problems palette of David Rock and Mary K. Porter 6. A local musician comes to your school to give a performance


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1 palette of problems Davd Rock and Mary K. Porter 1. If n represents an nteger, whch of the followng expressons yelds the greatest value? n,, n, n, n n. A 60watt lghtbulb s used for 95 hours before t burns out. What s the lfe of the lghtbulb n klowatthours? 3. How many multples of 4 from 1 through 1000 do not contan any of the dgts 6, 7, 8, 9, or 0? 4. Fnd the dfference between the sum of the frst 1,000,000 postve even numbers and the sum of the frst 1,000,000 postve odd numbers. 5. Whle fshng, you and a frend dscover that your boat has sprung a leak. You estmate that the leak lets n water at a rate of 8 gallons per mnute. Whle you row, your frend can dump water out of the boat at a rate of 3 gallons per mnute. The boat wll float as long as there are fewer than 40 gallons of water n t. If you are 600 feet from the dock, how fast, n mles per hour, do you need to row the boat (on average) to make t back to shore before the boat snks? 6. A local muscan comes to your school to gve a performance n the gym. The gym s arranged wth chars so that the 150 members of your grade are all seated n rows. Each row contans the same number of chars, and every char s occuped. If 10 more chars were added to each row, everyone could be seated n 4 fewer rows, allowng the backrow students to be closer to the performance. How many chars are n each row n the orgnal confguraton? 7. A mathematcs teacher collects old books. One day, a student asked hm how many old math books he has. He repled, If I dvde the books nto two unequal whole numbers, then 64 tmes the dfference between the two numbers equals the dfference between the squares of the two numbers. How many old math books does the teacher have? 8. Usng an analog clock, fll n the blank to complete the pattern of tmes: p.m., 1:37 a.m., 11:05 a.m., 8:33 p.m. 3 Mathematcs Teachng n the Mddle School Vol. 14, No. 1, August 008 Copyrght 008 The Natonal Councl of Teachers of Mathematcs, Inc. All rghts reserved. Ths materal may not be coped or dstrbuted electroncally or n any other format wthout wrtten permsson from NCTM.
2 Prepared by Davd Rock, Columbus State Unversty, 45 Unversty Ave., Columbus, GA 31907, and Mary K. Porter, Sant Mary s College, Notre Dame, IN MTMS readers are encouraged to submt sngle problems or groups of problems by ndvduals, student groups, or mathematcs clubs to be consdered for publcaton. Send to the Palette edtor, Davd Rock, at rock_davd@colstate.edu. MTMS s also nterested n students creatve solutons to these problems. Send to The Thnkng of Students edtor, Edward S. Mooney, at mooney@lstu.edu. Both problems and solutons wll be credted. For addtonal problems, see the NCTM publcaton, Menu Collecton: Problems Adapted from Mathematcs Teachng n the Mddle School (stock number 76). 9. Fnd the tmes that fall exactly between the other two tmes: 10:49 a.m.,, 1:57 p.m. 10:49 p.m.,, 1:57 p.m. 10. In a quz game, a player earns 5 ponts for every correct answer and loses 3 ponts for every wrong answer. If Steve s score s 0 after answerng 40 questons, how many questons dd he answer correctly? 13. What are the next three letters n ths sequence? c, a, x, f, b, v,, c, t, l, d, r, o, e,,, 14. How many squares (of any sze) are n the fgure below? 11. What percentage of the fgure shown here s shaded? 1. What are the next three letters n ths sequence? 15. How many nonsquare rectangles (of any sze) are n the fgure shown n queston 14? c, a, f, b,, c, l, d, o, e,,, (Solutons on pages 34 36) Vol. 14, No. 1, August 008 Mathematcs Teachng n the Mddle School 33
3 solutons to palette (Alternatve approaches to those suggested here are encouraged.) (Contnued from pages 3 33) 1. If n 0, then n yelds the greatest value. If n = 1, then n and /n both yeld the greatest (same) value. If n >, then n yelds the greatest value. Makng a table helps to determne the greatest value n each case. (See table 1 at rght.). 5.7 klowatthours; 60 watts 95 hours reveals the number of watt hours used, whch s 5700 watt hours. To convert watts to klowatts, multply by 1/1000 snce a watt s 1/1000 of a klowatt. Therefore, 5700 watt hours yelds /1000 = 5.7 klowatt hours All multples of 4 are even. The only sngledgt multples of 4 are 4 and 8. Therefore, 4 would be the only onedgt value that does not contan a 6, 7, 8, 9, or 0. Next, examne the twodgt numbers. Snce we cannot use the dgts 6, 7, 8, 9, we only look for even numbers less than 60 wth unts dgts of and 4. The twodgt multples of 4 would be restrcted to 1, 4, 3, 44, and 5. Next, examne threedgt numbers. They would have to end n 1, 4, 3, 44, and 5 but could only start (hundreds dgt) wth 1,, 3, 4, and 5. Therefore, there are 5 5 = 5 threedgt possbltes. There s 1 onedgt number, 5 twodgt numbers, and 5 threedgt numbers, whch sum to = 31 possbltes. Extend the problem to more than three dgts. Snce the same 5 twodgt numbers must be present n all the numbers wth the condtons mposed, a pattern s created for the total number of possbltes. For example, wth the numbers 1 through 10,000, there are = possbltes. Table 1 A tabular soluton to problem 1 n n / /n n n n undefned ,000,000. One strategy would be to try a smpler problem. The sum of the frst 10 postve odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19) s 100, and the sum of the frst 10 postve even numbers (, 4, 6, 8, 10, 1, 14, 16, 18, 0) s 110. The dfference s 10, whch s the number of odd and even numbers. We can generalze as follows: The frst odd number s 1, and the frst even number s. The dfference between the sums s 1. If we look at the sums of the frst three odd numbers ( = 9) and the frst three even numbers ( = 1), we notce that the dfference s 3, the number of even and odd numbers. Therefore, the even numbers wll be 1 greater for each number. Snce we are workng wth the frst 1,000,000 even and the frst 1,000,000 odd numbers, the dfference s 1,000, Approxmately 3.7 mph. Snce per mnute the leak lets n 8 gallons and your frend can remove 3 gallons, the water comes n at a rate of 5 gallons per mnute. Therefore, you have 8 mnutes before you reach the 40gallon lmt, because 40/5 = 8. You must travel 600 feet n 8 mnutes; 600/8 = 35 feet per mnute. Convert 35 feet per 1 mnute to mles per hour. Multply 35 ft. 1 mn. 60 mn. 1 hr. = 19,500 ft. per hr. Next, multply 19,500 ft./1 hr. by 1 m./580 ft. (or dvde 19,500 by 580, snce there are 580 feet n a mle) to yeld approxmately 3.7 mles per hour. 600 ft. 8 mn. 60 mn. 1 hr. 1 m. 3.7 mph 580 ft. 34 Mathematcs Teachng n the Mddle School Vol. 14, No. 1, August 008
4 Let the number of chars n each row be represented by c and the number of rows be represented by r. There are 150 seats, whch yelds rc = 150, or r = 150/c. Four fewer rows would be r 4, and 10 more chars n each row would be c From ths, we can construct the equaton (r 4)(c + 10) = 150, whch can be rewrtten as rc 4c + 10r 40 = 150. Snce r = 150/c, we can use substtuton n the second equaton to yeld the followng: c 150 c c c 40 = c + 40 = 150 c c + 40 = 0 c Multplyng both sdes by c yelds: 4c c = 0 4c 40c = 0 c + 10c 375 = 0 Factorng yelds (c 15)(c + 5) = 0, and we see that 15 and 5 are the solutons, but only 15 s a possble number of chars per rows. The problem can also be solved wthout algebra by examnng the combnatons of factor pars that produce 150. Students can dscover that = 150 and that (10 4) ( ) = 6 5 = Let a and b represent the two unequal whole numbers. Then 64(a b) = a b. Factorng the rght sde yelds 64(a b) = (a b)(a + b). Snce a and b are not equal, (a b) does not equal zero. Therefore, you can dvde both sdes by (a b), whch yelds 64 = a + b. To solve the problem, you do not have to know the values of a and b, just the sum of a and b, whch s 64, the total number of books. 8. 4:09. Subtract to fnd the dfference between one term and the next. For example, we can compute 11:05 1:37 n ths way: 11:05 s 11 hours and 5 mnutes, or 10 hours and 65 mnutes. Now subtract 1 hour and 37 mnutes to get 9 hours and 8 mnutes, or 9:8. Ths tme s also the dfference between the thrd and fourth terms: 11:05 + 9:8 = 0:33, whch s 8:33 on a 1hour clock. Thus, to fnd the mssng frst term n our lst, we must fnd the tme that s 9 hours and 8 mnutes before 1:37. To do ths, frst note that on a 1hour clock, addng 1 hours to any tme brngs us back to the same tme on the clock. Thus, 1:37 + 1:00 (or 13:37) s stll 1:37; so 1:37 9:8 s the same as 13:37 9:8, whch s 4: :3 p.m.; 6:3 a.m. In the frst problem, determne the number of mnutes between the two tmes: 11 mnutes to 11:00 a.m., 60 mnutes to noon, 60 mnutes to 1:00 p.m., and 57 mnutes to 1:57 p.m. The total mnutes are 188; 188/ = 94 mnutes. Therefore, the mdpont s 10: mnutes = 1:3 p.m. To fnd the mdpont n the second problem, compute 10:49 + 1:57. Note that 10:49 + 1:57 s 11 hours and 106 mnutes, whch s 11 hours plus 1 hour and 46 mnutes, or 1 hours and 46 mnutes. Thus, 10:49 + 1:57 = 1:46 = 6:3 a.m. We can check ths answer by dong two computatons (addng 1 hours and/or convertng hours to mnutes as needed): (1) 6:3 10:49 = 18:3 10:49 = 17:83 10:49 = 7:34 (so the dfference between 10:49 and 6:3 s 7:34). () 6:3 + 7:34 = 13:57, whch s 1:57, as expected questons. Let x represent the number of questons Steve answered correctly, and let y represent the number of questons he answered ncorrectly. Then x + y = 40. Steve earned 5 ponts for every correct answer, so 5x s the total number of ponts he earned from hs correct answers on the quz. Steve loses 3 ponts for every ncorrect answer, so 3y s the total number of ponts he lost from hs ncorrect answers on the quz. Overall, Steve s score was 0, so 5x 3y = 0. Snce x + y = 40, then y = 40 x. Substtutng ths nto the equaton 5x 3y = 0, we fnd that 5x 3(40 x) = 0, so 5x x = 0. Thus, 8x 10 = 0, so 8x = 10. Therefore, x = 10/8 = 15. Thus, Steve answered 15 questons correctly (and = 5 questons were answered ncorrectly). Ths gave hm an overall score of 5(15) 3(5) = = /3 percent. Note that the fgure has dmensons 3 unts by 5 unts and can be dvded nto 15 squares of equal sze. Because 3.5 of these squares are shaded, the fracton of the fgure that s shaded s = 7 30 = Thus, 3 1/3 percent of the fgure s shaded. 1. r, f, u. If we examne the sequence (c, a, f, b,, c, l, d, o, e), we notce that t s actually made up of two sequences: The underlned terms (c, f,, l, o) and the ones that are not underlned (a, b, c, d, e). The terms that are not underlned follow a famlar pattern. The underlned terms (c, f,, l, o) also follow a pattern: Each term s the letter n the alphabet that s found three letters after the prevous term n ths sequence. For Vol. 14, No. 1, August 008 Mathematcs Teachng n the Mddle School 35
5 example, f s found three letters after c n the alphabet. Thus, the sequence formed by the underlned terms looks lke ths: c, f,, l, o, r, u. Therefore, the orgnal sequence contnues n ths way: c, a, f, b,, c, l, d, o, e, r, f, u, g,..., so the next three terms n the orgnal sequence are r, f, u. 13. p, r, f. If we examne ths sequence (c, a, x, f, b, v,, c, t, l, d, r, o, e), we notce that t s actually made up of two sequences: The underlned terms (x, v, t, r) and the ones that are not underlned (c, a, f, b,, c, l, d, o, e). The terms that are not underlned comprse the sequence n the prevous problem, so the next terms of that sequence, as determned n the soluton to the prevous problem, are r, f. The underlned terms (x, v, t, r) also follow a pattern: Each term s the letter n the alphabet that s found two letters before the prevous term. For example, v s found two letters before x n the alphabet. Thus, the sequence formed by the underlned terms looks lke ths: x, v, t, r, p. Therefore, the orgnal sequence contnues n ths way: c, a, x, f, b, v,, c, t, l, d, r, o, e, p, r, f,..., so the next three terms n the orgnal sequence are p, r, f squares. There are 4 squares of sze 1 1 (1 unt by 1 unt). There are 3 squares of sze. There s only 1 square of sze 3 3 and just 1 square of sze 4 4. Thus, there are 9 squares n ths fgure rectangles. As shown n the prevous problem, 9 squares are n ths fgure. Count the number of nonsquare rectangles n ths fgure: 7 rectangles are 1, 3 rectangles are 1 3, 1 rectangle s 1 4, 4 rectangles are 3, rectangles are 4, and only 1 rectangle s 3 4. Ths gves us a total of 1 nonsquare rectangles. Thus, we have = 7 rectangles n the fgure. l Soluton to August s Solve It (Contnued from page 31) The number of cows n pen C s four tmes the number of cows n pen B. There are multple ways to solve ths problem, ncludng by example and algebracally, as shown below. The frst condton ndcates that the number of anmals n each pen was the same before and after the anmals escaped. Let varables represent the number of anmals n each pen. The varable wll be the same value before and after the anmals escaped. Let A represent the number of anmals n each pen. The second condton ndcates that the number of pgs n pen A s the same as the number of pgs n pen B. Snce these numbers are equal, let one varable represent both values. Let x represent both the number of pgs n pen A and n pen B. Smlarly, snce the number of sheep n pen A s the same as the number of sheep n pen C, use the same expresson to represent each value. From the fourth condton, the number of sheep n pen A s /3 the number of pgs n pen A; the number of sheep n pen A and pen C can each be expressed by (/3)x. Table 1 lsts Table 1 The anmals after they escaped Number of cows Table The updated table Number of cows the anmals n each pen after they escaped. There are (/3)x sheep n pen A and (/3)x sheep n pen C. Orgnally, there were A sheep n pen B. Ths means that there are A (/3)x (/3)x, or A (4/3)x, sheep stll n pen B. There are x pgs n pen A and x pgs n pen B. Orgnally, there were A pgs n pen C. Ths means that there are A x x, or A x, pgs n pen C. See table for these updates. We know how many sheep and pgs are n each pen, and we know the total anmals n each pen. We can fnd the number of cows n each pen by subtractng the number of both sheep and pgs from the orgnal number of anmals n the pen. By examnng pen B, we know that the number of anmals s A. The number of sheep n the pen s A (4/3)x, and the number of pgs s x. So the number of cows n pen B s A (A (4/3)x) x, or (1/3)x. In pen C, the number of anmals s A; the number of pgs s A x, and the number of sheep s (/3)x. So the number of cows n pen C s A (A x) (/3)x, or (4/3)x. Ths means that the number of cows n pen C, (4/3)x, s four tmes the number of cows n pen B, (1/3)x. l Pen A Pen B Pen C Number of sheep (/3)x (/3)x Number of pgs x x Pen A Pen B Pen C Number of sheep (/3)x A (4/3)x (/3)x Number of pgs x x A x 36 Mathematcs Teachng n the Mddle School Vol. 14, No. 1, August 008
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