# 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

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?

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?

(60+40)÷2=50 miles per hour

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

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.

# A mouse taught a cat how to solve a math problem

A cat sneaked out of the animal school and looked very unhappy. Just at that time, he saw a mouse. He got hold of the mouse and was about to eat the mouse. The mouse was very smart. He asked the cat: “You look very unhappy. Is there anything I can help you with before you have me for lunch?” The cat thought for a while and said: “My teacher gave me a math problem and I did not know how to solve it. “

“What was it?”

“In the equation: __________÷ 10=15….. (_______), what is the largest possible value for the dividend? What is the smallest possible value for the dividend?” The cat continued:” If I cannot solve it before school is over, the teacher will keep me after school for tutoring. I want to go home after school.”

The mouse said: “If I can help you to solve the problem, do you promise to let me go?”

“Well….yes.”

“Listen: in a division equation, the remainder must be smaller than the divisor. In this equation, the divisor is 10 so the largest possible value for the remainder is 9. When the remainder is 9, the dividend is:

15 x 10 +9=159;

When the remainder is 1, the dividend has the smallest possible value:

15x 10+1=151

Now, Mr. Cat, do you think that I am right?”

The cat thought for a while and said:” You are right. I need to remember the formulas to find the dividend and also the rules in a division equation. Thank you. You can go now.”

“Mr. Cat, do not sneak out of your class any more. Study hard and do not give up.”

The cat let go of the mouse and returned to school.

If you need more problems on division equations, please get the books:

Word Problems-Detailed Explanations of Reasoning and Solving Strategies Volumes 5 and 6

# Share a watermelon

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?

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.

# Lily has \$50. Tom has 4 times as much as Lily. How much does Tom have?

Lily has \$50. Tom has 4 times as much as Lily. How much does Tom have?

Times means groups of.  The amount of money that Tom has is 4 groups of the amount of money that Lily has.

The total amount found in 4 groups is found by using multiplication:

50 x 4=\$200.00

.