Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 23 By Herbert I. Gross and Richard A. Medeiros next.

Answer: 225 Solution for #1: Method 1 If we wanted to use the most primitive approach we might make a chart that would look like… next © 2007 Herbert I. Gross next Number of Marbles Bill Has Number of Marbles John Has Total Number of Marbles Difference 0500 14995004982 500496349750049444965004925495500490

Solution for # 1: However, using this approach we can see that at the rate we’re going it will take “a lot of rows” before we get to the row in which the difference is 50. next © 2007 Herbert I. Gross next Number of Marbles Bill Has Number of Marbles John Has Total Number of Marbles Difference 0500 14995004982 500496∙ ∙ ∙ 500∙ ∙ ∙ 500∙ ∙ ∙??50050

Solution for # 1: Nevertheless, the chart does give us a few hints. Perhaps the most important one is that as we go down from row to row, the difference column decreases by 2 each time. next © 2007 Herbert I. Gross next Number of Marbles Bill Has Number of Marbles John Has Total Number of Marbles Difference 0500 1499500 2498500 3497500 4496500 5495500 498 496 494 492 490

Solution for # 1: We want to get to the row in which the difference is 50. That is, we want the difference to decrease from 500 in the first row to the row in which the difference is 50. This means we want a total decrease of 450; and since the decrease from row to row is 2, it means that we have to go down to the 225 th row below the first row. next © 2007 Herbert I. Gross As a check we see… next 22527550050

Solution for # 1: Method 2 If we took away 50 marbles from John, he would then have the same number of marbles that Bill has. In that case, John and Bill together would have only 450 marbles with each of them having 225. next © 2007 Herbert I. Gross And if we now give John back the 50 marbles we took away from him, John would have 275 marbles, and Bill would still have 225 marbles. next

Solution for # 1: Method 3 We can also represent Method #2 more visually by using the “corn bread” model. Namely, let’s assume that the corn bread has been sliced into 500 equally sized pieces with each piece representing 1 marble. next © 2007 Herbert I. Gross Since John has 50 pieces more than Bill, if B represents the size of Bill’s share then B + 50 represents John’s share. next

Solution for # 1: Method 4 Let’s suppose Bill and John had an equal amount of marbles. Then each would have had 250 marbles. next © 2007 Herbert I. Gross 250 0 Bill’s AmountJohn’s AmountDifference – 25+ 25 225 27550 To obtain the difference of 50, we take 25 marbles from Bill and give them to John. next

Solution for # 2: next © 2007 Herbert I. Gross The fact that Herb has three times as many marbles as Ben means that we multiply the number of marbles Ben has by 3 to find the number of marbles Herb has. next Method 1 Don’t confuse “three times as many” with “three more than”.

Solution for # 2: For example, if Ben has 20, Herb has 3 × 20 or 60, and if Ben has 40, Herb has 3 × 40 or 120. next © 2007 Herbert I. Gross Namely… next 206080 (less than 120) Ben’s AmountHerb’s AmountTotal Amount 40120160 (more than 120) These two estimates already give us a hint as to how the trial-and-error process might proceed. next

Solution for # 2: Because 120 is between 80 and 160, Ben’s amount must be between 20 and 40. Moreover, the fact that the relationship is linear tells us that since 120 is halfway between 80 and 160, Ben’s amount must be halfway between 20 and 40. Hence, Ben must have 30 marbles. next © 2007 Herbert I. Gross next 3090120 Ben’s AmountHerb’s AmountTotal Amount As a check, we see that…

next Note Reading correctly is always an important component of mathematics. © 2007 Herbert I. Gross next “30” would have been the correct answer if the problem had asked for the number of marbles Ben had. For example, notice that the problem asks for the number of marbles Herb has. If you write “30” as the answer, it is incorrect.

Solution for # 2: Method 2 (A more primitive method) If we didn’t notice that the relationship is linear, we could guess at any number between 20 and 40 and see what happens. next © 2007 Herbert I. Gross For example… next 2678104 (too few) Ben’s AmountHerb’s AmountTotal Amount 27 81108 (still too few)2884112 (still too few) 29 87116 (still too few) 3090120 (just right) 30 3090 120 (just right)

next Note In addition to giving us the answer, the chart shows us why the relationship is linear. © 2007 Herbert I. Gross next Namely, we see from the chart that every time Ben’s amount increases by 1, the total amount increases by 4.

Solution for # 2: Method 3 (An even more primitive method) Keep in mind that we didn’t have to recognize that Ben’s amount was between 20 and 40. We could have started with Ben and Herb each having no marbles and proceeded one marble at a time from there. next © 2007 Herbert I. Gross For example… next 000 Ben’s AmountHerb’s AmountTotal Amount 1342683912

Solution for # 2: Method 3 While it would take a while to get the “Total Amount” to equal 120 this way, we would still notice from the chart that every time “Ben’s Amount” increases by 1, the “Total Amount” increases by 4. next © 2007 Herbert I. Gross next We want the “Total Amount” to increase from 0 to 120; and since 120 ÷ 4 = 30, we want to increase “Ben’s Amount” by 30. This tells us that Ben must have 30 marbles; and hence that Herb has 90.

Solution for # 2: Method 4 If your arithmetic skills are good, there is no real need to even use a chart. Namely, the fact that Herb has three times as many marbles as Ben means that Herb has 3 / 4 of the marbles and Ben has 1 / 4 of the marbles. Since 3 / 4 of 120 is 90, we see that Herb must have 90 marbles. next © 2007 Herbert I. Gross

Solution for # 2: Method 5 (Corn Bread method) In this case, the corn bread is pre-sliced into 120 equally sized pieces and each piece represents 1 marble. In the diagram below, B denotes Ben’s share of the corn bread and H (which in this exercise is equal to 3B) represents Herb’s share. next © 2007 Herbert I. Gross next BBBB9030 H 120 30 next

Problem #3 © 2007 Herbert I. Gross next The only money Mary has with her is 30 coins consisting of dimes and quarters. If she has twice as many quarters as dimes, how much money does she have with her? next Answer: $6

Solution for #3: We can start by using a chart that is based on the fact that there are twice as many quarters as dimes. That is, to find the number of quarters, we multiply the number of dimes by 2. next © 2007 Herbert I. Gross next 000 # of Dimes# of QuartersTotal # of Coins 1 23246 3 69 In terms of a chart…

Solution for # 3: We want to get from the top row in which the total number of coins is 0 to the row in which the total number of coins is 30. We see from the chart that whenever the number of dimes increases by 1, the number of coins increases by 3. next © 2007 Herbert I. Gross next Since 30 ÷ 3 = 10, we must get to the 10th row; and since the number of dimes increases by 1 as we go from one row to the next, in the 10th row the number of dimes is 0 + 10 or 10.

Answer: $6 Solution for #3: In this case, since she has 10 dimes and she has twice as many quarters as dimes, she then has 20 quarters, and we see… next © 2007 Herbert I. Gross next 000 # of Dimes# of QuartersTotal # of Coins 1 23 246 3 69 4812 5 1015 61218 7 1421 8 1624 91827 10 2030

Solution for # 3: However, the problem does not ask us to find the number of coins. Rather we are asked to find how much money Mary has. next © 2007 Herbert I. Gross next Since each dime is worth 10 cents, 10 dimes are worth $1; and since each quarter is worth 25 cents, 20 quarters are worth $5. Altogether, then, she has $6. That is… # Dimes # Quarters 1020$1.00$5.00 ValueValue Total Value $6.00 next

Note on #3 The reason for starting the first row of the chart with 0’s is merely to simplify the arithmetic. That is, by starting with 0, the row number is equal to the number of dimes. © 2007 Herbert I. Gross next This would not have been the case, if we elected to start with 4 dimes in the first line. In that case, the chart would look like… # of Dimes# of QuartersTotal # of Coins 4812 5 1015 61218 7 1421

next Note on #3 We now want to get from a total of 12 coins in the first row to the row in which the total is 30 coins. Since 30 = 12 + 18, we need 18 more coins. For every dime we add, the total number of coins increases by 3; and since 18 ÷ 3 = 6, it means we must go 6 rows below the first row. © 2007 Herbert I. Gross next Because the number of dimes in the first row is 4; on the 6th line below the first line, the number of dimes will be 4 + 6 or 10.

Note on #3 If you understand fractions the quickest way to solve this problem may be to notice that 2 / 3 of the 30 coins are quarters and the other 1 / 3 are dimes. © 2007 Herbert I. Gross This tells us that we have 20 quarters and 10 dimes. next

© 2007 Herbert I. Gross next L W L W L L W W ++ + = 2(L + W) next = 2L + 2W Recall that the perimeter (P) of a rectangle is the distance around its four sides. That is, the perimeter is equal to 2(L + W), where L represents the length and W represents the width of the rectangle. Preface to Problem #4

Problem #4 © 2007 Herbert I. Gross next A rectangle’s length (L) is 22 inches more than three times the rectangle’s width (W). If the perimeter (P) of the rectangle is 180 inches, what is the length of the rectangle? next Answer: 73 inches

Solution for # 4: Based on what we are told, if we knew the width of the rectangle, we would obtain its length by multiplying the width by 3 and then adding 22 inches. next © 2007 Herbert I. Gross next We could then add the length and the width, multiply the sum by 2, and thus obtain the perimeter of the rectangle.

Solution for # 4: We can always begin by using a table of values and making guesses. For example, suppose we begin with the guess that the width is 5 inches; (W = 5). next © 2007 Herbert I. Gross next In this case, 3W will be 15 and the length, L, which is 3W + 22, will be 37. Hence, the length plus the width will be 52 and twice this sum (which is the perimeter of the rectangle) will be 104. We could keep making guesses until we got to the row in which the perimeter was 180.

Solution for # 4: The chart below illustrates this as we let W increase by 1 (inch) each time… next © 2007 Herbert I. Gross next W3WL = 3W + 22L + WP = 2(L + W)515374284618404692721435010082446541089274958116

Solution for # 4: We could continue this way until we got to the row in which P was 180, but it is quicker to use the fact that P increases by 8 every time W increases by 1. To get from P = 84 to P = 180 means that P has to increase by 96. Since 96 ÷ 8 = 12, we know that we have to go down 12 more rows; and 12 rows below the row in which W = 5 is the row in which W = 17. next © 2007 Herbert I. Gross next W3WL = 3W + 22L + WP = 2(L + W)17517390180 This row shows us that the length (L) is 73 inches. next

Problem #5 © 2007 Herbert I. Gross next Mary has twice as much money as Jane. If Mary had $4 more and Jane had $3 less, Mary would then have 4 times as much money as Jane. How much money does Jane have? next Answer: $8

Solution for # 5: If we let J denote the amount of money Jane has and M the amount of money Mary has then M = 2J. next © 2007 Herbert I. Gross next Moreover, M + 4 represents the amount of money Mary would have if she had an additional $4, and J – 3 represents the amount of money Jane would have if she had had $3 less.

Solution for # 5: The fact that if Mary had an additional $4 and Jane had $3 less, Mary would have had 4 times as much money as Jane tells us that we want to find the row in which the entry in the “M + 4” column is 4 times as much as the entry in the “J – 3” column. next © 2007 Herbert I. Gross next This occurs in the shaded row, and in this row we see that J = 8. 6123 7144 8165 9186 JMJ – 3M + 4 510214 16 18 20 22 48112 816520 next

Solution for # 6: In our chart, B will represent Bill’s present age; J ( = 3B), John’s present age; B + 7, Bill’s age in 7 years; and J + 7 will represent John’s age in 7 years. next © 2007 Herbert I. Gross next Thus, our chart might look like… 3910 41211 51512 61813 BJB + 7J + 7 26913 16 19 22 25 13810 72114 82415 28 31

Solution for # 6: The fact that in 7 years John will be only twice as old as Bill means that we want the row in which the entry in The “J + 7”column is twice the entry in the “B + 7”column. next © 2007 Herbert I. Gross next We have shaded this row in the above chart. 3910 41211 51512 61813 BJB + 7J + 7 26913 16 19 22 25 13810 72114 82415 28 31 7211428 next

Problem #7 © 2007 Herbert I. Gross next A car went from A to B at a constant rate of 30 mph and made the return trip at a constant rate of 20 mph. If the round trip took a total of 30 hours, what is the distance between A and B? next Answer: 360 miles

Solution for # 7: Since 60 is a common multiple of 30 and 20, our chart might look like… next Distance between A & B Time at 30 mphTime at 20 mphTotal Time 60 miles2 hours3 hours5 hours120 miles4 hours6 hours10 hours180 miles6 hours9 hours15 hours240 miles8 hours12 hours20 hours300 miles10 hours15 hours25 hours360 miles12 hours18 hours30 hours420 miles14 hours21 hours35 hours © 2007 Herbert I. Gross 360 miles12 hours18 hours30 hours The row in which the “Total Time” is 30 hours is shaded in above. next

Notes on #7 Once we completed the first row of the above chart, we could have concluded that for every 60 mile segment between A and B; the time for a round trip would be 5 hours. © 2007 Herbert I. Gross next Since the round trip took 30 hours, it meant that there were six 5 hour segments, and since each segment consisted of 60 miles, the total distance between A and B was 6 × 60 miles or 360 miles.

Notes on #7 Notice that the average speed for the round trip was not 25 hours. It would have been if the same amount of time was spent at each speed. However, even though the distance from A to B is the same as the distance from B to A, more time was spent at the slower speed. © 2007 Herbert I. Gross next In fact, notice that for the round trip it takes 5 hours to travel 120 miles. 120 miles ÷ 5 hours = 24 miles per hour.

next Notes on #7 A “cute” way to see the above is in terms of what happens for each mile. Namely, at 30 mph it takes 2 minutes to travel 1 mile while at 20 mph it takes 3 minutes to travel 1 mile. That is, traveling 1 mile in one direction takes 3 minutes while in the other direction it takes 2 minutes. So for each round trip mile the car traveled 2 miles in 5 minutes. Since there are twelve 5 minute segments in 1 hour, in 1 hour the car traveled 12 × 2 miles. In other words, the average speed for the round trip is 24 miles per hour. © 2007 Herbert I. Gross

Problem #8 © 2007 Herbert I. Gross next John has three times as many marbles as Bill, and Tom has four more marbles than John. How many marbles does Tom have if altogether they have 214 marbles? next Answer: 94 marbles

Solution for # 8: Once again we could make a table such as the one below. The table is based on the information that for every marble Bill has, John has 3; and since Tom has 4 more marbles than John it means that for every marble Bill has, Tom has 7 (that is 4 more 3). next # of Marbles Bill has # of marbles John has (3B) # of marbles Tom has (J+4) Total # of Marbles 1371126101839132541216325151939 © 2007 Herbert I. Gross

Solution for # 8: We want to find the row in which the total number of marbles is 214. Our chart shows us that each time Bill has 1 more marble the total number of marbles increases by 7 marbles. next © 2007 Herbert I. Gross The difference between 214 and 11 is 203, and 203 ÷ 7 = 29. So to get from 11 marbles (the entry in the first row) to 214 (the entry on the row we are looking for), we have to move 29 rows.

Solution for # 8: Hence, we have to move 29 rows beyond the first row, which means we are then at the 30 th row in our chart. The 30th row is… next # of Marbles Bill has # of marbles John has # of marbles Tom has Total # of Marbles 267882186278185193288488200298791207309094214 © 2007 Herbert I. Gross 309094214

next Notes on #8 If Tom had had 4 less marbles, he would have had the same number of marbles as John had. In that case, there would have been only 210 marbles, and both Tom and John would each have 3 times as many marbles as Bill. © 2007 Herbert I. Gross

next Notes on #8 In other words, the ratio would have been 1:3:3; or in terms of fractions, Bill would have had 1 / 7 of the marbles, and John and Bill would each have had 3 / 7 of the marbles. © 2007 Herbert I. Gross 1 / 7 of 210 marbles is 30 marbles (which is how many marbles Bill has). 3 / 7 of 210 marbles is 90 marbles (which is how many marbles John has). And since Tom has 4 more marbles than John, he has 94 marbles. next

next Notes on #8 © 2007 Herbert I. Gross next Since 210 ÷ 7 = 30, each piece is worth 30 marbles, and if we now return the 4 marbles to Tom, we see that… BJT 30 4 Bill has 30 marbles, John has 90 marbles, and Tom has 94 marbles for a total of 214 marbles.

Problem #9 © 2007 Herbert I. Gross next A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the second piece. How long is the first piece? next Answer: 8 inches

Solution for # 9: One way to approach this problem is to pretend that the third piece is 5 inches shorter than it actually is. In that case, the total length of the piece of string would be 40 inches, and the third piece is then the same length as the second piece and each of these two pieces is twice as long as the first piece. next © 2007 Herbert I. Gross

Solution for # 9: In this case, the ratio of the lengths of the three pieces is 1:2:2. Thus, the first piece is 1 / 5 of the total length (which is assumed to be 40 inches) or 8 inches; and the length of each of the other 2 pieces is 2 / 5 of the total length or 16 inches. next © 2007 Herbert I. Gross However, this solution assumed that the third piece was 5 inches shorter than it actually was. Hence, the length of the third piece is 21 inches.

next Notes on #9 © 2007 Herbert I. Gross next where f = the length of the first piece The corn bread version of the above solution might look like… s = the length of the second piece t = the length of the third piece fstt 88888 16 8 1016 40 inches

next Notes on #9 © 2007 Herbert I. Gross next If we then restore the 5 inches we took away from the third piece, we see that… Hence, the third piece is 16 + 5 = 21 inches; which checks with the fact that the total length is 45 inches. fstt 8161016 40 inches 5 5

next First Length Second Length × 2 Third Length × 2 + 5 Total Length 1 inch2 inches7 inches10 inches2 inches4 inches9 inches15 inches3 inches6 inches11 inches20 inches4 inches8 inches13 inches25 inches5 inches10 inches15 inches30 inches6 inches12 inches17 inches35 inches7 inches14 inches19 inches40 inches © 2007 Herbert I. Gross 8 inches16 inches21 inches45 inches next 8 inches16 inches21 inches45 inches Notes on #9 Again, keep in mind that making a table can often supply us with patterns we might not have otherwise noticed. For example…

next Notes on #9 From the chart, we see several patterns; one of which is that every time the length of the first piece increases by 1 inch the total length increases by 5 inches. © 2007 Herbert I. Gross First Length Second Length × 2 Third Length × 2 + 5 Total Length 1 inch2 inches7 inches10 inches 2 inches4 inches9 inches15 inches 3 inches6 inches11 inches20 inches 4 inches8 inches13 inches25 inches 5 inches10 inches15 inches30 inches 6 inches12 inches17 inches35 inches 7 inches14 inches19 inches40 inches 8 inches16 inches21 inches45 inches8 inches16 inches21 inches45 inches

next Notes on #9 Therefore, if we didn’t want to spend too much time entering new rows on the chart, we need only notice that to get from the total length in the first row (10 inches) to the row in which the total length is 45 inches the total length has to increase by 35 inches. Since it increases by 5 every time we go to the next row, we have to move 7 rows, which is the row in which f = 8. © 2007 Herbert I. Gross

Problem #10 © 2007 Herbert I. Gross next A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the first piece. How long is the first piece? next Answer: 10 inches

next Caution Problems #9 and #10 are, among other things, designed to emphasize the importance of reading comprehension. Namely, at a quick first glance the two exercises seem to be the same. © 2007 Herbert I. Gross In fact, the wording is exactly the same in both exercises; except that in Problem #9, the third piece is 5 inches longer than the second piece; while in Problem #10 the third piece is 5 inches longer than the first piece. next

Answer: 10 inches Solution for # 10: Once again we can pretend that the third piece is 5 inches shorter than it is. In that case the total length of the string is 40 inches (just as it was in Problem #9). next © 2007 Herbert I. Gross However, the third piece is now the same length as the first piece. In this case the first and third pieces have the same length and the second piece is twice the length of either the first or the third piece. In other words, the pieces are now in the ratio of 1:2:1. next

© 2007 Herbert I. Gross next Solution for # 10: Thus, this time the first piece is 1/4 of the total length (which is 40 inches) or 10 inches. In terms of our corn bread model… fst 10 20 102010 40 inches where…f = the length of the first piece s = the length of second piece t = the length of the third piece

next © 2007 Herbert I. Gross next Solution for # 10: If we then restore the 5 inches we took away from the third piece, we see that… Note that the change of just one word in going from Problem #9 to Problem #10 changes the answer from… f = 8, s = 16 and t = 21 to f = 10, s = 20 and t = 15 5 5 10 fsst 40 inches 20 s

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 23 By Herbert I. Gross and Richard A. Medeiros next.

Similar presentations

Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 23 By Herbert I. Gross and Richard A. Medeiros next."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 23 By Herbert I. Gross and Richard A. Medeiros next.

Similar presentations

Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 23 By Herbert I. Gross and Richard A. Medeiros next."— Presentation transcript:

Similar presentations

About project

Feedback