# 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)

# a math problem from SAT

The following is a problem from a SAT practice test:

If 6x + 1/x = 5, then x = ?

1. -1/6
2. 1/6
3. 1/4
4. 1/2
5. 2

You can try each number but you can also solve the problem on your own first and find the match for your answers.

How is how to solve it:

Multiply both sides of the equal sign with x so the equation becomes:

6x+1=5x

6x2-5x+1=0

Factor:

(2x-1)(3x-1)=0

X can be either ½ or 1/3.

The following is a question from a SAT practice test:

The average (arithmetic mean) of 3 numbers is 22 and the smallest of these numbers is 2. If the two remaining numbers are equal, what is the value of each of the remaining numbers?

1. 22
2. 32
3. 40
4. 64
5. 66

Here is how to solve for the answer:

The average of the 3 numbers is 22 so the sum of them should be:

22 x 3=66

One of the numbers is 2 so the sum of the other two numbers should be:

66-2=64

Since the other two numbers have the same value the value of each number should be:

64÷2=32

# 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.”

# cutting a stick

Can you solve this problem?

1/8 of a stick is cut off. If If another stick that is 15 inches long is glued to the remaining part of the first stick, the first stick will be 1/2 longer than its original length. What is the original length of the first stick?

In this problem, you need to find the match between the fractional part and the amount in it ( if you do not know about these terms, please go to Word Problem: Detailed Explanations of Reasoning and Solving Strategies Volumes 11-A, 11-B and 12)

15 inches  not only has covered  1/8 of the stick that is cut off but also makes the  the stick 1/2 longer so we can say that 15 inches match the sum of the fractional parts: 1/8 and 1/2.

We divide the sum of the fractional parts: 1/8+1/2=5/8 into the amount that matches it: 15 inches to find the amount in 1, which is the length of the original stick: ( please go to the books Word Problem: Detailed Explanations of Reasoning and Solving Strategies Volumes 11-A, 11-B and 12 for more detailed explanation):

15÷5/8=24 inches

The original stick is 24 inches.

# 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.

# 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.