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]]>Yesterday we solved this problem:
Div’s bicycle tour consists of three legs of equal length. For the first leg Div averaged 16 kilometers per hour. For the second leg he averaged 24 kilometers per hour. What speed must Div average for the final leg in order to average 24 kilometers per hour for the entire tour?
A) 20 kilometers per hour
B) 28 kilometers per hour
C) 32 kilometers per hour
D) 40 kilometers per hour
E) 48 kilometers per hour
We made a pretty good use of the RTD table, but we still took quite a while to arrive at our answer, and we did a lot of computation and algebra, every step inviting some sort of error.
There are two shortcuts that could save us some time and energy while reducing opportunities for computational error.
Notice that this problem doesn’t specify the lengths of the legs, but that every answer is a constant. That implies that so long as the legs are all of the same length as required by the problem, any leg length will yield the same answer.
Let’s take advantage of that fact to avoid some unnecessary algebra. Let’s stipulate an easy leg length. What leg length would be easiest to work with? 48 kilometers, since 48 is the least common multiple of 16 and 24.
Now we can put constants in the cells for leg lengths as well as in some of the cells for rates.
We can also determine a time for the bottom row—a combined time for all three legs—by dividing the combined distance by the average rate.
That allows us to determine the time for the third leg, since the sum of the times for the three legs is the combined time for the tour.
Finally, we can determine the rate for the third leg by dividing the distance by the time.
The little shortcut isn’t as powerful as the big one, and it’s peculiar to average problems, so you won’t get much use out of it, but here it is.
Since the rate for the second leg is equal to the average rate, we can ignore that leg. Because Div averages 24 kph for the entire tour and 24 kph for the second leg, he must also average 24 kph for the balance of the tour. So we could leave the second leg out altogether.
What would the RTD table for that amended problem look like? Well, if we use the big shortcut, we’ll still probably make each leg 48 kilometers, since 48 is the least common denominator of the rate in kph of the first leg and the rate in kph of the entire tour.
We can complete the top and bottom rows by dividing distance by the time.
Next, we can complete the “time” column.
Finally, we can divide the length of the third leg, 48 kilometers, by the time for the third leg, 1 hour, to determine the rate for the third leg.
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]]>If you’d like to review before moving on, check out my previous posts on the topic:
We used the RTD (Rate, Time, and Distance) table to manage these problems, but some test-takers probably found that method to be overkill. Those readers judged that they could translate the problems directly from English to algebra without benefit of any table.
Today I’m going to give you a more complicated problem, one in which the travelers don’t move simultaneously:
Jan’s house and Cindy’s house are joined by a straight road 24 miles long. Jan and Cindy agree to meet at a restaurant along that road, twice as far from Jan’s house as from Cindy’s. Beginning at noon, Cindy walks to the restaurant at a constant speed of 3 miles per hour. Later, Jan drives to the restaurant at a constant speed of 30 miles per hour. If they arrive at the restaurant simultaneously, at what time did Jan begin her drive?
A) 12:32
B) 1:25
C) 1:52
D) 2:08
E) 2:40
This is likely to be a frustrating problem. The algebra—if we can get to the algebra—probably won’t be too complicated, but the translation into algebra looks daunting, if only because we haven’t likely seen a very close model of this problem before. Let’s try our good friend the RTD table, and see if it eases the translation.
We’ll set the table up as we have before, with columns for rate, time, and distance, and with rows for Jan, Cindy, and their combined distance:
We’re given Jan’s and Cindy’s rates directly, so let’s put those in.
We are given the combined distances for Jan and Cindy, 24 miles, so we can include that. We’re told that the restaurant is “twice as far from Jan’s house as from Cindy’s.” If we call the distance from Cindy’s house d, then the distance from Jan’s house must be 2d.
d+2d=24
3d=24
d=8.
So the restaurant is 8 miles from Cindy’s house and 16 miles from Jan’s.
Let’s complete the rows for Cindy and Jan. Since we know that Cindy’s rate is 3 mph and that her distance is 8 miles, we can conclude that her time was 8/3 hours. Since we know that Jan’s rate is 30 mph and that her distance is 16 miles, we can conclude that her time was 16/30 hours.
But now what? Well, if Cindy took 8/3 of an hour, then she took hours, or 2 hours and 40 minutes. Since she started at noon, she arrived at the restaurant at 2:40.
Since Cindy and Jan arrived simultaneously, Jan too arrived at 2:40. At what time did she leave? Well, she took 16/30 of an hour. That’s 32/60, or 32 minutes. Since she arrived at 2:40 after 32 minutes of travel, she must have started her drive at 2:08.
If you’ve read my other blogs on the use of this table, then you know that the answer is “yes,” at least for some people. The table is an aid to translation, and a few people are able to translate even very complicated rate problems directly into algebra without benefit of the table.
Still, I think that the RTD table is especially useful for this problem, even though as we used it here it turned out to just be two equations stacked one on the other, and even though it left us with a fair bit of work still to do. Disciplined well-practiced use of the RTD table allows us to chip away at the familiar parts of the problem and to reserve our cognitive resources for the less familiar parts.
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]]>We wanted to solve for Mary’s time, t.
In every row the relationship among rate, time, and distance is the same: RT=D. In this diagram the bottom row looks the most promising, since it alone contains only the variable for which we’re solving. (Why mess around with d if we don’t need to?) So let’s look at the equation implicit in that bottom row:
Cross-multiplying to get the difference between those fractions yields:
or:
Yes. I wanted to show an efficient use of the table, but I didn’t want to insist on the optimal use.
You could, for instance, have represented Mary’s rate as 200 rather than as 1000/5, and Kate’s rate as 500/3 rather than as 1000/6. Doing that in the diagram or simplifying the moment you’d pull the equation out of the diagram would have yielded instead of That would have allowed you to multiply through by 3 rather than to cross-multiply:
That definitely saves some time and effort, but if our first, efficient-but-not-optimal use of the table comes more easily, that’s OK. Sometimes there’s a little trade-off between the ease of translation (English-into-algebra) and the ease of solution (algebraic manipulation).
By the way, I wouldn’t take it a step further and represent Kate’s rate as . Mixed numbers are usually more difficult to clear than are improper fractions.
For instance, many people build RTD tables with no bottom row for the sum or the difference of the rates and distances. This is just inviting complexity and computational error. Such a table often yields a system of two equations and two variables rather than a single equation with a single variable.
Consider what our table would have looked like with no bottom roappw:
This would have yielded a system of equations:
If you distribute those fractions right away you’re in for a lot of work. It might occur to you to instead isolate t in each equation, then to solve for d, and finally to solve for t. It turns out that that, too, is a lot of work:
But 3750 isn’t one of our answers! That’s because we’re solving for t rather than for d. We have to return to one of our equations that related t to d, substitute 3750 for d, and solve for t:
Bottom-line? Always include an extra row for simultaneous-movement problems. You won’t need to use it every time, but it will you save you a lot of trouble when you do need it.
Hey, I think that I could answer this problem without the table!
Well, then you probably could.
You might remember this formula from my earlier post on the RTD table: (combined rate)(time)=(combined distance). We used that for travelers moving simultaneously in opposite directions.
Here’s a similar, and similarly simple, formula for travelers moving simultaneously in the same direction: (difference in rates)(time)=(difference in distances).* Applying that formula to this problem directly yields our equation:
You’ve still got to do the math, of course.
*In fact, some people point out that the second formula is just a special case of the first, since subtraction is one way to combine values. I find that confusing, since for non-mathematicians like us “combined” usually means “added.”
Just in case you’ve patiently read these posts on the RTD table and you still haven’t seen a problem that you couldn’t translate directly from English to Algebra, my next post will feature a more complicated rate problem.
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]]>Today and tomorrow we’re going to use the table just a tiny bit differently to solve another common sort of rate problem, one in which two travelers move in the same direction simultaneously. We’ll set up the table today and use it to solve the problem tomorrow.
Many problems in which two travelers move in the same direction simultaneously involve a faster traveler starting behind a slower traveler, and catching up to or passing him.
Mary and Kate are running clockwise around a circular track with a circumference of 500 meters, each at her own constant speed. Mary runs 1000 meters every five minutes and Kate runs 1000 meters every six minutes. If Mary and Kate start opposite one another on the circular track, how many minutes must Mary run in order to pass Kate and catch her again?
We’ll begin with our familiar layout: three columns (for rate, time, and distance) and three rows (for Mary, Kate, and the difference between them). A lot of people mark the bottom row in such a table as “C,” for “combined,” just as they would if Mary and Kate were traveling in opposite directions, though “combined” in this case doesn’t mean “added together.” Just to be clear, I’m going to label that bottom row with a delta, to remind myself to determine the difference between Mary’s and Kate’s rates and distances. I’m also going to circle the cell representing Mary’s time, since that’s what we’re solving for.
Before we talk about how best to manage this problem and table, go ahead and copy the table above, then add the information from the problem.
Now let’s add the information that we’ve been given. Begin with the rates. Mary’s rate is not 5 minutes and Kate’s rate is not 6 minutes. Speeds are always D/T, so Mary’s and Kate’s rates are 1000/5 and 1000/6 respectively.
Similarly, Mary’s time and distance are not 5 minutes and 1000 meters, and Kate’s time and distance are not 6 minutes and 1000 meters. Those figures are given to you only to determine Mary’s and Kate’s rates. NEVER ASSUME THAT THE TIME AND DISTANCE THAT FIX THE RATE ARE THE APPROPRIATE FIGURES FOR THE TIME AND DISTANCE COLUMNS!
In fact, Mary’s and Kate’s distances are so far unknown to us. Let’s represent them with variables. Or better, let’s represent them with a single variable, using d for Kate’s distance. If Kate ran d meters, how far would Mary run? Well, she’d have to run an additional 250 meters to catch Kate, and then an additional 500 meters to lap her. That means that if Kate’s distance were d meters, Mary’s distance would be d+750 meters.
Since Kate and Mary run simultaneously, a single time t will suffice for all three rows.
Finally, subtract Kate’s rate and distance from Mary’s to fill in the bottom row.
As we saw in our earlier posts on the RTD table, diagrams generally produce equations which we then solve algebraically. Diagrams are usually tools to help us translate word problems into algebra; they are not usually substitutes for manipulating equations.
What equation can you draw from this table that would allow you to solve for t?
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]]>1) Line A has the equation 3x + y = 7. Which of the following lines is perpendicular to Line A?
2) Line P has a positive slope and a positive y-intercept. Line Q also has a positive slope and a positive y-intercept. The two slopes and the two y-intercepts are four different numbers, none equal. Lines P & Q have a single intersection point — what is the full set of possible quadrants in which the intersection point could be?
3) The median of a triangle is the line from any vertex to the midpoint of the opposite side. Triangle ABC has vertices A = (0, 5), B = (–1, –1), and C = (5, 2). What is the slope of the median from A to the midpoint of BC?
In the previous post, I discuss the idea of slope, but I didn’t talk about the special cases of horizontal and vertical lines.
A horizontal line is all run and no rise: therefore, its slope, its rise/run, is zero. By contrast, a vertical line is all rise and no run — when we calculate its rise over run, we get a mathematical error, because we can’t divide by zero. This means, vertical lines have undefined slope; very roughly, we can think of vertical lines as having “infinite” slope, although technically, that way of interpreting vertical lines does not bear rigorous analysis of the idea of infinity —- which, of course, is well beyond anything on the GMAT.
Horizontal lines have a slope equal to zero. Vertical lines have an undefined slope: the value of their slope does not make any mathematical sense. I would strongly recommend against saying that either has “no slope”, because that is a term ripe for confusion: does “no slope” mean a slope with a value of zero or does it mean not being able to compute the value of the slope? I recommend that you absolutely banish the term “no slope” from your vocabulary.
Parallel lines in the x-y plane have equal slope. If the think about what the slope means geometrically, you will see that this has to be true. The GMAT expects you to know this.
The case with perpendicular lines is a little trickier. Let’s think about this. Consider any slanted line with a positive slope (i.e. moving up to the right) — then the line perpendicular to it will have to have a negative slope (i.e. moving down to the right, moving up to the left). Similarly, if the original line has a negative slope, the perpendicular would have to have a positive slope. Thus, the slope of a perpendicular line is always has a sign opposite of the sign of the original line. That’s part one of the idea.
Now, let’s think about if we take a “rise over run triangle” and rotate it 90°. Let’s ignore ± signs for a moment, and just think about absolute values.
So, we have two perpendicular lines, the original line AB and the perpendicular line AE, intersecting at point A. For each, we have constructed “rise over run” triangle for the slope. The slope of line AB = (original rise)/(original run) = BC/AC, and the slope of line AE = (new rise)/(new run) = AD/DE. But notice that these two triangles are congruent! Triangle ADE is just Triangle ACB rotated 90° around point A. Because the triangle are congruent, the corresponding lengths are equal: AC = AD and BC = DE. In other words, the rise of one is the run of the other! When we go from the original line to a line perpendicular to it, the rise and the run switch places in the slope fraction. When the numerator and denominator of a fraction switch places, that’s called a reciprocal. For example, 2/5 and 5/2 are reciprocals; 3/11 and 11/3 are reciprocals; 7 and 1/7 are reciprocals.
Now, put these two facts together — when line #1 has a slope, and line #2 is perpendicular to it, the slope of line #2 must have both the opposite sign from and must be a reciprocal of the original slop of line #1. More elegantly, perpendicular slopes are opposite reciprocals. The GMAT expects you to know this.
If one line has a slope = +4/7, then the perpendicular line has a slope = –7/4. If one line has a slope of –3, the perpendicular line has a slope = +1/3. If you know the slope of one line, you flip that fraction over and change its sign to the opposite to make the slope of the perpendicular line. If this is a new idea to you, or one which learned but forget a long time ago, then I strongly recommend you actually get some graph paper and actually graph sets of perpendicular lines and find their slopes, so you have a visual understanding of this idea.
Occasionally, the GMAT will ask to you to find the coordinates of the midpoint of a segment. You will be given the two endpoints of the segment — for example, (4, 1) and (10, 15). You find the coordinates of the midpoint by averaging the values they give you. The x-coordinate of the midpoint is the average of the x-coordinates of the end points; here, this average is (4 + 10)/2 = 7. Similarly, the y-coordinate of the midpoint is the average of the y-coordinates of the end points; here, this average is (1 + 15)/2 = 8. Therefore, the coordinates of the midpoint are (7, 8). If you can take an average —- twice —- then you can find the coordinates of a midpoint.
Any question about any points or lines in the x-y plane is a visual question. If you are not given a diagram, always sketch a diagram. You open up a whole new level of understanding when you add a visual approach.
If the two practice questions at the top gave you some difficulty when you first looked at them, now that you have read this article, take another look at them before you jump into the explanations below.
1) What’s tricky about this problem: we have to begin by solving the given equation for y, so that we know its slope.
The slope of the original line is m = –3, and the negative reciprocal of that is +1/3, so the perpendicular line must have a slope of +1/3. Among the answer choices, the only line with a slope of +1/3 is (C). Just so you have a visual for this, here’s a diagram of these line (whenever possible, always verify questions about the x-y plane visually, with at least a quick sketch!)
2) If you forget quadrants, take a look at this post. This is a tricky one, because you have to be careful to think about all cases. The first important point to realize is: any line with both a positive slope and a positive y-intercept goes through quadrants I & II & III, but never quadrant IV. Neither of these lines is ever in Quadrant IV, so they can’t intersect there. As it turns out, for different values of the slope, they can intersect in any of the other three quadrants. Here are visual examples.
Therefore, the answer = (D)
3) For your visual understanding, here’s a diagram of the situation.
First to find the midpoint of B & C — average the x-coordinates: (–1+ 5)/2 = 2; and average the y-coordinates: (–1 + 2)/2 = 1/2. Thus, the midpoint has coordinates (2, 1/2). We want the slope from A = (0, 5) to (2, 1/2). The rise is the change in the y-coordinates: 1/2 – 5 = -9/2. The run is the change in the x-coordinates: 2 – 0 = 2. Slope = rise/run = [–9/2]/2 = –9/4. Answer = (E)
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]]>
First, try these challenging practice questions.
1) Of the 80 houses in a development, 50 have a two-car garage, 40 have an in-the-ground swimming pool, and 35 have both a two-car garage and an in-the-ground swimming pool. How many houses in the development have neither a two-car garage nor an in-the-ground swimming pool?
A. 10
B. 15
C. 20
D. 25
E. 30
2) A certain school has three performing arts extracurricular activities: Band, Chorus, or Drama. Students must participate in at least one, and may participate in two or even in all three. There are 120 students in the school. There are 70 students in Band, 73 in the Chorus, and 45 in the Drama. Furthermore, 37 students are in both the Band and Chorus, 20 are in both the Band and the Drama, and 8 students are in all three groups. Twenty-five students are just in the chorus, not in anything else. How many students participate in only the drama?
A. 11
B. 12
C. 14
D. 17
E. 21
The idea of a set is, in some sense, the most fundamental idea in all of mathematics. Nonetheless, it’s a very simple idea. A set is simply a collection of objects or elements. When the members of a set are all numbers, then we use “set notation”, which consists of brackets. For example:
A = {2, 3, 3, 5, 7}
That notation denotes set A with five numerical members. When all the members of the set are numbers, typical questions involve computations like the mean or the median: here, the mean (or average) of set A is 4, and the median of A is 3. You can read more about those calculations at this post.
Numerical sets can be handled with statistical calculations. Non-numerical sets, sets in which the members are people or cars or companies, are the stuff of tricky word problems. Ordinarily, there’s nothing particular challenging if there’s only one set: some of the people are in that set, whatever it is, and the rest aren’t. No challenge.
Things get more interesting if there are two or more overlapping sets. For example, in the SF Bay Area, many adult residents were born out of state — many, but not all; many adult residents have a college education — many, but not all; and many adult residents are SF Giants fans — many, but not all. Those are three overlapping sets —- any particular adult resident of the SF Bay Area many be a member of none, one, two, or all three of those three categories. As it happens, I am a member of exactly two of those categories. This is exactly the situation of the practice questions posed above.
Venn diagrams are the best method for untangling overlapping sets. If you have two overlapping sets, you need a two-circle Venn diagram:
This diagram contains three discrete regions:
A = those elements in just the left circle
B = those element in both categories, in the overlap
C = those elements in just the right circle
There may also be a fourth discrete region, those elements that are not members of any set. Typically, the problem will only give us information about totals — the total number of elements altogether, the total number in each circle, and the overlap. If you are told there are 70 members in the right circle, and 20 members in the overlap, then you would know B + C = 70 and B = 20, so from that you could deduce C = 50, the number of elements that are just in the portion of the circle labeled C. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.
If there are three overlapping categories, we use a three-circle Venn diagram:
This diagram has at least seven discrete regions
A = members of all three circles
B = members of the green and blue circles, but not the red circle
C = members of the green and red circles, but not the blue circle
D = members of the blue and red circles, but not the green circle
E = members of the green circle but of neither the blue nor the red circles
F = members of the blue circle but of neither the green nor the red circles
G = members of the red circle but of neither the green nor the blue circles
Depending on context, there may also be a eighth discrete region, those elements that are not members of any of the three set. Typically, the problem will only give us information about totals. This gets very tricky. If we are told the total in any one circle, that includes four discrete regions; for example, the green circle includes A + B + C + E. Similarly, the overlap of two circles contains two discrete regions: for example, the overlap of the blue and red circles includes A + D. The problem will always tell you how many elements are in the central region (A), and will often tell you how many are in each circle, and how many in each overlap of two circles. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.
With these strategies, you may find the practice problems at the beginning somewhat more approachable. Try them again, before reading the explanations below.
In the next post, we will look at set problems in which each element is categorized according to two different variables at once.
1) Here, we have two categories: (a) with or without two-car garage, and (b) with or without an in-the-ground pool. Houses can be members of either, both, or neither category. We will use a two circle Venn diagram:
We know the total of the group is 80 —– A + B + C + D = 80. We know the green circle, two-car garages, has 50 members, so A + B = 50. We know the blue circle, in-the-ground pool, has 40 members, so B + C = 40. We also know the crucial overlap region, B = 35. If B = 35, in the green circle, we can deduce that A = 15, and in the blue circle, we can deduce that C = 5. Then
A + B + C + D = 15 + 35 + 5 + D = 80
D = 25
Thus, 25 houses in this development have neither a two-car garage nor an in-the-ground swimming pool. Answer = D.
2) Here, we have three categories, so we need three circles. Every student must take at least one of these three performing arts extracurricular activities, so there will be no one outside the three circles.
The sum of all seven = 120 (we never use this number in this question)
The totals for the band (70), the chorus (73), and the drama (45) each involve the sum of four discrete regions. We will have to find other information before we can employ them.
“8 students are in all three groups”
N = 8.
“37 students are in both the Band and Chorus”
37 = K + N = K + 8 —> K = 29
“20 are in both the Band and the Drama”
20 = M + N = M + 8 —> M = 12
“twenty-five students are just in the chorus, not in anything else”
L = 25
We now have identified three of the regions in the Chorus circle, so we can solve for P.
chorus = 73 = K + L + N + P
73 = 29 + 25 + 8 + P
P = 11
Now, we have identified three of the regions in the Drama circle, so we can solve for Q.
drama = 45 = M + N + P + Q
45 = 12 + 8 + 11 + Q
Q = 14
This is precisely what the question was asking: how many students are only in drama? There are 14 students who take only drama.
Answer = C
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]]>Fact: All lines with slopes of 1 make 45º angles with both the x- and y-axes.
Conversely, if a line makes a 45º angles with either the x- of y-axes, you know immediately its slope must be . This first fact is true, not only for y = x and y = –x, for all lines of the form y = mx + b in which m equals either 1 or –1. If the slope is anything other than , you would need trigonometry to figure out the angles, and that’s beyond the scope of GMAT math. The GMAT could expect you to know this one fact about these special lines, especially on Data Sufficiency.
Fact: Suppose we treat the line y = x as a mirror line. If you take any point (a, b) in the coordinate plane, and reflect it over the line y = x, the result is (b, a). It reverses the x- and y-coordinates!
The corollary of this is that if we compare any two points with reversed coordinates, say (2, 7) and (7, 2), we automatically know that each is the image of the other by reflection over the line y = x. Add now the geometry fact that a mirror line is the set of all points equidistant from the original point and its image. This means that the midpoint of the segment connect (2, 7) and (7, 2) must lie on the line y = x. In fact, any point on the line y = x will be equidistant from both (2, 7) and (7, 2). Without doing a single calculation, we know, for example, that the triangle formed by, say, (2, 7) and (7, 2) and (8, 8 ) must be an isosceles triangle. (See the diagram below.)
When we reflex over the line y = –x, the coordinate are reversed and made their opposite sign: e.g. (2, 7) reflect to (–7, –2), and (–5, 3) reflects to (–3, 5). The other conclusions, about equidistance, remain the same.
Fact: Any point (x, y) in the coordinate plane that is above the line y = x has the property that y > x. Any point (x, y) in the coordinate plane that is below the line y = x has the property that y < x.
Can you sense the veritable cornucopia of Data Sufficiency questions that could arise from this fact? If you every see a question about the coordinate plane asking whether y > x or y < x, chances are very good that the line y = x is hidden somewhere in the question.
1) Is the slope of Line 1 positive?
Statement #1: The angle between Line 1 and Line 2 is 40º.
Statement #2: Line 2 has a slope of 1.
(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question
(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.
2) Point (P, Q) is in the coordinate plane. Is P > Q?
Statement #1: P is positive.
Statement #2: Point (P, Q) above is on the line y = x + 1
(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question.
(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.
3) A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
(A) (1, 1)
(B) (1, 7)
(C) (–1, 9)
(D) (–3, –2)
(E) (–9, 1)
1) A straightforward prompt.
Statement #1 is intriguing: it gives us a specific angle measure. This is tantalizing, but unfortunately, it is only the angle between Line 1 and Line 2, and that angle could be oriented in any direction. Therefore, we can draw no conclusion about the prompt from this statement alone. Statement #1, by itself, is insufficient.
Statement #2 is also tantalizing, because it’s numerically specific. But, unfortunately, this tells us a lot about Line 2 and zilch about Line one, so this statement is, by itself, is also insufficient.
Now, combine the statements. From statement #2, we know Line 2 has a slope of 1, which means the angle between Line 2 and the positive x-axis is 45º. We know, from statement #1, that Line #1 is 40º away from Line 2. We don’t know which way, above or below Line 2. If Line 1 is steeper than Line 2, it makes an angle of 45º + 40º = 85º with the positive x-axis. If Line 1 is less steep than Line 2, it makes an angle of 45º – 40º = 5º with the positive x-axis. Either way, its angle above the positive x-axis is between 0º and 90º, which means it has a positive slope. The combined statements allow us to give a definitive answer to the prompt question. Answer = C.
2) We see the x > y type question in the prompt, which makes us suspect that the line y = x will play an important part at some point.
Statement #1 just tells us P is positive, nothing else. The point (P, Q) = (4, 2) has the property that P > Q, but the point (P, Q) = (4, 5) has the property that P < Q. Clearly, just knowing P is positive does nothing to help us figure out whether P > Q. Statement #1, by itself, is wildly insufficient.
Statement #2 is intriguing. It discusses not the line y = x but the line y = x + 1. What is the relationship of those two lines? First of all, they are parallel: they have the same slope. The line y = x has a y-intercept of zero (it goes through the origin), while the line y = x + 1 has a y-intercept of 1. This means: any point on the line y = x + 1 must be above the line y = x. If (P, Q) is on y = x + 1, then it is above y = x, which automatically means Q > P. We can give a definite “no” answer to the question. By itself, Statement #2 is sufficient. Answer = B.
3) For this problem, there’s a long tedious way to slog through the problem, and there’s a slick elegant method that gets to the answer in a lightning fast manner.
The long slogging approach — first, calculate the distance from (–4, 4) to (2, 3). As it happens, that distance, the radius, equals . Then, we have to calculate the distance from (–4, 4) to each of the five answer choices, and find which one has also has a distance of —- all without a calculator. 🙁
The slick elegant approach is as follows. The point (–4, 4) is on the line y = –x, so it is equidistant from any point and that point’s reflection over the line y = –x. The reflection of (2, 3) over the line y = –x is (–3, –2). Since (–3, –2) is the same distance from (–4, 4) as is (2, 3), it must also be on the circle. Answer = D.
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