Rates questions and their close cousins dealing with proportions always have to do with a relationship between numbers. Since a similar relationship can also be expressed by a fraction, a decimal, or a percent, it’s important to know the differences between these equivalent expressions. Percent questions are quite frequent on the SAT and have their own dedicated module in this course.
Rates, specifically, involve relationships between different quantities that can be modeled with an algebraic equation. More specifically, proportions relate two rates (expressed as fractions) to each other and can typically be solved through cross-multiplication. You will likely see 1-2 questions in these two categories on test day.
On her regular long-distance runs, Kirsten finds that she runs at an average speed of 6 miles per hour. Which of the following expressions models how long, in minutes, it would take Kirsten to run x miles at this pace?
A.
B.
C.
D.
Rate questions like this one can be deceptively difficult. We not only have to decide whether our answer should represent multiplication and division, but we also have to accomplish a unit conversion. Let’s take the tasks one at a time. With a rate question, you can clear up much possible confusion if you consistently use the rate formula: x . This formula is effective for questions involving any sort of rate; you simply need to assign either variables or constants to each of the three quantities in the formula. Let’s do so in this case:
Distance: miles
Rate: miles/hour
Time: ?
If we label time as , we can set up the equation: . Solving for , we get . From this we can conclude that it would take Kirsten hours to run miles. But the question asks for minutes. We know there are minutes in an hour, but does this mean we should multiply by or divide by ? It can be hard to tell. The best approach is to set up a product of fractions that demonstrate the units, like this:
hours minutes. (Notice that, with the fractions set up this way, the “hours” unit cancels, leaving us with units in minutes, which is what we want. The answer is 10x.
Though questions like this are typically best solved through algebra, they can often be checked by common sense, real-life reasoning. We can plug in a number of miles Kirsten runs and see which answer choice fits. In this case, if Kirsten runs miles per hour, it follows that she can run mile in minutes. We want the choice that gives an output of when we input for All we’re really saying in the final answer is that we need to multiply the number of miles she runs, , by minutes, to get the total time it takes for her run.
A similar concept to the ones found in this chapter is rate of change, which expresses how much a quantity changes per unit of time. We will address this concept in the unit on problem-solving and data.
The algebraic skill most helpful for proportion problems is cross-multiplication. The “cross” refers to the fact that you (twice) multiply the numerator of one fraction by the denominator of the other. You then set the products equal to each other. For a flashcard, you can represent cross-multiplication as follows:
-one side of the card: a/b = c/d
-the other side: ad=bc
The most important formula by far for rates, used many times in this lesson, is distance = rate x time (some know it as distance = speed x time). Make sure you’ve got that one down cold!
The average speed of a snail’s movement is mph. At this glacial pace, how many hours will it take a snail to cover miles?
A. 2.7
B. 30
C. 300
D. 3,000
The answer is 300. Consider, again, distance = rate x time. We can plug in for distance and for rate to give us the equation , where is time. Dividing both sides by , we arrive at our answer of . Note: the units naturally cancel out with this calculation; that is typically the case on the SAT, but make sure to use the UnCLES method and read carefully for exceptions such as this lesson’s Approach question, which required a conversion from hours to minutes.
Ivy surveys her closet and calculates that the ratio of blue items of clothing to green items is . If Ivy has total items of clothing in her closet that are either blue or green, how many clothing items in the closet are blue? (You may assume that this question disregards any items that are partially blue or partially green.)
A. 4
B. 10
C. 20
D. 25
The answer is 10. This is a great opportunity to set up a proportion, but we need to do some analysis first. Be careful when creating a proportion that the same kind of element is in both numerators and the same kind is in both denominators. If we’re going to use 14 in the proportion, we need to arrive at a total figure in the ratio, since 14 represents the total number of actual items. This means we need to add together the parts of the ratio to arrive at a total of 7 ratio “parts”.
Now we are ready to set up our proportion: (or ). Either of these ways of writing the ratio properly ensures that both numerators describe the same sort of quantity, as do both denominators. (If you accidentally flip a fraction, you will arrive at the wrong answer.)
We now may cross-multiply, arriving at the equation 7x=70. Thus our answer.
Some students may notice right away that there are total parts in the ratio () and will see to divide by to get the basic multiplier for a ratio “part”. There are “parts” that correspond to blue, so . This is likely the quickest way to solve the problem, but it may not be intuitive for all students. The algebraic method presented above is reliable.
If and , what is the value of ?
A. 7
B. 14
C. 28
D. 49
The answer is 49. Although it’s possible to plug in numbers to arrive at the answer here (say, assuming that and in the first equation, then plugging those values into the second equation), students may find it more algebraically straightforward to solve the first equation for a and substitute that value for a in the second equation. Multiplying both sides by in the first equation, we find that . Substituting for a in the second equation leads us here:
A teenager mowing lawns makes an algebraic calculation that he can mow acres of grass in minutes. Which expression represents the number of acres the teenager can mow, at the same rate, in minutes?
A.
B.
C.
D.
The answer is . This problem is rated at higher difficulty because the language may make it difficult to see how best to solve it. But the best strategy is a simple proportion relating the rate (ratio) given to the minutes given later. It looks like this:
It looks like we have a lot of unknowns, but we can use any variable we want for the question mark, then cross-multiply. Let’s use and cross-multiply, giving us . Dividing both sides by y gets rid of the y4$'s, then dividing by isolates the . , but since both and are divisible by , that fraction can be reduced to the final answer of .
Newton’s inverse square law states that the force of gravity between two objects is inversely proportional to the square of the distance between them. If, during the course of their natural orbits, two objects move five times further apart at time B than they were at time A, by what factor does the gravitational force between them exist at time B, compared to time A?
A. It is 25 times as great.
B. It is 5 times as great.
C. It is 1/5 times as great.
D. It is 1/25 times as great.
The answer is It is 1/25 times as great. This question requires careful interpretation of the inverse square law. The first keyword is inverse, which expresses a ratio in which, when one quantity goes up, the other goes down, and vice versa. Let’s use that definition combined with common sense: if two planets move further apart from each other, what will happen to their gravitational attraction? Surely it will decrease. It already appears that one of the two answers with fractions must be correct.
Here we consider the other keyword in the law and its definition: “square”. Newton discovered that when an object moves, say, times further away, the gravitational force does not change by a factor of but rather a factor of squared. Our answer becomes clear: the attraction is now smaller by a factor of , or, in other words, times as great.
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