Master gmat 2010 part 19 pot

In the question, you started with six terms. Let a through f equal those six terms: 19 5 a 1 b 1 c 1 d 1 e 1 f 6 114 5 a 1 b 1 c 1 d 1 e 1 f f 5 114 2~a 1 b 1 c 1 d 1 e! Letting f 5 the number removed, here’s the arithmetic-mean formula, applied to the remaining five numbers: 21 5 a 1 b 1 c 1 d 1 e 5 105 5 a 1 b 1 c 1 d 1 e Substitute 105 for (a 1 b 1 c 1 d 1 e) in the first equation: f 5 114 2 105 f 5 9 Step 5: If you have time, check to make sure you got the formula right, and check your calculations. Also make sure you didn’t inadvertently switch the numbers 19 and 21 in your equations. (It’s remarkably easy to commit this careless error under time pressure!) If you’re satisfied that your analysis is accurate, confirm your answer and move on to the next question. The correct answer is (C). QUESTION 3 Question 3 is moderately difficult. Approximately 50% of test takers respond correctly to questions like it. Here’s the question again: 3. If p pencils cost 2q dollars, how many pencils can you buy for c cents? [Note: 1 dollar 5 100 cents] (A) pc 2q (B) pc 200q (C) 50pc q (D) 2pq c (E) 200pcq Step 1: The first step is to recognize that this question involves a literal expression. Although it probably won’t be too time-consuming, it may be a bit confusing. You should also recognize that the key to this question is the concept of proportion. It might be appropriate to set up an equation to solve for x. Along the way, expect to convert dollars into cents. ALERT! Careless errors, such as switching two numbers in a problem, is by far the leading cause of incorrect GMAT responses. Chapter 7: Problem Solving 163 www.petersons.com Step 2: The five answer choices provide a couple of useful clues: • Notice that each answer choice includes all three letters (p, q, and c). So the solution you’re shooting for must also include all three letters. • Notice that every answer choice but (E) is a fraction. So anticipate building a fraction to solve the problem algebraically. Step 3: Is there any way to answer this question besides setting up an algebraic equation? Yes. In fact, there are two ways. One is to use easy numbers for the three variables—for example, p 5 2, q 5 1, and c 5 100. These simple numbers make the question easy to work with: “If 2 pencils cost 2 dollars, how many pencils can you buy for 100 cents?” Obviously, the answer to this question is 1, so you can plug the numbers into each answer choice to see which choice provides an expression that equals 1. Only choice (B) fits the bill: ~2!~100! ~200!~1! 5 1 Another way to shortcut the algebra is to apply some intuition to this question. If you strip away the pencils, p’s, q’s and c’s, in a very general sense the question is asking: “If you can buy an item for a dollar, how many can you buy for one cent?” Since one cent (a penny) is 1 100 of a dollar, you can buy 1 100 of one item for a cent. So you’re probably looking for a fractional answer with a large number such as 100 in the denominator (as opposed to a number such as 2, 3, or 6). Answer choice (B) is the only choice that appears to be in the correct ballpark. Choice (B) is indeed the correct answer. Step 4: You can also answer the question in a conventional manner using algebra. (This is easier said than done.) Here’s how to approach it: 1. Express 2q dollars as 200q cents (1 dollar 5 100 cents). 2. Let x equal the number of pencils you can buy for c cents. 3. Think about the problem “verbally,” then set up an equation and solve for x: “p pencils is to 200q cents as x pencils is to c cents” “The ratio of p to 200q is the same as the ratio of x to c” (in other words, the two ratios are proportionate) p 200q 5 x c pc 200q 5 x Step 5: Our solution, pc 200q , is indeed among the answer choices. If you arrived at this solution using the conventional algebraic approach (step 4), you can verify your solution by substituting simple numbers for the three variables (as we did in step 3). Or if you arrived at your solution by plugging in numbers, you can check your work by plugging in a different set 164 PART IV: GMAT Quantitative Section www.petersons.com of numbers or by thinking about the problem conceptually (as in step 3). Once you’re confident you’ve chosen the correct expression among the five choices, confirm your choice, and then move on to the next question. The correct answer is (B). SOME ADVANCED TECHNIQUES Now let’s take a look at some more advanced methods of Problem Solving. In this next section, you’ll: • Apply the success keys you learned earlier to more challenging Problem Solving questions. • Learn additional success keys that apply to certain types of Problem Solving questions and apply these keys to example questions. The first thing you’ll want to do is scan the answer choices to see what all or most of them have in common, such as radical signs, exponents, factorable expressions, or fractions. Then try to formulate a solution that looks like the answer choices. 4. If a Þ 0or1,then 1 a 2 2 2 a 5 (A) 1 2a 2 2 (B) 2 a 2 2 (C) 1 a 2 2 (D) 1 a (E) 2 2a 2 1 The correct answer is (A). Notice what all the answer choices have in common: Each one is a fraction in which the denominator contains the variable a, and no fractions appear in the numerator or denominator. That’s a clue that your job is to manipulate the expression given in the question so that the result includes these features. First, place the denominator’s two terms over the common denominator a. Then, divide a from the denominators of both the numerator fraction and the denominator fraction (this is a shortcut to multiplying the numerator fraction by the reciprocal of the denominator fraction): 1 a 2 2 2 a 5 1 a 2a 2 2 a 5 1 2a 2 2 Chapter 7: Problem Solving 165 www.petersons.com USE COMMONSENSE “GUESSTIMATES” TO NARROW THE FIELD If the question asks for a numerical value, you can probably narrow the answer choices by estimating the value and type of number you’re looking for. Use your common sense and real-world experience to formulate a rough estimate for word problems. However, don’t expect to eliminate all answer choices except the correct one by using common sense alone. 5. A spinner containing seven equal regions numbered 1 through 7 is spun two times in a row. What is the probability that the first spin yields an odd number and the second spin yields an even number? (A) 2 7 (B) 12 49 (C) 5 14 (D) 1 2 (E) 4 7 The correct answer is (B). This problem involves the concept of probability. Common sense about basic probability should tell you that, with odds of close to 50% of spinning the desired type of number on each of the two spins, the odds of spinning such a number twice in a row should be less than 50%. So you can eliminate choices (D) and (E). Your odds of answering the question correctly are now 1 in 3. But notice that the remaining choices—(A), (B), and (C)—are closely grouped in value. Also notice that, in each of these remaining choices, the denominator contains the sort of number you could end up with when you apply a mathematical operation to the numbers given in the question. Conclusion: You’ve probably reached the limits of applying common sense, and you’ll need to solve the problem mathematically to find the correct choice. Here’s how to do it. There are four odd numbers (1, 3, 5, and 7) and three even numbers (2, 4, and 6) on the spinner. So the chances of yielding an odd number with the first spin are 4 in 7, or 4 7 . The chances of yielding an even number with the second spin are 3 in 7, or 3 7 . To determine the probability of both events occurring, combine the two individual probabilities by multiplication: 4 7 3 3 7 5 12 49 Notice the “sucker” answer choice in this question: Answer choice (D) provides the simple average of the two individual probabilities: 4 7 and 3 7 . Aside from the fact that 1 2 , or 50%, is too high a probability from a commonsense viewpoint, (D) should strike you as too easy a solution to what appears to be a complex problem. 166 PART IV: GMAT Quantitative Section www.petersons.com WHEN TO PLUG IN NUMBERS FOR VARIABLES If the answer choices contain variables (like x and y), the question might be a good candidate for the “plug-in” strategy. Pick simple numbers (so the math is easy) and substitute them for the variables. You’ll definitely need your pencil for this strategy. 6. If a train travels r 1 2milesinh hours, which of the following represents the number of miles the train travels in 1 hour and 30 minutes? (A) 3r 1 6 2h (B) 3r h 1 2 (C) r 1 2 h 1 3 (D) r h 1 6 (E) 3 2 ~r 1 2! The correct answer is (A). This is an algebraic word problem involving rate of motion (speed). You can solve this problem either conventionally or by using the plug-in strategy. The conventional way: Notice that all of the answer choices contain fractions. This is a clue that you should try to create a fraction as you solve the problem. Here’s how to do it. Given that the train travels r 1 2milesinh hours, you can express its rate in miles per hour as r 1 2 h .In 3 2 hours, the train would travel this distance: S 3 2 DS r 1 2 h D 5 3r 1 6 2h The plug-in strategy: Let r 5 8 and h 5 1. Given these values, the train travels 10 miles (8 1 2) in 1 hour. Obviously, in 1 1 2 hours the train will travel 15 miles. Start plugging these r and h values into the answer choices. You won’t need to go any further than choice (A): 3r 1 6 2h 5 3~8!16 2~1! 5 30 2 ,or15 Plugging the values into choice (E) also gives an answer of 15, but you should eliminate choice (E) because it omits h. Common sense should tell you that the correct answer must include both r and h. Chapter 7: Problem Solving 167 www.petersons.com WHEN—AND WHEN NOT—TO WORK BACKWARD FROM NUMERICAL ANSWER CHOICES If a Problem Solving question asks for a number value and if you draw a blank as far as how to set up and solve the problem, don’t panic. You might be able to work backward by testing the answer choices, each one in turn. 7. A ball is dropped 192 inches above level ground, and after the third bounce, it rises to a height of 24 inches. If the height to which the ball rises after each bounce is always the same fraction of the height reached on its previous bounce, what is this fraction? (A) 1 8 (B) 1 4 (C) 1 3 (D) 1 2 (E) 2 3 The correct answer is (D). The fastest route to a solution is to plug in an answer. Try choice (C) and see what happens. If the ball bounces up 1 3 as high as it started, then after the first bounce it will rise up 1 3 as high as 192 inches, or 64 inches. After a second bounce, it will rise 1 3 as high, or about 21 inches. But the problem states that the ball rises to 24 inches after the third bounce. Obviously, if the ball rises less than that after two bounces, it’ll be way too low after three. So choice (C) cannot be the correct answer. We can see that the ball must be bouncing higher than one third of the way; so the correct answer must be a greater fraction, either choice (D) or choice (E). You’ve already narrowed your odds to 50%. Try plugging in choice (D), and you’ll see that it works: 1 2 of 192 is 96; 1 2 of 96 is 48; and 1 2 of 48 is 24. Although it would be possible to develop a formula to answer the question, doing so would be senseless, considering how quickly and easily you can work backward from the answer choices. Working backward from numerical answer choices works well when the numbers are easy and when few calculations are required, as in the preceding question. In other cases, however, applying algebra might be a better way. 168 PART IV: GMAT Quantitative Section www.petersons.com 8. How many pounds of nuts selling for 70 cents per pound must be mixed with 30 pounds of nuts selling at 90 cents per pound to make a mixture that sells for 85 cents per pound? (A) 10 (B) 12 (C) 15 (D) 20 (E) 24 The correct answer is (A). Is the easiest route to the solution to test the answer choices? Let’s see. First of all, calculate the total cost of 30 pounds of nuts at 90 cents per pound: 30 3 0.90 5 $27. Now, start with choice (C). 15 pounds of nuts at 70 cents per pound costs $10.50. The total cost of this mixture is $37.50, and the total weight is 45 pounds. Now you’ll need to perform long division. The average weight of the mixture turns out to be between 83 and 84 cents—too small valued for the 85 cent average given in the question. At least you can eliminate choice (C). You should realize by now that testing the answer choices might not be the most efficient way to tackle this question. Besides, there are ample opportunities for calculation errors. Instead, try solving this problem algebraically—by writing and solving an equation. Here’s how to do it. The cost (in cents) of the nuts selling for 70 cents per pound can be expressed as 70x, letting x equal the number that you’re asked to determine. You then add this cost to the cost of the more expensive nuts (30 3 90 5 2700) to obtain the total cost of the mixture, which you can express as 85(x 1 30). You can state this algebraically and solve for x as follows: 70x 1 2700 5 85~x 1 30! 70x 1 2700 5 85x 1 2550 150 5 15x 10 5 x 10 pounds of 70-cents-per-pound nuts must be added in order to make a mixture that sells for 85 cents per pound. FIND THE EASIEST ROUTE TO THE ANSWER If the question asks for an approximation, then you know that precise calculations won’t be necessary and you can safely “round off” the numbers as you go. But even in other questions, you can sometimes eliminate all but the correct answer without resorting to precise calculations. 9. What is the difference between the sum of all positive odd integers less than 102 and the sum of all positive even integers less than 102? (A) 0 (B) 1 (C) 50 (D) 51 (E) 101 Chapter 7: Problem Solving 169 www.petersons.com The correct answer is (D). To see the pattern, compare the initial terms of each sequence: odd integers: {1,3,5 99,101} even integers: {2,4,6 100} Notice that, for each successive term, the even integer is one more than the corresponding odd integer. There are a total of 50 corresponding integers, so the difference between the sums of all these corresponding integers is 250. But the odd-integer sequence includes one additional integer: 101. So the difference is (250 1 101), or 51. SEARCH GEOMETRY FIGURES FOR CLUES Most geometry problems are accompanied by figures. The pieces of information a figure provides can lead you, step-by-step, to the answer. 10. If O is the center of the circle in the figure above, what is the area of the shaded region, expressed in square units? (A) 3 2 p (B) 2p (C) 5 2 p (D) 8 3 p (E) 3p The correct answer is (E). This question asks for the area of a portion of the circle defined by a central angle. To answer the question, you’ll need to determine the area of the entire circle as well as what percent (portion) of that area is shaded. This multi-step question is as complex as any you might encounter on the GMAT. But there’s no need to panic; just start with what you know, then move step-by-step toward the answer. Mine the figure for a piece of information that might provide a starting point. DOCD is your first “stepping stone.” Here are the steps to the answer: 170 PART IV: GMAT Quantitative Section www.petersons.com You know that OC and OD are congruent (equal in length) because each one is the circle’s radius. In any triangle, angles opposite congruent sides are also congruent (the same size, or degree measure). Thus, ∠ODC must measure 60°—just like ∠OCD. For any triangle, the sum of the measures of all three interior angles is 180°. Thus, ∠COD measures 60°, just like the other two angles. Vertical angles created by two intersecting lines are congruent. Thus, ∠AOB also measures 60°. By the same reasoning as in steps 1 and 2, each angle of DABO measures 60°. Notice that the length of AB is given as 3. Accordingly, the length of each and every side of both triangles is 3. Since this length (3) is also the circle’s radius (the distance from its center to its, circumference), you can determine the circle’s area. The area of any circle is pr 2 , where r is the circle’s radius. Thus, the area of the circle is 9p. Now determine what portion of the circle’s area is shaded. The four angles formed at the circle’s center (O) total 360°. You know that two of these angles account for 120°, or 1 3 of those 360°. ∠AOC is supplementary to ∠DOC ; that is, the two angles combine to form a straight line, and so their measures total 180°. Therefore, ∠AOC measures 120°. 120° is 1 3 of 360°. Thus, the shaded portion accounts for 1 3 the circle’s area, or 3p. If you look at the 60° angle in the figure, you might recognize right away that both triangles are equilateral and, extended out to their arcs, form two “pie slices,” each one 1 6 the size of the whole “pie” (the circle). What’s left are two big slices, each of which is twice the size of a small slice. So the shaded area must account for 1 3 the circle’s area. With this intuition, the problem is reduced to the simple mechanics of calculating the circle’s area, then dividing it by 3. SKETCH A GEOMETRY FIGURE TO SOLVE A PROBLEM A geometry problem that does not provide a diagram might cry out for one. That’s your cue to take pencil to scratch paper and draw one yourself. 11. A rancher uses 64 feet of fencing to create a rectangular horse corral. If the ratio of the corral’s length to width is 3:1, which of the following most closely approximates the minimum length of additional fencing needed to divide the rectangular corral into three triangular corrals, one of which is exactly twice the area of the other two? (A) 24 feet (B) 29 feet (C) 36 feet (D) 41 feet (E) 48 feet Chapter 7: Problem Solving 171 www.petersons.com The correct answer is (B). Your first step is to determine the dimensions of the rectangular corral. Given a 3:1 length-to-width ratio, you can solve for the width (w) of the field using the perimeter formula: 2~3w!12~w!564 8w 5 64 w 58 Accordingly, the length of the rectangular corral is 24 feet. Next, determine how the rancher must configure the additional fencing to meet the stated criteria. This calls for a bit of sketching to help you visualize the dimensions. Only two possible configurations create three triangular corrals with the desired ratios: The top figure requires less fencing. You can determine this fact by calculating each length (using the Pythagorean theorem). Or you can use logic and visualization. Here’s how. As a rectangle becomes flatter (“less square”), the shorter length approaches zero (0), at which point the minimum amount of fencing needed in the top configuration would decrease, approaching the length of the longer side. However, in the bottom design, the amount of fencing needed would increase, approaching twice the length of the longer side. Your final step is to calculate the amount of fencing required by the top design, applying the theorem (let x 5 either length of cross-fencing): 8 2 1 12 2 5 x 2 64 1 144 5 x 2 208 5 x 2 x 5 = 208 ' 14.4 Thus, a minimum of approximately 28.8 feet of fencing is needed. Answer choice (B) approximates this solution. Since the question asks for an approximation, it’s a safe bet that estimating = 208 to the nearest integer will suffice. If you learned your “times table,” you know that 14 3 14 5 196, and 15 3 15 5 225. So = 208 must be between 14 and 15. That’s close enough to zero in on choice (B), which provides twice that estimate. 172 PART IV: GMAT Quantitative Section www.petersons.com . know that 14 3 14 5 196 , and 15 3 15 5 225. So = 208 must be between 14 and 15. That’s close enough to zero in on choice (B), which provides twice that estimate. 172 PART IV: GMAT Quantitative. viewpoint, (D) should strike you as too easy a solution to what appears to be a complex problem. 166 PART IV: GMAT Quantitative Section www.petersons.com WHEN TO PLUG IN NUMBERS FOR VARIABLES If the. the preceding question. In other cases, however, applying algebra might be a better way. 168 PART IV: GMAT Quantitative Section www.petersons.com 8. How many pounds of nuts selling for 70 cents