0.1 SolvingEquationsWithPicturesandWith Algebramath.uga.edu/~sybilla/5003Fa03/algebra.pdf0.1 SolvingEquationsWithPicturesandWith Algebra A major part of algebra is solving equations

0.1. SOLVING EQUATIONS WITH PICTURES AND WITH ALGEBRA1

The following is an excerpt from Mathematics for Elementary Teachers

first edition by Sybilla Beckmann, copyright Addison-Wesley, 2003.

0.1 Solving Equations With Pictures and With

Algebra

A major part of algebra is solving equations. In this section, we will studythe fundamental ideas we need to solve equations and we will apply theseideas to solve some of the simplest kinds of equations. We will also seehow drawing diagrams can make some algebra story problems easy to solvewithout using variables. We will then relate the diagram-drawing method ofsolving algebra story problems to standard algebraic methods. The diagram-drawing method of solving problems we will study is shown in the textbooksused in grades 3 – 6 in Singapore. Of the 38 nations studied in the ThirdInternational Mathematics and Science Study, children in Singapore scoredhighest in math (see [?]).

Solutions of Equations

We must first say what it means to solve an equation. Some equations aretrue; for example,

3 + 2 = 5

6 × 3 = 3 × 6

10 = 2 × 5

are true. Some equations are false; for example

1 = 0

3 + 2 = 6

are false. Most equations with variables are true for some values of thevariable and false for other values of the variable. For example,

3 + x = 5

is true when x = 2 but is false for all other values of x.

2

To solve an equation involving variables means to determine those val-ues for the variables which make the equation true. The values for the vari- solve an equa-

tionables which make the equation true are called the solutions to the equation.To solve the equationsolution

3x = 12

means to find all those values for x for which 3x is equal to 12. Only 3 · 4 isequal to 12; 3 times any number other than 4 is not equal to 12. So there isonly one solution to 3x = 12, namely x = 4.

Even young children in elementary school learn to solve simple equations.For example, first or second graders could be asked to fill in the box belowto make the equation true:

5 + = 7

One source of difficulty in solving equations is understanding that theequal sign does not mean “calculate the answer.” For example, when childrenare asked to fill in the box to make the equation

5 + 3 = + 2

true, many will fill in the number 8 because 5 + 3 = 8. In order to under-stand how to solve equations, we must understand that we want to makethe expressions to the left and right of the equal sign equal to each other.One helpful piece of imagery for understanding equations is a pan balance,as shown in Figure 1. If view the equation 5 + 3 = + 2 in terms of a panbalance, each side of the equation corresponds to a side of the pan balance.To solve the equation means to make the pans balance rather than tilt toone side or the other.

Solving Equations Using Algebra

How do we solve equations? The equation

2x = 6

is easy to solve because it asks:

2 times what number is 6?

which we can solve by division:

x = 6 ÷ 2 = 3


5 + 3 is not equal to 8 + 2

5 + 3 = ? + 2

what will make the pans balance?

?

5 + 3 = 6 + 2

Figure 1: Viewing the Equation 5 + 3 =? + 2 in Terms of a Pan Balance

The equation

x + 3 = 7

is also easy to solve because it asks:

what number plus 3 is 7?

which we can solve by subtraction:

x = 7 − 3 = 4

But how do we solve a more complex equation like

5x + 2 = 3x + 8?

The strategy for solving complex equations is to change the equation intoa new equation that has the same solutions and is easy to solve. We canchange an equation into a new equation that has the same solutions by doingany of the following:

4

1. Use properties of arithmetic or valid ways of operating with numbers(including fractions) to change an expression on either side of the equalsign into an equal expression. For example, we can change the equation

3x + 4x + 2 = 1

to the equation7x + 2 = 1

because according to the distributive property,

3x + 4x = (3 + 4)x = 7x

2. Add the same quantity to both sides of the equation or subtract thesame quantity from both sides of the equation. For example, we canchange the equation

6x − 2 = 3

to the equation6x − 2 + 2 = 3 + 2

which becomes6x = 5

by using item (1).

3. Multiply both sides of the equation by the same non-zero number ordivide both sides of the equation by the same non-zero number. Forexample, we can change the equation

5x = 3

to the equation

x =3

5

by dividing both sides of the equation by 5. Remember that 5x standsfor 5 · x, so when we divide this by 5, we get x.

Why do items (1), (2), and (3) change an equation into a new equationthat has the same solutions as the original one? When we use item (1), wedon’t really change the equation at all, we just write the expression on one


side of the equal sign in a different way. Therefore when we use item (1), thenew equation must have the same solutions as the original equation.

To see why item (2) should produce a new equation that has the samesolutions as the original one, think in terms of a pan balance. If the two pansare balanced and you add the same amount to both sides or take the sameamount away from both sides, then the pans will remain balanced. Likewisewith equations.

To see why item (3) should produce a new equation that has the samesolutions as the original one, think again in terms of a pan balance. If thetwo pans are balanced and you put twice as much or three times as much,and so on, on each pan, then the pans will remain balanced. Similarly, if thetwo pans are balanced and you take 1

2or 1

3, and so on, of each pan, then the

pans will remain balanced.

The strategy then for solving an equation such as

5x + 2 = 3x + 8

is to use items (1), (2), and (3) above to change the equation into an equationof the form

ax = b

where a and b are numbers, and which has the same solutions as the original.We can do this by getting all the terms with x on one side of the equationand getting all the numbers on the other side. Figure 2 shows how to solve5x+2 = 3x+8 this way. Each step is also illustrated in terms of a pan balance.Each small square on the pan balance represents 1. Each larger square with aquestion mark inside represents an x (because this x is the unknown amountwe are looking for). To solve the equation means to determine the value ofx that will make the equation true. In terms of the pan balance, we want todetermine how many small squares will balance a box with a question mark.We see from Figure 2 that the solution is x = 3, in other words, 3 smallsquares balance a square with a question mark.

6

? ?

? ?

? ? ?

?

? ?

? ?

? ? ?

?

? ? ?

? ?

?

5x + 2 = 3x + 8

- 2 - 2

5x = 3x + 6

- 3x - 3x

2x = 6

x = 3

take 2

away from both sides

take 2

away from

both sides

take 3x

away from

both sides

divide

both

sides

by 2

take 3

away from both sides

divide each side

in half, i.e., divide

each side by 2

with equations: with a pan balance:

Figure 2: Solving 5x + 2 = 3x + 8 With Equations and With a Pan Balance


Class Activity 0A: Solving Equations Algebraically and

With a Pan Balance

Solving Algebra Story Problems With Singapore-Style

Strip Diagrams

The pan balance view of equations presented above provides an excellentconcrete way to solve equations. However, many algebra story problems,especially those involving fractions or percentages, are not readily formulatedin terms of a pan balance. We will now study a way to use diagrams to solvesome algebra and other story problems.

In the elementary school mathematics textbooks used in Singapore (see[?]), narrow strips, such as the ones shown in Figure 3, are often used torepresent quantities. We will solve three different problems from the Singa-pore textbooks with the aid of such strip diagrams. For the second and thirdproblem, we will also see how to formulate the problem algebraically and howthe algebraic solution method mirrors the strip diagram solution method.

Problem 1 from a 3rd grade workbook

There were 120 boys at a concert. There were 15 more girls thanboys. How many girls were there? How many children were therealtogether? (From [?], volume 3A, part 1, page 20.)

Accompanying the problem is a diagram similar to Figure 3. This strip

Boys:

120 15

Girls:

Figure 3: A Strip Diagram to Determine How Many Girls and How ManyChildren Were at a Concert When There Were 120 Boys and 15 More GirlsThan Boys

diagram makes it clear that we should add 120 + 15 = 135 to find the

8

number of girls at the concert and then add 120 + 135 = 255 to find thetotal number of children at the concert. Notice also that we could calculatethe total number of children by adding 120 + 120 + 15, as we see from thediagram.

Problem 2 from a 3rd grade workbook

Rani had $47. After paying for 3 kg of prawns, she had $20 left.Find the cost of 1 kg of prawns. (From [?], volume 3A, part 1,page 55.)

This problem is accompanied by a strip diagram like Figure 4. We can see

$47

$20?

Figure 4: A Strip Diagram to Determine How Much 1 kg of Prawns Cost ifRani Started With $47 and Had $20 Left After Paying for 3kg of Prawns

from the diagram that 3 kg of prawns must cost $47− $20 = $27. Therefore1 kg of prawns must cost $27 ÷ 3 = $9.

We can also formulate this problem algebraically with an equation. Letx be the cost of 1 kg of prawns. Then 3 kg of prawns cost 3x dollars. Soafter paying for the 3 kg of prawns, Rani has

47 − 3x

dollars left, which the problem tells us is $20. Therefore

47 − 3x = 20

If we add 3x to both sides we obtain the equation

47 = 20 + 3x


which fits nicely with the strip diagram in Figure 4, where one small whitestrip represents x. If we subtract 20 from both sides of the equation, weobtain the equation

47 − 20 = 3x after simplifying: 27 = 3x

Dividing both sides by 3 we have

27

3= x after simplifying: 9 = x

So x = 9 is the solution and each kilogram of prawns costs $9. Notice thatthe calculations we performed when solving the problem algebraically are thesame as those we performed when using the strip diagram: in both cases wesubtracted 20 from 47 to obtain 27 and then we divided 27 by 3 to obtain 9.

Problem 3 from the 4th grade textbook

Although the previous two problems can be considered arithmetic problems,most math teachers would probably view the following 4th grade problem asan algebra problem rather than an arithmetic problem:

300 children are divided into two groups. There are 50 morechildren in the first group than in the second group. How manychildren are there in the second group? (From [?], volume 4A,page 40)

The Singapore textbook has not introduced algebra with variables at thispoint, so the children are expected to solve this problem with the aid of adiagram. They might draw a diagram like the one in Figure 5. From thediagram we see that if we subtract 50 from 300 and then divide the resultingamount by 2, we will have the number of children represented by each of thelong strips. Since 300 − 50 = 250 and 250 ÷ 2 = 125, the second group has125 children and the first group has 125 + 50 = 175 children.

We can also formulate this problem algebraically with equations. If welet A be the number of children in the first group and B be the number ofchildren in the second group, then, because there are 300 children in all, wehave

A + B = 300

10

50

300

Group 1:

Group 2:

Figure 5: A Strip Diagram to Determine the Number of Children in EachGroup if 300 Children are Divided into 2 Groups With 50 More Children inthe First Group Than the Second Group

We are given that A is 50 more than B, so

A = B + 50

Notice that we can also get these equations from Figure 5, where the bottomstrip represents B and the top, combined strip represents A. Since the secondequation tells us that A is equal to B + 50, we can replace A in the firstequation with B + 50 to obtain the equation

B + 50 + B = 300

Combining the Bs we have

2B + 50 = 300

We can see this equation in Figure 5 because the 2 long strips representingB and the strip representing 50 combine to make 300. Subtracting 50 fromboth sides we have

2B = 300 − 50 after simplifying: 2B = 250


B =250

2after simplifying: B = 125

So there are 125 children in the second group.We can also solve this problem algebraically by using only one variable

instead of two variables. Let C be the number of children in the smaller


group, which is the second group. Then since there are 50 more children inthe first group, the number of children in the first group is

C + 50

Since the total number of children is 300, we have

C + 50 + C = 300

Combining the Cs we have

2C + 50 = 300

We solve this equation in the same way that we solved 2B +50 = 300 above,using the letter C instead of B.

Notice that the calculations we performed when solving the problem alge-braically are the same as those we performed when using the strip diagram:in both cases we subtracted 50 from 300 to obtain 250 and then divided 250by 2 to obtain 125.

Class Activity 0B: Solving Story Problems With Singapore-

Style Strip Diagrams and With Algebra

Class Activity 0C: Solving Story Problems

Exercises for Section 0.1 on Solving Equations With Pic-

tures and With Algebra

1. Dante has twice as many jelly beans as Carlos. Natalia has 4 more jellybeans than Carlos and Dante together. All together, Dante, Carlos, andNatalia have 58 jelly beans. How many jelly beans does Carlos have?

(a) Solve this problem with the aid of a diagram. Explain your solu-tion.

(b) Now solve the problem algebraically as follows. Let x stand forthe number of jelly beans that Carlos has.

i. Write an expression in terms of x for the number of jelly beansthat Dante has. Relate your expression to your diagram inpart (a).

12

ii. Write an expression in terms of x for the number of jelly beansthat Natalia has. Relate your expression to your diagram inpart (a).

iii. Write an equation involving x which states that the total num-ber of jelly beans is 58. Relate your equation to your diagramin part (a).

iv. Solve your equation in part (iii). Relate your work to thework you did with your diagram in part (a).

2. A 6th grade problem: A tank is 2

5filled with water. When another 26

liters of water are poured in, the tank becomes 5

6full. Find the capacity

of the tank. (From [?] volume 6A, p. 71.) Solve this problem in twoways: with the aid of a diagram and with equations. Discuss how thetwo solution methods are related.

3. A 6th grade problem: Samy had 130 stickers and Devi had 50 stickers.After Samy gave Devi some stickers, Samy had twice as many stickersas Devi. How many stickers did Samy give Devi? (From [?] volume 6B,p. 34.) Solve this problem in two ways: with the aid of a diagram andwith equations. Discuss how the two solution methods are related.

Answers to Exercises for Section 0.1 on Solving Equa-

tions With Pictures and With Algebra

1. (a) We see from Figure 6 that the total number of jelly beans consistsof 6 equal groups of jelly beans and 4 more jelly beans. If wetake the 4 away from 58, we have 54 jelly beans that must bedistributed equally among the 6 groups. Therefore each grouphas 54 ÷ 6 = 9 jelly beans. Since Carlos has one group of jellybeans, he has 9 jelly beans. (Dante has 2 · 9 = 18 jelly beans, andNatalia has 9 + 18 + 4 = 31 jelly beans.)

(b) i. Since Dante has twice as many jelly beans as Carlos, Dantehas 2x jelly beans. In Figure 6, Carlos’s strip represents xjelly beans and since Dante has two such strips, he has 2xjelly beans.

ii. Since Natalia has 4 more than Carlos and Dante combined,Natalia has x + 2x + 4 jelly beans, which we can also write as


Carlos:

4

58Dante:

Natalia:

Figure 6: A Strip Diagram for Determining How Many Jelly Beans CarlosHas

3x + 4 jelly beans. We see in Figure 6 that Natalia’s portionconsists of a strip which represents x jelly beans, two morecopies of this strip which represent 2x jelly beans, and anotherstrip representing the additional 4 jelly beans. So all together,Natalia’s portion is 3x + 4 jelly beans.

iii. We are given that the total number of jelly beans is 58. Onthe other hand, the total number of jelly beans is Carlos’s,Dante’s, and Natalia’s combined, which is

x + 2x + 3x + 4

By combining the xs, we can rewrite this expression as

6x + 4

Since the two ways of expressing the total number of jellybeans must be equal, we have the equation

6x + 4 = 58

We can also deduce this equation from Figure 6 because thereare 6 strips which each represent x jelly beans and anothersmall strip representing 4 more jelly beans; the combinedamount must be 58, so 6x + 4 = 58.

iv. To solve 6x+4 = 58, first subtract 4 from both sides to obtain

6x = 58 − 4 after simplifying: 6x = 54

14

Then divide both sides by 6 to obtain

x =54

6after simplifying: x = 9

Therefore Carlos has 9 jelly beans. Notice that the arithmeticwe performed in order to solve the equation 6x + 4 = 58is the same as the arithmetic we performed in part (a): wefirst subtracted 4 from 58 and then we divided the resultingamount, 54, by 6.

2. As we see in Figure 7, 26 liters fill the difference between 2

5of the tank

and 5

6of the tank. Since

5

6−

2

5=

5 · 5

6 · 5−

2 · 6

5 · 6=

25

30−

12

30=

13

30

There are 13 parts (each of which is 1

30of the tank) that the 26 liters

fill. So each of these 13 parts holds 2 liters. The full tank consists of30 parts, so the full tank holds 30 · 2 = 60 liters.

2/5 = 12/30 26 liters fills 13 parts

5/6 = 25/30

Figure 7: A Strip Diagram for Determining the Capacity of a Tank if 26Liters Added to a 2

5Full Tank Makes the Tank 5

6Full

To solve the problem with equations, let x be the number of liters thatthe tank holds. Then 2

5x is the number of liters in 2

5of the tank and

5

6x is the number of liters in 5

6of the tank. We are given that

2

5x + 26 =

5

6x

Subtracting 2

5x from both sides we have

26 =5

6x −

2

5x


so

26 = (5

6−

2

5)x after simplifying: 26 =

13

30x

After multiplying both sides by 30

13(or dividing both sides by 13

30) we

have30

13· 26 =

30

13·13

30x after simplifying: 60 = x

Therefore x = 60 and the full capacity of the tank is 60 liters.

We perform the same arithmetic using both solution methods. Witheach method, we calculated 5

6−

2

5. When we solved the problem using a

diagram, we divided 26÷13 to find the number of liters in each 1

30part

of the tank. Then we multiplied that result, 2, by the total number ofparts in the tank, 30. Similarly, using the algebraic solution method,we had to calculate

30

13· 26

Which we can do by canceling 13s:

30

13/·

2

26/= 30 · 2 = 60

This way we again divide 26 by 13 and multiply the result, 2, by 30.

3. In Figure 8, focus on the amount Samy has after he gives some stickersto Devi. In Samy’s strip, we see that if we take away the amount in2 shaded strips from the 130 stickers, then the remaining amount willbe 3 times the number of stickers that Samy gave Devi. The 2 shadedstrips make 100 stickers, so taking these 100 stickers away from the130 stickers leaves 30. These 30 stickers must be 3 times the numberof stickers that Samy gave Devi is 30. Therefore Samy gave Devi 10stickers.

To solve the problem with algebra, let S be the number of stickers thatSamy gave Devi. Then after Samy gives away these stickers, he has130 − S stickers and Devi has 50 + S stickers. We are told that Samyhas 2 times as many as Devi, so

130 − S = 2 · (50 + S)

Therefore130 − S = 100 + 2S

16

Samy:

130

50

Devi:

Samy:

Devi:

Before:

After:

Figure 8: A Strip Diagram for Determining How Many Stickers Samy GaveDevi

Adding S to both sides we have

130 = 100 + 3S

Subtracting 100 from both sides we have

30 = 3S


10 = S

Therefore S = 10 and Samy gave Devi 10 stickers.

We perform the same arithmetic using both solution methods. In bothcases we multiplied 50 by 2 and then subtracted the resulting amount,100, from 130. We then divided the result, 30, by 3. We can seethe amount, S in the diagram as the small box. We can also see theequation 130 − S = 2 · (50 + S) in the strip representing the amountSamy has after giving stickers to Devi.


Problems for Section 0.1 on Solving Equations With

Pictures and With Algebra

1. Solve4x + 5 = x + 8

in two ways: with equations and with pictures of a pan balance. Relatethe two methods.

2. Solve3x + 2 = x + 8

in two ways: with equations and with pictures of a pan balance. Relatethe two methods.

3. A 4th grade problem: 1650 pupils took part in a parade. There weretwice as many boys as girls. How many boys were in the parade? (From[?] volume 4B part 1, p. 38.)


(b) Now solve the problem algebraically as follows. Let x stand forthe number of girls in the parade.

i. Write an expression in terms of x for the number of boys inthe parade. Relate your expression to your diagram in part(a).

ii. Write an expression in terms of x for the total number of chil-dren in the parade. Relate your expression to your diagramin part (a).

iii. Write an equation involving x which states that the total num-ber of children in the parade is 1650. Relate your equation toyour diagram in part (a).

iv. Solve your equation in part (iii) and use the solution to de-termine the number of boys in the parade. Relate your workto the work you did with your diagram in part (a).

4. A 5th grade problem: After spending 2

5of his money on a toy car, Samy

had $42 left. How much money did he have at first? (From [?] volume5A, p. 60.)

18


(b) Now solve the problem algebraically as follows. Let x stand forthe amount of money Samy had to start with.

i. Write an expression in terms of x for amount of money Samyhad left after buying the toy car. Relate your expression toyour diagram in part (a).

ii. Write an equation involving x which states that Samy has$42 left after buying his toy. Relate your equation to yourdiagram in part (a).

iii. Solve your equation in part (ii). Relate your work to the workyou did with your diagram in part (a).

5. A 5th grade problem: Mrs Chen bought some eggs. She used 1

2of them

to make tarts and 1

4of the remainder to make a cake. She had 9 eggs

left. How many eggs did she buy? (From [?] volume 5A, p. 60.)


(b) Now solve the problem algebraically as follows. Let x stand forthe number of eggs that Mrs Chen bought.

i. Write an expression in terms of x for the number of eggs MrsChen had left after making tarts. Then write an expressionin terms of x for the number of eggs Mrs Chen used to makea cake. Relate your expressions to your diagram in part (a).

ii. Write an expression in terms of x for the total number ofeggs Mrs Chen used. Write another expression in terms of xfor the total number of eggs Mrs Chen has left. Relate yourexpressions to your diagram in part (a).

iii. Write an equation involving x which states that Mrs Chenhas 9 eggs left. Relate your equation to your diagram in part(a).

iv. Solve your equation in part (iii). Relate your work to thework you did with your diagram in part (a).

6. A 4th grade problem: 3000 exercise books are arranged into 3 piles.The first pile has 10 more books than the second pile. The number of


books in the second pile is twice the number of books in the third pile.How many books are there in the third pile? (From [?] volume 4A, p.41.) Solve this problem in two ways: with the aid of a diagram andwith equations. Discuss how the two solution methods are related.

7. A 4th grade problem: 2500 people took part in a cross-country race.The number of adults were 4 times the number of children. If there were1200 men, how many women were there? (From [?] volume 4A part 1,p. 62.) Solve this problem in two ways: with the aid of a diagram andwith equations. Discuss how the two solution methods are related.

8. A 4th grade problem: A piece of rope 3 m 66 cm long was cut into 2pieces. The longer piece was twice as long as the shorter piece. Whatwas the length of the longer piece? (From [?] volume 4B, p. 73.) Solvethis problem in two ways: with the aid of a diagram and with equations.Discuss how the two solution methods are related.

9. A 4th grade problem: 2

3of a sum of money is $18. Find the sum of

money. (From [?] volume 4B, p. 100.) Solve this problem in two ways:with the aid of a diagram and with equations. Discuss how the twosolution methods are related.

10. A 5th grade problem: A hawker sold 2

3of his curry puffs in the morning

and 1

6in the afternoon. He sold 200 curry puffs altogether. How many

curry puffs had he left? (From [?] volume 5A, p. 60.) Solve this problemin two ways: with the aid of a diagram and with equations. Discusshow the two solution methods are related.

11. A 5th grade problem: Minghua bought a bag of marbles. 1

4of the

marbles were blue, 1

8were green and 1

5of the remainder were yellow.

If there were 24 yellow marbles, how many marbles did he buy? (From[?] volume 5A, p. 60.) Solve this problem in two ways: with the aid ofa diagram and with equations. Discuss how the two solution methodsare related.

12. A 5th grade problem: Ali saved twice as much as Ramat. Maria saved$60 more than Ramat. If they saved $600 altogether, how much didMaria save? (From [?] volume 5A, p. 90.) Solve this problem in twoways: with the aid of a diagram and with equations. Discuss how thetwo solution methods are related.

20

13. A 6th grade problem: 1

3of the beads in a box are red, 2

3of the remainder

are blue and the rest are yellow. If there are 24 red beads, how manyyellow beads are there? (From [?] volume 6B, p. 34.) Solve this problemin two ways: with the aid of a diagram and with equations. Discusshow the two solution methods are related.

14. (See Problem ?? in Section ??) Bonnie had some stickers. First, Bonniegave away 1

6of her stickers. Then when Bonnie gets 12 more stickers

she has 47 stickers. Determine how many stickers Bonnie had at thestart. Explain your solution.

15. (See Problem ?? in Secion ??) There was some flour in a bag. First,3

8of the flour in the bag was used. Then 1

5of the remaining flour was

used. At that point, there were 10 cups of flour in the bag. Determinehow much flour was in the bag at the start. Explain your solution.

16. Not every algebra story problem is naturally modeled with either a panbalance or with a strip diagram. Solve the following problem in anyway that makes sense and explain your solution.

Kenny has stickers to distribute among some goodie bags. When Kennytries to put 5 stickers in each goodie bag he is 2 stickers short. Butwhen Kenny puts 4 stickers in each bag he has 3 stickers left over. Howmany goodie bags does Kenny have? How many stickers does Kennyhave?

17. Not every algebra story problem is naturally modeled with either a panbalance or with a strip diagram. Solve the following problem in anyway that makes sense and explain your solution.

Frannie has no money now but she plans to save $D every week fromnow on so that she can buy a toy she would like to have. Frannie figuresthat if she saves her money for 3 weeks she will have $1 less than sheneeds to buy the toy but if she saves her money for 4 weeks she willhave $3 more than she needs to buy the toy. How much is Franniesaving every week and how much does the toy Frannie plans to buycost?

18. A 5th grade problem: David cuts a rope 60 m long into two pieces inthe ratio 2 : 3. What is the length of the shorter piece of rope? (From[?] volume 5A, p. 79.) Solve this problem and explain your solution.


19. A 5th grade problem: The ratio of Samy’s weight to John’s weight is6 : 5. If Samy weighs 48 kg, find John’s weight. (From [?] volume 5A,p. 79.) Solve this problem and explain your solution.

20. A 5th grade problem: The ratio of the number of boys to the numberof girls is 2 : 5. If there are 100 boys, how many children are therealtogether? (From [?] volume 5A, p. 79.) Solve this problem andexplain your solution.

21. A 6th grade problem: Susan and Sally had an equal amount of moneyeach. After Sally spend $15 and Susan spent $24, the ratio of Sally’smoney to Susan’s money was 4 : 3. How much money did each girlhave at first? (From [?] volume 6A, p. 38.) Solve this problem andexplain your solution.

22. A 6th grade problem: There are 10% more boys than girls in a choir.If there are 4 more boys than girls, how many children are there alto-gether? (From [?] volume 6A, p. 67.) Solve this problem and explainyour solution.

23. A 6th grade problem: 60% of the books in a library are for adults,5% are for young people and the rest are for children. If there are 280books for children, how many books are there altogether? (From [?]volume 6A, p. 67.) Solve this problem and explain your solution.

24. A 6th grade problem: At a sale, Mrs Li bought a fan for $140. Thiswas 70% of its usual price. What was the usual price of the fan? (From[?] volume 6A, p. 67.) Solve this problem and explain your solution.

25. A 6th grade problem: Rahim has 30% more books than Gopal. IfRahim has 65 books, how many books does Gopal have? (From [?]volume 6B, p. 64.) Solve this problem and explain your solution.

26. A 6th grade problem: John and Mary had $350 altogether. After Johnspent 1

2of his money and Mary spent 1

3of her money, they each had

an equal amount of money left. How much did they spend altogether?(From [?] volume 6B, p. 58.) Solve this problem and explain yoursolution.

22

27. A 6th grade problem: If 2

3of a number is 12, what is the value of 1

2

of the number? (From [?] volume 6B, p. 63.) Solve this problem andexplain your solution.

28. A 6th grade problem: 10 glasses of water can fill 5

8of a bottle. How

many more glasses of water are needed to fill up the bottle? (From [?]volume 6B, p. 63.) Solve this problem and explain your solution.

29. Jane had a bottle filled with juice. At first, Jane drank 1

5of the juice

in the bottle. After 1 hour, Jane drank 1

4of the remaining juice in

the bottle. After another 2 hours, Jane drank 1

3of the remaining juice

in the bottle. At that point, Jane checked how much juice was left inthe bottle: there was 2

3of a cup left. No other juice was added to or

removed from the bottle. How much juice was in the bottle originally?Solve this problem and explain your solution.

30. A flock of geese on a pond were being observed continuously. At 1:00P.M., 1

5of the geese flew away. At 2:00 P.M., 1

8of the geese that

remained flew away. At 3:00 P.M., 3 times as many geese as had flownaway at 1:00 P.M. flew away, leaving 28 geese on the pond. At no othertime did any geese arrive or fly away. How many geese were in theoriginal flock? Solve this problem and explain your solution.

Documents

0.1 SolvingEquationsWithPicturesandWith Algebramath.uga.edu/~sybilla/5003Fa03/algebra.pdf0.1 SolvingEquationsWithPicturesandWith Algebra A major part of algebra is solving equations