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M16-18

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M16-18 [#permalink]

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New post 15 Sep 2014, 23:59
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  95% (hard)

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35% (01:01) correct 65% (01:24) wrong based on 100 sessions

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Official Solution:


Question asks: is \(2*x*5*y=even\)? Now, since there is 2 as a multiple, then this expression will be even if \(5xy=integer\). Basically we are asked: is \(5xy=integer\)?

Notice that \(x\) and \(y\) may not be integers for \(2*x*5*y\) to be even (example \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\)) BUT if they are integers then \(2*x*5*y\) is even.

(1) \(2 + x + 5 + y\) is an even integer. Given: \(2+x+5+y=even\). So, \(7+x+y=even\), which means that \(x+y=odd\). Not sufficient. (Consider \(x=1\) and \(y=2\) for an YES answer and \(x=1.3\) and \(y=1.7\) for a NO answer).

(2) \(x-y\) is an odd integer. Given: \(x-y=odd\). Not sufficient. (Consider \(x=1\) and \(y=2\) for an YES answer and \(x=1.3\) and \(y=0.3\) for a NO answer).

(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\), so \(2x=even\). This implies that \(x=integer\) and therefore \(y=integer\). Hence sufficient.


Answer: C
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Re: M16-18 [#permalink]

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New post 18 Mar 2016, 02:37
Hi,

Hope zero can be considered rite?

If its taken, then the answer is E. Kindly debrief?

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Re M16-18 [#permalink]

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New post 27 Aug 2016, 04:43
I think this is a high-quality question and I agree with explanation.

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Re: M16-18 [#permalink]

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New post 21 Nov 2016, 07:22
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Great Question
Here we need to see if 10xy is even or not
Let P=10xy
Note => If x and y are integers => 10xy will be always even irrespective of the value of x and y
Lets look at statements
Statement 1
x+y+7=even
hence x+y=odd
hmm
x=1,y=2 => P = even
x=1.2,y=1.8 => P = 18*12/10 => not an integer(we don't have to perform the division here, as 5 is in the denominator and no 5 is in the numerator => This wont be an integer)
Hence not sufficient
Statement 2
x-y=odd
hmm
x=10,y=1=> P will be even
if x=10.1,y=1.1 => p will not be an integer.
Hence not sufficient
Combining the two statements

x+y=odd
x-y=odd
adding them up
2x=odd+odd = even
hence x => integer
now y=odd-x=> integer too
Hence x and y are both integers
hence P will always be even

Hence C
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Re: M16-18 [#permalink]

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New post 07 Jul 2017, 02:31
amazing questions

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Re: M16-18 [#permalink]

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New post 21 Nov 2017, 12:39
(1)+(2) Sum (1) and (2) (x+y)+(x−y)=odd1+odd2(x+y)+(x−y)=odd1+odd2, so 2x=even2x=even. This implies that x=integerx=integer and therefore y=integery=integer. Hence sufficient.

Can someone explain how 2x= even implies , X is an integer

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Re: M16-18 [#permalink]

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New post 21 Nov 2017, 20:08
HariharanIyeer0 wrote:
(1)+(2) Sum (1) and (2) (x+y)+(x−y)=odd1+odd2(x+y)+(x−y)=odd1+odd2, so 2x=even2x=even. This implies that x=integerx=integer and therefore y=integery=integer. Hence sufficient.

Can someone explain how 2x= even implies , X is an integer


2x = even;

x = even/2 = integer.
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New to the Math Forum?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Kudos [?]: 135577 [0], given: 12701

Re: M16-18   [#permalink] 21 Nov 2017, 20:08
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