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In an integer division operation, the divisor is x, the quotient is y,
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22 Apr 2015, 03:30
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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.
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
_________________
Re: In an integer division operation, the divisor is x, the quotient is y,
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24 Apr 2015, 07:31
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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
Kudos if its a correct solution
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?
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.
_________________
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
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 ?
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.
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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53% (02:13) correct 
