Algebra - GMAT Quantitative

Card 0 of 3752

Question

Which of the following would be a valid alternative definition of the function

$f(x) = \left | x + 1 \right | - \left | x - 1\right |$ ?

Answer

If $x \geq 1$ , then and are both positive, so

$f(x) = \left | x + 1 \right | - \left | x - 1\right |$

If , then , then is positive and is negative, so

$f(x) = \left | x + 1 \right | - \left | x - 1\right |$

$=x+1 - \left [-\left ( x-1\right ) \right ]$

$=x+1 - \left (-x+1\right )$

If $x \leq -1$ , then and are both negative, so

$f(x) = \left | x + 1 \right | - \left | x - 1\right |$

$=-\left ( x + 1 \right )- \left [-\left ( x - 1\right ) \right ]$

$=- x - 1- \left (- x + 1\right )$

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Question

Solve for :

Answer

and

and

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Question

Give the -intercept(s), if any, of the graph of the function $f (x) = \left | 4x + B \right | - 8$ in terms of .

Answer

Set and solve for :

$\left | 4x + B \right | - 8 = 0$

$\left | 4x + B \right | = 8$

Rewrite as a compound equation and solve each part separately:

$4x + B = -8 \textrm{ or } 4x + B = 8$

$4x \div 4 =\left ( -B -8 \right ) \div 4$

$x = \frac{-B -8 }{4}$

$4x \div 4 =\left ( -B +8 \right ) \div 4$

$x = \frac{-B +8 }{4}$

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Question

What is the -intercept of ?

Answer

To solve for the -intercept, you set the to zero and solve for :

-intercept:

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Question

Which of the following numbers is in scientific notation?

Answer

The number in front must have an absolute value greater than or equal to 1, and less than 10. Of these choices, only $6.7 \times 10^{-9}$ qualifies.

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Question

Let be the piecewise-defined function graphed above. Define a function .

Evaluate $\left (f \circ g^{-1} \right ) (6)$ .

Answer

$\left (f \circ g^{-1} \right ) (6) = f [g^{-1}(6)]$

$g^{-1}(6) = c$ if , so, since

Therefore, $g^{-1}(6) = 10$ , and

$\left (f \circ g^{-1} \right ) (6) = f (10)$

As can be seen from the diagram, however, the domain of is . 10 is not in the domain of . Therefore, is not in the domain of $f \circ g^{-1}$ .

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Question

A number is divided by 4; its decimal point is then moved to the right 3 places. This is the same as doing what to the number?

Answer

The best way to illustrate the answer to this question is to do these operations to the number 1.

First, divide by 4:

$1 \div 4 = 0.250$

Now move the decimal point right three spaces:

$0.250\Rightarrow 250.$

This has the effect of multiplying the number by 250.

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Question

Simplify

$\frac{x^{2}-5x+6}{x^{2}+6x-16}$ .

Answer

In order to simplify $\frac{x^{2}-5x+6}{x^{2}+6x-16}$ , we need to factor the numerator and denominator and then cancel out factors as needed:

$\frac{x^{2}-5x+6}{x^{2}+6x-16}$

Factoring the numerator and the denominator we are left with two binomials on both the top and the bottom.

$=\frac{(x-2)(x-3)}{(x-2)(x+8)}$

Since (x-2) appears in both the numerator and denominator they cancel eachother out.

This results in the final solution:

$=\frac{x-3}{x+8}$

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Question

Which equation is linear?

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

is a rational number. True or false:

Statement 1: $x^{2} - 3x - 10 = 0$

Statement 2: $1 - \frac{7}{x+2} = 0$

Answer

Assume Statement 1 alone and solve for :

$x^{2} - 3x - 10 = 0$

Either:

, in which case ,

or

, in which case .

Therefore, Statement 1 alone is inconclusive.

Assume Statement 2 alone and solve for :

$1 - \frac{7}{x+2} = 0$

$1 =\frac{7}{x+2}$

$1 \cdot (x+ 2) =\frac{7}{x+2} \cdot (x+ 2)$

The only solution to this equation is , so Statement 2 alone answers the question.

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Question

True or false: .

Statement 1: $(N+2)^{2} = 2N + 19$

Statement 2: $1 + \frac{10}{N-5} = 0$

Answer

Substitute 5 for in both statements, and it becomes immediately apparent that it is a solution of neither statement:

$(N+2)^{2} = 2N + 19$

$(5+2)^{2} = 2 \cdot 5 + 19$

$7^{2} = 10+19$

, a false statement.

$1 + \frac{10}{N-5} = 0$

$1 + \frac{10}{5-5} = 0$

$1 + \frac{10}{0} = 0$ , a false statement since the left quantity, having a zero denominator, is undefined.

Therefore, it follows from either statement alone that is false.

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Question

Solve the system of equations.

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

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

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

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

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.

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:

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

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.

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

$\frac{6^{3}}{36} + \frac{3^{68}}{3^{67}}=$

Answer

$\frac{6^{3}}{36} = \frac{6^{3}}{6^{2}} = 6$

$\frac{3^{68}}{3^{67}} = 3^{68-67} = 3$

Putting these together,

$\frac{6^{3}}{36} + \frac{3^{68}}{3^{67}}= 6 + 3 = 9$

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Question

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

Give the solution set for .

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).

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

Answer

Plug in the given value and then isolate .

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