Equations - GMAT Quantitative

Card 1 of 576

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

What is the -intercept of ?

Tap to reveal answer

Answer

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

-intercept:

← Didn't Know|Knew It →

Question

is a rational number. True or false:

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

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

True or false: .

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

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\dpi{100} \small 9\times 4=36$ .

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

← Didn't Know|Knew It →

Question

A tractor spends 5 days plowing $\dpi{100} \small x$ number of fields. How many days will it take to plow $\dpi{100} \small y$ number of fields at the same rate?

Tap to reveal answer

Answer

The equation that will be used is (rate * number of days = number of fields plowed). From the first part of the question, number of fields plowed $\dpi{100} \small (x)$ is calculated as:

$rate \cdot 5 = x$

To solve for rate both sides are divided by 5.

$rate = \frac{x}{5}$

This rate is used for the second part of the problem. $\frac{x}{5} * days = y$ . To solve for days, both sides are divided by $\frac{x}{5}$ , which is the same as multiplying by $\frac{5}{x}$ , cancelling out the $\frac{x}{5}$ and giving the answer of $days = \frac{5y}{x}$ .

← Didn't Know|Knew It →

Question

A 70 ft long board is sawed into two planks. One plank is 30 ft longer than the other, how long (in feet) is the shorter plank?

Tap to reveal answer

Answer

Let $\dpi{100} \small x$ = length of the short plank and $\dpi{100} \small x+30$ = length of the long plank.

$\dpi{100} \small x+x+30=70$ is the length of the pre-cut board, or combined length of both planks.

$\dpi{100} \small 2x+30=70$

$\dpi{100} \small 2x=40$

$\dpi{100} \small x=20$

← Didn't Know|Knew It →

Question

Solve $2x^{2} - 8x - 24 = 0.$

Tap to reveal answer

Answer

$2x^{2} - 8x - 24 = 2(x^{2}-4x-12)=0$

Divide both sides by 2: $x^{2}-4x-12=0$ . We need to find two numbers that multiply to $\dpi{100} \small -12$ and sum to $\dpi{100} \small -4$ . The numbers $\dpi{100} \small -6$ and $\dpi{100} \small 2$ work.

$x^{2}-4x-12= (x-6)(x+2)=0$

$\dpi{100} \small x-6=0$

$\dpi{100} \small x=6$

$\dpi{100} \small or\ x+2=0$

$\dpi{100} \small x=-2$

← Didn't Know|Knew It →

Question

If $1-\frac{4}{a}=2-\frac{7}{a}$ then $a=$

Tap to reveal answer

Answer

Multiply both sides of the equation by a: $a-4=2a-7$

Then, solve for $a$ .

← Didn't Know|Knew It →

Question

$\dpi{100} \small 4y+7=3x+5$

Solve for x.

Tap to reveal answer

Answer

We need to solve for x in terms of y by isolating x.

$x = \frac{4}{3}y + \frac{2}{3}$

← Didn't Know|Knew It →

Question

Which of the following is a solution to the equation ?

Tap to reveal answer

Answer

We need to plug in the answer choices and see which produce the value 4.

1. , correct

2. , incorrect

3. , correct

4. , incorrect

Therefore two of the answer choices are correct.

← Didn't Know|Knew It →

Question

Which is NOT a solution to the equation

Tap to reveal answer

Answer

To solve this, we need to plug the answer choices into the equation and see which choice does NOT work. Let's go through the answer choices.

$\dpi{100} (2,4,6):2(2)-4(4)+6=-6$

This is the correct answer. If this had been a solution to the equation, the equation would have produced 8.

$\dpi{100} (\frac{11}{2},1,1):2(\frac{11}{2}) -4(1)+1)=8$

$(\frac{7}{2},0,1):2(\frac{7}{2})-4(0)+1=8$

$(4+2a-\frac{b}{2}):$

Let's let and . Then this ordered pair becomes .

. Any other answer choices of this form will also work.

← Didn't Know|Knew It →

Question

Solve for :

Tap to reveal answer

Answer

$4 \cdot x +4 \cdot 8 + 13 = 6 \cdot x + 6 \cdot 9 - x$

← Didn't Know|Knew It →

Question

The volume of a fixed mass of gas varies inversely as the atmospheric pressure, as measured in millibars, acting on it, and directly as the temperature, as measured in kelvins, acting on it.

A balloon is filled to capacity at a point in time at which the atmospheric pressure is 104 millibars and the temperature is 295 kelvins. Six hours later, the temperature has increased to 305 kelvins, but the volume of the gas has not changed at all. What is the current atmospheric pressure?

Tap to reveal answer

Answer

The following variation equation can be set up:

$\frac{V_{1} P _{1}}{T_{1}} = \frac{V_{2} P _{2}}{T_{2}}$

But since the initial volume and the current volume are equal, or, equivalently, $V_{1} = V_{2}$ ,

$\frac{V_{1} P _{1}}{T_{1}} \div V_{1} = \frac{V_{2} P _{2}}{T_{2}}\div V_{2}$

so

$\frac{P _{1}}{T_{1}}= \frac{P _{2}}{T_{2}}$

We substitute $P _{1}= 104, T_{1}=295 ,T_{2}=305$ , and solve for $P _{2}$ :

$\frac{104}{295}= \frac{P _{2}}{305}$

$\frac{104}{295}\cdot 305 = \frac{P _{2}}{305} \cdot 305$

$P_{2} = \frac{104}{295}\cdot 305 \approx 108 \textrm{ mbar}$

← Didn't Know|Knew It →

Question

Find all real solutions to the following equation:

$\frac{1}{x^{2}} - \frac{8}{x} +12 = 0$

Tap to reveal answer

Answer

This can be best solved by substituting $u = \frac{1}{x}$ , and, subsequently, $u ^{2}=\left ( \frac{1}{x} \right ) ^{2} =\frac{1}{x ^{2}}$ , then solving the resulting quadratic equation.

$\frac{1}{x^{2}} - \frac{8}{x} + 12 = 0$

$\frac{1}{x^{2}} - 8 \cdot \frac{1}{x} + 12 = 0$

$u^{2} - 8u + 12 = 0$

Factor the expression on the left by finding two integers whose product is 12 and whose sum is :

Set each linear binomial factor to 0, solve separately for , and substitute back:

$\frac{1}{x} = 6$

$x = \frac{1}{6}$

or

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

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

← Didn't Know|Knew It →

Question

The period of a pendulum - that is, the time it takes for the pendulum to swing once and back - varies directly as the square root of its length.

The pendulum of a giant clock is 18 meters long and has period 8.5 seconds. If the pendulum is lengthened to 21 meters, what will its period be, to the nearest tenth of a second?

Tap to reveal answer

Answer

The variation equation for this situation is

$\frac{P_{1}}{\sqrt{L_{1}}} = \frac{P_{2}}{\sqrt{L}_{2}}$

Set $P_{1} = 8.5 , L_{1} = 18, {L}_{2}= 21$ , and solve for $P_{2}$ ;

$\frac{8.5}{\sqrt{18}} = \frac{P_{2}}{\sqrt{21}}$

$\frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} = \frac{P_{2}}{\sqrt{21}} \cdot{\sqrt{21}}$

$P_{2} = \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} \approx 9.2 \textrm{ sec}$

← Didn't Know|Knew It →

Question

Solve for .

Tap to reveal answer

Answer

$-x-3xy+4>0\Rightarrow 4>x+3xy$

$4>x(1+3y) \Rightarrow \frac{4}{1+3y}>x$

← Didn't Know|Knew It →

Question

$x = 8^{N}$

Which of these expressions is equal to $\log_{32} x$ ?

Tap to reveal answer

Answer

$2^{5} = 32$

$2 = \sqrt[5]{32}$

$8 = 2^{3} = \left ( \sqrt[5]{32} \right ) ^{3} = 32^{ \frac{3}{5} }$

$x = 8^{N}= \left (32^{ \frac{3}{5} } \right \)^ {N } =32^{ \frac{3}{5} N }$

$\log_{32} x =\log_{32}32^{ \frac{3}{5} N } = \frac{3}{5} N$

← Didn't Know|Knew It →

Question

What is the equation of the line that goes through and is parallel to ?

Tap to reveal answer

Answer

First, we have to find the slope from using the general form Therefore

Parallel lines have the same slope, so the new line has a slope of and point .

Use the point-slope equation: $y-y_{1}=m(x-x_{1})$

← Didn't Know|Knew It →

Question

$125^{N} = x$

Which of the following expressions is equal to $\log_{25}x$ ?

Tap to reveal answer

Answer

$5 ^{2} = 25$ , so $5 = \sqrt{25}$ , and $125 = 5 ^{3}= \left (\sqrt{25} \right )^{3} = 25^\frac{3}{2}$ .

$x= 125^{N} = \left ( 25^\frac{3}{2} \right )^{N} =25^{\frac{3}{2}N} \right$

$\log_{25} x = \frac{3}{2}N$

← Didn't Know|Knew It →

Question

A factory makes barrels of the same shape but different sizes; the amount of water they hold varies directly as the cube of their height. The four-foot-high barrel holds 20 gallons of water; how much water would the six-foot-high barrel hold?

Tap to reveal answer

Answer

Let be the height of a barrel and be its volume. Since varies directly as the cube of , the variation equation is

$V = k h^{3}$

for some constant of variation .

We find by substituting from the smaller barrels:

$20 = k \cdot 4^{3}$

$20 = k \cdot 64$

$k = \frac{20 }{64}= \frac{5 }{16}$

Then the variation equation is:

$V = \frac{5 }{16} h^{3}$

Now we can substitute to find the volume of the larger barrel:

$V = \frac{5 }{16} \cdot 6^{3} = \frac{5 }{16} \cdot 216 = \frac{5 }{16} \cdot \frac{216}{1} = \frac{5 }{2} \cdot \frac{27}{1} = \frac{135 }{2}= 67 \frac{1 }{2}$

The larger barrel holds $67 \frac{1 }{2}$ gallons.

← Didn't Know|Knew It →

0

Knew It