GMAT Math : Solving equations

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Example Question #1 : Solving Equations

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

Possible Answers:

$\dpi{100} \small 31$

$\dpi{100} \small 139$

$\dpi{100} \small 3$

$\dpi{100} \small 10$

$\dpi{100} \small 19$

Correct answer:

$\dpi{100} \small 31$

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

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

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Example Question #1 : Equations

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?

Possible Answers:

$\frac{5y}{x}$

$\frac{xy}{5}$

$\frac{y}{5x}$

$\frac{5x}{y}$

$\frac{5}{xy}$

Correct answer:

$\frac{5y}{x}$

Explanation:

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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Example Question #3 : Solving Equations

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?

Possible Answers:

$\dpi{100} \small 20$

$\dpi{100} \small 50$

$\dpi{100} \small 30$

$\dpi{100} \small 15$

$\dpi{100} \small 40$

Correct answer:

$\dpi{100} \small 20$

Explanation:

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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Example Question #2 : Equations

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

Possible Answers:

$\dpi{100} \small x=-6\ or\ x=2$

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

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

$\dpi{100} \small x=6\ or\ x=-2$

$\dpi{100} \small x=-12\ or\ x=-4$

Correct answer:

$\dpi{100} \small x=6\ or\ x=-2$

Explanation:

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

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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Example Question #1 : Solving Equations

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

Possible Answers:

$2$

$\frac{1}{3}$

$\frac{2}{3}$

$-1$

$3$

Correct answer:

$3$

Explanation:

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

Then, solve for $a$ .

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Example Question #6 : Solving Equations

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

Solve for x.

Possible Answers:

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

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

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

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

not enough information

Correct answer:

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

Explanation:

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

3x=4y+2

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

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Example Question #1 : Equations

Which of the following is a solution to the equation x+2y-3z=4 ?

Possible Answers:

two of the answer choices are correct

x=1,y=2,z=3

x=2,y=-1,z=5

x=3,y=2,z=1

x=8,y=1,z=2

Correct answer:

two of the answer choices are correct

Explanation:

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

1. x=3,y=2,z=1:3+2(2)-3(1)=3+4-3=4 , correct

2. x=1,y=2,z=3:1+2(2)-3(3)=1+4-9=-4 , incorrect

3. x=8,y=1,z=2:8+2(1)-3(2)=8+2-6=4 , correct

4. x=2,y=-1,z=5:2+2(-1)-3(5)=2-2-15=-15 , incorrect

Therefore two of the answer choices are correct.

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Example Question #8 : Solving Equations

Which is NOT a solution to the equation 2x-4y+z=8

Possible Answers:

(6,1,0)

(2,4,6)

$(\frac{11}{2},1,1)$

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

$(\frac{7}{2},0,1)$

Correct answer:

(2,4,6)

Explanation:

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.

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

(6,1,0):2(6)-4(1)+0=8

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

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

Let's let a=1 and b=0 . Then this ordered pair becomes (6,1,0) .

2(6)-4(1)+0=8 . Any other answer choices of this form will also work.

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Example Question #9 : Solving Equations

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

Possible Answers:

Infinitely many

Correct answer:

Explanation:

$x\cdot(x+k)=-7$ is the same as x^2 + kx + 7 = 0 .

For solutions sets of integers, we are able to divide it as: $(x+\_\_)(x+\_\_) = 0$

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 k=8 . If they're both negative numbers then k= -8 .

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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Example Question #10 : Solving Equations

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?

Possible Answers:

The information provided is insufficient to answer this question.

$104 \textrm{ mbar}$

$101 \textrm{ mbar}$

$108 \textrm{ mbar}$

$112 \textrm{ mbar}$

Correct answer:

$108 \textrm{ mbar}$

Explanation:

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, $V_{1} = V_{2}$ ,

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

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

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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GMAT Math : Solving equations

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Solving Equations

Example Question #1 : Equations

Example Question #3 : Solving Equations

Example Question #2 : Equations

Example Question #1 : Solving Equations

Example Question #6 : Solving Equations

Example Question #1 : Equations

Example Question #8 : Solving Equations

Example Question #9 : Solving Equations

Example Question #10 : Solving Equations

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