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1) \(\frac{x}{7} +5y=109\) can be written as \(x+35y=763\). Not sufficient. 2)\(175 y=763(5-x)\) Still insufficient.

On combining the two equations, method I- we get \(175/(5-x)= x+35y\). This is a quadratic quation.

method II- even on solving by traditional approach, we would be getting a quadatic equation of X. Quadratic equation implies two roots of X. So E. A transition from SC to PS. _________________

Lets multiply statement 1 by 5: \(x+35y=763\) \(5x+175y=763*5\)

Substract this new equation from statement 2. 763x-5x = 0 x=0 Hence sufficient.

Ans C it is

In general a 15 sec approach tip for such questions: We dont need to solve these question to find ultimate solution. If you can visualize that each such equation represents a line and these lines will intersect certainly at some point provided we have diffferent slopes. So u need to only quickly identify slopes (m in y=mx+c form) and determine between C or E. _________________

Lets multiply statement 1 by 5: \(x+35y=763\) \(5x+175y=763*5\)

Substract this new equation from statement 2. 763x-5x = 0 x=0 Hence sufficient.

Ans C it is

In general a 15 sec approach tip for such questions: We dont need to solve these question to find ultimate solution. If you can visualize that each such equation represents a line and these lines will intersect certainly at some point provided we have diffferent slopes. So u need to only quickly identify slopes (m in y=mx+c form) and determine between C or E.

Hi Vips. X=0 is just the one value of X. The other value of the equation has to satisfy the equation \(x-5-35y=0\) for which we don't know the value of y. The two equations are: \(x(x-5-35y)=0\) _________________

Hi Vips. X=0 is just the one value of X. The other value of the equation has to satisfy the equation \(x-5-35y=0\) for which we don't know the value of y. The two equations are: \(x(x-5-35y)=0\)

Hi Marcab, the value of x=0 and y=109/5 (from statement 1 or 2) these values satisfy both equations and are only values to do so.

I'm not sure, how did you formulate: \(x(x-5-35y)=0\)

Best thing to do (while solving such linear equations) is to eliminate 1 term, as i did in the solution. _________________

First given eqn: x + 35y = 109*7........... (i) x 5 we get: 175y + 5x = 763*5.......(iii)

From the 2nd given equation: 175y + 763x = 763*5....(ii);

we say (ii) minus (iii), we get 763x – 5x = (763*5) – (763*5) = 0 x = 0; y = 109*7 / 35 = 21.8;.....(0, 21.8) How come E?

OFFICIAL EXPLANATION Solution: E If we have two variables (x and y), we can generally solve for both of them if we have two independent linear equations, i.e. two linear equations that are not multiples of each other. In a data sufficiency question like this, be skeptical as soon as you notice that 763 (in statement (2)) just happens to be 109 (in statement (1)) times 7. Manipulate statement (2) until you get x and y on the same side: 5x + 175y = 3815. A little mental math should show that statement (2) is just 35 times statement (1), so the two equations are NOT independent and we cannot solve for x or y; (E). _________________

KUDOS me if you feel my contribution has helped you.

The two given equations are: 1)\(x+35y=763\) 2)\(175y=763(5-x)\) which can be written as \(175y/(5-x)=763\)

since the values on Left Hand Side in both equations are equal to 763, therefore: \(x+35y=175y/(5-x)\) or \((5-x)(x+35y)=175y\)

On expanding this above equation, we'll get: \(5x+175y-x^2+35xy=175y\)

\(5x+175y-x^2+35xy=175y\)

Now we get:

\(x^2-5x-35xy=0\)

Taking x common, we get:

\(x(x-5-35y)=0\) One value of x=0 and the other should satisfy the equation \(x-5-35y=0\) and for this we need to know the value of Y.

When one considers the value of x=0, then he is neglecting the other value that satisfies the equation \(x-5-35y=0\). Hence E. Hope that helps _________________

combining both eqnts: 109*7 – x = 109*7 - (109*7/5)x x - (109*7/5)*x = 0 => x = 0. from 35y = 109*7 - x, we get y = 109*7/35 = 109/5...same as before No other value satisfies the equation.

Implying that the OA SHOULD HAVE BEEN C (E not correct): Agree with Vips;

Hii BULL. You and Vips are equally right in your approach but when you are considering one value of X, you are letting the other one to slip. The solution that you provided gives only the value of X which is other than zero, i.e. second value.

Moreover 2x+2y=10 is same as 4x+4y=20. I hope you are getting my point. Both the statements are the same, but written in different manner and one cannot deduce a particular value of x or y. _________________

If you divide this by 35, then you will get the same equation as statement 1.

Since it is an equation, we can divide by 35 on both sides comprehensively. Just because Statement 1 has been multiplied by 35 to yield statement 2 doesnt makes it a different equation. _________________

That came intuitively apart from that silly and disastrous mistake in second solution, I feel comfortable with this solution because it yields two values of x. I agree to your's, Bull's and Vips's solution but whats wrong in this method. _________________

That came intuitively apart from that silly and disastrous mistake in second solution, I feel comfortable with this solution because it yields two values of x. I agree to your's, Bull's and Vips's solution but whats wrong in this method.

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