Mathematics in the Modern World

Galzote, Carlos Verueco Monroy, Jayzell Alvaro Saraga, Jill Andaluz

Trusted by over 1 million members

Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.





Problem Set no. 1 Mathematics in the Modern World

1. Alice, Ben and Carl collect stamps. They exchange stamps among themselves according to the following scheme : Alice gives Ben as many m any stamps as Ben has and Carl as many m any stamps as Carl has. After that, Ben gives Alice and Carl as many stamps as each of them has, and then Carl gives Alice and Ben as many stamps as each has. If I f each finally has 64 stamps, with how many stamps does Alice start? (Working Backward) SOLUTION: Let A be Alice’s number of stamps Let B be Ben’s number of stamps Let C be Carl’s number of stamps Statement 1 “If each finally has 64 stamps” It is said that they all have 64 stamps s tamps at the end. So, A = 64 B = 64 C = 64 Statement 2 “Carl gives Alice and Ben as many stamps as each has” If Carl gives them as many stamps as they have they will have 64 stamps. The trick is, what number must the giver will give and an d add it to itself to get the amount of 64? The number is none other than 32. So, Before Carl gives them the stamps, Alice and Ben has 32 stamps each while Carl has 128. Why 128? Because, after Carl gives the stamp s tamp he must have 64, giving away 32 to Alice and 32 to Ben, adding them up 64 + 32 +32 = 128





BEFORE: Alice (Receiver) = 32 Ben (Receiver) = 32 Carl (Giver) = 128 Statement 3 “Ben gives Alice and Carl as many stamps as each of them has” AFTER: Alice (Receiver) = 16 + 16 (from Ben) = 32 Ben (Giver) = 112 – 112 – 64 64 (given to Carl) - 16(given to Alice) = 32 Carl (Receiver) = 64 + 64 (from Ben) = 128 BEFORE: Alice (Receiver) = 16 Ben (Giver) = 112 Carl (Receiver) = 64 Statement 4 “Alice gives Ben as many stamps as Ben has and Carl as many stamps as Carl has” AFTER: Alice (Give) = 104 – 104 – 56 56 (given to Ben) – Ben) – 32 32 (given to Carl) = 16 Ben (Receiver) = 56 + 56 (from Alice) = 112 Carl (Receiver) = 32 + 32 (from Alice) = 64 BEFORE: Alice (Give) = 104 Ben (Receiver) = 56 Carl (Receiver) = 32 ANSWER : Alice has 104 stamps to start with [Source:https://brainly.ph/question/1607197] Source:https://brainly.ph/question/1607197] 2. Solve the following cryptarithms. In each problem, letters represent a single digit only.





where: S H E

=6 =2 =5

b. WRONG + WRONG = RIGHT (24153) (24153) (48306) where: W =2 R =4 O =1 N =5 G =3 I =8 H =0 T =6 3. How many squares of all sizes are in an 8x8 checkerboard? SOLUTION: 8 x 8 squares 7 x 7 squares 6 x 6 squares 5 x 5 squares 4 x 4 squares 3 x 3 squares 2 x 2 squares 1 x 1 squares

=1 =4 =9 = 16 = 25 = 36 = 49 = 64

ANSWER: 1+4+9+16+25+36+49+64 = 204 squares [Source:https://www.teachingideas.co.uk/problem-solving/squares-on-aSource:https://www.teachingideas.co.uk/problem-solving/squares-on-achessboard] 4. In order to encourage his son in the study s tudy of algebra, a father promised the son P8 for every problem solved correctly and to fine him P5 for each incorrect solution. After 26 problems neither owed anything to the other. How many problems did the boy solved correctly?





8 = 8,16,24,32,40,48,56,64,72,80 10 5 = 5,10,15,20,25,30,35,40,45,50,55,60,65,7 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80 0,75,80 16 26 ANSWER: The boy solved 10 problems correctly. 5. If a pup is worth a pooch and mutt, m utt, and a pup and a pooch are worth one bird-dog, and two bird-dogs are worth three mutts, how many pooches is a pup worth? SOLUTION: The given are: pup one bird dog 2 bird dogs

= pooch + mutt = pup + pooch = 3 mutts

2(pup + pooch) 2pup + 2 pooch

= 3(pup – 3(pup – pooch) pooch) = 3 pup – pup – 3 3 pooch

2 pooch + 3 pooch 5 pooch

= 3 pup – pup – 2 2 pup = pup

ANSWER: Therefore, 1 pup is worth 5 pooches. 6. Mang Ruben has only an 11-liter can and a 5-liter 5-l iter can. How can he measure out exactly 7 liters of water? SOLUTION: Since Mang Ruben has only 11-liter can and a 5-liter 5-l iter can. Here are the steps in order to measure out exactly 7 liters. Step 1: Fill 1: Fill up the 11-liter can and transfer 5 liters into the 5-liter can. Empty the 5-liter can. Step 2: Pour 2: Pour the remaining 6 liters into the 5-liter 5 -liter can. Empty the 5 liter can. Step 3: Pour 3: Pour the remaining liter into the 5-liter can. Step 4: Fill 4: Fill up the 11-liter can again. Pour 4 liters of it into the 5-liter 5- liter can that has 1 liter in it. What’s left in the 11-liter 11-liter can is 7 liters.





7. An egg vendor broke all the eggs that he was delivering del ivering to a local store. He could not remember how many eggs there were in all. However, he did remember that when he tried to pack them into packages of 2, 3, 4, 5, and 6 he had one left over each time. When he packed them into packages of 7, he had none left over. What is the smallest number of eggs he could have in the shipment? SOLUTION: Look for a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6, but evenly divisible by 7. Look at multiples m ultiples of 7. 7 leaves a remainder of 1 when divided by 2, 3, or 6. Unfortunately it doesn't work for 4. Now, get a number which leaves a remainder of 1 when divide it by 4. Then, take a number that leaves a remainder of 1 when divided by 2 and add 2 to that number, there would be another number that leaves a remainder of 1. If we add any multiple of 2, we still get a number that leaves a remainder of 1 when divided by 2. Similarly, if we take a number that leaves a remainder of 1 when we divide it by 3 and add any multiple of 3 to it, we would get another number that leaves a remainder of 1 when divide it by 3. Anything that leaves a remainder of 1 when divided by 2 or 3 is automatically leaves a remainder of 1 when divided by 6. So if we take 7 and add anything which is a multiple of 2, 3 and 7 to it, we would get another multiple of 7 which leaves a remainder of 1 when we divide it by 2 or 3. So we don't lose anything if we add a multiple multiple of 42 (that's 2 x 3 x 7) to 7. 7 + 42 = 49. If we add a multiple of 4 we still get a number that leaves a remainder of 1 when divided by 4. So if we add a multiple of 3 x 4 x 7, we'll get another another number that satisfies all the conditions. 3 x 4 x 7 = 84 So,

49 + 84 = 133







[Source: http://mathforum.org/library/drmath/view/58849.html] 8. There was a jar of chocolate chip cookies on the table. James and Monica were very hungry because they hadn’t had anything to eat since breakfast, so they ate half the cookies. Then Victor came along and noticed the cookies. He ate a third of what was left l eft in the jar. Sharon, who was waiting around nearby, decided to take a fourth of the cookies left in the jar. Then Tiffany came rushing up and took one cookie to munch on in her class. When Valerie looked at the cookie jar, she saw there were two cookies l eft. How many cookies were in the jar to begin with? (Working Backward) SOLUTION: Statement 1 “Valerie looked at the cookie jar, she saw there were two cookies left.” 2 cookies Statement 2 “Tiffany came rushing up and took one cookie to munch on in her class .” 2 cookies + 1 cookie = 3 cookies Statement 3 “Sharon, who was waiting around nearby, decided to take a fourth of the cookies left in the jar.” 3 cookies ÷ 3 / 4

=4

cookies

Statement 4 Victor came along and noticed the cookies. He ate a third of what was “





9. Every GOOP is a GORP. Half of all GORGS are GORPS. Half of all GORPS are GOOPS. There are 40 GORGS and 30 GOOPS. No GORG is a GOOP. How many GORPS are neither GOOPS nor GORGS? SOLUTION:

GORGS

20

GORPS

20

GOOPS

30

10 10. On a balance scale, two spools and one thimble balance 8 buttons. Also, one spool balances one thimble and one button. How many buttons will balance one spool? SOLUTION: 2 pool + 1 thimble 1button 1 thimble

= 8 buttons 1 spool





EXPERIENCES IN WORKING IN A GROUP OF PEOPLE THAT WE’RE WE’RE NOT FAMILIAR WITH GALZOTE, CARLOS VERUECO In solving a mathematics-based logical problem I always end up with a wrong answer, why? I always used a very simple approach wherein I do not apply the fundamentals of algebra but solving it with your groupmates I don't think it would be much difficult perhaps it would reduce its difficulty because different insights are used. I've learned that solving a difficult math problem with group would be easier than solving it by yourself. Every items that we've answered uses different concepts of algebra and technical math which indeed a quiet tricky and complex problem. My groupmates and I have not encountered such conflict because items were equally distributed. And so with the groupings of the whole class we were systematically grouped according to the availability of each student. Since we have enough time to do it, we took it up immediately and finished before the submission. In addition to this composition, I myself didn't have a hard time communicating with Jayzell and Jill since s ince they're approachable. Honestly, I would not be able to finish this activity without collaboration. So having a group study and group work means a lot. I've learned to be flexible





another. Honestly the problems are much difficult but I find it a little bit easier because of the help of my groupmates and also with the help of the internet. It is good that we have not encountered any conflict in answering the problems because the items were equally distributed. But I can say that those problems challenge my brain and it really needs a good analyzing skill. Meanwhile, choosing groupmates is not difficult because the whole class cooperate in in order to come up a suitable group group for each student. The entire class is divided equally so there would not be an issue of bias. Thus, it is truly indeed that working as a group can be the most effective way of solving complex problems.

SARAGA, JILL ANDALUZ Working in the group to solve the problem is very hard because for the first place I don’t don’t understand the problem. I am not also the type of person to solve some mathematics mathematics problem because I hate math.

I think the the positive side

working this problem by group is there are people to ask and there are many brains that is working on to solve, particularly the questions that I don’t know how. how. The topic that I really don’t understand even I already research it in google is the cryptarithms. I really don’t get the pattern that they are using. I read it so many times but I don’t understand a thing.





Mathematics in the Modern World

Recommend Documents