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 :–
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