Gruber's Essential Guide to Test Taking: Grades 6-9 - Gruber Gary R. 2019

Basic math strategies
Math strategies

A Note to Parents

You and your child will be presented with various strategies and critical-thinking skills to be used in solving math problems. If as you go along you begin to feel that your child is lacking certain basic information in an area of math, such as geometry, refer to the section entitled “Math Words, Concepts and Rules Your Child Should Know,” on pages 156—172. Your child can also look at “Math Shortcuts Your Child Should Know,” on pages 173—182. Finally, if you feel that your child should brush up on his or her basics before beginning to learn the strategies, have a look at “The Thirty-Two Key Basic Math Problems for Grades 6 • 7 • 8 • 9,” on pages 183—184. You may want to have your child work on them before teaching your child the beginning strategic material.

Before beginning to work on the math strategies presented in this part of the book, you should also review the four-step learning method described in the “Introduction to Parents,” on page 16.

Basic math strategies

The math strategies presented below will help your child to answer math questions more quickly and accurately by enabling your child to: first, focus his or her mind on each problem in the most appropriate way, and second, use a variety of strategic shortcuts in working out the solutions to problems.

MATH STRATEGY 1:

Know How to Approach Solutions to Math Questions

One of the most important things to know in answering any math question is how to start the solution. So many students rush into a solution without taking the time to think about what the question is really asking or aiming at.

The key to solving math questions is knowing how to extract from the question the useful pieces of information that will enable your child to solve it. Here are some examples:

EXAMPLE 1

Paul earns $22 in 5 days. How much does Paul earn in 8 days if he works at the same rate of pay?

(A) $88.00

(B) $40.00

(C) $35.20

(D) $35.40

Don’t just rush into the problem by multiplying the numbers given. Think about what piece or pieces of information can be extracted from this question that will be useful in finding a solution. If Paul earns $22 in 5 days, you can find out how much money he earns in 1 day. Just divide 22 by 5:

So Paul earns $4.40 in 1 day. Now you’ve got something very useful. If you want to find how much he earns in 8 days, just multiply $4.40

Choice C is correct.

Let’s look at another problem that has some piece of information that will start you off in the right direction toward solving the problem.

EXAMPLE 2

The number of pencils in 7 boxes, each containing 30 pencils, is the same number of pencils in

(A) 5 boxes containing 20 pencils in each box

(B) 10 boxes containing 15 pencils in each box

(C) 15 boxes containing 20 pencils in each box

(D) 21 boxes containing 10 pencils in each box

How do we approach this question? What do we look for first? Let’s look at the choices. They all state the number of boxes and how many pencils there are in each box. This gives us the total number of pencils in all the boxes for each choice. The question tells us the number of boxes and how many pencils there are in each box. This also gives us the total number of pencils. So we can find this total number by multiplying 7 by 30. This number is 210.

Now let’s look for a choice whose numbers (boxes × pencils) give you 210. Choice D has 21 boxes with 10 pencils (21 × 10 = 210). So Choice D is correct.

EXAMPLE 3

In the figure below, you are given a 6-minute clock. The clock starts at 0 and stops 243 minutes later. At which number does the clock stop?

(A)1

(B)2

(C)3

(D)4

How do we approach a problem like this? We have to think that the clock will reach the 0-minute mark again in 6 minutes, again in 12 minutes, again in 18, and so on. In other words, it will reach the 0 mark in every multiple of 6 minutes. So let’s find out how many 6’s go into the total time of 243 minutes: This will tell us how many times it hit the 0 mark and what was left over.

The clock hit the 0 digit 40 times and moved another 3 digits. So it ended up on the digit 3. Choice C is correct.

After you have shown your child the previous three problems with solutions, have him or her try the problems below and check to see whether he or she used the strategy explained in the solutions that follow the problems.

PROBLEMS

1The bar graph above describes the cost of various items used in a particular recipe. What item costs approximately 60 percent of the total cost for all the items?

(A)nuts

(B)apples

(C)potatoes

(D)other vegetables

2How many teachers teach at a high school if the ratio of teachers to students in that school is 1 to 5 and there are 200 students in the school?

(A)4

(B)40

(C)400

(D)1,000

3The dues for a certain club are $5 per month. How much would a person save in one year if he or she paid $50 for a yearly subscription to the club and did not have to pay dues for the year?

(A)$60

(B)$50

(C)$30

(D)$10

4A lemon syrup calls for 2 ounces of lemon juice to every 12 ounces of plain syrup. To make 42 ounces of lemon syrup, how many ounces of plain syrup must be used?

(A)14

(B)16

(C)36

(D)2

5The radius of Circle A is 10 and the radius of Circle B is 5. How much greater is the area of Circle A than the area of Circle B?

(A)25π

(B)50π

(C)75π

(D)100π

SOLUTIONS

1(D)What you want to know is what section of the graph looks as if it is 60% of the total graph (total cost). You can see that “Other Vegetables” make up a block that is about three fifths, or 60% of the total. All of the other sections make up less than one half (50 percent) of the total.

2(B)You want to find the number of teachers in the school. You are told that the ratio of teachers to students is 1 to 5 and there are 200 students. First use the fact that the ratio is 1 to 5.

Now you use the fact that there are 200 students in the school. We get:

Multiply both sides of the above equation by 200 to get rid of the 200 in the denominator of the right side of the equation:

3(D)You are told that the dues for each month are $5. You are looking for how much you will save in one year. So let’s find out how much it will cost per year at $5 per month. Just multiply $5 by 12 since there are 12 months in a year ($5 × 12 = $60 per year). If you pay $50 on a subscription, you save $10 ($60 — $50 = $10).

4(C)You are told that 2 ounces of lemon juice are used with 12 ounces of plain syrup. You want to find what you must use to make 42 ounces of lemon syrup. You must mix a certain amount of lemon juice with a certain amount of plain syrup to get a total amount of 42 ounces of lemon syrup. How do you approach this question? Try to figure out something from the information you are given. You know that 2 ounces of lemon juice are mixed with 12 ounces of plain syrup to make lemon syrup. Since 2 + 12 = 14, the 2 ounces of lemon juice + 12 ounces of plain syrup make 14 ounces of lemon syrup. You want 42 ounces of lemon syrup. To get 42 ounces, you will need times as much of each ingredient, since 14 × 3 = 42. So instead of 12 ounces of plain syrup, which would make 14 ounces of lemon syrup, you will need 36 ounces (12 × 3 = 36) of plain syrup to make 42 ounces (14 × 3 = 42) of lemon syrup.

5(C)You are told that the radius of Circle A is 10 and the radius of Circle B is 3. You want to know the difference between the areas. So what you have to think about is how to relate area to radius.

First find the areas of Circle A and Circle B:

area of Circle A = π × radius of Circle A × radius of Circle A

area of Circle B = π × radius of Circle B × radius of Circle B

We know that the radius of Circle A is 10 and the radius of Circle B is 5, so

area of Circle A = π × 10 × 10 = 100π

area of Circle B = π × 5 × 5 = 25π

The difference in areas is just

area of Circle A — area of Circle B

The difference is 100π — 25π = 75π.

MATH STRATEGY 2:

Use Math Symbols for Words

Math problems expressed in nonmathematical terms are the most difficult kinds of math questions. If your child does not know the right strategy, he or she could become exhausted trying to do some of these brain-racking problems. However, there is a strategy that can make these problems much simpler: substituting math symbols for words.

As early as possible, you should familiarize your child with math symbols, so that he or she can use them as a shortcut in solving those kinds of problems that are expressed mostly in words. Here’s an example:

EXAMPLE 1 :

12 is what percent of 6?

(A)72

(B)2

(C)20

(D)200

Here’s how to do the problem using the verbal-math translation method:

Translate:

So the problem becomes:

Divide both sides of the equation by 6:

Choice D is correct.

The table below will help your child in changing words into math symbols. Your child should gradually become familiar with this table.

Note: If a > b, and b > c, then a > c.

a > b is the same as b < a.

b > c is the same as c < b.

a > c is the same as c < a.

Let’s try two more examples using the above table.

EXAMPLE 2

If a sweater originally cost $21.50 and is reduced 40%, by how much was the sweater reduced?

(A)$12.90

(B)$8.60

(C)$21.40

(D)$4.50

Use the translation method:

Original cost

Reduced by

Thus Choice B is correct.

EXAMPLE 3

Mary is taller than John. Paul is shorter than John. Describe the heights of Mary, John, and Paul from tallest to shortest:

(A)Mary, John, Paul

(B)Paul, John, Mary

(C)John, Paul, Mary

(D)John, Mary, Paul

Translate from words to math:

Let Mary be M

John be J

Paul be P

Translate:

Mary is taller than John

M > J

Paul is shorter than John

P > J

Since M > J and P < J (J > P is the same as P < J) then M > J and J > P. And if you look at the translation table above, you’ll see that since M > J and J > P, M > P. So M > J > P. Choice A is correct.

Now have your child try these examples. Go over the solutions with your child, making sure he or she approaches the questions with the correct strategy.

PROBLEMS

1A television usually priced at $160 is reduced 20 percent. What is the new price?

(A)$128

(B)$145

(C)$132

(D)$32

The following two questions refer to the graph below:

2The contribution from Washington Junior High School was what part of the contributions from the other schools?

3What contribution did Lincoln Junior High School make?

(A)$24,000

(B)$12,000

(C)$6,400

(D)$4,800

4There are 24 girls at a party. The ratio of boys to girls at the party is 3 to 4. How many boys are there at the party?

(A)12

(B)18

(C)28

(D)36

515 is what percent of 45?

(A)20

(B)33%

(C)45

(D)50

625 is 150 percent of what number?

(A)8

(B)8⅓

(C)16⅔

(D)80

7Harry’s father gives Harry $2 for every log over 10 that Harry brings into the house. If Harry brings L logs into the house where L is greater than 10, how many dollars does Harry make?

(A)2L

(B)(L — 10)2

(C)(10 — L)2

(D)(L + 10)2

SOLUTIONS

1(A)Translate words into math:

2(B)Translate words into math:

Divide both sides by 75 to get x alone.

Reducing:

3(D)Translate words into math:

Lincoln Junior High School made 20% of total. 100% = $24,000 (given)

4(B)translate words into math:

Call the number of boys b. Now translate:

The ratio of boys to girls is 3 to 4: Note: 24 girls (given)

Multiply by 24 to get b alone:

Reduce:

Something you may want to note: Since it was stated that the ratio of boys to girls was 3 to 4, there must be fewer boys than girls. So there must be less than 24 boys because there are 24 girls. You can therefore rule out Choices C and D immediately.

5(B)Translate:

6(C)Translate:

7(B)Translate:

L is the number of logs Harry brings into the house.

He gets $2 for each log (over 10).

So L — 10 is the number of logs he gets paid for.

You can try an example to see if you’re right:

Suppose he carries 11 logs into the house. He gets paid for L — 10 = 11 — 10 = 1 log, which seems right.

Since he gets paid $2 for each log over 10, (L — 10) × 2 must be the amount of dollars he makes.

MATH STRATEGY 3:

Know When and How to Approximate

Many times your child will encounter a question that asks not for an exact answer but for an approximation to an answer. At other times your child may want to approximate to find an answer more rapidly. So it is important to know how and when to approximate. The following examples show typical problems where this strategy should be used.

EXAMPLE 1

The length of a fence is 32.96 meters. One fourth the length of the fence would be approximately how many meters?

(A)7

(B)8

(C)9

(D)10

Whenever you want to approximate an answer, always look at the answer choices just to see how far your approximation should go. In the question above, you can see that all of the answer choices are whole numbers. So let’s approximate 32.96 by the closest whole number 33. Now you’d like to take one fourth of 33. Take one fourth of 32—it’s easier. One fourth of 32 is 8. Choice B is correct.

EXAMPLE 2

15 is closest to which other number?

Looking at the choices, you can see that multiplying and dividing by 1/100 (Choices C and D) changes the number 15 very much. Subtracting only 1/100 from 15 does not change 15 as much as adding 15/100 to 15. So Choice A must be correct.

EXAMPLE 3

is closest to (A) 3.9 (B) 4.9 (C) 5.9 (D) 6.9

means that the number when multiplied by itself gives you 24. So which of the choices when multiplied by itself gives you a number closest to 24. Approximate all of the choices:

(A)3.9—4

(B)4.9—5

(C)5.9—6

(D)6.9—7

You can see that Choice B is closest since 5 × 5 = 25, which is very close to 24.

EXAMPLE 4

7.876 rounded to the nearest hundred is

(A)7,800

(B)7,870

(C)7,880

(D)7,900

Here you have to read the question carefully. It says you want to find 7.876 rounded to the nearest hundred. That means the number you are looking for must end in 00. So, although 7,880 (Choice C) is closest to 7,876, it is not rounded to the nearest hundred. The number to the nearest hundred is 7,900 (Choice D).

EXAMPLE 5

19 × 48 is closest to

(A)900

(B)1,000

(C)1,100

(D)1,200

Look at the choices. They all end in 00. So let’s approximate 19 by 20 and 48 by 50. Now multiply 20 × 50 — 1,000. Choice B is correct.

EXAMPLE 6

Approximate 397 as 400 and 2,401 as 2,400. So you want to get Reduce and cancel: Choice A is correct.

Have your child work on these questions. Go over his or her answers to see whether your child has approached the problems correctly by referring to the appropriate strategic explanations.

PROBLEMS

17.3 × .4 × .09 =

(A)26.28

(B)2.628

(C).2628

(D).02628

231 + 71 + 98 + 99 is closest to

(A)310

(B)300

(C)290

(D)280

3.46 × 32 =

(A)1.472

(B)14.72

(C)147.2

(D)1472

4 51,289 — 4,996 is closest to

(A)45,000

(B)46,000

(C)47,000

(D)48,000

5If x < , then which is true?

(A)8 < x < 9

(B)7 < x < 8

(C)6 < x < 7

(D)5 < x < 6

SOLUTIONS

1(C)You can see by looking at the choices that all you need to do is find where the decimal point should be placed. You can approximate very roughly:

Choice C is closest.

2(B)By looking at the choices, you can approximate:

Choice B is correct.

3(B)Look at the choices. You can see that it is possible to roughly approximate:

Choice B (15) is closest to 14.72.

4(B)Look at the choices. They all end in 000. So approximate 51,289 by 51,000 and 4,996 by 5,000. Now 51,000 - 5,000 = 46,000 (Choice B).

5(A) means that the number when multiplied by itself gives you 67. You know that 8 × 8 = 64, 9 × 9 = 81. So the number that gives you 67, when multiplied by itself, must be between 8 and 9. That is, it must be greater than 8 but less than 9 (Choice A).

MATH STRATEGY 4:

Reduce or Simplify Before Calculating

Often a child will go through a series of calculations without realizing that by reducing certain quantities first or simplifying others, he or she can arrive at a solution much more quickly. Here are some examples:

EXAMPLE 1

of (52 hours and 10 minutes) =

(A)12 hours and 24 minutes

(B)11 hours and 24 minutes

(C)11 hours

(D)10 hours and 26 minutes1

Don’t add 52 hours and 10 minutes yet.

of (52 hours and 10 minutes) =

Choice D is correct.

EXAMPLE 2

82 + (30% of 50) — (20% of 50) =

(A)87

(B)88

(C)89

(D)90

Don’t calculate 30% of 50 and 20% of 50. Look for a simpler way. 30% of 50 — 20% of 50 = (30% — 20%) of 50

All you have to do now is add 82 to 5 to get 87. Choice A is correct.

EXAMPLE 3

(A)155

(B)165

(C)175

(D)185

Don’t multiply 775 × 3 and then divide by 15! First reduce:

Choice A is correct.

EXAMPLE 4

Don’t multiply 3 × 4 × 5 × 6 and then divide that by 4 × 5 × 6 × 7!

Notice that a denominator of one fraction cancels with the numerator of the next fraction:

Choice D is correct.

EXAMPLE 5

(A)30

(B)32

(C)34

(D)36

Don’t divide

Get the value of □ by cross-multiplying, but don’t multiply out!

Now divide by 25:

Reduce:

Choice A is correct.

Have your child work on these questions, then check to see whether his or her method of approach is the same as the one used in the book for the solutions to these questions.

PROBLEMS

1of the total of 25 hours and 8 minutes is equal to

(A)6 hours and 23 minutes

(B)6 hours and 17 minutes

(C)8 hours and 15 minutes

(D)8 hours and 25 minutes

SOLUTIONS

1(B)Don’t add 25 hours + 8 minutes yet.

3(A)Don’t multiply 27 × 42 and then divide by 21 × 36!

Reduce:

4(C) First simplify by getting all fractions to have the same form. Try to get all fractions to be one type, like simple fractions.

So

Cancel numerators and denominators that are the same:

5(A)Don’t divide 420 by 75!

Cross-multiply:

Divide both sides of the equation by 10:

MATH STRATEGY 5:

Kart with Last Choice When Testing Choices

The test designer is a very clever person. The test maker expects that if your child has to try all the choices to see which one is correct, your child will start with Choice A. In order to weed out the poor students, the test maker hopes that if your child doesn’t know how to solve a problem, he or she will make a mistake before getting to the right choice. So for this particular type of example the test maker usually puts the right choice at the end of the string of choices: C or D (or D or E if there is an E choice). Thus, by the time your child gets to the right choice, he or she will usually have eliminated two or three of the incorrect ones. So here’s a useful strategy: For a four-choice question, always start with Choice D, and then go to C, B, and A

Practice using this strategy in the following examples.

EXAMPLE!

Which of the following sums will always be a whole number?

(A)a fraction added to a fraction

(B)a fraction added to a whole number

(C)a decimal fraction multiplied by a whole number

(D)a whole number added to a whole number

Look at Choice D first since you have to test all of the choices. A whole number added to a whole number is a whole number. To be sure, try some numbers: 1 + 2 = 3; 2 + 3 = 5. Choice D is correct. There is no need to look at the other choices.

EXAMPLE 2

All numbers are equal to except

Here you have to examine all of the choices, so start with Choice D. Compare so it is not equal to . Choice D is correct. There is no need to look at any of the other choices.

EXAMPLE 3

Which of the following can be the dimensions of a rectangular box whose volume is 50 cubic centimeters?

(A)5 cm × 5 cm × 3 cm

(A)4 cm × 10 cm × 2 cm

(C)24 cm × 2 cm × 1 cm

(D)25 cm × 2 cm × 1 cm

Look at choice D first: 25 × 2 × 1 = 50. Choice D is correct. There is no need to look at the other choices.

EXAMPLE 4

If in the addition problem above, xand / represent digits from 0 to 9, what can x be?

(A)3

(B)4

(C)5

(D)6

This is an excellent illustration of the last-choice method because using it not only allows you to save a lot of time but eliminates a great deal of mental exhaustion. Try Choice D first:

Let x = 6: For the addition you get:

Using x = 6 satisfies the addition. Choice D is correct.

Have your child work on these problems. Go over your child’s answers and make sure he or she has used the strategies presented in the explanatory answers to these problems.

PROBLEMS

1Which is always an odd number?

(A)an odd number plus an odd number

(B)an even number plus an even number

(C)an odd number divided by an odd number

(D)an odd number times an odd number

2Which fraction is greater than

3If a square has a whole number for one of its sides, which of the following could be its area?

(A)80

(B)66

(C)49

(D)32

4

In the multiplication above, x and y are integers from 0 to 9 (inclusive). What can y be?

(A)8

(B)6

(C)5

(D)4

SOLUTIONS

1(D)Since you have to test all of the choices, start with the last one, Choice D: An odd number times an odd number. Try different examples if you’re not sure of this one:

3 × 1 = 3 (odd × odd = odd)

5 × 5 = 25 (odd × odd = odd, again)

You can be pretty sure an odd number times an odd number is odd. There is no need to look at the other choices.

2(D)Since you have to test all of the choices, start with Choice D. Refer to the shortcut for comparing fractions described in the “Math Shortcuts” section at the end of this book (see pages 173—182).

Comparing Choice D:

Choice D is correct.

3(C)You know that the area of a square is: side × side. So you are looking for a choice in which a whole number times the same whole number equals the number in the choice. Since you have to test out all of the choices, start with Choice D. What whole number times itself gives you 32? No number. Choice D is incorrect.

Now look at Choice C: What whole number times itself gives you 49. It’s 7. So the square could have 7 as its side. Choice C is correct. There is no need to look at the other choices.

4(D)Look at Choice D first. Let y = 4. Then

This satisfies the multiplication where x = 4 and y — 4. So Choice D is correct.

MATH STRATEGY 6:

You Don’t Always Need a Complete Solution

At times when working on the solution to a math problem, your child will get some result but not the complete answer. At that point it may still be possible to eliminate incorrect choices by comparing his or her result with the answer choices.

Here are some examples:

EXAMPLE 1

What is the number HIJK?

(A)2,184

(B)2,548

(C)1,456

(D)17,108

You should realize that ABCD is derived by multiplying 364 × 6, DEFG by multiplying 364 × 7, and HIJK by multiplying 364 × 4. So when you multiply 364 × 4, you end up with a number ending in a 6, since 4 × 4 = 16. The only choice that ends in a 6 is Choice C. So Choice C must be correct. Therefore you don’t have to multiply out the whole product of 364 × 4.

EXAMPLE 2

What is the product of 1 × 2 × 3 × 4 × 5 × 6 × 7?

(A)5,039

(B)5,040

(C)5,041

(D)5,042

Instead of multiplying out, just realize that the 2 in the problem times the 5 in the problem gives you 10. And 10 times the rest of the numbers must end in 0, because 10 times any integer ends in 0. Thus your correct choice must end in 0. The only one that does is Choice B. So Choice B is correct. Look at all the time you saved!

Have your child try these questions. Go over the solutions with your child to see whether he or she has used the strategies described in the book.

PROBLEMS

1A bike is priced normally $234.95 but is reduced by 20 percent. By how much is the bike reduced by?

(A)$46.99

(B)$49.95

(C)$55.88

(D)$44.67

2

(A)1,088

(B)1,089

(C)1,090

(D)1,091

SOLUTIONS

1(A)Multiply 234.95 by 20%:

Stop here. When you move the decimal point over two places, you end up with a 9 as the last digit. So let’s look for a 9 in the answer. Only Choice A has a 9 in the answer.

There is no need to multiply the whole product out.

2(B)You should realize that these choices differ only in the last digit. Here you’re looking for a number that when multiplied by 4 will give you 4,356. That number cannot end in 8 (Choice A) because 8 × 4 = 32 and the last digit of 4,356 is not 2. The number cannot end in 0 (Choice C) because 0 × 4 = 0, and the last digit of 4,356 ends in 6, not 0. The number cannot end in 1 (Choice D) because 1 × 4 = 4, which is not the last digit of 4,356. Thus only Choice B remains.

MATH STRATEGY 7:

Know How to Work with Problems Dealing with Averages

Whenever your child sees a problem that says “average,” he or she should know what the word means. It is

As soon as you see the word “average,” translate the meaning as described above. Here are some examples:

EXAMPLE 1

The average of 7, 4, and — 3 is

Remember;

So

Choice C is correct.

EXAMPLE 2

The average of 6 numbers is 10. The sum of 5 of these numbers is 40. The remaining number is

(A)20

(B)30

(C)40

(D)50

Remember:

But what do you do since you don’t know the numbers? Call the numbers a, b, c, d, e, and f. Then

But you know that 5 of the numbers add up to 40. So let a + b + c + d + e = 40. Putting the value of a + b + c + d+ e in the equation above you get:

Multiply both sides of the equation by 6:

Now subtract 40 from both sides:

The remaining number is 20. (Choice A).

EXAMPLE 3

A car travels along a straight line from A to B to C. It travels from A to B in 2 hours and from B to C in 3 hours. What is the average speed of the car in kilometers per hour if the distance from A to B is 40 km and the distance from B to C is 70 km?

(A)55

(B)44

(C)33

(D)22

PROBLEMS

Have your child try these questions. Then check to see whether his or her answers match the ones that are given at the end of these questions. Make sure that your child uses the right strategies as explained in the explanatory answers.

1A four-sided figure has sides of 4 cm, 4 cm, 7 cm, and 8 cm. What is the average length per side of the figure?

(A)23 cm

(B)8½ cm

(C)5¾ cm

(D)7¾ cm

2The average of — 5, + 7, + 5, and — 7 is

(A)—2

(B)+2

(C)0

(D)—1

3The average score of 4 students on a test was 80. The lowest score was 60 and the highest score was 90. What was the sum of the remaining two scores on the test?

(A)140

(B)150

(C)160

(D)170

4A train travels from New York to Chicago in 24 hours and from Chicago to St. Louis in 16 hours. If the total distance from New York to St. Louis (straight line distance) is 1,200 miles, what was the average speed of the train for the complete trip from New York to St. Louis?

(A)30 mph

(B)40 mph

(C)50 mph

(D)cannot be determined

SOLUTIONS

1(C)Remember:

2(C)Remember:

Notice that the — 5 cancels with the + 5 and the — 7 cancels with the + 7 to give a 0 for the numerator of the average.

3(D)Remember:

The highest score was 90, the lowest 60. Let the other two scores be represented by a and b. So

Multiply both sides of the equation by 4:

Subtract 150 from both sides:

320 + 150 = 150 + a + b — 150

170 = a + b

But a + b are the sum of scores of the remaining students, so the answer is 170. (Choice D).

4(D)This is very tricky. The average speed is

The total time is 24 + 16 = 40 hours.

But the total distance is not 1,200 miles because the distance from New York to Chicago (as the train travels) plus the distance from Chicago to St. Louis (as the train travels) is going to be greater than the straight line distance from New York to St. Louis. (See the diagram below.) Thus you cannot determine an answer (Choice D).

MATH STRATEGY 8:

Label Sides, Angles, Etc., with Numbers or Letters

In questions with Diagrams, have your child label everything so that he or she can have as complete an understanding as possible of the problem. Here are some examples:

EXAMPLE 1

In the drawing below, in which all boxes are squares of equal area, the distance from B to C is 10 kilometers. What is the distance from A to B?

(A)3

(B)4

(C)5

(D)6

Since the distance from B to C is 10 kilometers, and there are 5 squares from B to D, each square has a side of 2. So label each side of the squares 2. Thus AB = 6 (Choice D).

EXAMPLE 2

In the figure below, CD and BC are the same length. ED and AE are also the same length. What is the perimeter of the figure?

(A)27

(B)32

(C)35

(D)40

Label CD = 5 (since CD = BC) and ED = 8 (since ED = AE).

This makes the perimeter 8 + 8+ 5 + 5 + 14 = 40 (Choice D).

EXAMPLE 3

The perimeter of an equilateral triangle is 21 y. What is the side of the triangle?

(A)7y

(B)By

(C)9y

(D)lOy

An equilateral triangle has equal sides. Label each side of the triangle s. Then perimeter = s + s + s = 3s. Set 3s = 21y. Divide by 3 to get s:

EXAMPLE 4

What is the perimeter of the parallelogram below:

(A)23

(B)38

(C)46

(D)61

Remember: In a parallelogram the opposite sides are equal. So label each of the sides:

Perimeter = 15 + 8+ 15 + 8 = 46. (Choice C).

Have your child try these questions. Then check to see whether his or her answers match the ones at the end of the examples. Make sure your child uses the right approach and strategy.

PROBLEMS

1An equilateral triangle has a perimeter of 32. What is its side?

(A)10

(B)16

(C)4

(D)10^⅔

2In this figure (right), AB = AE = ED = BC. DC = 5 and AE — 4. What is the perimeter of figure ABCDE?

(A)9

(B)17

(C)18

(D)21

3The distance around the circumference from A to B in the circle (right), is the same as the distance around the circumference from B to C and the same as the distance around the circumference from C to A. If the radius of the circle is 2, what is the circular distance from A to B?

(A)4π

(B)2π

4Each of the squares in the figure below has an equal area. If the area of the whole figure ABCD is 80, what is the area of each smaller square?

(A)2

(B)4

(C)8

(D)16

SOLUTIONS

1(D)All of the sides of an equilateral triangle are equal. Draw and label each side of the triangle s:

Perimeter = s + s + s = 3s.

But you were told that the perimeter = 32. So set 3s = 32.

Divide by 3:

2(D)Label all sides:

Perimeter = 4 + 4 + 4-1-4-1-5 = 21

3(C)Label the circular distance from A to B as a. Since the circular distance from B to C is the same as from C to A, label the other circular distances a also.

The radius of 2 tells us that the whole circumference = 2π(2) = 4π (since circumference = 2πr). Thus circumference = 4π. But we know that 3a is also the circumference.

So let 4π = 3a.

Divide by 3:

4(A)Label the width and length of the rectangle with number of units for length and width.

Area = length × width = 10 × 4 = 40.

So there are 40 little squares. But the area of the rectangle is 80, so each square must be 2 units in area if 40 squares have a total area of 80.

MATH STRATEGY 9:

Subtract Knowns from Knowns to Get Unknowns

This is a very useful strategy, especially when problems with figures are presented. Here are a few examples:

EXAMPLE!

The area of the shaded region is

(A)12 square meters

(B)17 square meters

(C)22 square meters

(D)34 square meters

You will find that it is almost impossible to find the shaded area by some direct calculation. So you should say to yourself, how can I get the area from the areas that I’m given?

STRATEGY: Subtract Known Areas from Known Areas to Get Unknown Areas. The area of the big rectangle minus the area of the smaller rectangle will give you the area of the shaded region, because that’s the area that is left over after subtracting.

The area of the big rectangle is length × width = 7 × 4 = 28

The area of the small rectangle is length × width = 3X2 = 6

The difference is 28 — 6 = 22. (Choice C)

Here’s another example with lines, not areas:

EXAMPLE 2

What is the distance of line AB?

(A)10.6

(B)9.5

(C)4.2

(D)cannot tell

Use the same strategy, but use it for lines. Subtract Knowns from Knowns to Get Unknowns:

The distance of line AB can be found by subtracting:

AC — BC = AB

AC = 7.4, BC = 3.2.

So AB = 7.4 - 3.2 = 4.2 (Choice C).

EXAMPLE 3

What is the area of the shaded region if the area of the small circle is 12 and the area of the large circle is 20?

(A)32

(B)8

(C)10

(D)cannot tell

The area of the shaded region is found by subtracting the area of the large circle from the area of the small circle. Thus the area of the shaded region is just 20 — 12 = 8. (Choice B).

Have your child try these questions. Check to see whether his or her answers match those in the book, making sure that your child used the same method of solution presented there.

PROBLEMS

1If the area of the large triangle ABD is 20 and the area of small triangle is 4, what is the area of shaded region?

(A)24

(B)16

(C)8

(D)cannot tell

2What is the distance x in the diagram below?

(A)1

(B)2

(C)3

(D)4

3If the side of the square is 4 and the radius of the inscribed circle is 2, what is the area of all the shaded regions?

(A)2

(B)8 — 2π

(C)4

(D)16 - 4π

SOLUTIONS

1(B)Subtract knowns from knowns to get unknowns: The area of the large triangle — the area of small triangle = the area of the shaded region. So 20 — 4 = 16.

2(B)Subtracting 4 from 5 (5 — 4) gives you the segment to the left of the segment of length x. Now the segment of length 3 minus the segment of length 1 (the segment to the left of x) gives you what the length of x is. (3 — 1 = 2).

3(D)The area of the shaded regions is equal to the area of the square minus the area of the circle:

area of shaded regions = area of square — area of circle.

So let’s find the area of the square and the area of the circle.

The area of the square = side × side = 4 × 4 = 16

The area of the circle = π × radius × radius = π(2 — 2) = 4π

So the area of shaded regions = 16 — 4π.