Are you want to prepare the ratio and proportion for class 7 in detail and score good marks in the exam. Then, you are in the right place. Here you will get a complete concept,ncert solution, worksheet, and their PDF

Ratio and proportion is an extremely simple chapter that is based on the fraction. This is the foundation of mathematic which we use in our day-to-day life and different fields of science. So, in this article, you will learn concept with the help of examples

## Ratio

It is a way of comparing two or more things of the same kind. We compare how much/many times is one thing of another, but that things should be the same. Let take an example to understand it clearly

Suppose Ram and Shyam are two bothers and their weight is 30 kg and 60 kg respectively. What is it mean?

It means Shyam’s weight is twice of ram weight. Here is the comparison between two things which is “weight”

let’s see how we represent it in the ratio’s

**There are three ways to represent a ratio**

(i) Use “:” to denote ratios

In the above example, one is double of others. So, it’s ratio will be 2:1 which denote first is two times of another.

(ii) Use the word ” is to” to denote the ratio

We can represent the above example as “2 is to 1”.

(iii) Or you can represent is in the form of a fraction such as \frac{2}{1}

**The ratio can be scaled up without changing its value**. Let’s see how

When the weight of Shyam and ram is 60kg and 30kg respectively. Then, the ratio was 2:1

But when the weight of Shyam and ram becomes 120kg and 60kg respectively. What will the ratio

The ratio will remain the same as 2:1 even the value gets the increased.

**Always Remember: The weight of both people should increase by the same amount of time.**

If one weight increase by 2 times, others weight should also increase by two times for the same ratio

Let’s take another example to understand this concept clearly

Suppose I have to prepare a delicious sweet in which sugar and water are 2kg and 4kg respectively.

So, the ratio of sugar to water will be 1:2

But what happens when I prepare the same sweet for more number of peoples. Simply I will multiply both quantities by 2,3,4…more, how much I required for people. In this way, I can easily know the quantity of sugar and water both accurately.

### Properties of ratio

(i) In the ratio a:b, the first term known as Antecedent, and the second term b known as consequent.

(ii) The value of a ratio remains unchanged if each one of its terms is multiplied or divided by the same non-zero number

(iii) a²:b² is the duplicate ratio of a:b.

(iv) a³:b³ is the triplicate ratio of a:b

(v) √a:√b is the sub-duplicate ratio of a:b

## Proportion for class 7

The proportion is nothing but a comparison between two ratios. We use”::” to denote proportionality. Let’s understand it with an example

Let you have two ratios 2:3 and 4:6 respectively.Now when we divide 4:6 or convert it into fraction form \frac{4}{6},it will becomes of form\frac{2}{3}after division .As you can see both ratios are equal. So, we can say 2:3 is proportional to 4:6 or 2:3:: 4:6

That’s means if both ratios are equal then, they are proportional

Now you might be thinking, why we don’t use = sign to denote this?

we don’t use” = “sign to denote proportional because ratio 8:12,16:24, etc are also proportional to 2:3.

So,when we will write \frac{2}{3}=\frac{8}{12},\frac{2}{3}=\frac{16}{24} etc

It is not equal to any particular number, so we simply denote it by proportionality sign

**Note: In the proportion a:b:: c:d,a and d are extreme values and b and c are mean values.**

Product of means=Product of extreme

**Properties of proportion**

(i) If x is the third proportional to a,b,then a:b :: b:x .

(ii) The mean proportional between a and b is √ab

### Application of proportion

- A fixed ratio of length and breadth is always maintained while making a national flag. The ratio may be different for different countries but it is generally 1.5:1 or 1.7:1. The approximate value of this ratio can be taken as 3:2.
- The drawings or the pictures look good when appropriate relative proportions are maintained.

## Ncert solution of ratio and proportion for class 7

Here you will complete solution for class7 ratio and proportions .After reading this ,you can check your concept through some problems which are given at the end for ratio and proportion class 7

Q.1 Express each of the following ratios in the simplest form

(i) 24:40

Solution:

Converting ratio in the simplest form means”Converting fraction into simplest form”

\frac{24}{40}=\frac{12}{20}=\frac{3}{5}(ii) 13.5:15

Solution(ii) \frac{13.5}{15}=\frac{135}{150}=\frac{9}{10}

(iii) 6\frac{2}{3} :7\frac{1}{2}

Solution:

\frac{20}{3}:\frac{15}{3}=\frac{4}{3}

(iv) \frac{1}{6}:\frac{1}{9}

Solution: \frac{1}{6}:\frac{1}{9}=\frac{\frac{1}{6}}{\frac{1}{9}}=\frac{3}{2}

(v) 4:5:\frac{9}{2}

Sol. Take the LCM of denominator i.e 2

Multiply all ratios by 2

8:10:9

Q.2 Express each of the ratios in the simplest form

(i) 75paise:3rupees

Solution: We know that 1rupee=100paise

So, 3rupees=300paise

75paise:300paise (Both ratios must have same unit)

\frac{75}{300}=\frac{1}{4}=1:4

(ii) 1m 5cm:63cm

Solution: Convert 1m 5cm into cm =105cm:63cm (1m=100cm)

\frac{105}{63}=\frac{5cm}{3cm}

Q.3 If A:B =7:5 and B:C=9:14,find A:C

Solution(3)

Since ,it is given

\frac{A}{B}=\frac{7}{5} ……(i)

\frac{B}{C}=\frac{9}{14} …….(ii)

simply multiply equation(i) and (ii)

\frac{A}{B}×\frac{B}{C}=\frac{7}{5}×\frac{9}{14}

\frac{A}{C}=\frac{9}{10}Q.4 If A:B=5:8 and B:C=16:25 ,find A:C

Sol. This is same type of problem which we solved earlier

Since ,it is given

\frac{A}{B}=\frac{5}{8} ……(i)

\frac{B}{C}=\frac{16}{25}. ……..(ii)

Multiply equation (i) and (ii)

\frac{A}{B}×\frac{B}{C}=\frac{5}{8}×\frac{16}{25}

\frac{A}{C}=\frac{2}{5}Q.5 If A:B=3:5 and B:C=10:13 then find A:B:C

Solution:

It is given A:B=3:5

\frac{A}{B}=\frac{3}{5}

So, A=\frac{3B}{5} ……….(i)

similarly B:C=10:13

\frac{B}{C}=\frac{10}{13}Value of C=\frac{13B}{10} …….(ii)

Subtitute the value from equation (i) and (ii) in A:B:C

\frac{3B}{5} :B:\frac{13B}{10}6:10:13

### Ratio and proportion for class 7 word problem

Q.7 Divide Rs.360 between kunal and rohit in ratio 7:8

Solution:

The sum of given ratio will be 8+7=15

Kunal have=\frac{7}{15}×360=168Rs

Rohit have=\frac{8}{15}×360=Rs 192

Q.8 Divide Rs 880 between rajan and kamal in ratio \frac{1}{5}:\frac{1}{6}

Solution: This is same problem as we solved above

The sum of given ratio will be=\frac{1}{5}+\frac{1}{6}=\frac{11}{30}

Rajan have=\frac{\frac{1}{5}}{\frac{11}{30}}×880=\frac{6}{11}×880=480

Kamal have =\frac{\frac{1}{6}}{\frac{11}{30}}×880=\frac{5}{11}×880=400

Q.9 Divide 5600Rs between A,B and C in ratio 1:3:4

Solution :The sum of given ratio are 1+3+4=8

A’s have =5600×\frac{1}{8}=Rs 700

B’s have=5600×\frac{3}{8}=Rs2100

C’s have =5600×\frac{4}{8}=Rs 2800

Q.10 What number must be added to each term in the ratio 9:16 to make the ratio 2:3?

Sol. Let x should be added to the given ratio

A/Q

(9+x):(16+x)=2:3

\frac{9+x}{16+x}=\frac{2}{3}

27+3x=32+2x

x=5

Q.11 What number must be subtracted from each term of ratio 17:33 so that ratio becomes 7:15?

Sol. Again, let x should be added to the given ratio

A/Q

(17-x):(33-x)=7:15

255-15x=231-7x

8x=24

x=3

Q.12 Two numbers are in the ratio 7:11. If 7 is added to each number, the ratio becomes2:3.Find the numbers

Sol. Let the numbers are 7x and 11x (Ratio can be scaled , so we can let it in multiple of some particular number).

A/Q, 7 is added to both numbers, and the ratio of both numbers becomes 2:3

(7x+7):(11x+7)=2:3

\frac{7x+7}{11x+7}=\frac{2}{3}

21x+21=22x+14

x=7

∴ Both numbers are 7x=7×7=49 and 11x=11×7=77

Q.13 Two numbers are in the ratio 5:9.On subtracting 3 from each number, the ratio becomes 1:2. Find the numbers

Sol. Again, let both numbers are 5x and 9x

A/Q

(5x-3):(9x-3)=1:2

10x-6=9x-3

x=3

∴ Both numbers are 5x=5×3=15 and 9x=9×3=27

Q.14 Two numbers are in the ratio 3:4. If their LCM is 180, find the numbers

Sol. Let both numbers are 3x and 4x

SO, their LCM will 12x

A/Q, LCM=180

12x=180

x=15

Hence both numbers are 3x=3×15=45 and 4x=4×15=60

Q.15 The ages of A and B are in the ratio 8:3. Six years hence, their ages will be in the ratio 9:4. Find their present ages

Sol. Let the present ages of A and B are 8x and 3x.

A/Q, Six years hence their ages will be in the ratio 9:4

(8x+6):(3x+6)=9:4

32x+24=27x+54

5x=30

x=6

The present age of A and B are 8x=8×6=48 years and 3x=3×6=18 years

Q.16 The ratio of copper and zinc in an alloy is 9:5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.

SOl. Let the weight of copper and zinc in the alloy is 9x and 5x respectively

Since it is given weight of copper =48.6 grams

9x=48.6grams

x=\frac{48.6}{9}

x= 5.4

Weight of zinc=5x=5×5.4=27 grams

Q.17 The ratio of boys and girls in the school is 8:3. If the total number of girls be 375, find the number of boys in the school.

Sol. Let the numbers of boys and girls in the school are 8x and 3x respectively

Since number of girls=375

3x=375

x=125

so, the number of boys will be 8x=8×125=1000

Q.18 The ratio of monthly income to the savings of a family is 11:2. If the savings be Rs 2500, find the income and expenditure

SOl. Let the monthly income and saving of a family are 11x and 2x respectively

Saving=Rs.2500

2x=2500

x=1250

so, the monthly income of the family will be 11x=11×1250=13750

Expenditure of family =monthly income -saving

Expenditure of family = 13750-2500

Expediture=112500

Q.19 A bag contains Rs.750 in the form of rupee 1 Rs,50 paise, and 25 paise coins in the 5:8:4.Find the numbers of the coin of each type

Sol.

Let number of coin of 1 Rs ,50 paise and 25 paise coin are 5x .8x and 4x respectively

Total amount of all coin (in paise)=(5x×100+8x×50+4x×25)=(500x+400x+100x)=1000x

Total amount of money in the bag(given)=Rs.750=75000 paise

1000x=75000

x=75

Number of coin of each types=5x ,8x and 4x =5×75,8×75,4×75=375 ,600 and 300

Q.20 If (4x + 5) : (3x + 11) = 13 : 17, find the value of x

Sol. \frac{4x + 5}{3x + 11}=\frac{13}{17}

68x+85=39x+143

29x=58

x=2

Q.21 If x:y =3:4 ,find (3x+4y):(5x+6y)

Sol. It is given \frac{x}{y}=\frac{3}{4}

so , x=\frac{3y}{4}

(3x+4y):(5x+6y)=( 3.\frac{3y}{4}+4y):(5.\frac{3y}{4}+6y)

(3x+4y):(5x+6y)=(\frac{9y+16y}{15y+24y})

(3x+4y):(5x+6y)=\frac{25y}{39y}

(3x+4y):(5x+6y)=\frac{25}{39}

Q.22 Two numbers are in the ratio 5:7 .If the sum of the numbers is 720 ,find the numbers

Sol. Let the two numbers are 5x and 7x respectively

Sum of the numbers=720

5x+7x=720

12x=720

x=60

∴ Two numbers are 5x=5×60=300 and 7x=7×60=420

Exercise 1.2

Q.1 show that 30 ,40 ,45 and 60 are in propertion

Sol. We know in prortion ,product of extremes=product of means

Product of extremes=30 × 60=1800

Product of products =40×45=1800

So, 30 ,40 ,45 and 60 are in proprtion .

Q.2 show that 36,49 ,6, and 7 are not in proportion.

Sol.

Product of extreme=36 ×7=252

Product of means=49×6=294

Since Product of extreme≠Product of means

So, given numbers are not proportional

Q.3 If 2:9:: x:27 ,find the value of x

Sol. Since 2,9,x, and 27 are in proportion

So, the product of extremes=product of means

2×27=9×x

54=9x

x=6

Q.4 If 8:x::16:35 ,find the value of x

Sol.

Product of means=product of extreme

8×35=16×x

280=16x

x=17.5

Q.5 If x:35::48:60 ,find the value of x

Sol.

Product of means=product of extreme

x×60=35×48

60x=1680

x=28

Q.6 Find the fourth proprtional to the numbers

(i) 8,36,6

(ii) 5,7,30

(iii) 2.8,14,3.5

Sol . (i) Let the fourth proportional is x

8:36:: 6:x

Product of extreme=product of means

8x=36×6

8x=216

x=27

(ii) Let the fourth proportional is x

5:7::30:x

product of extreme =Product of means

5×x=7×30

5x=210

x=42

(iii)Let the fourth proportional is x

2.8:14::3.5:x

product of extreme =Product of means

2.8×x=14×3.5

2.8x=49

x=17.5

Q.7 If 36,54,x are in continued proportion, find the value of x

Sol. It is given,

36,54,x are in continued proportion

36:54::54:x

Product of means=Product of extreme

54×54=36×x

36x=2916

x=81

Q.7 If 27,36,x is in continued proportion, find the value of x

Sol. It is given,

27,36,x are in continued proportion

27:36::36:x

Product of means=Product of extreme

36×36=27×x

27x=1296

x=48

Q.8 Find the third proportional

(i) 8 and 12

(ii) 12 and 18

(iii) 4.5 and 6

Sol. Let the third proportional is x

∴ 8:12::12:x

Product of means=Product of extreme

12×12=8×x

8x=144

x=18

(ii)

Let the third proportional is x

∴ 12:18::18:x

Product of means=Product of extreme

18×18=12×x

12x=324

x=27

(iii)

Let the third proportional is x

∴ 4.5:6::6:x

Product of means=Product of extreme

6×6=4.5×x

4.5x=36

x=8

Q.9 If the third proportional to 7 and x is 28, find the value of x

Sol . It is always of form 7:x::x:28

Product of means =product of extreme

x×x=7×28

x²=196

x=14

Q.10 Find the mean proportional of following

(i) 6 and 24

(ii) 3 and 27

(iii) 0.4 and 0.9

Sol. If x is the mean proportional of a and b

then, x=√ab

(i) Mean proportional of 6 and 24 =√6×24=√144=12

(ii) Mean proportional of 3 and 27 =√3×27=√81=9

(iii) Mean proportional of 0.4 and 0.9 =√0.4×0.9=√0.36=0.6

Q.11 What number must be added to each of the number 5,9,7,12 to get the number which are in proportion

Sol. Let we need to add a number which is x

So , (5+x):(9+x)::(7+x):(12+x)

Product of extreme=Product of means

(5+x)(12+x)=(9+x)(7+x)

60+5x+12x+x²=63+9x+7x+x²

60+17x=63+16x

x=3

Hence, the required number is 3

Q.12 What number must be subtracted from each of the numbers 10,12,19 and 24 to get the numbers that are in proportion?

Sol . Let the required number is x

A/Q

(10-x):(12-x)::(19-x):(24-x)

Since, product of extreme=Product of means

(10-x)×(24-x)=(12-x)×(19-x)

240-10x-24x+x²=228-12x-19x+x²

240-34x=228-31x

12=3x

x=4

So , required number is 4

Q.13

Q.14 At a certain time a tree 6 m high cast a shadow of length 8 meters. At the same time, a pole cast a shadow of length 20 meters. Find the height of the pole

Sol.

Length of the tree and their shadow is 6m and 8m respectively

At the same time, a pole shadow is 20 m

Let the length of pole is x

so , 6:8:: x:20

Product of extreme=Prioduct of means

6×20=8×x

120=8x

x=15

The height of the pole is 15m

### Ratio and proportion for class 7 MCQ with solution

Here are some MCQ questions for class 7 ratio and proportion with the solutions.

Q.15 Mark(**✓) against the correct answer **

If a:b=3:4 and b:c=8:9 ,then a:c =?

(i) 1:2

(ii) 3:4

(iii) 1:3

(iv) 2:3

Solution: It is given a:b=3:4 and b:c=8:9

Means \frac{a}{b}=\frac{3}{4} ….(i) , \frac{b}{c}=\frac{8}{9} …….(ii)

multiply equation(i) and (ii)

\frac{a}{c}=\frac{27}{32}Q.16 Mark (**✓) against the correct answer**

If A:B=2:3 and B:C=4:5 ,find C:A=?

(i) 15:8

(ii) 6:5

(iii) 8:5

(iv) 8:15

Solution: Since , \frac{A}{B}=\frac{2}{3} ……(i) ,\frac{B}{C}=\frac{4}{5} …..(ii)

multiply equation(i) and (ii)

\frac{A}{C}=\frac{8}{15}So,\frac{C}{A}=\frac{15}{8}

Q.17 Mark (**✓) against the correct answer**

If 2A=3B and 4B=5C ,then A:C=?

(i) 4:3

(ii) 8:15

(iii) 3:4

(iv) 15:8

Solution:

It is given 2A=3B

so, \frac{A}{B}=\frac{3}{2} ……….(i)

similarly \frac{B}{C}=\frac{5}{4} ……(ii)

Multiply both equation

\frac{A}{C}=\frac{15}{8}Q.18 Mark (**✓) against the correct answer**

If 15% of A=20% of B ,then A:B=?

(i) 3:4

(ii) 4:3

(iii) 17:16

(iv) 16:17

Solution:

According to question , 15% of A=20% of B

\frac{15}{100}×A=\frac{20}{100}×B

15A=20B

\frac{A}{B}=\frac{4}{3}

Q.19 Mark (**✓) against the correct answer**

If A=\frac{1}{3}B and B=\frac{1}{2}C ,then A:B C

(i) 1:3:6

(ii) 2:3:6

(iii) 3:2:6

(iv) 3:1:2

Solution: Since,

A=\frac{1}{3}B .. ……..(i)

and

B=\frac{1}{2}C ………… (ii)

Then value of C from equation(ii)

C=2B

Then , A:B:C =\frac{1}{3}B :B :2B=1:3:6

Q.20 If A:B=5:7 and B:C=6:11 ,find A:B:C

(i) 30:42:55

(ii) 30:42:77

(iii) 35:49:66

(iv) None of above

Solution: If A:B=5:7 and B:C=6:11

We will apply same above procedure

It is given, \frac{A}{B}=\frac{5}{7}

A=\frac{5B}{7} ……(i)

simiarly

\frac{B}{C}=\frac{6}{11}

C=\frac{11B}{C}

then, A:B:C= \frac{5B}{7} : B : \frac{11B}{C}= 30:42:77

Q.21 If 2A=3B=4C ,then A:B:C =?

(i) 2:3:4

(ii) 4:3:2

(iii) 6:4:3

(iv) 3:4:6

Solution:

2A=3B=4C

So, we have 2A=3B ……(i) and3B=4C …(ii)

Apply same procedure which we applied above

∴ A:B:C=\frac{3B}{2}:B:\frac{3B}{4}

A:B:C=6:4:3

Q.22 If \frac{A}{3}=\frac{B}{4}=\frac{C}{5} ,then A:B:C=?

(i) 3:4:5

(ii) 4:3:5

(iii) 5:4:3

(iv) 20:15:12

Solution:

It is given ,

\frac{A}{3}=\frac{B}{4}4A=3B

A=\frac{3B}{4}

similarly , \frac{B}{4}=\frac{C}{5}

C=\frac{5B}{4}

Then , A:B:C

\frac{3B}{4}:B:\frac{5B}{4}3:4:5

Q.24 If x:y=3:4 ,then(7x+3y):(7x-3y)=?

(i) 4:3

(ii) 5:2

(iii) 11:3

(iv) 37:39

Solution: It is given that \frac{x}{y}=\frac{3}{4}

Subtitute x=\frac{3y}{4} in (7x+3y):(7x-3y)

It becomes \frac{21y+12}{21y-12}=\frac{33y}{9y}=\frac{11}{3}

Q.25 If (3a+5b):(3a-5b)=5:1 ,then a:b=?

(i) 2:1

(ii) 3:2

(iii) 5:2

(iv) 5:3

Solution: It can be written as \frac{3a+5b}{3a-5b}=\frac{5}{1}

(3a+5b)=5(3a-5b)

3a+5b=15a-25b

30b=12a

\frac{a}{b}=\frac{30}{12}=\frac{10}{4}=\frac{5}{2}Q.26 If 7:x::35:45 ,then x=?

(i) 7

(ii) 15

(iii) 9

(iv) 5

Solution: As given ,these numbers are proprtional

Product of means=Product of extreme

x×35=7×45

35x=315

x=9

**More Ratio and proportion for class 7 word problems.**

Here are some word problems to strengthen your concept. In the end, you will get a practice problem for class 7 ratio and proportion.

Q.27 What number must be added to each number 3:5 so that the ratio become 5:6

(i) 6

(ii) 7

(iii) 12

(iv) 11

Solution: Let x should be subtracted from each number

A/Q \frac{3-x}{5-x}=\frac{5}{6}

6(3-x)=5(5-x)

18-6x=25-5x

-6x+5x=25-18

-x=7

x=-7

Q.28 Two numbers are in the ratio 3:5. If each number increased by 10, the ratio becomes 5:7. The sum of the numbers is

(i) 8

(ii) 16

(iii) 35

(iv) 40

Solution: let two numbers are 3x and 5x

A/Q

\frac{3x+10}{5x+10}=\frac{5}{7}7(3x+10)=5(5x+10)

21x+70=25x+50

20=4x

x=5

Q.29 What is the least should be subtracted from each 15:19 so that ratio becomes 3:4

(i) 3

(ii) 5

(iii) 6

(iv) 9

Solution: Let x is least required number

\frac{15-x}{19-x}=\frac{3}{4}4(15-x)=3(19-x)

60-4x=57-3x

x=3

So, option(i) is correct

Q.30 If 420 is divided between A and B in the ratio of 3:4 then, A’s share

(i) Rs 180

(ii) Rs 140

(iii) Rs 270

(iv) Rs 210

Solution: The sum of given ratio is 3+4=7

A have=\frac{3}{7}×420= 180

So, A’s share is 180

Q.31 The boys and girls in the school are in the ratio 8:5 . If the number of girls is 160 ,what is the total strebght of the school

(i) 250

(ii) 260

(iii) 356

(iv) 416

Solution: Let the number of boys and girls in school are 8x and 5x respectively

Since, number of girls=160

So, 5x=160

x=32

Therefore,number of boys =8x=8×32=256

Total number of students in school =160+256=416

Option(iv) is correct

Q.32 Which one is greater out of 2:3 and 4:7?

(i) 2:3

(ii) 4:7

(iii) both are equal

Solution: Take the LCM of 3 and 7

LCM of 3 and 7 will be 21

Now multiply 7 in \frac{2}{3} and 3 in \frac{4}{7}in numerator and denominator so that denominator becomes equal

\frac{14}{21} and \frac{12}{21}

As you see \frac{12}{21}<\frac{14}{21}

So , (4:7)<(2:3)

Q.33 The third proportional to 9 and 12

(i) 10.5

(ii) 8

(iii) 16

(iv)21

Solution: Let x is the third proportional of 9 and 12

So, 9:12::12:x

Product of extreme=product of means

9x=144

x=16

So, the third proportional of 9 and 12 is 16

Q.34 The mean proportional to 9 and 16 is

(i) 12.5

(ii) 12

(iii) 5

(iv) None of these

Solution: We know that mean proportional of a and b is given by √ab

SO, the mean proportional of 9 and 12 is √9×12=√144=12

Option(ii) is correct

Q.35 The ages of A and B are in the ratio 3:8. Six years hence, their ages will be in the ratio 4:9. Find the present age of A

(i) 18 years

(ii) 15 years

(iii) 12 years

(iv) 21 years

Solution: Let the present age of A and B are 3x and 8x respectively

Six years hence,ratio of their ages are 4:9

\frac{3x+6}{8x+6}=\frac{4}{9}

9(3x+6)=4(8x+6)

27x+54=32x+24

5x=10

x=2

Q.36 Compare 4:5 and 7:9

Solution: LCM of 5 and 9 is 45

So multiply 4:5 and 7:9 by 9 and 5 respectively

It become \frac{36}{45} and \frac{35}{45}

\frac{36}{45}>\frac{35}{45}

Q.37 Divide Rs 1100 among A, B, and C in the ratio of 3:4:5

Solution: The sum of the ratio is 13

A’s share =\frac{3}{13}×1100=220

B’s share =\frac{4}{13}×1100=330

C’s share =\frac{5}{13}×1100=550

Q.38 Show that the numbers 25,36,5, and 6 are not in proportion

Solution: We know that, if numbers are in proportion so

Product of extreme=Product of means

Here,

Product of extreme= 25×6=150

Product of means= 36×5=180

Q.39 If x.18 ,108 are in continued fraction .find the value of x

Solution:

x:18::18:108

108x=324

x=3

Q.40 Two numbers are in the ratio 5:7. If the sum of these numbers is 84, find the numbers

Solution: Let the two are 5x and 7x respectively

Since,sum of these numbers =84

12x=84

x=7

So, numbers are 5x and 7x =5×7,7×7=35,49

Q.41 The ages of A and B are in the ratio 4:3. Eight years ago, their ages are in the ratio 10:7. Find their present ages

Solution: Let the present age of A and B are 4x and 3x respectively

Eight years ago, the ratio of their ages are in 10:7

\frac{4x-8}{3x-8}=\frac{10}{7}

7(4x-8)=10(3x-8)

28x-56=30x-80

-56+80=30x-28x

24=2x

x=12

Q.42 If a car cover 54 km in an hour, how much distance will it cover in 40 minutes?

Solution:

Distance covered by car in 1 hour =54km

Distance covered by car in 1 minute=\frac{54}{60}

So,distance covered by car in 40 min =\frac{54}{60}×40=36km

Q.43 Find the third proportional to 8 and 12

Solution:

Let the third proportional of 8 and 12 is x

8:12::12:x

Product of means=Product of extreme’

8x=144

x=16

So, the third proportional of 8 and 12 is 144

Q.44 If 40 men can finish a piece of work in 60 days, in how many days will 75 men finish the work?

Solution:

40 men finish a piece of work =60 days

1 man finish that works in =\frac{60}{40}

So, 75 men will finish that work in \frac{60}{40}×75=32

In 32 days,75 men will finish that piece of work

Q.45 Fill in the blanks

(i) If A:B=2:3 and B:C=4:5 ,then C:A…….

Solution:

\frac{A}{B}=\frac{2}{3} ……(i) ,

\frac{B}{C}=\frac{4}{5} .. ………(ii)

multiply equation(i) and (ii)

\frac{A}{C}=\frac{8}{15}

So, value of C:A=15:8

(ii) If 16% of A= 20% of B ,then A:B=?

Solution: It is given that 16% of A= 20% of B,

\frac{16}{100}×A=\frac{20}{100}×B

16A=20B

4A=5B

\frac{A}{B}=\frac{5}{4}## Ratio and proportion for class 7 worksheet and extra questions

Here , you will get 20 questions for class 7 ratio and proportion. If you have any doubt regarding any question , comment below without any hesitation

Q.1 The fourth proportion to 3,5, and 21 is

Q.2 What must be added to each term of the ratio 49:68 so that it becomes 3:4

Q.3 A bag contains Rs 600 in the form of one rupee,50 paise, and 25 paise coins in the ratio 3:4:12. The number of 25 paise coins is

Q.4 Out of the ratios 7:20;13:25 ;17:30 and 11:15 ,the smallest one is

Q.5 20L of a mixture contains milk and water in the ratio 5:3. If 4L of the mixture is replaced by 4L of the milk, the ratio of milk to water in the new mixtures will become

Q.6 If 2x=3y=4z ,then x:y:z is

Q.7 The ratio of the number of boys and girls in a school of 720 students is 7:5. How many more girls should be admitted to make the ratio 1:1?

Q.8 The income of A and B are in the ratio 3:2 and their expenditure in the ratio 5:3. If each saves Rs 1500, then B’s income is

Q.9 The third proportional to 9 and 12 is

Q.10 The ratio of zinc and copper in a brass piece is 13:7. How much zinc will be there in 100kg of such a piece?

Q.11 What should be added to the ratio 5:11, so that the ratio becomes 3:4

Q.12 Two numbers are in the ratio 3:7. If their sum is 710, find the numbers

Q.13 The angles of a triangle are in the ratio 1:2:3. Find the angles of the triangle.

Q.14 A building of height 8m casts its shadow of the lengths 6m at the point of time during the day.If exactly the same of the day, the length of the shadow of a tower is 5m, what is the height of the tower?

Q.15 During the same time of the day, a pole casts a shadow of the length 10 m whereas a tree of 3m height casts a shadow of length 6m. Find the height of the pole.

Q.16 A person pays Rs 180 to buy 5kg of rice. How much will he pay to buy 20kg of rice?

Q.17 Anjali got 24 marks out of 30 in a test and Fatima got 28 out of 40. Whose score is better?

Q.18 The ratio of the monthly income to the saving of a person is 6:5. If the saving is Rs 2000, find the monthly income

Q.19 Three players scored 250 runs in a one-day cricket match in the ratio 2:3:5. Find the score of each player

Q.20 The ratio of a father’s age to that of his so is 3:1. If the sum of their ages is 60, find their ages

**Answer **

1.35

2.8

3.900

4.7:20

5.7:3

6.6:4:3

7.120

8.Rs6000

9.16

10.65kg

11.13

12.213,597

13.30º,60º,90º

14.6\frac{2}{3}m

15.5m

16.Rs.720

17.Anjali

18.Rs.2400

19.50,75,125

20.Father’s age=45 year and son’s age=15 years

### Ratio and proportion for class 7 worksheet PDF

ratio and proportion class 7 worksheet PDF

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