## NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion

Here at MentorAtHome, We provide NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion. NCERT solutions help students to improve their performance, and attain better results. In NCERT Solutions Class 6 Maths Chapter 12 there is 3 exercise.

### NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion Exercises 12.1

Class 6th Maths Chapter 12 Solutions ex 12.1

**Q.1. There are 20 girls and 15 boys in a class.****(a) What is the ratio of the number of girls to the number of boys?****(b) What is the ratio of th**e** number of girls to the total number of students in the class?**

Ans:-**(a)** Number of girls = 20

Number of boys = 15

Total number of students = 20 + 15 = 35

∴ Ratio of the number of girls to the number of boys

= Number of girls / Number of boys

=20 / 15 = 4 : 3**(b)** Ratio of the number of girls to the number of students

= Number of girls / Number of students

=20 / 35 = 4 : 7

**Q.2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of****(a) Number of students liking football to the** **number of students liking tennis.****(b) Number of students liking cricket to the total number of students**

Ans:-

Number of students in the class = 30

Number of students liking football = 6

Number of students liking cricket = 12

Number of students liking tennis

= 30 – (6 + 12) = 30 – 18 = 12**(a)** Ratio of the number of the students liking football to the number of students liking tennis

= 6 / 12 = 1 : 2**(b)** Ratio of the number of students liking cricket to the total number of students

= 12 / 30 = 2 : 5

**Q.3. See the figure and find the ratio of****(a) Number of triangles to the number of circles inside the rectangle.****(b) Number of squares to all the figures inside the rectangle.****(c) Number of circles to all the figures inside the rectangle.**

Ans:-**(a)** Number of triangles 3

Number of circles = 2

Number of squares = 2

Total number of figures = 7

∴ Ratio of number of triangles to the number of circles

= 3 / 2 = 3 : 2**(b**) Ratio of number of squares to the number of all the figures

= 2 / 7 = 2 : 7**(c)** Ratio of number of circles to the number of all the figures

= 2 / 7 = 2 : 7

**Q.4. Distance traveled** **by Hamid and Akhtar in an hour is 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar.**

Ans:-

Distance traveled by Hamid = 9 km.

Distance traveled by Akhtar = 12 km.

Speed of Hamid = 9 km/hr

Speed of Akhtar = 12 km/hr

∴ Ratio of the speed of Hamid to the speed of Akhtar

= 9 / 12 = 3 : 4

**Q.5. Fill in the following blanks:**

Ans:-

The cross multiplication

(15 x 6) / (18 x ☐)

☐ = 90 / 18

☐** ** = 5

The cross multiplication

(5 x ☐) / (6 x 10)

☐ = 60 / 5

☐ = 12

The cross multiplication

(10 x 30) / (12 x ☐)

☐ = 300 / 12

☐ = 25

Hence, 5, 12, and 25 are the numbers that come in the blanks, respectively.

Yes, all are equivalent ratios

**Q.6. Find the ratio of the following:****(a) 81 to 108****(b) 98 to 63****(c) 33 km to 121 km****(d) 30 minutes to 45 minutes**

Ans:-**(a)** 81 / 108

81 ÷ 27 / 108 ÷ 27

= 3 / 4

The ratio = 3 : 4**(b)** 98 / 63

98 ÷ 7 / 63 ÷ 7

= 14 / 9

The ratio = 14 : 9**(c)** 33 / 121

33 ÷ 11 / 121 ÷ 11

= 3 / 11

The ratio = 3 : 11**(d)** 30 / 45

30 ÷ 15 / 45 ÷ 15

= 2 / 3

The ratio = 2 : 3

**Q.7. Find the ratio of the following:****(a) 30 minutes to 1.5 hours****(b) 40 cm to 1.5 m****(c) 55 paise to ₹ 1****(d) 500 ml to 2 liters**

Ans:-**(a)** 30 minutes to 1.5 hours

1 hour = 60 minutes

∴ 1.5 hours = 60 x 1.5 minutes = 90 minutes

Ratio of 30 minutes to 90 minutes = 30 / 90

30 ÷ 30 / 90 ÷ 30

= 1 / 3

The ratio = 1 : 3**(b)** 40 cm to 1.5 m

1 m = 100 cm

∴ 1.5 m = 1.5 x 100 cm = 150 cm.

The ratio of 40 cm to 150 cm.=40 / 150

40 ÷ 10 / 150 ÷ 10

= 4 / 15

The ratio = 4 : 15**(c)** 55 paise to ₹ 1

₹1 = 100 paise

Ratio of 55 paise to 100 paise = 55 / 100

55 ÷ 5 / 100 ÷ 5

= 11 / 20

The ratio = 11 : 20**(d)** 500 ml to 2 liters

1 litre = 1000 mL

∴ 2 litres = 2 x 1000 mL = 2000 mL

Ratio of 500 mL to 2000 mL = 500 / 2000

500 ÷ 500 / 2000 ÷ 500

= 1 / 4

The ratio =1 : 4

**Q.8. In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of****(a) Money that Seema earns to the money she saves****(b) Money that she saves from the money she spends.**

Ans:-**(a) **Money earned by Seema = ₹ 1,50,000

Money saved by her = ₹ 50,000

∴ Money spent by her = ₹ 1,50,000 – ₹ 50,000 = ₹ 1,00,000

∴ Ratio of money earned by Seema to the money saved by h

= 150000 / 50000

150000 ÷ 50000 / 50000 ÷ 50000

= 3 / 1

The ratio = 3 : 1**(b)** Ratio of money saved by Seema to the money spent

=50000 / 100000

50000 ÷ 50000 / 100000 ÷ 50000

= 1 / 2

The ratio = 1 : 2

**Q.9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.**

Ans:-

Number of teachers = 102

Number of students = 3300

∴ Ratio of number of teachers to the number of students

= 102 / 3300

102 ÷ 6 / 3300 ÷ 6

=17 / 550

The ratio = 17 : 550

**Q10. In a college, out of 4320 students, 2300 are girls. Find the ratio of****(a) Number of girls to the total number of students.****(b) Number of boys to the number of girls.****(c) Number of boys to the total number of students.**

Ans:-

Total number of students = 4320

Number of girls = 2300

∴ Number of boys = 4320 – 2300 = 2020**(a)** Ratio of number of girls to the total number of students

= 2300 = 4320

2300 ÷ 20 / 4320 ÷ 20

=115 / 216

The ratio = 115 : 216**(b)** Ratio of number of boys to the number of girls

= 2020 / 2300

2020 ÷ 20 / 2300 ÷ 20

= 101 / 115

The ratio = 101 : 115**(c)** Ratio of number of boys to the total number of students

= 2020 / 4320

2020 ÷ 20 / 4320 ÷ 20

=101 / 216

The ratio = 101 : 216

**Q.11. Out of 1800 students in a school, 750 opted for basketball, 800 opted for cricket, and the remaining opted table tennis. If a student can opt only one game, find the ratio of(a) Number of students who opted for basketball to the number of students who opted for table tennis.(b) Number of students who opted for cricket to the number of students opting for basketball.(c) Number of students who opted for basketball to the total number of students.**

Ans:-

Total number of students = 1800

Number of students opting for basketball = 750

Number of students who opted for cricket = 800

Number of remaining students who opted table tennis

= 1800 – (750 + 800) = 1800 – 1550 = 250

**(а)**Ratio of number of students who opted for basketball to the number of students who opted for table tennis = 750 / 250

750 ÷ 250 / 250 ÷ 250

= 3 / 1

The ratio = 3 : 1

**(b)**Ratio of the students who opted for cricket to the number of students opting for basketball

= 800 / 750

800 ÷ 50 / 750 ÷ 50

=16 / 15

The ratio = 16 : 15

**(c)**Ratio of number of students who opted fo

**r**basketball to the total number of students

= 750 / 1800

750 ÷ 150 / 1800 ÷ 150

= 5 / 12

The ratio = 5 : 12

**Q.12. The cost of a dozen pens is ₹ 180 and the** **cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.**

Ans:-

Cost of a dozen pens** = **₹ 180

Cost of 1 pen = 180 / 12 = ₹ 15

Cost of 8 ball pens = ₹ 56

Cost of 1 ball pen = 56 / 8 = ₹ 7

The ratio of cost of 1 pen to the cost of 1 ball pen

= 15 : 7

**Q.13. Consider the statement: Ratio of breadth and length of a hall is 2: 5. Complete the following table that shows some possible breadths and lengths of the hall.**

Breadth of the hall (in meters) | 2 | ☐ | 40 |

Length of the hall (in meters) | 5 | 50 | ☐ |

Ans:- (i)

We have 2 : 5 :: ☐ : 50

The cross multiplication

= 2 x 50 / 5 x ☐

☐ = 100 / 5

☐ = 20

The breadth = 20 m**(ii)**

We also have 2 : 5 :: 40 : ☐

The cross multiplication

= 2 x ☐ / 5 x 40

☐ = 200 / 2

☐ = 100

The length = 100 m

**Q.14. Divide 20 pens between Sheela and Sangeeta in the ratio of 3: 2.**

Ans:-

Sum of these terms = 3 + 2 = 5

Total number of pen = 20

Number of pens having with Sheela = 3 / 5 × 20

= 3 x 4 = 12

Number of pens having with Sangeeta = 2 / 5 × 20

= 2 x 4 = 8

Thus Sheela gets 12 pens and Sangeeta gets 8 pens.

**Q.15. Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.**

Ans:-

Ratio of ages = 15 : 12 = 5 : 4

mother wants to divide ₹ 36 in the ratio of 5: 4

∴ Sum = 5 + 4 = 9

The amount Shreya get** = **5 / 9 × 36

= 5 x 4 = 20

The amount Sangeeta get = 4 / 9 × 36

= 4 x 4 = 16

Therefore Shreya will get ₹ 20 and Sangeeta will get ₹ 16

**Q.16. The present age of the father is 42 years and that of his son is 14 years. Find the ratio of****(a) Present age of father to the present age of the son****(b) Age of the father to the age of the son, when the** **son was 12 years old.****(c) Age of father after 10 years to the age of son after 10 years.****(d) Age of father to the age of son when father was 30 years old.**

Ans:-

The present age of the father = 42 years.

The present age of his son = 14 years.

(a) Ratio of the present age of father to the present age of the son

= 42 / 14

42 ÷ 14 / 14 ÷ 14

= 3 / 1

The ratio = 3 : 1**(b) **The son was 12 years old 2 years ago.

So the age of father 2 years ago will be

= 42 – 2 = 40 years

The ratio of the father’s age to the son’s age

= 40 / 12

40 ÷ 4 / 12 ÷ 4

= 10 / 3

The ratio = 10 : 3**(c) **After ten years age of father = 42 + 10 = 52 years

After 10 years age of son = 14 + 10 = 24 years

The ratio of the father’s age to the son’s age

= 52 / 24

52 ÷ 4 / 24 ÷ 4

= 13 / 4

The ratio = 13 : 4**(d)** 12 years ago, age of father was 30

At that time age of son = 14 – 12 = 2 years

The ratio of the son’s age to the age

= 30 / 2

30 ÷ 2 / 2 ÷ 2

= 15 / 1

The ratio = 15 : 2

### NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion Exercises 12.2

NCERT Class 6 Maths Chapter 12 Solutions ex 12.2

**Q.1. Determine if the following are in proportion.****(a) 15, 45, 40, 120****(b) 33, 121, 9, 96****(c) 24, 28, 36, 48****(d) 32, 48, 70, 210****(e) 4, 6, 8, 12****(f) 33, 44, 75, 100**

Ans:-**(a) **15, 45, 40, 120

= 15 / 45

15** ÷ **15 / 45 ÷ 15 = 1 / 3

=40 / 120

40 ÷ 40 / 120 ÷ 40 = 1 / 3

∴ 15 : 45 :: 40 : 120

∴ 15, 45, 40 and 120 are in proportion**(b)** 33, 121, 9, 96

= 33 / 121

33 ÷ 11 / 121 ÷ 11 = 3 / 11

9 / 96

9 ÷ 3 / 96 ÷ 3 = 3 / 32

∴ 33:121 ≠ 9: 96

∴ 33, 121, 9 and 96 are in proportion.**(c)** 24, 28, 36, 48

= 24 / 28

24 ÷ 4 / 28 ÷ 4 = 6 / 7

= 36 / 48

36 ÷ 12 / 48 ÷ 12 = 3 / 4

∴ 24: 28 ≠ 36:48

∴ 24, 28, 36 and 48 are not in proportion**(d)** 32, 48, 70, 210

= 32 / 48

32 ÷ 16 / 48 ÷ 16 = 2 / 3

= 70 / 210

70 ÷ 70 / 210 ÷ 70 = 1 / 3

∴ 32: 48 ≠ 70: 210

∴ 32, 48, 70 and 210 are not in proportion**(e)** 4, 6, 8, 12

= 4 / 6

4 ÷ 2 / 6 ÷ 2 = 2 / 3

= 8 / 12

8 ÷ 4 / 12 ÷ 4 = 2 / 3

∴ 4 : 6 :: 8 : 12

∴ 4, 6, 8 and 12 are in proportion.**(f)** 33, 44, 75, 100

= 33 / 44

33 ÷ 11 / 44 ÷ 11 = 3 / 4

= 75 / 100

75 ÷ 25 / 100 ÷ 25 = 3 / 4

∴ 33 : 44 : : 75 : 100

∴ 33, 44, 75 and 100 are in proportion.

**Q.2. Write True (T) or False ( F ) against each of the following statements :****(a) 16 : 24 :: 20 : 30****(b) 21: 6 :: 35 : 10****(c) 12 : 18 :: 28 : 12****(d) 8 : 9 :: 24 : 27****(e) 5.2 : 3.9 :: 3 : 4****(f) 0.9 : 0.36 :: 10 : 4**

Ans:-**(a) **16: 24 :: 20: 30

= 16 / 24

16 ÷ 8 / 24 ÷ 8 = 2 / 3

= 20 / 30

20 ÷ 10 / 30 ÷ 10 = 2 / 3

∴ The given statement (a) → (T)**(b)** 21: 6:: 35: 10

= 21 / 6

21 ÷ 3 / 6 ÷ 3 = 7 / 2

= 35 / 10

35 ÷ 5 / 10 ÷ 5 = 7 / 2

∴ The given statement (b) → (T)**(c) **12: 18 :: 28: 12

= 12 / 18

12 ÷ 6 / 18 ÷ 6 = 2 / 3

= 28 / 12

28 ÷ 4 / 12 ÷ 4 = 7 / 3

∴ The given statement (c) → (F)**(d)** 8: 9:: 24: 27

= 8 / 9

8 ÷ 1 / 9 ÷ 1 = 8 / 9

= 24 / 27

24 ÷ 3 / 27 ÷ 3 = 8 / 9

The given statement (d) → (T)**(e)** 5.2: 3.9:: 3: 4

= 5.2 / 3.9

5.2 ÷ 13 / 3.9 ÷ 13 = 0.4 / 0.3

= 3 / 4

3 ÷ 1 / 4 ÷ 1 = 3 / 4

The given statement (e) → (F)**(f)** 0.9: 0.36:: 10: 4

= 0.9 / 0.36 or 90 /36

90 ÷ 18 / 36 ÷ 18 = 5 / 2

= 10 / 4

10 ÷ 2 / 4 ÷ 2 = 5 / 2

∴ The given statement (f) → (T)

**Q.3. Are the following statements true?****(a) 40 persons : 200 persons = ₹ 15 : ₹ 75****(b) 7.5 litres : 15 litres = 5 kg : 10 kg****(c) 99 kg : 45 kg = ₹ 44 : ₹ 20****(d) 32 m : 64 m = 6 sec : 12 sec****(e) 45 km : 60 km = 12 hours : 15 hour**

Ans:-**(a) **40 persons : 200 persons = ₹ 15 : ₹ 75

= 40 / 200

40 ÷ 40 / 200 ÷ 40 = 1 / 5

= 15 / 75

15 ÷ 15 / 75 ÷ 15 = 1 / 5

∴ Statement (a) is true.**(b)** 7.5 litres : 15 litres = 5 kg : 10 kg

= 7.5 / 15 or 75 / 150

75 ÷ 75 / 150 ÷ 75 = 1 / 2

= 5 / 10

5 ÷ 5 / 10 ÷ 5 = 1 / 2

∴ Statement (b) is true.**(c) **99 kg : 45 kg = ₹ 44 : ₹ 20

= 99 / 45

99 ÷ 9 / 45 ÷ 9 = 11 / 5

= 44 / 20

44 ÷ 4 / 20 ÷ 4 = 11 / 5

∴ Statement (c) is true.**(d)** 32 m : 64 m = 6 sec : 12 sec

= 32 / 64

32 ÷ 32 / 64 ÷ 32 = 1 / 2

= 6 / 12

6 ÷ 6 / 12 ÷ 6 = 1 / 2

∴ Statement (d) is true.**(e)** 45 km : 60 km = 12 hours : 15 hours

= 45 / 60

45 ÷ 15 / 60 ÷ 15 = 3 / 4

= 12 / 15

12 ÷ 3 / 15 ÷ 3 = 4 / 5

∴ Statement (e) is not true.

**Q.4. Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.****(a) 25 cm : 1 m and ₹ 40 : ₹ 160****(b)39 litres : 65 litres and 6 bottles : 10 bottles****(c) 2 kg : 80 kg and 25 g : 625 g****(d) 200 mL : 2.5 litre and ₹ 4 : ₹ 50**

Ans:-**(a) **25 cm : 1 m and ₹ 40 : ₹ 160

25 cm : 1 m = 25 cm : 100 cm [∵ 1 m = 100 cm]

= 25 / 100

25 ÷ 25 / 100 ÷ 25 = 1 / 4

= 40 / 160

40 ÷ 40 / 160 ÷ 40 = 1 / 4

∴ The given ratios are in proportion.

Extreme terms are 25 cm and ₹ 160.

Middle terms are 1 m and ₹40.**(b)** 39 litres : 65 litres and 6 bottles : 10 bottles

= 39 / 65

39 ÷ 13 / 65 ÷ 13 = 3 / 5

= 6 / 10

6 ÷ 2 / 10 ÷ 2 = 3 / 5

∴ The given ratios are in proportion.

Extreme terms are 39 litres and 10 bottles.

Middle terms are 65 litres and 6 bottles.**(c)** 2 kg : 80 kg and 25 g : 625 g

= 2 / 80

2 ÷ 2 / 80 ÷ 2 = 1 / 40

= 25 / 625

25 ÷ 25 / 625 ÷ 25 = 1 / 25

∴ The given ratios are not in proportion.**(d)** 200 mL : 2.5 litre and ₹ 4 : ₹ 50

200 mL : 2.5 litres = 2.5 litres = 2.5 x 1000 mL = 2500 mL

= 200 / 2500

200 ÷ 100 / 2500 ÷ 100 = 2 / 25

= 4 / 50

4 ÷ 2 / 50 ÷ 2 = 2 / 25

∴ The given ratios are in proportion.

Extreme terms are 200 mL and ₹ 50

Middle terms are 2.5 liters and ₹ 4.

### NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion Exercises 12.3

Ch 12 Class 6 Maths Solutions ex 12.3

**Q.1. If the cost of 7 m of cloth is ₹ 1470, find the cost of 5 m of cloth.**

Ans:-

cost of 7 m of cloth = ₹294

Cost of 1 m of cloth = ₹ 294 / 7

Cost of 5 m of cloth = ₹(294 / 7 x 5)

= ₹(42 x 5) = ₹ 210

Thus, the required cost = ₹ 210

**Q.2. Ekta earns ₹ 3000 in 10 days. How much will she earn in 30 days?**

In 10 days Ekta earn ₹ 3000

In 1 day Ekta will earn ₹ 3000 / 10

In 30 days Ekta will earn ₹(3000 / 10 x 30)

= ₹(300 x 30) = ₹ 9000

Thus the money earned by Ekta in 30 days = ₹9000

**Q.3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.**

Ans:-

In the last 3 days the rain falls = 276 mm.

In 1 day the rain falls = (276 / 3) mm.

in 7 days the rain will fall = (276 / 3 x 7) mm.

= (92 x 7) = 644 mm or 64.4 cm [∵ 1 cm = 10 mm]

Thus, the rain fall in week = 64.4 cm.

**Q.4. The cost of 5 kg of wheat is ₹ 91.50.****(a) What will be the cost of 8 kg of wheat?****(b) What quantity of wheat can be purchased in ₹ 183?**

Ans:-**(a)** Cost of 5 kg of wheat = ₹ 91.50

Cost of 1 kg of wheat = ₹ 91.50 / 5

Cost of 8 kg of wheat = ₹( 91.50 / 5 x 8)

= ₹(18.30 x 8) = ₹ 146.40

Thus, the required cost = ₹ 146.40**(b)**

₹ 91.50 Wheat purchased = 5 kg

₹ 1 Wheat purchased = (5 / 91.50) kg

₹ 183 wheat purchased = [(5 / 91.50) × 183]

= (0.054 x 183) = 10kg

**Q.5. The temperature dropped 15 degree celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?**

Ans:-

30 days temperature drop = 15^{0} C

1 day temperature drop = 15 / 30

10 days temperature drop in next = (15 / 30 × 10)

= (1 / 2) × 10 = 5^{0} C

**Q.6. Shaina pays ₹ 15000 as rent for 3 months. How much does she have** **to pay for a whole year, if the rent per month remains the same?**

Ans:-

3 months rent paid by Shaina = ₹ 15000

1 month rent paid = ₹ (15000 / 3)

12 months rent paid = ₹ (15000 / 3 x 12)**= **₹ (5000 x 12) = ₹ 60000

**Q.7. Cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹ 90?**

Ans:-

∵ 1 dozen = 12 units

∴ 4 dozen of bananas = 12 x 4 = 48 bananas

₹ 180 is the cost of = 48 bananas

₹ 1 is the cost of = (48 / 180) bananas

₹ 90 is the cost of = (48 / 180 x 90) bananas

= (4 / 15 x 90) = 10 bananas

**Q.8. The weight of 72 books is 9 kg. What is the weight of 40 such books?**

Ans:-

Weight of 72 books = 9 kg

Weight of 1 books = (9 / 72) kg

Weight of 40 books = (9 / 72 x 40) kg

= (1 / 8 x 40) = 5 kg

**Q.9. A truck requires 108 liters of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?**

Ans:-

To cover 594 km, the diesel required = 108 liters.

To cover 1 km, the diesel required = (108 / 594) liters

To cover 1650 km, the diesel required = (108 / 594 x 1650) liters

= (2 / 11 x 1650) = 300 liters

**Q.10. Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper?**

Ans:-

Cost of 10 pen = ₹150

Cost of 1 pen = ₹ (150 / 10) = ₹ 15

For Manish,

Cost of 7 pens = ₹ 84

Cost of 1 pen = ₹ (84 / 7) = ₹12

∴ Pens purchased by Manish are cheaper than Raju

**Q.11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?**

Ans:-

runs made by Anish in 6 overs = 42

runs made by him in 1 over = (42 / 6) = 7 runs.

runs made by Anup in 7 overs = 63

runs made by him in 1 over = (63 / 7) = 9 runs.

∴ 9 runs > 7 runs.

Thus, Anup has made more runs.

Class 6 Science NCERT Notes | Class 6 Complete Study Material |

NCERT Solutions for Class 6 Science | Class 6 Maths Chapter 6 Solutions |

NCERT Solutions For Class 6 Maths Chapter 12 Ratio and Proportion are based on NCERT Books. You can download ncert book for class 6 maths chapter 12. Ch 12 class 6 maths solutions are designed by our subject expert team.

## What We Learn?

- For comparing quantities of the same type, we commonly use the method of taking difference between the quantities.
- In many situations, a more meaningful comparison between quantities is made by using division, i.e. by seeing how many times one quantity is to the other quantity. This method is known as comparison by ratio. For example, Isha’s weight is 25 kg and her father’s weight is 75 kg. We

say that Isha’s father’s weight and Isha’s weight are in the ratio 3 : 1. - For comparison by ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit before the ratio is taken.
- The same ratio may occur in different situations.
- Note that the ratio 3 : 2 is different from 2 : 3. Thus, the order in which quantities are

taken to express their ratio is important. - A ratio may be treated as a fraction, thus the ratio 10 : 3 may be treated as10/3.
- Two ratios are equivalent, if the fractions corresponding to them are equivalent. Thus, 3 : 2 is equivalent to 6 : 4 or 12 : 8.
- A ratio can be expressed in its lowest form. For example, ratio 50 : 15 is treated as50/15; in its lowest form 50/15=10/3 . Hence, the lowest form of the ratio 50 : 15 is 10 : 3.
- Four quantities are said to be in proportion, if the ratio of the first and the second quantities is equal to the ratio of the third and the fourth quantities. Thus, 3, 10,15, 50 are in proportion, since 3/10=15/50 We indicate the proportion by 3:10 :: 15:50, it is read as 3 is to 10 as 15 is to 50. In the above proportion, 3 and 50 are the extreme terms and 10 and 15 are the middle terms.
- The order of terms in the proportion is important. 3, 10, 15 and 50 are in proportion,but 3, 10, 50 and 15 are not, since 3/10 is not equal to 50/15.
- The method in which we first find the value of one unit and then the value of the required number of units is known as the unitary method. Suppose the cost of 6 cans is 210rs.To find the cost of 4 cans, using the unitary method, we first find the cost of 1 can. It is 210/6= 35. From this, we find the price of 4 cans as 35*4=140.

###### What are Ratio and Proportion?

The ratio is used for comparing two quantities of the same kind. The ratio formula for two numbers says a and b is given by a : b or a/b. When two or more such ratios are equal, they are said to be in proportion.

###### What are the 4 types of proportion?

**There are four types of proportion.**

Direct Proportion.

Inverse Proportion.

Compound Proportion.

Continued Proportion.