Ratio and Proportion Class 6 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8. In this article you will learn about RATIO . Visit official Website **CISCE** for detail information about ICSE Board Class-6 Mathematics.

Board | ICSE |

Subject | Maths |

Class | 6th |

Ch-8 | Ratio and Proportion |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-8A |

Academic Session | 2024 – 2025 |

### RATIO

**Definition:** The fraction a/b is the ratio of two non-zero numbers a and b, and it is written as a:b, or ‘a to b where a is the antecedent and b is the consequent

- The ratio is the comparison of a quantity with respect to another quantity.
- It is denoted by symbol ( : )
- Two quantities can be compared only if they are in the same unit
- Ratio must be in simplest form

**Exercise- 8A **

(Ratio and Proportion Class 6 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8)

**Que-1: Find the ratio of each of the following in simplest form :**

(i) Rs7.80 to 65 paise (ii) 15 hours to 1 day (iii) 750gm to 2kg (iv) 64ml to 1 litre (v) 18mm to 3cm (vi) 80kg to 2 quintals.

**Sol: **(i) 65 paise = Rs0.65

7.80/0.65 = 780/65

= 12/1 = 12:1

(ii) 1 days = 24 hours

15/24 = 5/8 = 5:8

(iii) 2kg = 1000gm

750/2000 = 3/8 = 3:8

(iv) 1 litre = 1000ml

64/1000 = 8/125 = 8:125

(v) 3cm = 30mm

18/30 = 3/5 = 3:5

(vi) 2 quintals = 200kg

80/200 = 2/5 = 2:5

**Que-2: Express each of the following ratios in simplest form :**

(i) 15:25 (ii) 49:35 (iii) 90:72 (iv) 146:365 (v) 111:259 (vi) (1/6):(1/8) (vii) (3/4):(4/5):(5/6) (viii) {2*(1/3)}:{3*(1/4)}:{5*(1/6)}

**Sol: **(i) 15:25 = 3:5

(ii) 49:35 = 7:5

(iii) 90:72 = 5:4

(iv) 146:365 = 2:5

(v) 111:259 = 3:7

(vi) (1/6):(1/8)

Taking LCM of 6,8 = 24

= (24/6) : (24/8)

= 4:3

(vii) (3/4):(4/5):(5/6)

Taking the LCM of 4,5,6 is 60

Multiplying the numerator by 60

= (60×3)/4 : (60×4)/5 : (60×5)/6

=(15×3) : (12×4) : (10×5)

= 45 : 48 : 50

(viii) {2*(1/3)}:{3*(1/4)}:{5*(1/6)}

= (7/3):(13/4):(31/6)

Taking the LCM of 3,4,6 is 12

Multiplying the numerator by 60

= (12×7)/3 : (12×13)/4 : (12×31)/6

= (4×7) : (3×13) : (2×31)

= 28 : 39 : 62.

**Que-3: In a school, there are 420 boys and 180 girls. Find the ratio of**

(i) girls to boys (ii) girls to total number of students (iii) boys to total number of students.

**Sol: **Total number of students = number of boys + number of girls

= 420 + 180 = 600.

(i) girls/boys

= 180/420 = 3:7

(ii) girls/total number of students

= 180/600 = 3/10

(iii) boys/total number of students

= 420/600 = 7/10

**Que-4: The total population of a village is 3540, out of which 2065 are males. Find the ratio of males to females.**

**Sol: **Given, total population of a district is 3,540 out of which 2,065 are males.

Numbers of females = (3,540−2.065)

= 1,475

∴ The ratio of males to females = 2,065/1,475

= 7/5 = 7:5.

**Que-5: Find the ratio of the price of coffee to that of tea, if coffee costs Rs48 per 100gm and tea costs Rs280 per kg.**

**Sol: **The price of coffee is ₹48 per 100 gm.

The price of tea is ₹280/10= ₹28 per 100 gm.

So the ratio of price of coffee to price of tea is

₹48 : ₹28

= 48/28

= 12/7

= 12:7

**Que-6: The ratio of tin and zinc in an alloy 3:4. In 21 gm of alloy, find the quantity of (i) tin and (ii) zinc**

**Sol: **Let T/Z = 3/4

T = No. of gms of Tin in alloy

Z = No. of gms of zinc in alloy

T+Z = 21 gms

(3Z/4) + Z = 21

7Z/4 = 21

z = 21×(4/7)

Z = 12 gms

T = (3/4)×12 = 9 gms

(i) Quantity of Tin in alloy = 9 gms.

(ii) Quantity of zinc in alloy = 12 gms.

**Que-7: Divide Rs1155 between Amit and Sanya in the ratio 7:4.**

**Sol: **We should divide it i.e 7+4=11 parts

1155/11=105

1 part is 105

7 parts is equal to 7(105) = 735

4 parts is equal to 4(105) = 420

**Que-8: Divide Rs3500 among A, B and C in the ratio 3:4:7.**

**Sol: **A’s share = 3x

B’s share = 4x

C’s share = 7x

Total money = 3500

3x+4x+7x = 3500

14x = 3500

x = 250

Hence A’s share = 3x = ₹750

B’s share = 4x = ₹1000

C’s share = 7x = ₹1750

**Que-9: Divide Rs3010 among A, B and C in such a way that A gets double of what B gets and B gets double of what C gets.**

**Sol: **Let the amount of money C gets be Rs. x

∴ Amount of money B gets = Rs. 2x

Amount of money A gets = Rs.4x

According to question,

4x+2x+x = 3010

⇒ 7x = 3010

⇒ x = 3010/7 = 430

∴ Amount of money A gets = Rs.(4×430)

= Rs.1720

Amount of money B gets = Rs.(2×430)

= Rs.860

And amount of money C get = Rs.430.

**Que-10: Divide 182 in three parts in the ratio (1/10):(1/15):(1/20).**

**Sol: **The ratios in which 182 is divided are 1:10, 1:15 and 1:20.

1:10 =1/10, 1:15 = 1/15, 1:20 = 1/20

1/10+1/15+1/20 = (6+4+3)/60 = 13/60

(1/10)×(60/13)×182 = 84

(1/15)×(60/13)×182 = 56

(1/20)×(60/13)×182 = 42

**Que-11: The sides of a triangle are in the ratio 2:3:4. If its perimeter is 54cm, find the lengths of the sides of the triangles.**

**Sol: **Let the number be x

2x+3x+4x = 54

9x = 54

x = 54/9 = 6

The sides of the triangle are

2x = 2×6 = 12

3x = 3×6 = 18

4x = 4×6 = 24

Then,

We get the sides of the triangle are 12,18 and 24.

**Que-12: The angles of a triangle are in the ratio 3:5:7. Find the measure of each angle of the triangle.**

**Sol: **Ratio in angles of a triangle is 3: 5: 7

But sum of angles of a triangle = 180°

3x+5x+7x = 180∘ (Angle sum property of a triangle)

⇒ 15x = 180°

⇒ x = 12°

∴ The angles of the triangle are 3× 12=36∘,

5× 12 = 60° and

7× 12 = 84°

**Que-13: A ratio in simplest form is 11:15. If its antecedent is 121, find the consequent.**

**Sol: **The ratio = 11:15

Let the numbers be 11x and 15x.

Given : antecedent = 121

Then,

11x = 121

x = 121/11

x = 11

Therefore, the consequent = 15x = (15*11) = 165

**Que-14: A ratio in the simplest form is 7:12. If its consequent is 144, find its antecedent.**

**Sol: **The ratio is 7:12

Let the common ratio be x

The antecedent and consequent are 7x, 12x respectively

By the problem,

=>12x = 144

=> x = 144/12

= 12

∴ Antecedent is 7x = 7×12

= 84

Hence the antecedent is 84.

**–: Ratio and Proportion Class 6 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8 :–**

**Return to :- ICSE Class -6 RS Aggarwal Goyal Brothers Math Solutions**

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