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# A jar contains x red marbles, y white marbles and z blue marbles wherE

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A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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15 May 2018, 01:41
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Difficulty:

95% (hard)

Question Stats:

47% (01:45) correct 53% (02:10) wrong based on 132 sessions

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FRESH GMAT CLUB TESTS' 700 LEVEL QUESTION

A jar contains x red marbles, y white marbles, z blue marbles, where x > y > z, and no other marbles. How many red marbles are there in the jar?

(1) To ensure that at least one marble of each color is removed from the jar, minimum 55 marbles must be removed
(2) To ensure that a red marble is removed from the jar, the smallest number of marbles that must be removed is 51

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A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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Updated on: 15 May 2018, 06:09
hi all,

first statement gives: x+y+1=55, insufficient
2nd one gives: y+z+1=51, insufficient,

Both statments give 2 equations, 3 unknowns, thus insufficient.

Ans E. Am i right ?

Originally posted by wysiwyg on 15 May 2018, 02:52.
Last edited by wysiwyg on 15 May 2018, 06:09, edited 1 time in total.
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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15 May 2018, 06:06
2
statement 1... X+Y+1=55.....X+Y=54
statement 2...Y+Z+1=51......Z+Y=50
NOW X>Y>Z
This gives y=26,x=28,z=24
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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15 May 2018, 06:34
1
Bunuel wrote:

FRESH GMAT CLUB TESTS' 700 LEVEL QUESTION

A jar contains x red marbles, y white marbles, z blue marbles, where x > y > z, and no other marbles. How many red marbles are there in the jar?

(1) To ensure that at least one marble of each color is removed from the jar, minimum 55 marbles must be removed
(2) To ensure that a red marble is removed from the jar, the smallest number of marbles that must be removed is 51

Statement 1: x+y+1 = 55

NOT SUFFICIENT

Statement 2: y+z+1 = 51

NOT SUFFICIENT

Combining the two statements

x+y = 54
y+z = 50

i.e. x - z = 4

Case 1: x =28, y =26, z = 24

SUFFICIENT

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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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15 May 2018, 21:59
Can somebody please explain how you got x+y+1=55 from the 1st statement. There is no mention anywhere that we are taking the blue marble out. 1 could signify x,y or z.

Thank you.
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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15 May 2018, 22:20
StrugglingGmat2910 wrote:
Can somebody please explain how you got x+y+1=55 from the 1st statement. There is no mention anywhere that we are taking the blue marble out. 1 could signify x,y or z.

Thank you.

Check that there are total x red, y white, and z blue marbles.
As per statement 1, minimum 55 marbles must be removed to ensure at least 1 marble of each color is removed

Let's assume we are to ensure that we will pick 1 blue marble
Now, in the worst case scenario, we will pick all the red and white marbles first; and once there are no more red or white marbles, the next pick will be by default blue marble
This is exactly what is given in statement 1

It is also mentioned that x > y > z, that is why when we consider the worst case scenario, we form the equation as x + y + 1 = 55
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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03 Jun 2018, 18:09
My understanding is that the reason why
Statement 1: x+y+1 = 55
Statement 2: y+z+1 = 51
is sufficient even though we have 3 variables and 2 equations is because of the additional constraint that 0 > z > y > x. Is there an easy way to confirm that y=26,x=28,z=24 is the ONLY answer? What's a good way to solve this question?
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A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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08 Jun 2018, 17:44
1
dracobook wrote:
My understanding is that the reason why
Statement 1: x+y+1 = 55
Statement 2: y+z+1 = 51
is sufficient even though we have 3 variables and 2 equations is because of the additional constraint that 0 > z > y > x. Is there an easy way to confirm that y=26,x=28,z=24 is the ONLY answer? What's a good way to solve this question?

The constraint that z < y < x tells us that x,y,z are distinct integers.
Per Statement 2, y+z=50. So y and z cant both be 25. The next logical case we can examine is if y=26 and z=24.
Per Statement 1, x+y=54. Since y=26, that makes x=28. This gives us the answer everyone's posting, (x,y,z) = (28,26,24).

Let's take the next case, y=27 and z=23. This fulfills Statement 2. To fulfill Statement 1, x has to equal 27. But this violates the constraint that x,y,z are distinct integers. And for all other increasing values of y, x will be less than y, which violates the constraint that y<x.

Another way to think about it is to subtract Statement 2 from Statement 1. What you get: x-z=4. This means that the range of the set is 4. We found that (x,y,z) = (28,26,24) satisfies both statements. If we try other values of y and z that satisfy Statement 2, there's no way the set will have a range of 4.

Hope that helps.
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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01 Jul 2018, 06:42
from statement 1, how do we know if x+y=54, wouldn't y+z or x+z be possibilities to consider as well? Thanks.
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Re: A jar contains x red marbles, y white marbles and z blue marbles wherE  [#permalink]

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01 Jul 2018, 17:56
Andy24 wrote:
from statement 1, how do we know if x+y=54, wouldn't y+z or x+z be possibilities to consider as well? Thanks.

(1) To ensure that at least one marble of each color is removed from the jar, minimum 55 marbles must be removed

We know that the most amount of marbles of a given color is x, since x > y > z, and then next greatest is y. Therefore, in order to guarantee that you have one of each color, you assume the worse case scenario, which is that you first pick out all x of the red marbles, then all y of the white marbles, then FINALLY you pick 1 of the blue marbles. Hence, you get $$x + y + 1 = 55 \implies x + y = 54$$
Re: A jar contains x red marbles, y white marbles and z blue marbles wherE &nbs [#permalink] 01 Jul 2018, 17:56
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