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M05-28

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New post 16 Sep 2014, 00:26
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A
B
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New post 16 Sep 2014, 00:26
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D
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New post 06 Jan 2015, 18:04
The answer is simply B!
y=4--->z=6-----> x=3
so x=-1 is clearly unacceptable. It is assumed y=4, how it is possible that y=-4?
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New post 07 Jan 2015, 07:21
arashroghani wrote:
The answer is simply B!
y=4--->z=6-----> x=3
so x=-1 is clearly unacceptable. It is assumed y=4, how it is possible that y=-4?


Sorry, but I don't understand what is confusing there...

Does not x = -1, y = -4 and z = -2 satisfy the given equations?
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New post 07 Jan 2015, 13:35
Bunuel wrote:
arashroghani wrote:
The answer is simply B!
y=4--->z=6-----> x=3
so x=-1 is clearly unacceptable. It is assumed y=4, how it is possible that y=-4?


Sorry, but I don't understand what is confusing there...

Does not x = -1, y = -4 and z = -2 satisfy the given equations?


Dear Bunuel
y=-4 is contradictory with statement 2!
2)y=4
so x=-1 is not possible and as a result statement 2 is sufficient.
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New post 08 Jan 2015, 06:45
arashroghani wrote:
Bunuel wrote:
arashroghani wrote:
The answer is simply B!
y=4--->z=6-----> x=3
so x=-1 is clearly unacceptable. It is assumed y=4, how it is possible that y=-4?


Sorry, but I don't understand what is confusing there...

Does not x = -1, y = -4 and z = -2 satisfy the given equations?


Dear Bunuel
y=-4 is contradictory with statement 2!
2)y=4
so x=-1 is not possible and as a result statement 2 is sufficient.


The correct answer to the question is D, not B! The answer choice D means that EACH statement ALONE is sufficient to answer the question asked. So, the second statement as well as the first one is sufficient ALONE.

As for your remark that y=-4 contradicts statement 2: yes, it does but we got those values based only on the info given in the stem. When then we proceed to the second statement we can reject this set (x = -1, y = -4 and z = -2) and we'll be left with the second set, which we also got from the set: x = 3, y = 4, z = 6.

Please re-read the question and the solution.
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New post 13 May 2015, 08:19
Hi Bunuel,
I tried solving in another way and I think B is the answer. Please check and confirm if this makes sense.

Given,

x^2=y+5
y=z-2
z=2x

I tried to get everything under one variable "x"

y=2x-2
y^2=(2x-2)^2 = 4x^2+4-8X

x^2=2x-2+5=2x+3
X^3 = x(2x+3) = 2x^2+3X


Eqn is x^3+y^2+z

Subsituting values of y and z in terms of x

=2x^2+3x+4x^2+4-8X+2x
=6x^2-3x+4

Subsituting X>0 values, this eqn is satisfied both for x=1 & x=3.

If x=1, then y=-4, z=-6, eqn value is -1+16-6=9/7 - no
If x=3, then y=4, z=6 eqn value is 27+16+6 = 49 - yes

We have 2 options with x>0 and so (A) is not sufficient

with option (B), it is clear that the eqn is divisible by 7.

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New post 13 May 2015, 08:36
dpbe25 wrote:
Hi Bunuel,
I tried solving in another way and I think B is the answer. Please check and confirm if this makes sense.

Given,

x^2=y+5
y=z-2
z=2x

I tried to get everything under one variable "x"

y=2x-2
y^2=(2x-2)^2 = 4x^2+4-8X

x^2=2x-2+5=2x+3
X^3 = x(2x+3) = 2x^2+3X


Eqn is x^3+y^2+z

Subsituting values of y and z in terms of x

=2x^2+3x+4x^2+4-8X+2x
=6x^2-3x+4

Subsituting X>0 values, this eqn is satisfied both for x=1 & x=3.

If x=1, then y=-4, z=-6, eqn value is -1+16-6=9/7 - no
If x=3, then y=4, z=6 eqn value is 27+16+6 = 49 - yes

We have 2 options with x>0 and so (A) is not sufficient

with option (B), it is clear that the eqn is divisible by 7.

Deep


x=1 is not a solution. x=-1 is.

Your set x=1, y=-4, and z=-6 is NOT correct. We are told that z=2x, which is not satisfied with your set of numbers.
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New post 13 May 2015, 10:25
Thanks Bunuel. I missed rechecking on z=2x parameter. Now it makes sense.
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New post 18 Aug 2015, 08:51
Not able to understand while using the statement (2) i.e. y=4, using this value if we put in equation x^2 = y+5then x=+3 and -3 whereas z = 6 and z = -6,
using x=3, z=6 and y=9 answer from final equation is 49, which is divisible by 7
using x=-3, z=-6 and y=9 answer from final equation is 17, which is not divisible by 7
so pls let me know can answer to the question ( A )
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New post 18 Aug 2015, 09:53
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kshitijkhurana3010 wrote:
Not able to understand while using the statement (2) i.e. y=4, using this value if we put in equation x^2 = y+5then x=+3 and -3 whereas z = 6 and z = -6,
using x=3, z=6 and y=9 answer from final equation is 49, which is divisible by 7
using x=-3, z=-6 and y=9 answer from final equation is 17, which is not divisible by 7
so pls let me know can answer to the question ( A )


z = -6 is not possible (it violates y = z - 2 for y = 4). The same way x = -3 is not possible.
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New post 21 Sep 2015, 20:06
A very good question. Fell for easy B trap.
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New post 21 Sep 2015, 20:29
Here,
x^2=y+5 , y=z-2, z=2x
Putting 2nd and 3rd eqn in 1 we get => x^2-2x-3=0=> x=3 or x=-1
for x=3 we get y=4 and z=6 => x^3+y^2+z=49
for x=-1we get y=-4 and z=-2
since X>0 hence x can only be 3 , also since y=4 is given only 1 st condition is possible.
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New post 29 Sep 2015, 07:57
Rahulnair2802 wrote:
Here,
x^2=y+5 , y=z-2, z=2x
Putting 2nd and 3rd eqn in 1 we get => x^2-2x-3=0=> x=3 or x=-1
for x=3 we get y=4 and z=6 => x^3+y^2+z=49
for x=-1we get y=-4 and z=-2
since X>0 hence x can only be 3 , also since y=4 is given only 1 st condition is possible.

What I did was almost the same as Bunuel. First I solved the system of equations, getting to \(x=3\) or \(x=-1\).
Then, for statement i) \(x>0\), I used \(x=3\), getting \(y=4\) and \(z=6\): \(x^3+y^2+z = 27+16+6=49\), divisible by \(7\)
Now, for statement ii) \(y=4\), I just assumed that \(y=4\) is possible when \(x=3\), so \(z=6\), then the statement will be sufficient. Is my assumption correct or should I have used the \(x=-1\) for the 2nd triplet?

EDIT: The best approach was what Bunuel did. Get the two solutions and test them to see if they are divisible by 7. It could have happened that the two triplets weren't divisible by 7 (using \(x=3\) or \(x=-1\)). In this case, what option do we choose? D?
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New post 09 Dec 2015, 10:44
2
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange:

\(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D



Dear Bunuel

It was an interesting solution how you have inferred even before putting pen to paper " on of the values of the expression will be divisible by 7 & other will be not" . If this was known before setting out to solve the question then I think in this DS question knowing that any pointers to establish / confirm the nature of the triplets will be sufficient.

My point here is how do we know that both the values will not be divisible by 7. Had the question asked divisible by 5 or 19 for that matter. Does the inference remains same that one value of the expression will remain divisible & other will not in each of the case?
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New post 25 Sep 2016, 14:13
1
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D



Bruno,

Thanks for the great response. I understand this is a 700 level problem, but is there a quicker way to "rephrase" the question. As I was making the formula translations, and then calculating the values of x, y, and z for each set, I had already spent more than 2 minutes.

Thanks,

DDB
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New post 27 Jul 2018, 14:05
ankushbagwale wrote:
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange:

\(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D



Dear Bunuel

It was an interesting solution how you have inferred even before putting pen to paper " on of the values of the expression will be divisible by 7 & other will be not" . If this was known before setting out to solve the question then I think in this DS question knowing that any pointers to establish / confirm the nature of the triplets will be sufficient.

My point here is how do we know that both the values will not be divisible by 7. Had the question asked divisible by 5 or 19 for that matter. Does the inference remains same that one value of the expression will remain divisible & other will not in each of the case?


have the same question! Please help with the explanation
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New post 28 Jul 2018, 04:14
grsm wrote:
ankushbagwale wrote:
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange:

\(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D



Dear Bunuel

It was an interesting solution how you have inferred even before putting pen to paper " on of the values of the expression will be divisible by 7 & other will be not" . If this was known before setting out to solve the question then I think in this DS question knowing that any pointers to establish / confirm the nature of the triplets will be sufficient.

My point here is how do we know that both the values will not be divisible by 7. Had the question asked divisible by 5 or 19 for that matter. Does the inference remains same that one value of the expression will remain divisible & other will not in each of the case?


have the same question! Please help with the explanation


The first paragraph is just a summary of what we THEN got by algebraic manipulations.
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Re M05-28  [#permalink]

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New post 03 Aug 2018, 19:50
I think this is a high-quality question. Yes
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M05-28  [#permalink]

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New post 16 Aug 2018, 12:15
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D

Hello Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.
Can you please clarify my answer.
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M05-28 &nbs [#permalink] 16 Aug 2018, 12:15

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