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Free Sample Math Question Paper for GMAT 2014

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by ravikashyap1196 » Tue Oct 29, 2013 11:36 pm
High difficulty level math question that has high probability of appearing in your GMAT exam.

Get the Answes - https://admissiontimes.com/gmat-math
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by theCodeToGMAT » Tue Oct 29, 2013 11:57 pm
x^2 - 6x + 5 = 0

x = 5 & 1

{A} x^2 = -1
{B} x^2 - x - 2 = 0 ==> x = 2 & -1
{C} 2X^2 = 2 => X = +1,-1
{D} x^2 - 2x - 3 = 0 ==> X = 3, -1
{E} x^2 - 10x - 5 = 0 ==> [10 +- sqrt(100-4(1)(-5)]/2a = We wont have Integer roots.

Answer [spoiler]{C}[/spoiler]
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by Brent@GMATPrepNow » Wed Oct 30, 2013 6:16 am
Which of the following equations has a root (solution) in common with x² - 6x + 5 = 0?

{A} x² + 1 = 0
{B} x² - x - 2 = 0
{C} 2x² - 2 = 0
{D} x² - 2x - 3 = 0
{E} x² - 10x - 5 = 0
Factor x² - 6x + 5 = 0 to get (x - 1)(x - 5) = 0
So, the roots are x = 1 and 5

At this point, we aren't necessarily required to solve each equation. We can simply plug one of the roots into each of the five equations and see if it works.

Since it's pretty easy to plug a 1 into an equation, let's plug x = 1 into the answer choices.

{A} x² + 1 = 0: we get 1² + 1 = 2 [nope, we wanted 0]
{B} x² - x - 2 = 0: we get 1² - 1 - 2 = -2 [nope, we wanted 0]
{C} 2x² - 2 = 0: we get 2(1)² - 2 = 0 perfect!!
STOP

So, x = 1 is a solution to the equation in answer choice C, so this is the correct answer.

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by Matt@VeritasPrep » Wed Oct 30, 2013 10:16 pm
Just as a word of caution to students getting excited about solving this "high level" question rather easily ... this is NOT a high level question (so don't slack off just yet :))
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