Equations / Inequalities - GRE Quantitative Reasoning

If each pencil sells at 50 cents, x pencils will sell at 50x. The smallest value of x such that

50xgeq 1200+20x

xgeq 40

The answer is:

y-7< x+2y-14

-y-7< x-14

-1(-y< x-7)

y > -x+7

From the equation we can see that the slope is –1.

This question is not as hard as it seems. Remember that for real numbers, square roots cannot be taken of negative numbers. Therefore, we know that this function is undefined for:

This is simple to solve. Merely add to both sides:

Then, divide by :

Therefore, quantity A is less than quantity B. This means that quantity B is greater than it.

Recall that when you have an absolute value and an inequality like

,

this is the same as saying that must be between and . You can rewrite it:

To solve this, you merely need to subtract from all three values:

Since is between and , it could be both larger or smaller than . Therefore, you cannot determine the relationship based on the given information.

Take the values of y that are possible, i.e. 2 and 3, and plug them into the first inequality. First, plug in 2. 2 – 3x > 21. Subtract 2 from both sides, and then divide by –3. Don't forget that when you divide or multiply by a negative number in an inequality you must flip the inequality sign. Thus, x < –19/3. Now plug in 3. We find, following the same steps, that when y=3, x < –6. Thus –7 is the correct answer.

If each pencil sells at 50 cents, x pencils will sell at 50x. The smallest value of x such that

50xgeq 1200+20x

xgeq 40

The answer is:

y-7< x+2y-14

-y-7< x-14

-1(-y< x-7)

y > -x+7

From the equation we can see that the slope is –1.

This question is not as hard as it seems. Remember that for real numbers, square roots cannot be taken of negative numbers. Therefore, we know that this function is undefined for:

This is simple to solve. Merely add to both sides:

Then, divide by :

Therefore, quantity A is less than quantity B. This means that quantity B is greater than it.

Recall that when you have an absolute value and an inequality like

,

this is the same as saying that must be between and . You can rewrite it:

To solve this, you merely need to subtract from all three values:

Since is between and , it could be both larger or smaller than . Therefore, you cannot determine the relationship based on the given information.

The quantity can be regrouped to be –3 = a – b. Thus, (a – b)2 = (–3)2 = 9.

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

Start by setting up an equation using Quantity A and Quantity B. In other words, you can solve an inequality where Quantity A > Quantity B. You would have one of four outcomes:

1. Quantity A = Quantity B: the two quantities are equal.
2. The inequality is always satisfied: Quantity A is always larger.
3. The inequality is never satisfied (but the two are unequal): Quantity B is always larger.
4. The inequality is not always correct or incorrect: the relationship cannot be determined.

So solve:

–5x + 4 > 8 – 2x (Quantity A > Quantity B)

+2x +2x

–3x + 4 > 8

–4 –4

–3x > 4 or x < –4/3

*remember to switch the direction of the inequality when you divide by a negative number

As the inequality \[x < –4/3\] is always false for \[x>0\], Quantity B is always larger.

First we need the area or the rectangle. 24 * 8 = 192. So now we know that 10 gallons will cover 75 ft2 and x gallons will cover 192 ft2. We set up a simple ratio and cross multiply to find that 75_x_ = 1920.

x = 25.6

First, solve for x:

11 + 3_x_ = 29

29 – 11 = 3_x_

18 = 3_x_

x = 6

Then, solve for 2_x_:

2_x_ = 2 * 6 = 12

For this type of problem, we use the formula:

distance = ratetimes time

When Karen catches up with Jen, their distances are equivalent. Thus,

rate (Jen) times time (Jen)=rate(Karen)times time(Karen)

We then make a variable for Jen's time, t. Thus we know that Karen's time is t-.5 (since we are working in hours).

Thus,

There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:

15H = 22.5

So H = 1.5.

Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.

Start by combining like terms.

4_x_ + 5 = 13_x_ + 4 – x – 9

4_x_ + 5 = 12_x_ – 5

–8_x_ = –10

x = 5/4

First we solve for x.

Subtracting 3 from both sides gives us –3_x_ < 17.

Dividing by –3 gives us x > –17/3.

–6 is less than –17/3.

In order to solve for y, place x in terms of y in the first equation and then substitute that for x in the second equation.

The first equation would yield: .

Substituting into the second equation, we get: .

Simplify: