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Eq 1 and 2 are insuff alone as they inclide two variables and its impossible to solve for one variable i.e x. Combined they provide the same info, so nothing new. Answer should be E.
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(1) 2x - 5y + 6 = 12 => 2x - 5y = 6

(2) 8x - (4x + 10y) + 27 = 39 => 4x - 10y = 12 or 2 (2x -5y ) = 2* 6

Since both equation are same, x can not be determined, hence E
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Bunuel

Tough and Tricky questions: Sequences.



What is the value of x?

(1) 2x - 5y + 6 = 12

(2) 8x - (4x + 10y) + 27 = 39

Kudos for a correct solution.


S1 : 2x-5y = 6....can have many different values of x & Y e.g. X=8, Y=2 is one solution ; x=13, Y=4 is another one. Hence S1 is insufficient.

S2 : 4x-10y = 12 ....the equation is the same as S1 [multiplied by 2]. Hence S2 is insufficient.

Since both equations are same, we can't solve for X & Y using these two equations. Hence taking S1 & S2 together is also insufficient.

Hence Answer E.
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Bunuel

Tough and Tricky questions: Sequences.



What is the value of x?

(1) 2x - 5y + 6 = 12

(2) 8x - (4x + 10y) + 27 = 39

Kudos for a correct solution.

I agree it is E as well. Both equations are same so x cannot be solved.
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IMO E

both are same equations when simplified

OA?
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Bunuel

Tough and Tricky questions: Sequences.



What is the value of x?

(1) 2x - 5y + 6 = 12

(2) 8x - (4x + 10y) + 27 = 39

Kudos for a correct solution.

The parentheses is just here to confuse you

8x-4x +10y= 12
4x-2x =10y =6
2x -5y =6

redundant info

E
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Bunuel

Tough and Tricky questions: Sequences.



What is the value of x?

(1) 2x - 5y + 6 = 12

(2) 8x - (4x + 10y) + 27 = 39

Kudos for a correct solution.


The subtle trick played here is redundant information- the questions asks for the value of X, specifically.

St 1

2x-5y= 6

insuff

St 2

8x-4x-10y =12
4x- 10y= 12
2(2x-5y) =12
(2x-5y) =6

No better than st 1

insuff

St 1 & 2

Insuff because both statements give us the same thing

E
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Bunuel

Tough and Tricky questions: Sequences.



What is the value of x?

(1) 2x - 5y + 6 = 12

(2) 8x - (4x + 10y) + 27 = 39

Kudos for a correct solution.

Target question: What is the value of x?

Statement 1: 2x - 5y + 6 = 12
Subtract 6 from both sides to get: 2x - 5y = 6
Since we have one equation with 2 variables, there are infinitely many solutions.
In other words, x can have infinitely many values, which means statement 1 is NOT SUFFICIENT.

ASIDE: If you're not 100% convinced that there are infinitely many solutions, you can always test some values
There are several values of x and y that satisfy the equation 2x - 5y = 6. Here are two:
Case a: x = 3 and y = 0, in which case x = 3
Case b: x = 8 and y = 2, in which case x = 8

Statement 2: 8x - (4x + 10y) + 27 = 39
Subtract 27 from both sides to get: 8x - (4x + 10y) = 12
Expand: 8x - 4x - 10y = 12
Simplify: 4x - 10y = 12
Here we have another equation with 2 variables, which means there are infinitely many solutions.
In other words, x can have infinitely many values, which means statement 2 is NOT SUFFICIENT.

Statements 1 and 2 combined
Statement 1 tells us that 2x - 5y = 6
Statement 2 tells us that 4x - 10y = 12
IMPORTANT: the two equations above are EQUIVALENT equations.
If we take the equation 4x - 10y = 12 and divide both sides by 2 we get the EQUIVALENT equation 2x - 5y = 6
Since the two statements provide the same information, and since each statement alone is NOT SUFFICIENT, we can conclude that the COMBINED statements are NOT SUFFICIENT

Answer: E
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What happen to the plug in method? Each statement is useless on its on but if you use the solution for y in the first statement and plug it in to the second equation you would find the value of x. So what if the two statements of x is the same...i honestly didn't event get there becuause now that we are using both equations, I would find the value of one variable to find the other....so confused here
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What happen to the plug in method? Each statement is useless on its on but if you use the solution for y in the first statement and plug it in to the second equation you would find the value of x. So what if the two statements of x is the same...i honestly didn't event get there becuause now that we are using both equations, I would find the value of one variable to find the other....so confused here

If the two linear equations are the SAME, then there will be infinitely many solutions.
For example, let's say the two equations are:
x + y = 10
2x + 2y = 20
There are infinitely many solutions that satisfy BOTH equations:
x = 1 and y = 9
x = 4 and y = 6
x = -3 and y = 13
x = -6.7 and y = 16.7
etc
So, if the target question asks "What is the value of x?", there's no way to answer that question.



Conversely, if the two linear equations are DIFFERENT, then there will be either zero or one solution.
For example, let's say the two equations are:
x + 2y = 10
x - y = 4
The one solution is x = 6 and y = 2
So, if the target question asks "What is the value of x?", we can say, "x MUST equal 6"

Does that help?

For more see:
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