the greatest common factor—question and answer

Three  ropes  are  120  inches  long,  90  inches  long  and  150 inches  long.  They are cut into sections with no remaining parts.  If each section has the same length, what is the largest possible length for each section?  How  many  sections  can  we  get  altogether?

Worksheets for this type of questions

Answer to the question:

Since each section has the same length, that means we need to find the common factors of 120, 90 and 150. Since we need to find the largest possible length of each section, we need to find the greatest common factor. That is 30. So the largest possible length of each section is 30 inches.

The  rope  that  is  120  inches  long  is  cut  into  4  sections. The  rope  that  is  90  inches  long  is  cut  into  3  sections. The  rope  that  is  150  inches  long  is  cut  into  5  sections. Therefore, all the ropes are cut into a total of 12  sections.

the least common multiple and the greatest common factor

The  product  of  two  natural  numbers  is  420.  If  their greatest  common  factor  is  12,what  is  their  least  common multiple?

To work on similar questions on your own, please click here

 

Answer:

 

According  to  this  rule:  The  product  of  the  greatest  common factor  and  the  least  common  multiple  of  the  two  numbers  equals to  the  product  of  the  two  numbers

The  least  common  multiple  of  the  two  numbers  is:  420÷12=35

 

Average Speed

Lisa and Sarah were given a word problem to work on:

The distance between City A and City B is 360 miles. A car was going at 60 miles per hour from City A to City B; on the way back, the car was going at  40 miles  per hour because it was raining really hard so that driver could not see very well. What was the average speed of the car for the round trip?

This was Lisa’s answer:

(60+40)÷2=50 miles per hour

 

Sarah said: “No, you cannot solve the problem like that. This is how to solve it.”

This was Sarah’s answer:

The total distance for the round trip: 360×2=720 miles

The amount of time the car used to travel from City A to City B:

360÷60=6 hours

The amount of time the car used to travel back to City A from City B:

360÷40=9 hours

The total amount of time for the round trip:

6+9=15 hours

The average speed for the round trip= the total distance for the round trip÷ the total amount of time spent on the round trip= 720÷15=48 miles per hour

 

Friends, I believe that you all think that Sarah is right and she is. Remember: when you solve a word problem on travelling, finding the average speed is not the same as finding the average age, the average height, etc.

 

 

cutting a stick

A stick is 1 yard long. If 1/5 is cut off at first, 1/6 of the remaining section is cut off secondly, 1/7 of the last remaining section is cut off thirdly and the cutting keeps on until 1/10 of the last remaining section is cut off finally, what is the length of the remaining stick?
The solution to this type of question usually has a pattern:

When 1/5 is cut off, 4/5 yard is left;

When 1/6 of the remaining section is cut off, 5/6 of 4/5 yard is left:

4/5 yard × 5/6 =4/6 yard

When 1/7 of the last remaining section is cut off, 6/7 of 4/6 yard is left:

6/7  × 4/6 = 4/7 yard

In this pattern, when 1/10 of the last remaining part is cut off, the amount left should be:

4/10 yard= 2/5 yard ( simplest term)

the 3 elements you need to find to solve a word problem

To solve a whole number or a decimal word problem, 3 elements are essential: the number of groups (days, hours, rows, lines, etc.), the amount in each group, and the total amount in all groups. 

 

To solve for one element, you need to determine what the other two elements are. 

 

If the amount in each group is NOT the same, use addition or subtraction to solve for one of the 3 elements. 

 

Examples: 

 

  1. Tom has $2.00; Peter has $5.00. How much do they have altogether? 

 

In this example, there are two people (groups) and each person does not have the same amount of money. To find the total amount of money they have, we need to use addition. Therefore, the total amount of money they have is : 2+5=$7.00 

 

If the amount in each group is the SAME, use one of the following 3 elements formulas to solve for one of the 3 elements: 

 

  1. Tom has $2.00; Peter also has $2.00. How much do they have altogether? 

 

In this example, there are two people (groups) and each person has the same amount of money. 

 

To find the total amount of money they have, we can use addition but we can also use the following 3 elements formulas for whole number and decimal word problems: Total amount in all groups= the number of groups x the amount in each group

 

To find more on how to use 3 elements formulas to solve word problems with multiplication and division, please purchase word problems; detailed explanations of reasoning and solving strategies (volumes 3volume 4, volume 5volume 6, volume 7, volume 8, volume 9volume 10, volume 11 and volume 12)

 

Cutting a log

Peter asked John this question:” If it takes me 6 minutes to cut a log into 3 sections, how many minutes will it take for me to cut the log into 6 sections?”

John was very quick:” That’s easy. Since 3 sections take 6 minutes, 6 sections should take two groups of 6 minutes so it should be …..12 minutes.”

 Friends, do you think that John is right?

 

The answer is no. This is why: In 6 minutes, he does two cuttings and these two cuttings turn the logs into 3 sections so each cutting takes 3 minutes on the average. To cut the log into 6 sections, Peter will need to do 5 cuttings. Since each cutting takes 3 minutes, 5 cuttings will take 5 groups of 3 minutes: 5×3=15 minutes

It will take Peter 15 minutes to cut the log into 6 sections. 

Share a watermelon

Deuteronomy 6:1-9:6 (NIV): These are the commands, decrees and laws the Lord your God directed me to teach you to observe in the land that you are crossing the Jordan to possess, so that you, your children and their children after them may fear the Lord your God as long as you live by keeping all his decrees and commands that I give you, and so that you may enjoy long life. Hear, Israel, and be careful to obey so that it may go well with you and that you may increase greatly in a land flowing with milk and honey, just as the Lord, the God of your ancestors, promised you.Hear, O Israel: The Lord our God, the Lord is one.[a] Love the Lordyour God with all your heart and with all your soul and with all your strength. These commandments that I give you today are to be on your hearts. Impress them on your children. Talk about them when you sit at home and when you walk along the road, when you lie down and when you get up. Tie them as symbols on your hands and bind them on your foreheads. Write them on the door frames of your houses and on your gates.

Mom bought one big watermelon for the hot day. My brother was so thirsty that he wanted half of the watermelon. Mom cut the watermelon into two sections and gave my brother the larger section. Then she asked: ”Is this half of the watermelon that you wanted?” My brother said: “yes, yes. ” 

“No. When a watermelon is cut into halves, your section and your brother’s should be the same big.”

“Oh! Half a watermelon means that the two sections of the watermelon are the same. Brother, next time, I will make sure that mom cuts the watermelon into two same sections so that you and I each can have a section of the same big.”

Keys and locks

There are 6 keys and 6 locks. At most how many tries are required before it is certain that each lock has found its key?

Which answer is correct?

One answer is as follows: 6 x 6=36 tries ( Is this correct?)

The other answer is as this:

  1. Use the 6 keys to try 1 lock at a time and at most 5 tries are required before the lock finds its key. Move the key and lock away from the groups. Now there are 5 locks and 5 keys left. 
  2. Use the 5 keys to try another lock and at most 4 tries are required before the lock finds its key. Move the second set of key and lock out of the groups. Now there are 4 locks and 4 keys left. 
  3. Use 4 keys to try another lock and at most 3 tries are required before the lock finds its key. Move the third set of key and lock out of the groups. Now there are only 3 locks and 3 keys left. 
  4. Use the 3 keys to try another lock and at most 2 tries are required before the lock finds its key. Now more the fourth set of key and lock out of the groups. Now there are only 2 locks and 2 keys left.
  5. Use the 2 keys to try one of the remaining locks and at most 1 try is required for the lock to find its key.  Move the 5th set of key and lock out of the groups. Now there are only 1 key and one lock left and they should be one set. 

How many tries are there from the above steps?

5+4+3+2+1=15 tries

Of course, the second answer is correct. When you see a verbal problem, do not rush to get  the answer. Always think it through. 

 

Simple ways to calculate math problems

2 Corinthians 13:5: Test yourselves to see if you are in the faith; examine yourselves! Or do you not recognize this about yourselves, that Jesus Christ is in you–unless indeed you fail the test?

 

I was tutoring a 7th-grade student on fraction addition and subtraction and saw that she used her fingers to try to find the least common multiples and the largest common factors. 

I was tutoring a 5th-grade student to solve a word problem with “how many groups of …” in it: he was drawing pictures and using his fingers to try to get the answer. 

I was student teaching a 3rd-grade class and saw all the students using their fingers to do math. 

No! It does not have to be that way. The children can do better than just using their fingers. They can use their brains to do math.

Yes, they can be trained to use their brains to do math. 

Do you know how much 999+899+102 is by doing it in your head only?

How would you do it? Do you think that your 2nd or 3rd grade child can do it?

Yes! They can. This is how:

First, we need to know this: 102=100+1+1

Therefore:  999+899+102=999+1+899+1+100=1,000+900+100=2,000

 

In the blogs, I will show you how to solve different types of math problems using different types of skills. You will also be given worksheets practice the skills: both free and membership worksheets.