GMAT Quantitative - Algebra

If $\dpi{100} \small 5x+4=19$ , what is the value of $\dpi{100} \small 4x^{2}-5$ ?

$\dpi{100} \small 31$

$\dpi{100} \small 139$

$\dpi{100} \small 19$

$\dpi{100} \small 10$

$\dpi{100} \small 3$

Explanation

First, we need to solve for $\dpi{100} \small x$ from the first equation in order to calculate the second quadratic function. To solve for $\dpi{100} \small x$ , we need to subtract four on each side of the equation, then we will get

$\dpi{100} \small 5x=15$

The answer for $\dpi{100} \small x$ would be $\dpi{100} \small \frac{15}{5}$ , which is $\dpi{100} \small 3$ .

So now we can calculate the function by plugging in $\dpi{100} \small x=3$ .

$\dpi{100} \small 3^{2}=9$ , and $\dpi{100} \small 9\times 4=36$ .

$\dpi{100} \small 36-5=31$

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

$\dpi{100} \small 9$

$\dpi{100} \small 18$

$\dpi{100} \small 6$

$\dpi{100} \small 36$

cannot be determined

Explanation

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

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

Putting these together,

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

There is water tank already $\frac{4}{7}$ full. If Jose adds 5 gallons of water to the water tank, the tank will be $\frac{13}{14}$ full. How many gallons of water would the water tank hold if it were full?

$14$

$5$

$15$

$20$

$25$

Explanation

In this case, we need to solve for the volume of the water tank, so we set the full volume of the water tank as $x$ . According to the question, $\frac{4}{7}$ -full can be replaced as $\frac{4}{7}x$ . $\frac{13}{14}$ -full would be $\frac{13}{14}x$ . Therefore, we can write out the equation as:

$\frac{4}{7}x+5=\frac{13}{14}x$ .

Then we can solve the equation and find the answer is 14 gallons.

Solve $\left | 2x - 5 \right |\geq 3$ .

$x \leq 1, x\geq 4$

$x \leq -1, x\geq -4$

$1 < x < 4$

$-2 \leq x\leq 5$

$x < 1, x > 4$

Explanation

It's actually easier to solve for the complement first. Let's solve $\left | 2x-5 \right |<3$ . That gives -3 < 2x - 5 < 3. Add 5 to get 2 < 2x < 8, and divide by 2 to get 1 < x < 4. To find the real solution then, we take the opposites of the two inequality signs. Then our answer becomes $x\leq 1 \textsc{ or } x\geq 4$ .

Solve $\left | 2x - 5 \right |\geq 3$ .

$x \leq 1, x\geq 4$

$x \leq -1, x\geq -4$

$1 < x < 4$

$-2 \leq x\leq 5$

$x < 1, x > 4$

Explanation

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

What is ?

Explanation

From the second equation:

Substitute into the first, then solve:

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

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

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

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

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

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

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

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

For what value of $\small N$ would the following equation have no solution?

$\small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\small N= 2$

The equation must always have at least one solution regardless of the value of $\small N$ .

$\small \small N= -1$

$\small \small N= 1$

$\small N= -2$

Explanation

Simplify both sides of the equation as much as possible, and solve for $\small x$ in the equation in terms of $\small N$ :

$\small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\small 3 \cdot 4x - 3 \cdot 7 + 12 = 2 \cdot 5x - 2 \cdot 3 + N \cdot x - N \cdot 3$

$\small 12x - 21 + 12 = 10x - 6 + Nx - 3N$

$\small 12x - 9 = (10+N)x + (- 6 - 3N)$

$\small \small 12x - (10+N)x = (- 6 - 3N) + 9$

$\small (2-N)x = 3 - 3N$

$\small x = \frac{3 - 3N}{2 -N}$

$\small x$ has exactly one solution unless the denominator is 0 - that is, $\small N = 2$ . We make sure that this value renders no solution by substituting:

$\small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + N (x - 3)$

$\small \small \small \small 3 (4x - 7) + 12 = 2 (5x - 3) + 2 (x - 3)$

$\small 12x - 21 + 12 = 10x - 6 + 2x - 6$

$\small 12x - 9 = 12x - 12$

$\small - 9 = - 12$

The equation has no solution, and $\small N = 2$ is the correct answer.

Solve for :

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

$x = -29 \frac{1}{3}$

$x = -4\frac{2}{3}$

$x = -9\frac{7}{9}$

$x = -24\frac{1}{3}$

The equation has no solution.

Explanation

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

$\left (4x + 7 \right ) + (2x+30) = 3x-51$

$x = \frac{-88}{3}= -29 \frac{1}{3}$

Consider the incomplete equation

$10(N - 18) + 5N = \square (N-12 )$

What number replaces the box in order to form an equation with no solution?

None of the other choices yields a correct answer.

Explanation

Set to be the number that replaces the box.

Simplify first:

Now solve for in terms of :

$\left (15 -A \right )N= 180 - 12A$

$N=\frac{ 180 - 12A}{15 -A }$

The only possible value of that might preclude the existence of a solution is , since it makes the denominator 0. However, let us test this value in the original equation:

As it turns out, replacing the box with 15 yields an identity, not a contradiction, so the solution set is the set of all real numbers. There is no number that fits the description.

Students at a local college decide to make and sell t-shirts in order to raise money for their activities. They will pay the manufacturer per t-shirt made and a fixed fee of . If they sell each t-shirt for , how many t-shirts would they have to make and sell to raise ?