an integer division operation, the divisor is x, the quotient is y, the dividend is z, and remainder is r. Is x > y?

(1) z = 151

(2) r = 11



Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Quick hypothetical situations should prove statement 1 to be insufficient. If x is 151, then you'd be dividing the same thing by itself and y would be 1 with no remainder, and x would be much larger than y. But you could flip that and make x = 1, in which you'd just be dividing 151 by 1 and the quotient (y) would be 151, much larger than x.

Recognizing the algebra of division can be quite helpful here, too. Since Quotient = Dividend * Divisor + Remainder, just knowing statement 1 tells you that 151 = xy + r, and clearly that's not enough information to solve.

Statement 2 is also insufficient. Keep in mind here that, without knowing statement 1, you essentially just have an undefined problem with a remainder of 11. Algebraically that's:

z = xy + 11

And clearly not enough information to solve.

But taken together, the information is sufficient. Why, if it still leaves two variables?

151 = xy + 11, so:

140 = xy

The other crucial element of this being a division problem is that the divisor can't be less than the remainder! If you're dividing 151 by, say, 7, you'd never have 11 left over. Even though 7 * 20 = 140, you wouldn't stop there and say that the remainder is 11, because 11 can still divide one more time by 7 (making it 21, remainder 4). So since 140 = xy AND x > 11, there's no way for an integer y to be greater than x. Therefore the two statements together are sufficient.




How come x can't equal 70 and y equal 2? The remainder would be 11 is this case as well.
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Re: In an integer division operation, the divisor is x, the quotient is y,   [#permalink] 30 May 2019, 21:26
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