Inequalities

The answer is B. This inequality might look relatively simple, especially when you know that inequalities can be solved like equations. We divide both sides by $-6$, so the right side is $-11$. But wait! This is a problem where the “inequality exception” applies; because we are dividing by a negative value, we must change the direction of the inequality symbol. So the answer is not $A$ (trap answer), but $B$.

Since this problem is a bit tricky due to its trap answer, it’s worth taking a few more seconds to plug in something like $-12$. Seeing how $-12$ makes the inequality true confirms our confidence in our answer.

The answer is B. If the CLT presents you with an inequality-based word problem, it will often be a linear model with a variable aspect and a constant part. We need to identify which part is variable (which one will change) so we can attach our variable to the number involved. In this case, the changing part of the equation is the $\$0.15$ per mile cost, so we should plan to have a term of $0.15x$. Meanwhile, the $\$40$ only gets charged once, so $40$ can be added, making $0.15x+40$. We need to stay at or under a budget of $\$115$; this sounds like an inequality but with the equals sign included (since spending exactly $\$115$ is acceptable in this case). The inequality comes together as follows:

$$0.15x+40\le 115$$

Once we subtract $40$ from both sides, we can see we’ll need to divide $75$ by $0.15$–not an easy task without a calculator. You can use long division, but always seek ways to use $10$’s to help you, since our math system is based on tens. One aspect of “ten-ness” in math is the decimal; since the decimal has to do only with the place value, we can ignore it for the moment and think about how many times $15$ goes into $75$. If you see that the answer to that is $5$, you should see the right answer already. But in case you want to make sure, the two zeroes in the right answer correspond to the two places we have to move the decimal in $0.15$ to make $15$. A combination of math reasoning and common sense can often save you time in calculating!

For an alternate strategy, consider how well this problem works with the “B, then C” strategy of plugging in the answers. If we start with one of the answers in the middle, that answer will either be 1) correct, or 2) too large, or 3) too small. If it’s too large or too small, you can quickly arrive at the answer. Since $500$ (choice $B$) is the answer here, let’s pretend we started with choice $C$ instead. If we suppose that we drove the car 600 miles, that would accrue $(600)(\$0.15)=\$90$ in variable charges. Add the $\$40$ fixed charge and we have $\$130$, which is over our budget. You can probably tell already that an answer a little smaller, like $500$, is correct at this stage, but you could plug in either choice $A$ or choice $B$ to check.

The answer is D. To understand this question, especially given how the answers are organized, it’s most helpful to envision how three consecutive integers might be described algebraically. We can always label a quantity as $x$, so we might as well use $x$ to represent the smallest number in the sequence. Once we do that, it becomes apparent that the next highest number is $x+1$ and the one after that is $x+2$. The sum of the three, then, must be $x+x+1+x+2=3x+3$. There’s only one answer that fits that, and the “greater than” symbol makes sense in that context.

Plugging in could help in this case if you’re uncertain about mapping the algebra. Choose three numbers whose sum is greater than $50$, like $20-21-22$. Plugging in one of these numbers results in a false statement for choices $A$, $B$, and $C$. This process can give you choice $D$ as an educated guess and may well deepen your understanding of the problem so that you become more certain of the answer.

The answer is A. Our best bet seems to be solving both inequalities, but where to begin? The linear equality is more approachable than the cubic inequality (the one with $y^3$), so let’s begin there. Subtracting $6$ from both sides is the first step, but be careful: the negative remains in front of the 9y, so our result is this:

$-9y<-36$

Dividing both sides by $-9$ reverses the sign so we arrive at:

$y>4$.

Have you done the L step of the UnCLES method, looking carefully at the answers? If so, you might notice that we don’t need to deal with the other inequality at all!. There is only one answer in which $y>4$. However, here is a solution to the second inequality if you’re interested:

$14+y^3<-13$

$y^3<-27$

$y<-3$

(Note: we are allowed to take the cube root of both sides because a cube root of a negative number is legal; it’s only even powers, like the square root, where we run into problems with negative numbers.)

The answer is A. This is as much a number properties problem as it is an inequalities problem, so feel free to click on the cross-reference link to the Number Properties lesson. We have two inequalities; the more concrete of the two tells us that $b$ is a negative number unless it is either $0$ or a fraction less than $1$. We should therefore consider all three of those possible number categories for $b$ (negative, zero, or small positive fraction) as we relate $b$ to $c$. We should be careful not to assume too much about $c$: just because it is greater than $b$ does not mean it is positive; $b$ could be negative while $c$ is just a bit less negative.

With this in mind, we can address the statements given with the Roman numerals. Statement I can be disproved by considering that a number squared is positive or zero; if we choose a “negative enough” number, we can make $b^2$ larger than $c$. For example, if $b=-10$ and $c=-5$, then $b<c$ but $b^2>c$. Since Statement I is not always true, answer choice $C$ is eliminated.

Statement II changes the relationship because now the quantity that can never be negative ($c^2$) is on the “greater than” side of the inequality. This tells us that as long as $b$ is negative, the inequality will be true because $b^3$ will also be negative. But what if $b$ is $0$ or $0<b<1$? If $b=0$, then $b^3=0$ and in that instance $c$, and therefore $c^2$, must be positive. So that works. If $b$ is something like $\frac{1}{2}$, $b^3$ will actually be smaller than that. If $c>b$, then $c^2$ won’t decrease as much as $b^3$ does and therefore can never be smaller. Put in numerical terms, imagine that $b=\frac{1}{3}$ and $c=\frac{1}{2}$. In that case, $b^3=\frac{1}{27}$ while $c^2$ is much larger: $\frac{1}{4}$. Statement II always works.

Does Statement III also work? We can actually deal with this statement more quickly by applying the same logic as we did with Statement I. Since $b^4$ must always be at least $0$ and could in fact very positive, we cannot guarantee it will stay less than $c^3$. Statement III is not always true, therefore, and the answer is $A$.