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Ratio and Proportion Word Problems — Examples & Practice - Expii

WORD PROBLEMS ON RATIO AND PROPORTION

The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy.

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x + 5x + 7x)/3 = 25

Age of the first boy = 3x

Age of the first boy = 5x

Age of the first boy = 7x

So, the age of the youngest boy is 15 years.

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7 : 6, find his new weight.

Given : Original weight of John = 56.7 kg. He is going to reduce his weight in the ratio 7:6.

We can use the following hint to find his new weight, after it is reduced in the ratio 7 : 6.

problem solving ratio and proportion

His new weight is

= (6 ⋅ 56.7)/7

So, John's new weight is 48.6 kg.

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2 : 3.

Sum of the terms in the given ratio is

So, no. of boys in the school is

= 720 ⋅ (3/8)

No. of girls in the school is

= 720 ⋅ (5/8)

Given : Number of new girls admitted in the school is 18.

Let x be the no. of new boys admitted in the school.

After the above new admissions,

No. of boys in the school = 270 + x

No. of girls in the school = 450 + 18 = 468

Given : The ratio after the new admission is 2 : 3.

(270 + x) : 468 = 2 : 3

Use cross product rule.

3(270 + x) = 468 ⋅ 2

So, the number of new boys admitted in the school is 42.

The monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenditures are in the ratio 7 : 9. If each saves $50 per month, find the monthly income of the second person.

From the given ratio of incomes ( 4 : 5 ),

Income of the 1st person = 4x

Income of the 2nd person = 5x

(Expenditure = Income - Savings)

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

(4x - 50) : (5x - 50) = 7 : 9

9(4x - 50) = 7(5x - 50)

36x - 450 = 35x - 350

Then, the income of the second person is

So, income of the second person is $500.

The ratio of the prices of two houses was 16 : 23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11 : 20. Find the original price of the first house.

From the given ratio 16 : 23,

Original price of the 1st house = 16x

Original price of the 2nd house = 23x

After increment in prices,

Price of the 1st house = 16x + 10% of 16x

Price of the 2nd house = 23x + 477

After increment in prices, the ratio of prices becomes 11:20.

17.6x : (23x + 477) = 11 : 20

20(17.6x) = 11(23x + 477)

352x = 253x + 5247

Then, original price of the first house is

So, original price of the first house is $848.

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Worked out Problems on Ratio and Proportion

Worked out problems on ratio and proportion are explained here in detailed description using step-by-step procedure. Solved examples involving different questions related to comparison of ratios in ascending order or descending order, simplification of ratios and also word problems on ratio proportion. Sample questions and answers are given below in the worked out problems on ratio and proportion to get the basic concepts of solving ratio proportion.

1. Arrange the following ratios in descending order.

2 : 3, 3 : 4, 5 : 6, 1 : 5 Solution: Given ratios are 2/3, 3/4, 5/6, 1/5 The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60

Now, 2/3 = (2 × 20)/(3 × 20) = 40/60 3/4 = (3 × 15)/(4 × 15) = 45/60 5/6 = (5 × 10)/(6 × 10) = 50/60 1/5 = (1 × 12)/(5 × 12) = 12/60 Clearly, 50/60 > 45/60 > 40/60 > 12/60 Therefore, 5/6 > 3/4 > 2/3 > 1/5 So, 5 : 6 > 3 : 4 > 2 : 3 > 1 : 5

2. Two numbers are in the ratio 3 : 4. If the sum of numbers is 63, find the numbers. Solution: Sum of the terms of the ratio = 3 + 4 = 7 Sum of numbers = 63 Therefore, first number = 3/7 × 63 = 27 Second number = 4/7 × 63 = 36 Therefore, the two numbers are 27 and 36.

3. If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y) Solution: x : y = 1 : 2 means x/y = 1/2 Now, (2x + 3y) : (x + 4y) = (2x + 3y)/(x + 4y) [Divide numerator and denominator by y.] = [(2x + 3y)/y]/[(x + 4y)/2] = [2(x/y) + 3]/[(x/y) + 4], put x/y = 1/2 We get = [2 (1/2) + 3)/(1/2 + 4) = (1 + 3)/[(1 + 8)/2] = 4/(9/2) = 4/1 × 2/9 = 8/9 Therefore the value of (2x + 3y) : (x + 4y) = 8 : 9

More solved problems on ratio and proportion are explained here with full description.

4. A bag contains $510 in the form of 50 p, 25 p and 20 p coins in the ratio 2 : 3 : 4. Find the number of coins of each type.

Solution: Let the number of 50 p, 25 p and 20 p coins be 2x, 3x and 4x. Then 2x × 50/100 + 3x × 25/100 + 4x × 20/100 = 510 x/1 + 3x/4 + 4x/5 = 510 (20x + 15x + 16x)/20 = 510 ⇒ 51x/20 = 510 x = (510 × 20)/51 x = 200 2x = 2 × 200 = 400 3x = 3 × 200 = 600 4x = 4 × 200 = 800. Therefore, number of 50 p coins, 25 p coins and 20 p coins are 400, 600, 800 respectively.

5. If 2A = 3B = 4C, find A : B : C Solution: Let 2A = 3B = 4C = x So, A = x/2 B = x/3 C = x/4 The L.C.M of 2, 3 and 4 is 12 Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12 = 6x : 4x : 3x = 6 : 4 : 3 Therefore, A : B : C = 6 : 4 : 3

6. What must be added to each term of the ratio 2 : 3, so that it may become equal to 4 : 5? Solution: Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5 ⇒ (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 2

7. The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now? Solution: Length of ribbon originally = 30 cm Let the original length be 5x and reduced length be 3x. But 5x = 30 cm x = 30/5 cm = 6 cm Therefore, reduced length = 3 cm = 3 × 6 cm = 18 cm

More worked out problems on ratio and proportion are explained here step-by-step. 8. Mother divided the money among Ron, Sam and Maria in the ratio 2 : 3 : 5. If Maria got $150, find the total amount and the money received by Ron and Sam. Solution: Let the money received by Ron, Sam and Maria be 2x, 3x, 5x respectively. Given that Maria has got $ 150. Therefore, 5x = 150 or, x = 150/5 or, x = 30 So, Ron got = 2x = $ 2 × 30 = $60 Sam got = 3x = 3 × 60 = $90

Therefore, the total amount $(60 + 90 + 150) = $300

9. Divide $370 into three parts such that second part is 1/4 of the third part and the ratio between the first and the third part is 3 : 5. Find each part. Solution: Let the first and the third parts be 3x and 5x. Second part = 1/4 of third part. = (1/4) × 5x = 5x/4 Therefore, 3x + (5x/4) + 5x = 370 (12x + 5x + 20x)/4 = 370 37x/4 = 370 x = (370 × 4)/37 x = 10 × 4 x = 40 Therefore, first part = 3x = 3 × 40 = $120 Second part = 5x/4 = 5 × 40/4 = $50 Third part = 5x = 5 × 40 = $ 200

10. The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term. Solution: Let the fourth term be x. Thus 42, 36, 35, x are in proportion. Product of extreme terms = 42 ×x Product of mean terms = 36 X 35 Since, the numbers make up a proportion Therefore, 42 × x = 36 × 35 or, x = (36 × 35)/42 or, x = 30 Therefore, the fourth term of the proportion is 30.

More worked out problems on ratio and proportion using step-by-step explanation. 11. Set up all possible proportions from the numbers 8, 12, 20, 30. Solution: We note that 8 × 30 = 240 and 12 × 20 = 240 Thus, 8 × 30 = 12 × 20 ………..(I) Hence, 8 : 12 = 20 : 30 ……….. (i) We also note that, 8 × 30 = 20 × 12 Hence, 8 : 20 = 12 : 30 ……….. (ii) (I) can also be written as 12 × 20 = 8 × 30 Hence, 12 : 8 = 30 : 20 ……….. (iii) Last (I) can also be written as 12 : 30 = 8 : 20 ……….. (iv) Thus, the required proportions are 8 : 12 = 20 : 30 8 : 20 = 12 : 30 12 : 8 = 30 : 20 12 : 30 = 8 : 20

12. The ratio of number of boys and girls is 4 : 3. If there are 18 girls in a class, find the number of boys in the class and the total number of students in the class. Solution: Number of girls in the class = 18 Ratio of boys and girls = 4 : 3 According to the question, Boys/Girls = 4/5 Boys/18 = 4/5 Boys = (4 × 18)/3 = 24 Therefore, total number of students = 24 + 18 = 42.

13. Find the third proportional of 16 and 20. Solution: Let the third proportional of 16 and 20 be x. Then 16, 20, x are in proportion. This means 16 : 20 = 20 : x So, 16 × x = 20 × 20 x = (20 × 20)/16 = 25 Therefore, the third proportional of 16 and 20 is 25.

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Ratio and Proportion Questions & Word Problems | GMAT GRE Maths

Whether you are using units from the Metric system (as we do in this post) or US measurement system (the GMAT being an American test), the concepts don’t change.

Introduction to Ratios

Ratio is the quantitative relation between two amounts showing the number of times one value contains or is contained within the other.

( Reference : Oxford dictionary )

Notation : Ratio of two values a and b is written as a:b or a/b or a to b.

For instance, the ratio of number of boys in a class to the number of girls is 2:3. Here, 2 and 3 are not taken as the exact count of the students but a multiple of them, which means the number of boys can be 2 or 4 or 6…etc and the number of girls is 3 or 6 or 9… etc. It also means that in every five students, there are two boys and three girls.

Question : In a certain room, there are 28 women and 21 men. What is the ratio of men to women? What is the ratio of women to the total number of people?

Men : women = 21 : 28 = 3:4

Women : total number of people = 28 : 49 = 4 : 7

Question : In a group, the ratio of doctors to lawyers is 5:4. If the total number of people in the group is 72, what is the number of lawyers in the group?

Let the number of doctors be 5x and the number of lawyers be 4x.

Then 5x+4x = 72 → x=8.

So the number of lawyers in the group is 4*8 = 32.

Question : In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the ratio 4:7:3:1. If the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks, what is the number of ‘B’ blocks?

Let the number of the blocks A,B,C,D be 4x, 7x, 3x and 1x respectively

4x = 3x + 50 → x = 50.

So the number of ‘B’ blocks is 7*50 = 350.

If the ratio of chocolates to ice-cream cones in a box is 5:8 and the number of chocolates is 30, find the number of ice-cream cones.

Let the number of chocolates be 5x and the number of ice-cream cones be 8x.

5x = 30 → x = 6.

Therefore, number of ice-cream cones in the box = 8*6 = 48.

Introduction to Proportion

A lot of questions on ratio are solved by using proportion.

Definition & Notation

A proportion is a comparison of two ratios. If a : b = c : d, then a, b, c, d are said to be in proportion and written as a:b :: c:d or a/b = c/d.

a, d are called the extremes and b, c are called the means.

For a proportion a:b = c:d, product of means = product of extremes → b*c = a*d.

Let us take a look at some examples:

In a mixture of 45 litres, the ratio of sugar solution to salt solution is 1:2. What is the amount of sugar solution to be added if the ratio has to be 2:1?

Number of litres of sugar solution in the mixture = (1/(1+2)) *45 = 15 litres.

So, 45-15 = 30 litres of salt solution is present in it.

Let the quantity of sugar solution to be added be x litres.

Setting up the proportion,

sugar solution / salt solution = (15+x)/30 = 2/1 → x = 45.

Therefore, 45 litres of sugar solution has to be added to bring it to the ratio 2:1.

A certain recipe calls for 3kgs of sugar for every 6 kgs of flour. If 60kgs of this sweet has to be prepared, how much sugar is required?

Let the quantity of sugar required be x kgs.

3 kgs of sugar added to 6 kgs of flour constitutes a total of 9 kgs of sweet.

3 kgs of sugar is present in 9 kgs of sweet. We need to find the quantity of sugar required for 60 kgs of sweet. So the proportion looks like this.

3/9 = x/60 → x=20.

Therefore, 20 kgs of sugar is required for 60 kgs of sweet.

Problems on Mixtures / Blends

If a 60 ml of water contains 12% of chlorine, how much water must be added in order to create a 8% chlorine solution?

Let x ml of chlorine be present in water.

Then, 12/100 = x/60 → x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, 8/100 = 7.2/y → y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90-60 = 30 ml of water.

There is a 20 litres of a solution which has 20% of bleach. Extra bleach is added to it to make it to 50% bleach solution. How much water has to be added further to bring it back to 20% bleach solution?

This question has 3 parts.

In the first part, there is 20% of bleach in 20 L of solution → 4 L of bleach in 16 L of water = 20 L of solution. Let’s note the details in a table for better clarity and understanding.

In the second part, Extra bleach is added to bring it to 50% of total solution. Let the amount of bleach added be x litres.

Then, (4+x)/(20+x) = 50/100 → x = 12 L of bleach is added.

Now, there is 4+12 = 16 L of bleach in 16 L of water in a total of 32 L of solution.

Now, to bring the bleach percentage back to 20%, extra water is added and the amount of bleach remains the same. Let this extra amount of water be y litres.

16 L of bleach constitutes 20% of the solution →

16/(32+y) = 20/100 → y = 48.

Therefore, 48 litres of water has to be added to the solution if bleach has to be 20% of the whole solution.

1 kg of cashews costs Rs. 100 and 1 kg of walnuts costs Rs. 120. If a mixture of cashews and walnuts is sold at Rs. 105 per kg,then what fraction of the total mixture are walnuts?

For this type of problems, first step is to determine how much each of the items is above or below the target.

Our target price is Rs. 105. Cashews price is Rs. 5 below the target price and walnuts price is Rs. 15 above the target price.

So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

This means that adding 3 kgs of cashews and 1 kg of walnuts gives a mixture that can be sold at Rs. 105 per kg.

So, 3 kgs of cashews present for every 1 kg of walnuts. The ratio of cashews to walnuts is 3:1. Fraction of walnuts in the mixture = 1/(3+1) = 1/4 of the total mixture

Practice Questions in Ratio and Proportion

Problem 1: Click here

On a certain map, 1 cm = 12 km actual distance. If two places are 96 km apart, what is their distance on map?

A. 10 cm B. 12 cm C. 96 cm D. 8 cm

Answer 1: Click here

1cm/12 km = x cm/100 km → x = 8 cm

Problem 2: Click here

A person types 360 words in 4 minutes. How much time does he take to type 900 words? A. 15 B. 90 C. 10 D. 9

Answer 2: Click here

4/360 = x/900 → x=10

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12 thoughts on “Ratio and Proportion Questions & Word Problems | GMAT GRE Maths”

Hey can you please emphasize on the working of the cashews and walnuts question?

I’m stuck on how we’re assuming this: So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

Why do they have to be added to cancel out eaach other?

Would appreciate the help.

I think if they equal zero, then is the only time a ratio can be found. If it isn’t equal to zero you won’t get a ratio.

assume the following Lets the total Qty of the Mixture is 10 KG Lets assume the Qty of Cashew is X Lets assume the Qty of Walnut is 10-X As we know that the purchase qty and sold qty can be equal for instance 10 KG now put this into equation 100(X) + 120(10-X) = 10*105 ” Total purchase = total Sales” X = 7.5 , hence walnut = 2.5 now 2.5/10=25%

given 1 KG of cashews cost 100 and walnuts was 120 given their mixture costs 105 rupees this 105 rupees of mixture cost is for just 1 kg includes X grams of cashews and Y grams of walnuts so, X+Y=1 KG or X+Y=1000 grams——1 1 KG of cashews cost is 100 so ,1000 grams=100 rupees. Hence 1 gram=1/10 rupees same way 1 gram of walnuts costs 12/100 rupees so, X/10 +Y*(12/100)=105——-2 solve equation 1 and 2 . Quantity of X and Y will be 750 grams and 250 grams so their ratio is 3:1 walnuts to the total ratio will be 1/4

Hello I have this peoblem The ratio of the area of the dining room to the family room is 2 to 3. After remodeling the family room is now 1/2 as large as it used to be and has 60 Sq less than the dining room. How many Sq feet is the isning room?

Hi Let the area of the dining room be d and the family room be f. d/2 = b f/3 = k d= 2k f=3k Total area = 5k 5k=3k +60(1/2*3k +1/2*3k + 60) 2k= 60 k = 30 Area of family = 3*30 /2 = 45 Area of dining = 105 45+105 = 150 and total area remains the same

I had a look at Erica’s answer, and unfortunately it doesn’t add up. However, I will stick to the same format she used:

dining room = d family room = f

Lets make that ratio easier to handle: d : f = 4 : 6

Now what does 4 : 6 mean? It means that the dining room represents 4 parts whilst the family room represents 6 parts. A part can be any size in a ratio, what matters is the proportion that the ratio describes.

Lets say 1 part = k, therefore d : f = 4k : 6k

Now if the family room halves in size, the ratio becomes 4k : 3k because the family room used to be 6 parts but is now 3 parts.

It is now also 60sq ft less than the dining room, so ‘size of f’ – ‘size of d’ = 60 sq ft therefore 4k – 3k = 60 sq ft i.e 1k = 60sq ft

Remember that the dining room is 4k so its size is:

4k = 4 x 60 sq ft = 240 sq ft

For every 2 boy students there are 3 girl students and for every one teacher there are 10 students whereas for every 4 male teachers there are 5 female teachers. Which of the following is the ratio of number of boy students to the number of male teachers?

For 1 teacher, there are 10 students which contains 4 boys, and 6 girls(because ratio of boys to girls is 2:3)

Therefore ratio of boys to teacher is 4:1

Now for 9 (4 male+5 female) teachers, there are total 90 students which has 36(90*0.4) boys

This means for a group of 36 boys, there are 4 male teachers.

Hence the ratio of boys to male teacher is 36:4 or 9: 1

Three similar lamps use 4 liters of oil in 80 hours. How much oil will 6 lamps of the same kind use in 40 hours?

Worker*Rate*Time=Object

Worker(lamp)=3 Time=80 hours Object = 4 liters

3R80=4. R=4/240 or R=1/60

Worker(lamp)=6 Time=40 hours Object= x

6R40=X (Rate is 1/60 from previous). 6(1/60)40=x (240/60)=x and x=4

Object or Liters is 4 liters.

a farmer has a total yield of 42,000 bu of corn from a 350-acre farmer what total yield should he expect from a similar 560-acre farm?

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Ratios and proportions and how to solve them

Let's talk about ratios and proportions. When we talk about the speed of a car or an airplane we measure it in miles per hour. This is called a rate and is a type of ratio. A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours.

A ratio can be written in three different ways and all are read as "the ratio of x to y"

$$\frac{x}{y}$$

A proportion on the other hand is an equation that says that two ratios are equivalent. For instance if one package of cookie mix results in 20 cookies than that would be the same as to say that two packages will result in 40 cookies.

$$\frac{20}{1}=\frac{40}{2}$$

A proportion is read as "x is to y as z is to w"

$$\frac{x}{y}=\frac{z}{w} \: where\: y,w\neq 0$$

If one number in a proportion is unknown you can find that number by solving the proportion.

You know that to make 20 pancakes you have to use 2 eggs. How many eggs are needed to make 100 pancakes?

$$\frac{eggs}{pancakes}=\frac{eggs}{pancakes}\: \: or\: \: \frac{pancakes}{eggs}=\frac{pancakes}{eggs}$$

If we write the unknown number in the nominator then we can solve this as any other equation

$$\frac{x}{100}=\frac{2}{20}$$

Multiply both sides with 100

$${\color{green} {100\, \cdot }}\, \frac{x}{100}={\color{green} {100\, \cdot }}\, \frac{2}{20}$$

$$x=\frac{200}{20}$$

If the unknown number is in the denominator we can use another method that involves the cross product. The cross product is the product of the numerator of one of the ratios and the denominator of the second ratio. The cross products of a proportion is always equal

If we again use the example with the cookie mix used above

$$\frac{{\color{green} {20}}}{{\color{blue} {1}}}=\frac{{\color{blue} {40}}}{{\color{green} {2}}}$$

$${\color{blue} {1}}\cdot {\color{blue} {40}}={\color{green} {2}}\cdot {\color{green} {20}}=40$$

It is said that in a proportion if

If you look at a map it always tells you in one of the corners that 1 inch of the map correspond to a much bigger distance in reality. This is called a scaling. We often use scaling in order to depict various objects. Scaling involves recreating a model of the object and sharing its proportions, but where the size differs. One may scale up (enlarge) or scale down (reduce). For example, the scale of 1:4 represents a fourth. Thus any measurement we see in the model would be 1/4 of the real measurement. If we wish to calculate the inverse, where we have a 20ft high wall and wish to reproduce it in the scale of 1:4, we simply calculate:

$$20\cdot 1:4=20\cdot \frac{1}{4}=5$$

In a scale model of 1:X where X is a constant, all measurements become 1/X - of the real measurement. The same mathematics applies when we wish to enlarge. Depicting something in the scale of 2:1 all measurements then become twice as large as in reality. We divide by 2 when we wish to find the actual measurement.

Video lesson

$$\frac{x}{x + 20} = \frac{24}{54}$$

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Here is an example of calculating ratios and proportions in word problems.

If Arnold can type 600 pages of manuscript in 21 days, how many days will it take him to type 230 pages if he works at the same rate?

First, circle what you're asked to find— how many days. One simple way to work this problem is to set up a “framework” (proportion) using the categories given in the equation. Here the categories are pages and days.

Therefore, a framework may be

problem solving ratio and proportion

Note that you also may have used

problem solving ratio and proportion

The answer will be the same. Now simply plug into the equation for each instance.

problem solving ratio and proportion

Cross multiplying

problem solving ratio and proportion

Previous Compound Interest

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Ratio and Proportion Word Problems
This math video tutorial provides a basic introduction into ratio and proportion word problems. Here is a list of examples and practice
Solving Ratio & Proportion Problems
MathAndScience.com Twitter: https://twitter.com/JasonGibsonMath In this lesson, you will learn how to solve ratio and proportion ...
Ratio/Proportion Word Problems
Set up a proportion to solve each problem, show all work, and label all answers. 1. The ratio of boys to girls is 3 to 2. If there are 12 boys
Ratio and Proportion Word Problems
A proportion is two ratios that are set equal to one another. An example of a proportion would be: 1 dollar10 pesos=10 dollars100 pesos As you can see, they are
Word Problems on Ratio and Proportion
WORD PROBLEMS ON RATIO AND PROPORTION ... Problem 1 : The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of
Worked out Problems on Ratio and Proportion
Worked out Problems on Ratio and Proportion · 1. Arrange the following ratios in descending order. · 2. Two numbers are in the ratio 3 : 4. · 3. If x : y = 1 : 2
Ratio and Proportion Questions & Word Problems
Question: In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the
Ratios and proportions and how to solve them
A proportion on the other hand is an equation that says that two ratios are equivalent. For instance if one package of cookie mix results in 20 cookies than
How to Solve Ratio Word Problems
By itself, a ratio is limited to how useful it is. However, when two ratios are set equal to each other, they are called a proportion. For
Ratio and Proportion
Ratio and Proportion ... Here is an example of calculating ratios and proportions in word problems. ... If Arnold can type 600 pages of manuscript in 21 days, how