Math Series

7/21/2019 Math Series

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Math ser ies

S o l v i n g m a t h w o r d p r o b l e m sThere are two steps to solving math word problems:

1. Translate the wording into a numeric equationthat combines smaller "expressions"

2. Solve the equation!

Suggestions: Read the problem entirely

Get a feel for the whole problem List information and the variables you identify

Attach units of measure to the variables (gallons, miles, inches, etc.) Define what answer you need,

as well as its units of measure Work in an organized manner

Working clearly will help you think clearly Draw and label all graphs and pictures clearly Note or explain each step of your process;

this will help you track variables and remember their meanings Look for the "key" words (above)

Certain words indicate certain mathematical operations:

Math expressions (examples):after you review the keywords,test yourself

addition: 5+x subtraction: 5-x

multiplication: 5*x; 5x division: 5 x; 5/x

Exercise: ("mouse over" the block for answer)

Key words for addition +

increased by; more than; combined together; total of; sum; added to

What is the sum of 8 and y? 8 + y

Express the number (x) of applesincreased by two

x + 2

Express the total weight of

Alphie the dog (x) and Cyrus the cat (y)x + y

Key words for Subtraction -

less than, fewer than, reduced by, decreased by, difference of

What is four less than y y - 4

What is nine less than a number (y) y - 9

What if the number (x) of pizzaswas reduced by 6? x - 6

What is the difference of my weight (x)

and your weight (y)x - y

Key words for multiplication * x or integers next to each other (5y, xy):

of, times, multiplied by

What is y multiplied by 13 13yor 13 * y

Three runners averaged "y" minutes.

Express their total running time:3y

I drive my car at 55 miles per hour.How far will I go in "x" hours?

55x

Key words for division /

per, a; out of; ratio of, quotient of; percent (divide by 100)

What is the quotient of y and 3 y/3or y 3

Three students rent an apartment x/3or x 3
http://www.studygs.net/mathproblems3.htmhttp://www.studygs.net/mathproblems3.htm


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for $ "x" /month. What will each have to pay?

"y" items cost a total of $25.00.

Express their average cost:25/yor25 y

More vocabulary and key words: "Per" means "divided by"

as "I drove 90 miles on three gallons of gas, so I got 30 miles per gallon."(Also 30 miles/gallon)

"a" sometimes means "divided by"

as in "When I filled up, I paid $10.50 for three gallons of gasoline,so the gas was 3.50 a gallon, or $3.50/gallon

"less than"If you need to translate "1.5 less than x", the temptation is to write "1.5 - x". DON'T! Put a "real world" situationin, and you'll see how this is wrong: "He makes $1.50 an hour less than me." You do NOT figure his wage bysubtracting your wage from $1.50.Instead, you subtract $1.50 from your wage

"quotient/ratio of" constructionsIf a problems says "the ratio ofxandy",it means "xdivided by y" or x/yor x y

"difference between/of" constructionsIf the problem says "the difference ofxandy",it means "x- y"

What if the number (x) of children was reduced by six, and then

they had to share twenty dollars? How much would each get? 20/(x - 6)

What is 9 more than y? y + 9

What is the ratio of 9 more than y to y? (y + 9)/y

What is nine less than the

total of a number (y) and two(y + 2) - 9 or y - 7

The length of a football field is 30 yards more than its width "y".

Express the length of the field in terms of its width y y + 30

Maths Word Problems and Solutions

Problem 1In one store in the afternoon sold twice more pears, than in the morning. During the whole day

they sold 360 kg. pears. How many kilograms of pears are sold in the morning and how many in theafternoon?Solution:

Lets take that the sold in the morning pears are x kg, then in the afternoon are sold 2x kg. Their sum x +

2x = 3x kg is the whole quantity sold pears, 360 kg. So we get the following equation3x = 360 x = 360/3 x = 120Therefore in the morning they sold 120 kg pears, and in the afternoon 2.120 = 240 kg.

Problem 2Ivan gathered twice more chestnuts than Peter and Boris gathered 2kg. more than Peter.Together they gathered 26 kg. chestnuts. How many kilogrammes gathered each one of them?Solution:Lets take Peters chestnuts as x kg., then Ivan gathered 2x kg. and Boris (x +2) kg . All gathered chestnutsare: x + 2x +x +2 = 4x +2 and by condition they are 26 kg. We get the equation: 4x +2 = 26 4x =

24 x = 6Therefore Peter gathered 6 kg., Ivan 2.6 = 12kg, Boris 6 +2 = 8 kg chestnuts.

Problem 3Kamen read 2/3 of a book and calculated that the read part is with 90 pages more than the unread. Howmany pages is the whole book?Solution:Lets take the whole book as x pages. The part he read is 2/3 from x , i.e. 2/3.x We will get the unread partwhen from the whole book we subtract the read part, i.e. x - 2/3 . x = 3/3x - 2/3x = 1/3x The read part2/3x is with 90 pages more than the unread one, which is 1/3x Therefore2/3x 1/3x = 90 1/3x =90 x = 90.3 = 270 So the book is 270 pages.


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Problem 4One tract can be ploughed with 6 tractors for 4 days, if they plough 120hectares a day. Two of the tractors

were moved to another tract. The rest 4 ploughed the same tract for 5 days. How many decares average aday ploughed the 4 tractors?Solution:

If 6 tractors ploughed 120 hectares a day and finished the tract for 4 days, then the whole tract is: 120.6.4= 720.4 = 2880 hectaresLets take that every one of the four tractors for the 5 days ploughed x hectares.Therefore the finished work:5.4. x = 20 . x hectares and this is the whole tract 2880 hectaresSo we get 20x = 2880 x = 2880/20 = 144 decares a day ploughed every on of the four tractor-

drivers.

Problem 5One student thought of a number, multiplied it by 2. From the received product subtracted 138 and got 102. Which is the number the student thought of?Solution:

Lets take the thought number as x, when he multiplied it with 2 he got (2x); from which he subtracted 138i.e. 2*x - 138 and by condition received 102 2. x -138 = 102 We must solve this equation to find thethought number

2*x - 138 = 102 2x = 240 x = 240/2 x = 120

Problem 6

I thought of a number, divided it by 5, from the received quotient I subtracted 154 and got 6. Which is thenumber i thought of?Instruction: The thought number is x and the equation: x/5 -154 = 6Solve the equation by yourself. Answerx = 800

Problem 7

The distance between two towns is 380km. A car and a lorry started from the two towns at the same time.At what speed drove the two vehicles, if the speed of the car is with 5km/h faster than the speed of thelorry and we know that they met after 4 hours?

Solution:The basic dependence that is used in problems with movement is that the distance is equal to the speedmultiplied by the time S = V.t

V km./h. t h. S km.

Car x + 5 4 4(x +5)

Lorry X 4 4x

4(x + 5) + 4x = 380 4x + 4x = 380 - 20 8x = 360 x = 360/8 x = 45Therefore the lorry was driving with 45 km/h., and the car with 50 km/h.

Problem 8One of the sides of a rectangle is with 3cm. shorter than the other one. Find the sides of the rectangle if weknow that, if we increase every side with 1cm., the surface of the rectangle will be increased with 18cm

2

Solution:Lets take that one of the sides is x cm. (x > 3), then the other will be x 3 cm. For the surface we find S1=x(x - 3) cm

2. If we increase the proportions with 1cm. the sides will be (x + 1) cm. and (x - 3 + 1 ) = (x -

2) cm. and these are the new proportions of the rectangle so the surface is S2= (x + 1).(x - 2) cm2and by

condition it is with 18 cm2bigger than the first one. Therefore we get the following equation:

S1+18 = S2 x(x - 3) + 18 = (x + 1)(x - 2) x2- 3x + 18 = x2+ x - 2x - 2 2x = 20 x =

10 And so the sides of the rectangle are 10 cm. and (10 - 3) = 7cm.

Problem 9For one year from two cows were milked 8100l. The next year the first cow increased her yield of milk with15% and the second one with 10%. So they milked 9100 l. from the two cows. How many litres are milkedfrom every cow during the first and the second year?Solution:If during the first year the first cow gave x l., then the second one gave (8100 x) l. The increase in the

yield of milk is 15% of x, i.e. 15/100.x and 10% of (8100 x), i.e. 10/100 . (8100 x) . Then during thesecond year the two cows gave the amount milk from the first year + the increase of the second yearSo we get the following equation: 8100 + 15/100.x + 10/100 . (8100 x) = 9100


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Therefore 8100 + 3/20x + 1/10 (8100 x) = 9100 1/20 . x = 190 x = 3800And so for the first year the milked 3800 and 4300 l. from every cow and for the second year 4370 l and

4730 l.

Problem 10The distance between stations A and B is 148 km. From station A to station B leaves an express train whichproceeds with 80 km/h and at the same time from station B towards station A leaves a goods train with 36km/h. We know that before the two trains meet at station C the express train made a 10min and the goodstrain - 5min. Find:a) The distance between station C and station B

b) At what time the goods train left station B if the meeting with the express train at station C was at 12oclock.Solutiona) We mark the distance from station B to station C with x km. Then the distance from station C to station Ais (148 x)km. By the time of the meeting at station C the express train ran (148 x)/80 + 10/60 hoursand the goods train x/36 +5/60. Because the trains left at the same time these times are equal : (148

x)/80 + 1/6 = x/36 + 1/12 We reduce to a common denominator, which for 6, 12, 36, 80 is 720 We releasefrom denominator and we get:9(148 x) +120 = 20x +60 1332 9x + 120 = 20x + 60

29x = 1392 x = 48 Therefore the distance from station B to station C is 48 km.b) By the time of meeting at station C the goods train ran 48/36 + 5/60 hours, i.e. 1 hour and 25 min.Therefore he left station B in 12 - 1.25/60 = 10.35/60 oclock, i.e. in 10 h. and 35min.

Complexity=4, Mode=simple

Solve. Give the answer in simplest form.

1. Pearl practices one hour of soccer on Saturday

and two-thirds of an hour of soccer on Sunday.

How many more hours of soccer did she

practice on Saturday than on Sunday?

hours of

soccer

2. Mike has 2 pizzas and

eats 1 pizza. What

fraction of the pizzas

are left?

Complexity=8, Mode=simpleSolve. Give the answer in simplest form.

1. Steven has five-sixths of a pound of jelly

beans and a third of a pound of gummi

bears. How many more pounds of jelly beans

than gummi bears does he have?

pounds

2. Mary has 10 cookies and

eats 6 cookies. What

fraction of the cookies

are left?


1. There are 4 apples, 12 bananas, and

2 melons on a table. What fraction of

the pieces of fruit are apples?

2. Miguel has1/3cup of milk and

1/3cup

of brown sugar. How many cups of

ingredients does he have? cups


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Complexity=4, Mode=mixedSolve. Give the answer in simplest form.

1. Mike has 21/2cups of granolas and

11/2cups of dried cherries. How

many more cups of granolas than

dried cherries does he have?

cups

2. Peter runs3/4mile on Monday

and 11/4miles on Tuesday.

How many total miles did he

run?

miles


1. Peter has 21/3cups of

flour and 13/4cups of

sugar. How many more

cups of flour than sugar

does he have?

cups

2. Steven runs 23/4kilometers on Friday

and1/4kilometer on Saturday. How

many more kilometers did he run on

Friday than on Saturday?

kilometers


1. Jennifer has three and five-

ninths pounds of peanuts and

three and a quarter pounds of

cashews. How many pounds of

food does she have?

pounds

2. Jose runs 32/3miles on Tuesday

and 31/10miles on Wednesday.

How many more miles did he run

on Tuesday than on Wednesday?

miles

Complexity=12, Mode=3 mixedSolve. Give the answer in simplest form.

1. Miguel practices 11/9hours

of soccer on Wednesday,

12/9hours of soccer on

Thursday, and1/10hour of

soccer on Friday. How

many hours of soccer didhe practice altogether?

hours

of

soccer

2. Paul practices a tenth of an

hour of basketball on Friday,

two and a quarter hours of

basketball on Saturday, and

one and five-sixths hours of

basketball on Sunday. Howmany hours of basketball did

he practice altogether?

hours of

basketball

Answers

Complexity=4, Mode=simple

Solve. Give the answer in simplest form.

# Problem Correct AnswerYour

Answer

1 Pearl practices one hour of soccer on Saturday and two-thirds of an hour of

soccer on Sunday. How many more hours of soccer did she practice on

1hours of


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Saturday than on Sunday?

3

soccer

Solution

The question is

How many more hours of soccer did she practice on Saturday than on Sunday?

Which operation should you use?

The key words how many moreimply subtraction.Equation

1 -

2

3

=

1

3

# Problem Correct Answer Your Answer

2 Mike has 2 pizzas and eats 1 pizza. What fraction of the pizzas are left?

1

2

Solution

The fraction of pizzas that are left is

Total # of pizzas - # pizzas eaten

Total # of pizzas

=

2 - 1

2

=

1

2


# Problem Correct Answer YourAnswer

1

Steven has five-sixths of a pound of jelly beans and a third of a pound of

gummi bears. How many more pounds of jelly beans than gummi bears does

he have?

1

2

pounds

Solution

The question isHow many more pounds of jelly beans than gummi bears does he have?


The key words how many moreimply subtraction.

Equation

5 - 1 = 1


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6 3 2

# Problem Correct Answer Your Answer

2 Mary has 10 cookies and eats 6 cookies. What fraction of the cookies are left?

2

5

Solution

The fraction of cookies that are left is

Total # of cookies - # cookies eaten

Total # of cookies

=

10 - 6

10

=

4

10

Simplify the answer.

4

10

=

2

5


# ProblemCorrect

Answer

Your

Answer

1There are 4 apples, 12 bananas, and 2 melons on a table. What fraction of the

pieces of fruit are apples?

2

9

Solution

The fraction of pieces of fruit that are apples is

# apples

Total # of pieces of fruit

=

4

4 + 12 + 2

=

4

18

Simplify the answer.


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4

18

=

2

9


Answer

2Miguel has 1/3cup of milk and

1/3cup of brown sugar. How many cups of

ingredients does he have?

2

3

cups

Solution

The question is

How many cups of ingredients does he have?


The question implies addition.

Equation

1

3

+

1

3

=

2

3



Answer

1Mike has 2

1/2cups of granolas and 1

1/2cups of dried cherries. How many more

cups of granolas than dried cherries does he have?1

cups

Solution

The question is

How many more cups of granolas than dried cherries does he have?



Equation

2

1

2

- 1

1

2

= 1


Answer


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2Peter runs

3/4mile on Monday and 1

1/4miles on Tuesday. How many total

miles did he run?2

miles

Solution

The question is

How many total miles did he run?Which operation should you use?

The key word totalimplies addition.

Equation

3

4

+ 1

1

4

= 2



Answer

1Peter has 2

1/3cups of flour and 1

3/4cups of sugar. How many more cups of

flour than sugar does he have?

7

12

cups

Solution

The question is

How many more cups of flour than sugar does he have?



Equation

21

3

- 13

4

=7

12


Answer

2Steven runs 23/4kilometers on Friday and

1/4kilometer on Saturday. How

many more kilometers did he run on Friday than on Saturday? 2

1

2

kilometers

Solution

The question is

How many more kilometers did he run on Friday than on Saturday?


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Equation

2

3

4

-

1

4

= 2

1

2



Answer

1 Jennifer has three and five-ninths pounds of peanuts and three and a quarterpounds of cashews. How many pounds of food does she have?

6

29

36

pounds

Solution

The question is

How many pounds of food does she have?


The question implies addition.

Equation

3

5

9

+ 3

1

4

= 6

29

36


Answer

2Jose runs 3

2/3miles on Tuesday and 3

1/10miles on Wednesday. How many more

miles did he run on Tuesday than on Wednesday?

17

30

miles

Solution

The question is

How many more miles did he run on Tuesday than on Wednesday?



Equation

3

2

3

- 3

1

10

=

17

30


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Complexity=12, Mode=3 mixedSolve. Give the answer in simplest form.


Answer

1Miguel practices 11/9hours of soccer on Wednesday, 12/9hours of soccer onThursday, and 1/10hour of soccer on Friday. How many hours of soccer did

he practice altogether?

2

13

30

hours of

soccer

Solution

1

1

9

+ 1

2

9

+

1

10

= 2

13

30


Answer

2

Paul practices a tenth of an hour of basketball on Friday, two and a

quarter hours of basketball on Saturday, and one and five-sixths hours of

basketball on Sunday. How many hours of basketball did he practice

altogether?

4

11

60

hours of

basketball

Solution

1

10

+ 2

1

4

+ 1

5

6

= 4

11

60

What percent of 20is 30?

We have the original number (20)and the comparativenumber (30). The unknown in this problem is the rate orpercentage. Since the statement is "(thirty) is (some percentage) of(twenty)", then the variable stands for the percentage, and theequation is:

30 = (x)(20)

30 20 =x= 1.5

Sincexstands for a percentage, I need to remember to convertthis decimal back into a percentage:

1.5 = 150%

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Thirty is 150%of 20.

What is 35%of 80?

Here we have the rate (35%)and the original number (80); the unknown is the comparative number whichconstitutes 35%of 80. Since the exercise statement is "(some number) is (thirty-five percent) of (eighty)", then thevariable stands for a number and the equation is:

x= (0.35)(80)

x= 28

Twenty-eight is 35%of 80.

45%of what is 9?

Here we have the rate (45%)and the comparative number (9); the unknown is the original number that 9is 45%of. Thestatement is "(nine) is (forty-five percent) of (some number)", so the variable stands for a number, and the equation is:

9 = (0.45)(x)

9 0.45 =x =20

Nine is 45%of 20.

The format displayed above, "(this number) is (some percent) of (that number)", alwaysholds true for percents. In any givenproblem, you plug your known values into this equation, and then you solve for whatever is left.

Suppose you bought something that was priced at $6.95, and the total bill was $7.61. What is the sales tax ratein this city? (Round answer to one decimal place.)

The sales tax is a certain percentage of the price, so I first have to figure what the actual tax was. The tax was:

7.616.95 = 0.66

Then (the sales tax) is (some percentage) of (the price), or, in mathematical terms:

0.66 = (x)(6.95)

Solving forx, I get:

0.66 6.95 =x= 0.094964028... = 9.4964028...%

The sales tax rate is 9.5%.In the above example, I first had to figure out what the actual tax was. Many percentage problems are really "two-part-ers" likethis: they involve some kind of increase or decrease relative to some original value. Warning: Always figure the percentage ofchange relative to the originalvalue.

Suppose a certain item used to sell for seventy-five cents a pound, you see that it's been marked up to eighty-one cents a pound. What is the percent increase?

First, I have to find the absolute increase: Copyright Elizabeth Stapel 1999-2011 All Rights Reserved

8175 = 6

The price has gone up six cents. Now I can find the percentage increase over the original price.

Note this language, "increase/decrease overthe original", and use it to your advantage: it will remind you to put the increase ordecrease overthe original value, and then divide.

This percentage increase is the relative change:

6/75= 0.08


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...or an 8%increase in price per pound.

Sometimes geometry word problems wrap the geometry in a thick layer of "real life". You will need to be able to "see" thegeometry, and extract the relevant information.

Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet in diameter. Howmuch sheet metal was used in its construction?

What they are asking for here is the surface area of the water tank. The total surface area of the tank will be the sum ofthe surface areas of the side (the cylindrical part) and of the ends. If the diameter is eight feet, then the radius is four feet.

The surface area of each end is given by the area formula for a circle with radius r:A= (pi)r2

. (There are twoendpieces, so I will be multiplying this area by 2when I find my total-surface-area formula.) The surface area of the cylinder isthe circumference of the circle, multiplied by the height:A= 2(pi)rh.

Side view of thecylindrical tank,

showing the radius "r".

An "exploded" view ofthe tank, showing thethree separatesurfaces whose areas Ineed to find.

Then the total surface area of this tank is given by:

2 ( (pi)r2) + 2(pi)rh (the two ends, plus the cylinder)

= 2( (pi) (42

) ) + 2(pi) (4)(30)= 2( (pi) 16 ) + 240(pi)= 32(pi) + 240(pi)= 272(pi)

Since the original dimensions were given in terms of feet, then my area must be in terms of square feet:

the surface area is 272(pi)square feet.

By the way, this is one of those exercises that doesn't translate well into "real life". In reality, the sheet metal has thickness, andadjustments would be required in order to account for this thickness. For example, if you needed to figure out the amount of sheetmetal required to create a tank with a certain volume, you'd have to account for the fact that the volume is on the inside, while thesurface area is on the outside. Since the walls of a real-world tank have thickness, the real-world answer would not match the"ideal" mathematical one.

A piece of 16-gauge copper wire 42cm long is bent into the shape of a rectangle whose width is twice its length.Find the dimensions of the rectangle.

Do I care that the wire is made of copper, or that the wire is a length of sixteen-gauge? No; all I care is that the length isforty-two units, that the units are centimeters, that the rectangle is twice as long in one direction as the other, and that I'msupposed to find the values of each of these directions. I can ignore the other information.

Since the wire is42centimeters long, then the perimeter of the rectangle is 42centimeters. That is:

2L+ 2W= 42

I also know that the width is twice the length, so:

W= 2L

Then: Copyright Elizabeth Stapel 2000-2011 All Rights Reserved


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2L+ 2(2L) = 42 (by substitution for Wfrom the above equation)2L+ 4L= 426L= 42L= 7

Since the width is related to the length by W= 2L, then W= 14, and the rectangle is 7centimeters longand 14centimeters wide.

A circular swimming pool with a diameter of 28feet has a deck of uniform width built around it. If the area of the

deck is 60(pi)square feet, find its width.

I have this situation:

A pool issurrounded by a deck.

The pool has radius 14,and the deck has width "d ".

If the diameter of the pool is 28, then the radius is 14. The area of the pool is then:

(pi)r2= (pi)(14)2= 196(pi)

Then the total area of the pool plus the surrounding decking is:

196(pi) + 60(pi) = 256(pi)

Working backwards from the area formula, I can find the radius of the whole pool-plus-deck area:

256(pi) = (pi)r2256 = r216 = r

Since I already know that the pool has a radius of 14 feet, and I now know that the whole area has a radius of 16, thenclearly:

the deck is two feet wide.

If one side of a square is doubled in length and the adjacent side is decreased by two centimeters, the area of theresulting rectangle is 96 square centimeters larger than that of the original square. Find the dimensions of the

rectangle.

I'm starting from a square with sides of some unknown length. The sides of the rectangle are defined in terms of thatunknown length. So I'll pick a variable for the unknown side-length, create expressions for the rectangle's sides, and thenwork from there.

square's side length:xone side is doubled: 2xnext side is decreased by two:x2square's area:x2rectangle's area: (2x)(x2) = 2x24xnew area is 96more than old area: 2x24x=x2+ 96

2x24x=x2+ 96x

24x96 = 0


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(x12)(x+ 8) = 0x= 12 or x=8

I'm supposed to find the dimensions of a rectangle, so can I just erase that one "minus" sign and say that the rectangle

is 12by 8? No! I defined "x" as standing for the side length of the square, not as one of the sides of the rectangle.Looking back at my definitions, I see what "x= 12" (the only reasonable solution for the square) means that the sides ofthe rectangle have lengths 2(12)and (12)2:

The rectangle measures 24cm by 10cm.

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Math Series