Free Geometry Book Download: Amsco’s Geometry

Free Geometry Book Download: Amsco’s Geometry
Amsco’s Geometry is a comprehensive textbook custom designed for complete coverage of the New York State Core Curriculum for Geometry.

Description
    Geometry is a new text for high school geometry that continues the approach that has made Amsco a leader in presenting mathematics in a contemporary, integrated manner. Formal logic is presented as the foundation for geometric reasoning. A logical system of reasoning and proof is carefully built using appropriate language, postulates and theorems.

Table of Contents
    Essentials of Geometry
    Logic
    Proving Statements in Geometry
    Congruence of Line Segments, Angles, and Triangles
    Congruence Based on Triangles

Transformations and the Coordinate Plane     Geometric Inequalities     Slopes and Equations of Lines     Parallel Lines     Quadrilaterals     The Geometry of Three Dimensions     Ratio, Proportion, and Similarity     Geometry of the Circle     Locus and Construction

Book Details
Author(s): Ann Xavier Gantert
Publisher: Amsco School Publications
Format(s): PDF

Chapter 1	Essentials of Geometry
Chapter 2	Logic
Chapter 3	Proving Statements in Geometry
Chapter 4	Congruence of Line Segments, Angles, and Triangles
Chapter 5	Congruence Based on Triangles
Chapter 6	Transformations and the Coordinate Plane
Chapter 7	Geometric Inequalities
Chapter 8	Slopes and Equations of Lines
Chapter 9	Parallel Lines
Chapter 10	Quadrilaterals
Chapter 11	The Geometry of Three Dimensions
Chapter 12	Ratio, Proportion, and Similarity
Chapter 13	Geometry of the Circle
Chapter 14	Locus and Construction
Index	Index

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Free Geometry Book Download: The Geometry and Topology of Three-Manifolds

The Geometry and Topology of Three-Manifolds
This is an electronic edition of the 1980 lecture notes distributed by Princeton University. It is available in pdf and postscript formats.
Description
    These notes (through p. 9.80) are based on my course at Princeton in 1978–79. Large portions were written by Bill Floyd and Steve Kerckhoff. Chapter 7, by John Milnor, is based on a lecture he gave in my course; the ghostwriter was Steve Kerckhoff. The notes are projected to continue at least through the next academic year. The intent is to describe the very strong connection between geometry and lowdimensional topology in a way which will be useful and accessible (with some effort) to graduate students and mathematicians working in related fields, particularly 3-manifolds and Kleinian groups.

Contents     Geometry and three-manifolds     Geometric structures on manifolds     Hyperbolic Dehn surgery     Flexibility and rigidity of geometric structures     Gromov’s invariant and the volume of a hyperbolic manifold     Computation of volume     Kleinian groups

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Book Details
Author(s): William P. Thurston
Format(s): PDF, PostScript, DVI
Number of pages: 360
Link: Download.

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Problem and Solution: Part 5 (Ratio Exercise problem and solution, Proportion Exercise problem and solution, Rate & Partnership Exercise problem and solution)

Question: 27 Coffee A normally costs $ 100 per 1b. It is mixed with coffee B, which normally cost $ 70 per 1b, to form a mixture that costs $ 88 per 1b. If there are 10 1bs of the mix, how many pounds of coffee A is used in the mixture?
Option: (a) 4 (b) 5 (c) 6 (d) 7 (e) None

Explore: Let, a be the amount of coffee A present in per 1b of the mixture.
:. a x 100 + (1 - a)70 = 88
:. (100 - 70)a + 70 = 88
:. 30a = 18
:. a = 18/30 = 0.6 1b
:. In 10 1b, we have 6 1b of coffee A
Answer: (c)

Question: 28 Two blends of tea costing $ 2.80 and $ 3.20 per kg, respectively, are mixed in the proportion 2:3. The mixture is sold at $ 4.80 per kg, What is the percentage of profit?
Option: (a) 50.7 (b) 50.9 (c) 57.9 (d) 60

Explore: Let, tea of 1st kind and 2nd kind are mixed 2 kg & 3 kg, respectively. :. Total cost of 5 kg mixtrue = 2.8 x 2 + 3.2 x 3 = 5.6 + 9.6 =$ 15.2 Selling price of 5 kg = $ 4.8 x 5 = $ 24.0 :. Profit = $ 24 - $ 15.2 = $ 8.8 :. Percentage of profit = {(Profit)/(Original cost)} x 100% = {(8.8)/(15.2)} x 100% = 57.9 % Answer: (c)

Question: 29 Mr. John won an election where the ratio of his votes and those of his opponent, Mr. yunus was 4:3. The total number of voters was 581, of which 91 did not vote. Calculate the margin of votes by which Mr, Yunus was defeated.

Explore: No. of voters who applied their votes = 581 - 91 = 490
Ratio of votes of Mr. John  & Mr. Yunus = 4:3
Sum of the ratio = 4 + 3 = 7
:. Mr. John got = 490 x (4/7) = 280
Mr. Yunus got = 490 x (3/7) = 210
:. Margin of votes = 280 - 210 = 70
Answer: The margin of votes by which Mr, Yunus was defeated was 70.

Question: 30 A sum of money is distributed among A, B & C. The amount of money received by A & B were in the ratio 1:2. The amount of money received by B & C were in the ratio 3:4. How much money did A receive?
Option: (a) 989 (b) 999 (c) 968 (d) 990 (e) None

Explore: 5 + 7 + 10 = 22. So, the number must he divisible by 22, 968 & 990 are both divisible by 22. But since we are looking for the largest value, 990 is the answer.
Answer: (d)

Question: 31 $1105 was divided between A, B & C. The amount of money received by A & B were in the ratio 1:2. The amount of money received by B & C were in the ratio 3:4. How much money did A receive?
Option: (a) $ 205 (b) $ 195 (c) $ 1500 (d) $ 100 (e) None of these

Explore: A:B = 1:2 = 3:6
B:C = 3:4 = 6:8
:. A:B:C = 3:6:8
:. A got {(3)/(3+6+8)} x 1105 = (3/17) x 1105 = $ 195
Answer: (b)
Ratio Maths Questions - webcrawler.com?Adwww.webcrawler.com/?Search multiple engines for ratio maths questionsMath Research - intelligence.org?Adwww.intelligence.org/?MIRI researches mathematical logic, decision algorithms, and other mathMachine Intelligence Research Institute has 105 followers on Google+Search ResultsRatio word problem exercise example 1 - Khan Academywww.khanacademy.org/math/...math/...ratios...ratio.../ratio-word...?by Salman Khan - in 9,778 Google+ circlesThe given data in the question says the girl to boy ratio is 5 to 8. In other words, one may say that, for every 5 girls there are eight boys in the class. Now imagine ...Ratio word problems | Khan Academyhttps://www.khanacademy.org/exercise/ratio_word_problems?Practice ratio word problems with Khan Academy's free online exercises. ... Ratio word problems. 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A proportion is a set of 2 fractions ...Proportions 1 | Khan Academyhttps://www.khanacademy.org/math/.../ratio-proportion.../proportions _1?Proportions 2 exercise examples · Proportions 2 · Constructing proportions to solve application problems · Constructing proportions to solve application ...Percentage word problems 1 | Khan Academyhttps://www.khanacademy.org/exercise/percentage_word_problems_1?Practice percentage word problems 1 with Khan Academy's free online exercises.Ratio word problems | Khan Academywww.khanacademy.org/math/...proportions/e/ratio_word_problems?Proportion validity example · Solving ratio problems with tables exercise · Solving ratio problems with tables exercise 2 · Solving ratio problems with tables ...Proportions 2 exercise examples | Ratios and ... - Khan Academywww.khanacademy.org/...proportion.../proportions-2-exercise-e...?by Salman Khan - in 9,778 Google+ circlesFind an unknown in a proportion 2 · Proportions 1 · Proportions 2 exercise examples · Proportions 2 · Constructing proportions to solve application problems.Ratio and Proportion - Aptitude Questions and Answerswww.indiabix.com › Aptitude?In this section you can learn and practice Aptitude Questions based on "Ratio and Proportion" and improve your skills in order to face the interview, competitive ...Percent and Proportions - Math Goodieswww.mathgoodies.com/lessons/percent/proportions.html?by Gisele Glosser - in 99 Google+ circlesPercent and proportions are part of our Consumer Math unit by Math ... But what would you do if you given this problem: 8 is what percent of 20? .... Exercises ...Word Problem Exercises: Simple Proportions (5 problems)www.algebralab. org/practice/practice.aspx?...ProportionsSimpleWordPro...?The Ridilla family plan to drive to their vacation destination in 2 days. On day 1 the family traveled 324 miles in 4.5 hours. On day 2, how many hours will it take ...Solving Simple Proportions - Purplemathwww. purplemath. com/modules/ratio4.htm?through worked examples how to solve basic 'proportion' problems. ... Solving proportions is simply a matter of stating the ratios as fractions, setting th[PDF]7 Ratio and Proportionwww.cimt.plymouth.ac.uk/projects/mepres/book8/bk8_7.pdf?Exercises. 1. Write each of these ratios in its simplest form: (a) 2 : 6. (b) 4 : 20. (c) 3 : 15. (d) 6 : 2 ... Direct proportion can be used to carry out calculations like the one below: .... Give answers to the following questions correct to. 1 decimal place.Ratio And Proportion Problems?Adwww.webcrawler.com/?Search multiple engines for ratio and proportion problemsFree Math Worksheets?Adwww. education.com/Worksheets?Print Math Worksheets By Grade Help Kids Develop Essential Skills!Education.com has 219 followers on Google+Free Math Problem?Adwww.curriki.org /?Free Lessons, Tests, Ideas, Plans, Worksheets for Teaching Math.Searches related to proportion exercises problemspercentage practice problemspercentage problems workshee tsproportion worksheets word problemsexamples of proportion word problems examples of direct proportion problemsratio and proportion problems worksheets

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Divisibility Exercise problem and solution

Problem:	Is the number 621 prime or composite?
Method:	In the last lesson, we learned to find all factors of a whole number to determine if it is prime or composite. We used the procedure listed below.
	To determine if a number is prime or composite, follow these steps:
	Find all factors of the number. If the number has only two factors, 1 and itself, then it is prime. If the number has more than two factors, then it is composite.
The above procedure works very well for small numbers. However, it would be time-consuming to find all factors of 621. Thus we need a better method for determining if a large number is prime or composite. Every number has one and itself as a factor. Thus, if we could find one factor of 621, other than 1 and itself, we could prove that 621 is composite. One way to find factors of large numbers quickly is to use tests for divisibility.

Definition	Example
One whole number is divisible by another if, after dividing, the remainder is zero.	18 is divisible by 9 since 18 ÷ 9 = 2 with a remainder of 0.
If one whole number is divisible by another number, then the second number is a factor of the first number.	Since 18 is divisible by 9, 9 is a factor of 18.
A divisibility test is a rule for determining whether one whole number is divisible by another. It is a quick way to find factors of large numbers.	Divisibility Test for 3: if the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

We can test for divisibility by 3 (see table above) to quickly find a factor of 621 other than 1 and itself. The sum of the digits of 621 is 6+2+1 = 9. This divisibility test and the definitions above tell us that... 621 is divisible by 3 since the sum of its digits (9) is divisible by 3. Since 621 is divisible by 3, 3 is a factor of 621. Since the factors of 621 include 1, 3 and 621, we have proven that 621 has more than two factors. Since 621 has more than 2 factors, we have proven that it is composite.

Let's look at some other tests for divisibility and examples of each.

Divisibility Tests	Example
A number is divisible by 2  if the last digit is 0, 2, 4, 6 or 8.	168 is divisible by 2 since the last digit is 8.
A number is divisible by 3  if the sum of the digits is divisible by 3.	168 is divisible by 3 since the sum of the digits is 15 (1+6+8=15), and 15 is divisible by 3.
A number is divisible by 4  if the number formed by the last two digits is divisible by 4.	316 is divisible by 4 since 16 is divisible by 4.
A number is divisible by 5  if the last digit is either 0 or 5.	195 is divisible by 5 since the last digit is 5.
A number is divisible by 6  if it is divisible by 2 AND it is divisible by 3.	168 is divisible by 6 since it is divisible by 2 AND it is divisible by 3.
A number is divisible by 8  if the number formed by the last three digits is divisible by 8.	7,120 is divisible by 8 since 120 is divisible by 8.
A number is divisible by 9  if the sum of the digits is divisible by 9.	549 is divisible by 9 since the sum of the digits is 18 (5+4+9=18), and 18 is divisible by 9.
A number is divisible by 10  if the last digit is 0.	1,470 is divisible by 10 since the last digit is 0.

Let's look at some examples in which we test the divisibility of a single whole number.

Example 1:	Determine whether 150 is divisible by 2, 3, 4, 5, 6, 9 and 10.
	150 is divisible by 2 since the last digit is 0.
	150 is divisible by 3 since the sum of the digits is 6 (1+5+0 = 6), and 6 is divisible by 3.
	150 is not divisible by 4 since 50 is not divisible by 4.
	150 is divisible by 5 since the last digit is 0.
	150 is divisible by 6 since it is divisible by 2 AND by 3.
	150 is not divisible by 9 since the sum of the digits is 6, and 6 is not divisible by 9.
	150 is divisible by 10 since the last digit is 0.
Solution:	150 is divisible by 2, 3, 5, 6, and 10.

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Example 2:	Determine whether 225 is divisible by 2, 3, 4, 5, 6, 9 and 10.
	225 is not divisible by 2 since the last digit is not 0, 2, 4, 6 or 8.
	225 is divisible by 3 since the sum of the digits is 9, and 9 is divisible by 3.
	225 is not divisible by 4 since 25 is not divisible by 4.
	225 is divisible by 5 since the last digit is 5.
	225 is not divisible by 6 since it is not divisible by both 2 and 3.
	225 is divisible by 9 since the sum of the digits is 9, and 9 is divisible by 9.
	225 is not divisible by 10 since the last digit is not 0.
Solution:	225 is divisible by 3, 5 and 9.

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Example 3:	Determine whether 7,168 is divisible by 2, 3, 4, 5, 6, 8, 9 and 10.
	7,168 is divisible by 2 since the last digit is 8.
	7,168 is not divisible by 3 since the sum of the digits is 22, and 22 is not divisible by 3.
	7,168 is divisible by 4 since 168 is divisible by 4.
	7,168 is not divisible by 5 since the last digit is not 0 or 5.
	7,168 is not divisible by 6 since it is not divisible by both 2 and 3.
	7,168 is divisible by 8 since the last 3 digits are 168, and 168 is divisible by 8.
	7,168 is not divisible by 9 since the sum of the digits is 22, and 22 is not divisible by 9.
	7,168 is not divisible by 10 since the last digit is not 0 or 5.
Solution:	7,168 is divisible by 2, 4 and 8.

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Example 4:	Determine whether 9,042 is divisible by 2, 3, 4, 5, 6, 8, 9 and 10.
	9,042 is divisible by 2 since the last digit is 2.
	9,042 is divisible by 3 since the sum of the digits is 15, and 15 is divisible by 3.
	9,042 is not divisible by 4 since 42 is not divisible by 4.
	9,042 is not divisible by 5 since the last digit is not 0 or 5.
	9,042 is divisible by 6 since it is divisible by both 2 and 3.
	9,042 is not divisible by 8 since the last 3 digits are 042, and 42 is not divisible by 8.
	9,042 is not divisible by 9 since the sum of the digits is 15, and 15 is not divisible by 9.
	9,042 is not divisible by 10 since the last digit is not 0 or 5.
Solution:	9,042 is divisible by 2, 3 and 6.

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Example 5:	Determine whether 35,120 is divisible by 2, 3, 4, 5, 6, 8, 9 and 10.
	35,120 is divisible by 2 since the last digit is 0.
	35,120 is not divisible by 3 since the sum of the digits is 11, and 11 is not divisible by 3.
	35,120 is divisible by 4 since 20 is divisible by 4.
	35,120 is divisible by 5 since the last digit is 0.
	35,120 is not divisible by 6 since it is not divisible by both 2 and 3.
	35,120 is divisible by 8 since the last 3 digits are 120, and 120 is divisible by 8.
	35,120 is not divisible by 9 since the sum of the digits is 11, and 11 is not divisible by 9.
	35,120 is divisible by 10 since the last digit is 0.
Solution:	35,120 is divisible by 2, 4, 5, 8 and 10.

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Example 6:	Is the number 91 prime or composite? Use divisibility when possible to find your answer.
	91 is not divisible by 2 since the last digit is not 0, 2, 4, 6 or 8.
	91 is not divisible by 3 since the sum of the digits (9+1=10) is not divisible by 3.
	91 is not evenly divisible by 4 (remainder is 3).
	91 is not divisible by 5 since the last digit is not 0 or 5.
	91 is not divisible by 6 since it is not divisible by both 2 and 3.
	91 divided by 7 is 13.
Solution:	The number 91 is divisible by 1, 7, 13 and 91. Therefore 91 is composite since it has more than two factors.

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Problem and Solution: Part 4 (Ratio Exercise problem and solution, Proportion Exercise problem and solution, Rate & Partnership Exercise problem and solution)

Question: 22 The ratio of gold and silver in an ornament weighing 42 gm is 4:3. How much gold will need to be added for the ratio of gold and silver to be 5:3?
Explore: The ratio of gold & silver = 4:3
Sum of the ratio = 4+3 = 7
The amount of gold = 42 gm x 4/7 = 24 gm
The amount of silver = 42 gm x 3/7 = 18 gm
Let, X gm gold will need to be added.
According to the question,
(24 +X ): 18 = 5:3
=> (24+X)/18 = 5/3
=> 72 + 3X = 90
=> 3X = 90 - 72 = 18
=> X = 18/3 = 6
Answer: 6 gm gold will need to be added

Question: 23 The ratio of boys and girls in a class is 1:2 and the classroom has 24 students. How many boys would have to be admitted to make ratio of boys to girls 1:1?

Option: (a) 6 (b) 8 (c) 10 (d) 12 (e) 14 Explore: Total no. of students = 24 Boys: Girls = 1:2 Sum of the ratio = 1+2 = 3 No. of boys = 24 x 1/3 = 8 No. of girls = 24 x 2/3 = 16 Let, X boys have to be admitted. According to the question, (8+X): 16 = 1:1 => (8+X) / 16 = 1/1 =1 => 8+X = 16 => X = 16 - 8 => X = 8 Answer: (b)

Question: 24 A jar contains white, red & green marbles in the ratio 2:3:4. Five more green marbles are added to make the ratio 2:3:5. How many white marbles are there in the jar?
Option: (a) 4 (b) 6 (c) 8 (d) 10 (e) None

Explore: Initially, white: green = 2:4
Let, G be the number of green marbles.
So, (White)/(Green) = 2/4 => (White)/G = 2/4
:. White = G/2
So, after 5 green marbles are added,
(White)/(Green) = (G/2)/(G + 5)
According to the problem,
(G/2)/(G + 5)
According to the problem,
(G/2)/(G + 5) = 2/5
=> (G)/(G+5) = 4/5 => 5G = 4G + 20
=> G = 20
:. No. of white marbles = 10
Answer: (d)

Question: 25 In an MBA class the ratio of number of commerce graduates to the number of science graduates is 2 to 5. If 2 more commerce graduates enter the class the ratio become 1 to 2. How many commerce graduates are in the class?

Explore: Let, there are 2X commerce graduates in the class and 5X science graduates.
According to the question,
(2X +2): 5X = 1:2
=> (2X+2)/5X = 1/2
=> 4X + 4 = 5X
=> 5X - 4X = 4
=> X = 4
Answer: No. of commerce graduates is 2 x 4 = 8

Question: 26 2 partners X & Y have 60% & 40% shares in business. After sometime a 3rd partner Z joined the business by investing $ 5 Lack & thus having 20% of the share in the business. What is Y's share now in the business?
Option: (a) 32% (b) 48% (c) 36% (d) 50% (e) None

Explore: Let, $X,Y,Z be X's, Y's & Z's share.
:. X/Y = 60/40 = 3/2
:. X = Y(3/2)
After Z joined the business we have
(X+Y)/Z = 80/20 [Z holds 20%, X & Y 80%]
:. ((5/2)Y)/500000 = 4 => (5/2)Y = 20,00,000
:. Y = 8,00,000
:. X = (3/2)Y = (3/2) x 8,00,000 = 12,00,000
:. X:Y:Z = 12,00,000 : 8,00,000 : 5,00,000
= 12 : 8 : 5
:. Y's share = 8/(12+8+5) x 100%
= 8/25 x 100%
= 32%
Answer: (a)

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Ratio word problem exercise example 1 · Ratio word problem ...Ratio Worksheets | Ratio Worksheets for Teachers - Math Aidswww.math-aids.com/Ratios/‎These Ratio Worksheets are perfect for teachers, homeschoolers, moms, dads, and children looking for some practice in Ratio problems.Ratios and Rates Word ... - ‎Equivalent Ratio Worksheets - ‎Simple Ratio WorksheetsRatios - Maths Tutorww.mathtutor.ac.uk/arithmetic/ratios/exercise‎Ratios: Exercises. 1/6. 1. Attempt the following questions. Express the following ratios in their simplest form: 1. 2 to 10. 80 to 20. to 1. 3. 50p : £3.50. 6m : 30cm.Ratios, Proportions - Purplemath www.purplemath.com/modules/ratio.htm‎Explains the basic terminology and formatting of ratios, and demonstrates how to solve typical exercises. ... Return to the Purplemath home page.[PDF]Ratios - Mathcentrewww.mathcentre.ac.uk/resources/uploaded/mc-ty-ratios-2009-1.pdf‎xercises so that they become second nature. ... www.mathcentre.ac.uk. 1 ... Exercises. 1. Express these ratios in their simplest form: (a) 2 to 10. (b) 80 to 20 (c) 1.Math Exerciseswww.emathematics.net/‎Revise your understanding of proportional reasoning in maths. Here you will find exercises on ratios and proportions, direct proportions, inverse proportions, ...Ratios and Proportions Exercises - Mathematics - About.commath.about.com/od/Ratios-Proportions-Exercises/‎Exercises for Ratios and Proportions. Exercises involving Ratios and Proportions. Proportions Word Problems Worksheet 2. A proportion is a set of 2 fractions ...Ratios and Proportions Math Worksheets! - edHelper.comedhelper.com/ ratios.htm‎Proportions: State whether the ratios are proportional (fractions) · Proportions: State whether the ratios are proportional (mixed ways of writing ratios)BBC - GCSE Bitesize: Ratioswww.bbc.co.uk › Home › Maths › Number‎A secondary school revision resource for GCSE Maths about foundation level fractions, decimals and ratios.Searches related to ratio exercises mathhow to get ratiosmaths ratio and proportion exercisesratio worksheetsequivalent ratios worksheetratio and proportion worksheetratio problems worksheetratio sheetssimple ratio worksheetsRatios and Proportions Exercises - Mathematics - About.commath.about.com/od/Ratios-Proportions-Exercises/‎Exercises for Ratios and Proportions. Exercises involving Ratios and Proportions. Proportions Word Problems Worksheet 2. A proportion is a set of 2 fractions ...Proportions 1 | Khan Academyhttps://www.khanacademy.org/math/.../ratio-proportion.../proportions_1‎Proportions 2 exercise examples · Proportions 2 · Constructing proportions to solve application problems · Constructing proportions to solve application ...Percentage word problems 1 | Khan Academyhttps://www.khanacademy.org/exercise/percentage_word_problem _1‎Practice percentage word problems 1 with Khan Academy's free online exercises.Ratio word problems | Khan Academywww.khanacademy.org/math/...proportions/e/ratio_word_problems‎Proportion validity example · Solving ratio problems with tables exercise · Solving ratio problems with tables exercise 2 · Solving ratio problems with tables ...Proportions 2 exercise examples | Ratios and ... - Khan Academy www.khanacademy.org/ ...proportion.../proportions-2-exercise-e...‎by Salman Khan - in 9,778 Google+ circlesFind an unknown in a proportion 2 · Proportions 1 · Proportions 2 exercise examples · Proportions 2 · Constructing proportions to solve application problems.Ratio and Proportion - Aptitude Questions and Answerswww.indiabix.com › Aptitude‎In this section you can learn and practice Aptitude Questions based on "Ratio and Proportion" and improve your skills in order to face the interview, competitive ...Percent and Proportions - Math Goodieswww.mathgoodies.com/lessons/percent/proportions.html‎by Gisele Glosser - in 99 Google+ circlesPercent and proportions are part of our Consumer Math unit by Math ... But what would you do if you given this problem: 8 is what percent of 20? .... Exercises ...Word Problem Exercises: Simple Proportions (5 problems) www.algebralab.org/ practice/practice.aspx?...ProportionsSimpleWordPro...‎The Ridilla family plan to drive to their vacation destination in 2 days. On day 1 the family traveled 324 miles in 4.5 hours. On day 2, how many hours will it take ...Solving Simple Proportions - Purplemath www.purplemath.com/ modules/ratio4.htm‎through worked examples how to solve basic 'proportion' problems. ... Solving proportions is simply a matter of stating the ratios as fractions, setting th[PDF]7 Ratio and Proportionwww.cimt.plymouth.ac.uk/projects/mepres/book8/bk8_7.pdf‎Exercises. 1. Write each of these ratios in its simplest form: (a) 2 : 6. (b) 4 : 20. (c) 3 : 15. (d) 6 : 2 ... Direct proportion can be used to carry out calculations like the one below: .... Give answers to the following questions correct to. 1 decimal place.Ratio And Proportion Problems‎Adwww.webcrawler.com/‎Search multiple engines for ratio and proportion problemsFree Math Worksheets‎Adwww .education. com/Worksheets‎Print Math Worksheets By Grade Help Kids Develop Essential Skills!Education.com has 219 followers on Google+Free Math Problem‎Adwww.curriki.org/‎Free Lessons, Tests, Ideas, Plans, Worksheets for Teaching Math.Searches related to proportion exercises problemspercentage practice problemspercentage problems worksheets proportion worksheets word problemsexamples of proportion word problems examples of direct proportion problemsratio and proportion problems worksheets

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Problem and Solution: Part 3 (Ratio, Proportion, Rate & Partnership)

Question: 17 If Marie has twice as much money as Curie has, who has three times as much money as Sunny has. What is the ratio of the amount of money Sunny has to the amount of money Marie has?
 
Option: (a) 1:8 (b) 1:6 (c) 1:4 (d) 1:2 (e) 2:1

Explore : Let, Sunny hat $ X
:. Curie has $ 3X
:. Marie has $ (3X x 2) or $ 6X
:. The ratio of amount of money Sunny has to the amount of money Marie has
= X:6X = X/6X = 1/6 = 1:6
Answer: (b)

Question: 18Masum has twice as much money as Selim and Selim has 50% more money than what Babal has. If the average money with them is $ 110, then determine the amount of Masum's Money.

Explore: Let, Babal has $ X :. Selim has $ (X + X x 50%) = $ (X x 0.5%X) = $1.5X :. Masum has $ (1.5X x 2) = $3X According to the question,      X + 1.5X + 3X = 110 x 3 => 5.5X = 330 => X = 3.30/5.5 = 60 :. Masum has $ (3 x 60) = $ 180 Answer: $ 180

Question: 19 The cost of a book is 1.5 times that of a CD. The price of a pencil is 1/3 rd of that of a pen, The cost of the pen is twice as much as that of the CD. What is the ratio of cost of the book to cost of the pencil?
Explore: Let, the cost of CD =X
:. Cost of a book = 1.5X
Cost of a pen = 2X
?Cost of a pencil = 2X x 1/3 = 2X/3
:. The ratio of cost of the book to the cost of the pencil
= 1.5X :2X/3 = 1.5X/(2X/3) = 1.5X x 3/2X
= 4.5/2 = 9/4 = 9:4
Answer: 9 : 4

Question: 20 A room has a floor area of 150sq. feet. If the length & breath of the room are in the ratio 3:2, then what is the length of the floor in feet?
Option: (a) 10 (b) 15 (c) 20 (d) 9 (e) None

Explore: Let, L be the length of the room
According to the problem,
length/ width = 3/2
Or, L/width = 3/2 => width 2L/3
:. L x 2L/3 = 150
=> LxL = (3 x 150)/2 = 3 x 3 x 25
:. L 3 x 5 = 15
Answer: (b)

Question: 21 A gentleman spends 2/5th of his monthly income on food, 1/4th on education and the rest $ 200 is saved. What is his monthly income?
Option: (a) 2000 (b) 1500 (c) 1200 (d) 1000 (e) None

Explore: Let, X be his monthly income.
:. (2/5)X  + (1/4)X + 200 = X
:. 200 = X - (2/5)X - (1/4)X
= X (1-2/5 -1/4)
= X((20-8-5)/20) = X(7/.20)
:. X = (200 x 20)/7 571.43
Answer: (e)

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Ratio, Proportion & Partnership: (part 2 )Exercise Problem & solved examples

Question: 10 Given (P + 7Q)/4P = 19/20, what is the ratio of Q:P?
Option: (a) 1:2 (b) 1:3 (c) 2:4 (d) 2:5 (e) 2:7

Explore: (P + 7Q)/4P = 19/20
=> 20P + 140Q = 76P
=> 140Q = 56P
=> P/Q = 140/56 = 5/2
=> P:Q = 5:2
:. Q:P = 2:5

Answer: (d)

Question: 11 Nine years ago the age of P and Q| were in the ratio of 2:3. After 7 years, the ratio of their age will be 3:4. What is the present age of P?

Explore: Let 9 years ago the age of P and Q were 2x & 3x respectively.
:. Present age of P = 2x + 9 
Present age of Q = 3x + 9
According to the question,
(2x +9+7):(3x+9+7) = 3:4
=> (2x+16)/(3x+16) = 3/4
=> 4(2x+16) = 3(3x +16)
=> 8x + 64 = 9x +48
=> 9x-8x = 64 - 48
=> x = 16
:. Present age of P = 2 x 16 + 9 = 41
Answer: 41

Question: 12 Age of three persons are now in the proportion 2:3:4 and in 5 years from now, the proportion will be 5:7:8. What is the present age of the youngest person?
Option: (a) 30 (b) 25 (c) 20 (d) 15 (e) None

Explore: The ratio of the age of the three persons is 2:3:4. So, their age can be 2X, 3X, 4X, where X is any number.
After 5 years their age ratio is given by 5:7:8. So, their age may be 5Y, 7Y & 8Y where Y is any other number.
So, we can write
2X+5 = 5Y .......(1)
3X+5 = 7Y .......(2)

(2) - (1) gives X = 2Y
Or, Y = 1/2
Replacing this value in (1)
2X +5 = (5/2)X
Or, 5 = ((5/2) - 2)X = (1/2)X
:. X = 10
:. Required age 2X = 2x10 = 20
Answer: (c)

Question: 13 A fruit salad mixture consists of apples, peaches and grapes in the ratio 6:5:2 respectively, by weights. If 39 pounds of the mixture is prepared, the mixture includes how many more pounds of apples than grapes?

Explore: In the mixture of 39 pounds,
ratio of apples, peaches & grapes = 6:5:2
:. Weight of apples = 39 x (6/(6+5+2)) = 39 x (6/13) = 18 Pounds
Weight of peaches = 39 x (5/13) = 15 Pounds
Weight of grapes = 39 x (2/13) = 6 Pounds
:. Weight of apples is (18 - 6) or 12 pounds more than grapes.
Answer: 12 pounds

Question: 14 Jahirul and Jalil agree to form a partnership. The partnership agreement requires that Jahirul invest $7,000 less than one half of what Jalil invests. If the total investment is $ 1,25,000 how much does Jalil invest?

Explore: Let, Jalil invest $X
:. Jahirul invests $ (X/2) - 7000
According to the question,
X+(X/2) - 7000 = 125000
=> (2X+X)/2 = 125000+7000
=> 3X/2 = 132000
:. X = (132000 x 2)/3
=> X = 88000
Answer: Jalil invests $ 88000

Question: 15 Arif, Babu and Salm started a business jointly with a total amount of $28000. Arif paid $ 4500 more than Babu and Babu paid $7000 less than Salam. If the company made a profit of $5600, how much profit should Babu receive?

Explore: Let Babu paid $X
:. Arif paid $(X+4500)
Salam paid $(X+7000)
According to the question,
X+X+4500+X+7000 = 2800
=> 3X + 11500 = 28000
=> 3X = 28000 - 11500
=> X = (16500/3) = 5500
:. Babu paid $ 5500
Arif paid $ (5500+4500) = $10000
Salam paid $ (5500 + 7000) = $12500
They will share profit according to the money they paid.
:. Ratio of money paid by Babu, Arif and Salam
= 5500:10000:12500
= 55:100:125
= 11:20:25
Some of the ratio = 11+20+25 = 56
:. Babu received from the profit of $5600
= $5600 x (11/56)
= $1100
Answer: $1100

Question: 16 Arif, Babu and Salam started a business jointly with a total amount of $280. Arif paid $45 more than Babu and Babu paid $70 less than Salam. If the company made a profit of $56, how much profit should Babu receive?
Option: (a) 22 (b) 20 (c) 25 (d) 27 (e) None

Explore: Let, Babu's investment be $B
:. Arif's investment is $(B+45)
And Salam's investment $ (B+70)
According to the problem,
b+(b+45)+(b+70) = 280
=> 3B = 280 -115 = 165
:. B = 55
Since the profit is shared in proportion of capital investment, so,
Babu's profit = ((Babu's capital)/(Total capital)) x Total profit = ((55)/(250)) x 56 = $11
Answer: (e)

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Brain Waves Game

Instructions for Brain Waves

Challenge your brain with 7 fun mini games. Read the instructions before every mini game.
Wait for fully loaded, then click continue and then click continue and play.

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Ratio, Proportion & Partnership: Exercise Problem & solved examples

Question: 1 The ratio of 1/5 to 2/7 is:
Option: (a) 3:5 (b) 5:7 (c) 7:9 (d) 7:10 (e) 3:7

Explore: The ratio of 1/6 to 2/7
= 1/5:2/7
= (1/5)/()2/7
= 1/5 x 7/2
= 7/10
= 7: 10
Answer: (d)

Question: 2 The ratio of 1/4 to 3/5 is
Option: (a) 1 to 3 (b) 3 to 20 (c) 5 to 12 (d) 3 to 4 (e) 5 to 5

Explore: 1/4 to 3/5 = 1/4:3/5 = (1/4)/(3/5) = 5/12
Answer: (c)

Question: 3 A:B = 4:5, A:C = 10:9, then A:B:C = ?
Option: (a) 4:5:9 (b) 4:5:10 (c) 8:9:10 (d) 8:9:10 (e) 20:25:18

Explore: A:B = 4:5
=> B:A = 5:4
=> B:A = 5:5 x 4:5 = 25:20
A:C = 10:9
=> A:C = 10x2:9x2 = 20:18
:. B:A:C = 25:20:18
:. A:B:C = 20:25:18
Answer: (e)

Question: 4 If A:B = 1:2, B:C = 4:3, & A+B+C = 450. What is the value of B?
Explore: A:B = 1:2 = 1x2:2x2 = 2:4
B:C = 4:3
:.  A:B:C = 2:4:3
Sum of the numbers = 2+4+3 = 9
:. Value of B = 450x(4/9) = 200
Answer: 200

Question: 5 Let, X:Y = 3:4 and X:Z = 6:5, then Z:Y =?
Option: (a) 5:3 (b) 6:7 (c) 4:2 (d) 5:4 (e) 5:8

Explore: X:Y = 3:4
=> Y:X = 4:3 => Y:X = 4x2:2x3 = 8:6
Again, X:Z = 6:5
:. Y:X:Z =8:6:5
:. Z:Y = 5:8
Answer: (e)

Question: 6 If X:Y = Y:Z = 1.5 and Z = 2, what is the value of X?
Option: (a) 3 (b) 4 (c) 4.5 (d) 2.5 (e) 5

Explore: Y:Z = 1.5 or Y/Z = 1.5
Or, Y = 1.5Z = 1.5x2 = 3
Now, Z:Y = 1.5
:. X = 1.5Y = (3/2)3 = 9/2 = 4.5
Answer: (c)

Question: 7 If X is 2/5 of Y and Y is 5/7 of Z, what is the ratio of Z:X?
Explore: According to the question,
X = 2Y/5 & Y = 5Z/7
=> X = 2/5 x 5Z/7
=> X = 2Z/7
=>Z/X = 7/2 = 7:2
Answer: 7:2

Question: 8 In a school, the ratio of boys to girls is 3 to 7. If there are 150 boys and girls in the school, how many boys are there?
Option: (a) 45 (b) 75 (c) 90 (d) 105

Explore: The ratio of boys to girls = 3:7
Sum of the ratio = 3 + 7 = 10
No. of boys = 150 x 3/10 = 45
Answer: (45)

Question: 9 The ratio of girls to boys in a class is 3:4. The number of students in the class could be any of the following except:
Option: (a) 35 (b) 28 (c)48 (d) 42

Explore: Since ratio of girls to boys is 3:4 and no. of student can't be fraction and must be divisible by 7(3+4), so, 35, 28 & 42 are divisible by 7.
But 48 is not divisible by 7.

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