Remainders, Probability, and Number Property

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242. If x and y are positive integers, is x - y divisible by 4?
(1) xy is divisible by 16.
(2) x is divisible by 4.

[spoiler]Answer: E[/spoiler]
S1)
xy can be 16 = (1,16), (2,8), (4,4)
xy can be 32=(1, 32), (2, 16), (4, 8)
xy can be 48=(1, 48), (2, 24), (3, 16), (4, 8)

(x-y) can be divisible by 4, and not divisible by 4. Thus insufficient.

S2) x is divisible by 4. Then x must be 4a. No information on y. Thus insufficient.

If x is 4, then the pairs: (1, 16) and (2,8) are possibilities. Looking at (x-y): 16-1 is not divisible by 4, but (8-4) is divisible by 4. Thus insufficient.

Is there an easier way to do this other than listing the factor pairs?


There are 20 books in a bookcase. If one book is selected at random, what is the probability that the book is either a hardback or a novel?
(1) 8 books in the bookcase are novel books and 10 books are hardbacks.
(2) 3 books in the bookcase are hardback novels.

[spoiler]Answer: C[/spoiler]
S1: We know that there are 8 novels and 10 hardbacks, but we do not know how many books are both. Insufficient.
S2: We know there are 3 books which are both hardbacks and novels. Thus 20-3=17 must be the number of books which are either hardbacks or novels. Thus 17/20 is the answer. Sufficient.

Can anyone explain way I need statement 1?

256. If n is a positive integer between 30 and 60, inclusive, what is the value of n?
(1) When n is divided by 4, the remainder is 1.
(2) When n is divided by 5, the remainder is 2.

[spoiler]Answer: E[/spoiler]

30≤n≤60

S1: n=4a+1
4(8)+1=33
4(9)+1= 37
4(10)+1=41
4(11)+1=45
4(12)+1=49
4(13)+1=53
4(14)+1=57
Insufficient
S2: n= 5b+2
5(6)+2=32
5(7)+2=37
5(8)+2=42
5(9)+2=47
5(10)+2=52
5(11)+2=57
Insufficient.

Together: 37 and 57 is in both sets of numbers. Thus insufficient.
Again, does anyone have a better method than listing out the possibilities?
Source: — Data Sufficiency |

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by diebeatsthegmat » Mon Nov 15, 2010 6:26 pm
Cedagmat wrote:242. If x and y are positive integers, is x - y divisible by 4?
(1) xy is divisible by 16.
(2) x is divisible by 4.

[spoiler]Answer: E[/spoiler]
S1)
xy can be 16 = (1,16), (2,8), (4,4)
xy can be 32=(1, 32), (2, 16), (4, 8)
xy can be 48=(1, 48), (2, 24), (3, 16), (4, 8)

(x-y) can be divisible by 4, and not divisible by 4. Thus insufficient.

S2) x is divisible by 4. Then x must be 4a. No information on y. Thus insufficient.

If x is 4, then the pairs: (1, 16) and (2,8) are possibilities. Looking at (x-y): 16-1 is not divisible by 4, but (8-4) is divisible by 4. Thus insufficient.

Is there an easier way to do this other than listing the factor pairs?


There are 20 books in a bookcase. If one book is selected at random, what is the probability that the book is either a hardback or a novel?
(1) 8 books in the bookcase are novel books and 10 books are hardbacks.
(2) 3 books in the bookcase are hardback novels.

[spoiler]Answer: C[/spoiler]
S1: We know that there are 8 novels and 10 hardbacks, but we do not know how many books are both. Insufficient.
S2: We know there are 3 books which are both hardbacks and novels. Thus 20-3=17 must be the number of books which are either hardbacks or novels. Thus 17/20 is the answer. Sufficient.

Can anyone explain way I need statement 1?

256. If n is a positive integer between 30 and 60, inclusive, what is the value of n?
(1) When n is divided by 4, the remainder is 1.
(2) When n is divided by 5, the remainder is 2.

[spoiler]Answer: E[/spoiler]

30≤n≤60

S1: n=4a+1
4(8)+1=33
4(9)+1= 37
4(10)+1=41
4(11)+1=45
4(12)+1=49
4(13)+1=53
4(14)+1=57
Insufficient
S2: n= 5b+2
5(6)+2=32
5(7)+2=37
5(8)+2=42
5(9)+2=47
5(10)+2=52
5(11)+2=57
Insufficient.

Together: 37 and 57 is in both sets of numbers. Thus insufficient.
Again, does anyone have a better method than listing out the possibilities?
hmhmhm
for the second question ... we need statement 1 because you made a mistake...
you cant take 20-3
its must be 20=8+10-3+number of books which are not novel and hardback to find the number of book who are neither hardback or novel
and its = 5
P(not h and not N)=5/20=1/4
so P(both)=1-1/4=3/4

for the first and third question, actually yo dont need to make details. i think
question 1.
statement 2 is insufficient cos we dont know about y ( ignore it)
statement1 xy=16 x =4 y=4 so x-y=0 (yes)
x=2 y=8 so x-y=-6 and -6 cant be divisible by 4 ( no)
1+2) x=4 y=4 so x-y=0 (yes)
x=4 y=8 so x-y=-2 so no
E
question 3.
i did the same but i just use the detail in statement 1 to guess the answer

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by Geva@EconomistGMAT » Tue Nov 16, 2010 1:38 am
I think what you're missing for 1 and 3 is a more focused approach. You don't need to list out every possible scenario - you just need two counter examples:

In 1: You need two counter examples, one yes (x-y) is divisible by 4, on no (x-y is not divisible by 4).

Stat. (1) allows x and y to be 4 and 4, or 1 and 16. in the first case, the 4-4 =0 is divisible by 4 (zero is div by every integer except itself), so "yes". in the second case, 16-1=15 is not div by 4, so "no.

Stat. (2): you said it - we don't know anything about y.

combined: x is div by 4 doesn't mean x is divisible only by one 4: x could still be 4 or 16 (as 16 is still div by 4, it still satisfies stat. (2)), so the two cases we used for stat. (1) are still applicable when combined.


In Q3: Each statement is obviously insufficient, as it allows several values.
Combined: I'd list the terms of one of the statements, and just look among them for terms that also satisfy the other statement. For example:
multiples of 4 with rem of 1: 33, 37, 41, 45, 49, 53, 57 (just start with 33 and move in jumps of 4)
Out of these, are there any terms that, when reduced by 2, give a number divisible by 5?

Only 37 and 57 count - take 2 away, and you are left with a number that ends with a units digit of 5, which is a sign of divisibility by 5. But that's still one more than value, so answer is still E.
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