Algebra - GMAT Quantitative

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Question

True or false: $N^{2} < N + 6$

Statement 1:

Statement 2:

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Answer

First, solve the inequality:

$N^{2} < N + 6$

$N^{2} - N - 6 < N + 6 - N - 6$

$N^{2} - N - 6 < 0$

The boundary points of the intervals to be tested are the points at which:

; that is, , or

; that is, .

Therefore, the intervals $(-\infty, -2), (-2, 3), (3, \infty)$ should be tested; do so by testing one value in each interval in the inequality and testing its truth:

$(-\infty, -2)$ - test

$N^{2} < N + 6$

$(-3)^{2} < (-3 )+ 6$

False - reject this interval.

- test

$N^{2} < N + 6$

$0^{2} < 0 + 6$

True - accept this interval.

$(3, \infty)$ - test

$N^{2} < N + 6$

$4^{2} < 4 + 6$

False - reject this interval.

The solution set is the interval ; therefore, $N^{2} < N + 6$ if and only if .

From Statement 1 alone, we know that ; this provides insufficient information, since, for example, and both fall in this range, but only the former is a solution of $N^{2} < N + 6$ . For similar reasons, Statement 2 alone provides infufficient information.

Assume both statements are true. Together, the two statements are euivalent to saying that . Therefore, it holds that , or , and it can be established that $N^{2} < N + 6$ .

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Question

Which of the following is equal to $2 + \log 6$ ?

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Answer

$2 + \log 6 = \log \left ( 10^{2} \right ) + \log 6 = \log 100 + \log 6 = \log\left ( 100 \cdot 6 \right ) = \log 600$

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Question

Solve for :

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Answer

can be rewritten as the inequality

$\frac{-172}{-5} > \frac{- 5x }{-5} > \frac{148}{-5}$ (note the change in direction of the inequality symbols)

$34 \frac{ 2}{5} > x > -29 \frac{3}{5}$

This is the set $\left ( -29 \frac{3}{5}, 34 \frac{ 2}{5} \right )$ .

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Question

Divide:

$\left (56x^{3} +21x ^{2} -63x + 77 \right )\div 7 x ^{2}$

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Answer

$\left (56x^{3} +21x ^{2} -63x + 77 \right )\div 7 x ^{2}$

$= \frac{ 56x^{3} }{7 x ^{2}} + \frac{ 21x ^{2} }{7 x ^{2}}- \frac{ 63x }{7 x ^{2}}+ \frac{ 77 }{7 x ^{2}}$

$= \frac{ 56 }{7 } x^{3-2} + \frac{ 21 }{7 }- \frac{ 63 }{7}\cdot \frac{1}{ x ^{2-1}}+ \frac{ 77 }{7} \cdot \frac{1}{ x ^{2}}$

$= 8 x +3 -\frac{9}{ x }+ \frac{ 11 }{x ^{2}}$

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Question

Give the solution set of the inequality

$x^{2} + 6x + 2 > 6x - 7$

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Answer

To solve a quadratic inequality, move all expressions to the left first

$x^{2} + 6x + 2 > 6x - 7$

$x^{2} + 6x + 2 - 6x + 7 > 6x - 7 - 6x + 7$

$x^{2} + 9 >0$

$x^{2} > -9$

Since the square of any number must be nonnegative, it follows that for any ,

$x^{2} \ge 0 > -9$

and the solution set is the set of all real numbers, $(-\infty, \infty)$ .

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Question

Solve the system of equations.

$\dpi{100} \small 4x+3y=7$

$\dpi{100} \small 7x-14y=21$

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Answer

Let's first look at the 2nd equation. All three terms in $\dpi{100} \small 7x-14y=21$ can be divided by 7. Then $\dpi{100} \small x-2y=3$ We can isolate x to get $\dpi{100} \small x=3+2y$

Now let's plug $\dpi{100} \small x=3+2y$ into the 1st equation, $\dpi{100} \small 4x+3y=7:$

$\dpi{100} \small 4\left ( 3+2y \right )+3y=7$

$\dpi{100} \small 12+8y+3y=7$

$\dpi{100} \small 11y=-5$

$\dpi{100} \small y=\frac{-5}{11}$ Now let's plug our y-value into $\dpi{100} \small x=3+2y$ to solve for y:

$\dpi{100} \small x=3+2\left\left ( \frac{-5}{11} \right )=3-\frac{10}{11}=\frac{33}{11}-\frac{10}{11}=\frac{23}{11}$

So $\dpi{100} \small x=\frac{23}{11}\and\ y=\frac{-5}{11}$

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Question

Which equation is linear?

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Answer

Let's go through all of the answer choices.

1. $x+\Pi y+ez=log(5)$ : $\Pi$ and $e$ are both constants, so the equation is actually linear.

2. 5x + 7y - 8yz = 16: This is not linear because of the yz term.

3. $\frac{y+8}{x-2}=x+6$ : This can be transformed into y + 8 = (x + 6)(x - 2). Clearly when this is expanded, there will be an $x^{2}$ term, so this is not linear.

4. $y=x^{2}-55x+1$ : This is not linear either, also because of the $x^{2}$ term.

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Question

What is $x^{2} - y?$

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Answer

Solve the first equation to get

Substitute that into the second equation and get

Solve the equation to get , then substitute that into the first equation to get .

Plugging those two values into $x^{2} - y$ , gives

$5^{2}-7=18$

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Question

Solve.

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Answer

Solve for in the first equation:

$x=13-\frac{3}{2}y$

Substitute into the second equation:

$3(13-\frac{3}{2}y)-4y=5$

$39-\frac{9}{2}y-4y=-34$

$-\frac{9}{2}y-4y=-34$

$-\frac{17}{2}y=-34$

$y=(-34)(-\frac{2}{17})=4$

Solve for .

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Question

Solve the system of equations:

$\frac{x}{3}-\frac{y}{4} =1$

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Answer

Multiply both sides of the first equation by 12:

$\frac{x}{3}-\frac{y}{4} =1$

$12 \cdot \left (\frac{x}{3}-\frac{y}{4} \right ) =12 \cdot 1$

$\frac{12}{1} \cdot \frac{x}{3}-\frac{12}{1} \cdot \frac{y}{4} =12$

$4x-\3y =12$

Now, add both sides of the two equations:

$4x-\3y =12$

$\underline{-4x+3y = 6}$

Since this is impossible, the system of equations is inconsistent and thus has no solution.

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Question

Choose the statement that most accurately describes the system of equations.

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Answer

Subtract the first equation from the second:

$x+y = 4\Rightarrow y=4-x$

Now we can substitute this into either equation. We'll plug it into the first equation here:

$4x + 3(4-x) = 19\Rightarrow 4x+12-3x = 19\Rightarrow x+12 = 19$

Thus we get and .

Therefore is positive and is negative.

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Question

$x^{2} - y ^{2} = 125$

Give the solution set for .

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Answer

$x^{2} - y ^{2} = 125$

The expression on the left factors as the difference of squares:

Since , we can substitute:

$25(x-y ) \div 25 = 125\div 25$

We now have a system of linear equations to solve:

$\underline{x + y = 25}$

$2x \div 2 =30\div 2$

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Question

A company wants to ship some widgets. If the weight of the box plus one widget is 6 pounds, and the weight of the box plus two widgets is 10 pounds, then what is the weight of the box and the weight of the widget? Put the answer in an ordered pair such that the ordered pair is (box weight, widget weight).

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Answer

Let the weight of the box be represented by and the weight of the widget be represented by . Since the weight of the box plus the weight of one widget is 6 pounds, this can be represented by the equation

Since the weight of the box plus two widgets is 10 pounds, this can be represented by the equation

We now have two equations and two unknowns and we can now solve for and . To do this we solve the first equation for and substitute it into the second equation. Solving the first equation for we get

Substituting this into the second equation we get

$(6-W)+2W=10\Rightarrow 6-W+2W=10\Rightarrow$

$6+W=10\Rightarrow W=10-6=4.$

Using and substituting it into the first equation we get

$B+4=6\Rightarrow B=6-4=2.$ So the weight of the box is 2 pounds and the weight of the widget is 4 pounds. This gives us the ordered pair $\left ( 2,4 \right )$ .

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Question

Solve for when

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Answer

Plug in the given value and then isolate .

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Question

Whats the value of when :

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Answer

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Question

Solve for :

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Answer

$x=\frac{10-4y}{2}$

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Question

Define an operation $\bigstar$ as follows:

For any real , $a ; \bigstar ; b = a^{2} +b^{2} -8$ .

For what value or values of is it true that $N ; \bigstar ; N = 0$ ?

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Answer

Substitute into the definition, and then set the expression equal to 0 to solve for :

$a ; \bigstar ; b = a^{2} +b^{2} -8$

$N ; \bigstar ; N = N^{2} +N^{2} -8 = 2N^{2} -8$

$2N^{2} -8 = 0$

$2N^{2} =8$

$N^{2} =4$

$N = -2 \textrm{ or } N = 2$

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Question

$x^{2}+ y^{2} = 20$

What is ?

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Answer

From the second equation:

Substitute into the first, then solve:

$\left ( 10-2y\right )^{2}+ y^{2} = 20$

$10^{2} - 2 \cdot 10 \cdot 2y + \left ( 2y \right )^{2} + y^{2} = 20$

$4y ^{2} + y^{2} - 40y + 100 = 20$

$5 y^{2} - 40y + 100 = 20$

$5 y^{2} - 40y + 100 - 20= 20 - 20$

$5 y^{2} - 40y + 80= 0$

$5 \left ( y^{2} - 8y + 16 \right )= 0$

$5 \left ( y - 4 \right )^{2}= 0$

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Question

If and ; what is the value of ?

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Answer

For this problem we can use the elimination method to solve for one of our variables. We do this my multiplying our first equation by -2.

From here we can combine this equation with our second equation given in the question and solve for x.

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Now we plug 1 back into our original equation and solve for y.

Therefore,

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Question

Find the point of intersection of the two lines.

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Answer

The correct answer is $\left(2,-\frac{1}{6}\right)$

There are a few ways of solving this. The method I will use is the method of elimination.

(Start)

(Multiply the 2nd equation by -1 and add the result to the first equation, combining like terms. Now the top equation simplifies to

Now that we have one of the variables solved for, we can plug into either of the original equations, and we can get our , Let's use the 2nd equation.

$y=-\frac{1}{6}$

Hence the point of intersection of the two lines is $\left(2,-\frac{1}{6}\right)$ .

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