Solving equations - GMAT Quantitative Reasoning

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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

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

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

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

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

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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?

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 $days = \frac{5y}{x}$ .

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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?

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.

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

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

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

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Question

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

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.

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

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

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

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

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

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Question

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

Answer

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

Then, solve for $a$ .

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Question

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

Solve for x.

Answer

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

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

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Question

Define $f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of ?

Answer

By definition:

If $x \geq 0$ , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$

If , then ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Question

Which of the following is a solution to the equation ?

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.

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Question

Which is NOT a solution to the equation

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.

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Question

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Answer

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

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

Either

or

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

8 turns out to be the only solution.

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Question

Given the equation $x\cdot(x+k)=-7$ , if can be any integer, how many different sets of integer solutions are there for ?

Answer

$x\cdot(x+k)=-7$ is the same as .

For solutions sets of integers, we are able to divide it as:

When we multiply this, we notice that the constants must return 7. Since 7 is a prime number, we now know that these constants are 1 and 7. The signs of these can still switch however. If both of the numbers are negative, it will also return a product of positive 7. These are the only ways to make the constant term work. If they are both positive numbers, then we get . If they're both negative numbers then .

For the values of , we simply notice the only way for the overall product to be 0 is for one (or both) of the pieces to be zero. This is done by having equal the additive inverse of the constant piece.

Thus we have 2 integer solution sets: $\left{1,7\right}$ and $\left{-1,-7\right}$

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Question

Solve for :

Answer

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

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Question

Give all real solutions of the following equation:

$x^{4} - 10x^{2} + 9 = 0$

Answer

By substituting $u = x^{2}$ and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ , this equation be rewritten as a quadratic equation, and solved as such:

$x^{4} - 10x^{2} + 9 = 0$

$u^{2} - 10u + 9 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with a product of 9 and a sum of ; these integers are .

Substitute back:

$(x^{2}-9)(x^{2}-1) = 0$

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-1 = 0 \Rightarrow x = 1$

$x+1 = 0 \Rightarrow x = - 1$

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left { -3, -1, 1, 3\right }$ .

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Question

Solve for :

$x - 7 \sqrt{x} + 12 = 0$

Answer

Substitute $u = \sqrt{x}$ , and, subsequently, $u^{2} = x$ , to rewrite this equation as quadratic, then solve by factoring.

$x - 7 \sqrt{x} + 12 = 0$

$u^{2} - 7 u + 12 = 0$

We can rewrite the quadratic expression as , where the question marks are replaced with integers whose product is 12 and whose sum is ; these integers are $-3\ and -4$ .

Set each factor to zero and solve for ; then substitute back and solve for :

$\sqrt{x} =3$

$\sqrt{x} =4$

The solution set, which can be confirmed by substitution, is $\left { 9,16\right }$ .

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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?

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 $P _{2}$ :

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

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

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

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Question

Find all real solutions to the following equation:

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

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

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

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

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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?

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 $P_{2}$ ;

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

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

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

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