Samacheer Class 6 Maths - Ratio and Proportion

(i) 15 : 20
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
= 3 : 4(ii) 32 : 24
= \(\frac{32}{24}\)
= \(\frac{4}{3}\)
= 4 : 3
(iii) 7 : 15
(iv) 12 : 27
= \(\frac{12}{27}\)
= \(\frac{4}{9}\)
= 4 : 9
(v) 75 : 100
= \(\frac{75}{100}\)
= \(\frac{3}{4}\)
= 3 : 4

Total no of students = 50
No of boys = 30
No of girls = 50 – 30 = 20
(i) Ratio of boys to girls = 30 : 20
= \(\frac{30}{20}\)
= \(\frac{3}{2}\)
= 3 : 2
(ii) Ratio of girls to the total number of students = 20 : 50
= \(\frac{20}{50}\)
= \(\frac{2}{5}\)
= 2 : 5
(iii) Ratio of boys to the total no of students = 30 : 50
= \(\frac{30}{50}\)
= \(\frac{3}{5}\)
= 3 : 5Objective Type Questions

(i) 3 : 23 : 2 = 6 : 4 = 9 : 6
(ii) 1 : 61 : 6 = 2 : 12 = 3 : 18
(iii) 5 : 45 : 4 = 10 : 8 = 15 : 12

(i) 4 : 5 (or) 8 : 154 : 5 > 8 : 15
(ii) 3 : 4 (or) 7 : 87 : 8 > 3 : 4
(iii) 1 : 2 (or) 2 : 1
1 : 2 = \(\frac{1}{2}\)
2 : 1 = \(\frac{2}{1}\)
= 2
\(\frac{2}{1}\) > \(\frac{1}{2}\)
2 : 1 > 1 : 2

(i) Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = 20
1 part = \(\frac{20}{5}\) = 4
3 parts = 3 × 4 = 12
2 parts = 2 × 4 = 8
20 can be divided in the form as 12, 8.
(ii) Ratio = 4 : 5
Sum of the ratio = 4 + 5 = 9
9 parts = 27
1 part = \(\frac{27}{9}\) = 3
4 parts = 4 × 3 = 12
5 parts = 5 × 3 =15
27 can be divided in the form as 12, 15.
(iii) 40 in the ratio 6 : 14
Ratio = 6 : 14
Sum of the ratio = 6 + 14 = 20
20 parts = 40
1 part = \(\frac{40}{20}\) = 2
6 parts = 2 × 6 = 12
14 parts = 2 × 14 = 28
40 can be divided in the form as 12, 28.

Allotted amount = Rs 4000
Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = Rs 4000
1 part = Rs \(\frac{4000}{5}\) = Rs 800
Provisions : Vegetables = 3 : 2
3 parts = 3 × Rs 800 = Rs 2400
2 parts = 2 × Rs 800 = Rs 1600
Amount spent for provisions = Rs 2400
Amount spent for vegetables = Rs 1600

Total length = 63 cm Ratio = 3 : 4
Sum of the ratio = 3 + 4 = 7
7 parts = 63 cm
1 part = \(\frac{63}{7}\) = 9 cm
3 parts = 3 × 9 cm = 27 cm
4 parts = 4 × 9 cm = 36 cm
∴ 63 cm can be divided into the parts as 27 cm and 36 cm.
Objective Type Questions

(i) 3, 9, 4, 12
Here product of extremes = 3 × 12 = 36
Product of means = 9 × 4 = 36
3 : 9 : : 4 : 12
(ii) Also if we take 9, 3, 12, 4
Product of extremes = 9 × 4 = 36
Product of means = 3 × 12 = 36
9 : 3 : : 12 : 4

Yes, 12 : 24 : : 18 : 36
Because product of extremes 12 × 36 = 432
Product of means = 24 × 18 = 432
12 : 24 :: 18 : 36.

(i) 78 : 130 :: 12 : 20
Extreme terms are 78 and 20.
Mean terms are 130 and 12.
Product of Extremes = 78 × 20 = 1560
Product of Means = 130 × 12 = 1560
Product of Extremes = Product of means
It is in proportion.
(ii) 400 : 50 : : 25 : 625
Product of extremes = 400 × 625 = 2,50,000
Product of means = 50 × 25 = 1250
Here product of extremes ≠ product of means
400 : 50 and 25 : 625 are not in proportion.

6480 : 756, 60 : 7
Product of the means = 756 × 60 = 45360
Product of the extremes = 6480 × 7 = 45360
ad = bc
∴ They are in proportionObjective Type Questions

In 2 hours, pages read = 20
In 1 hour, pages read = \(\frac{20}{2}\) = 10
In 8 hours, pages read = 10 × 8
= 80 pages

In 1 hour (60 minutes), distance covered = 6 km
In 1 minute, distance covered = \(\frac{6 km}{60 min}\) = \(\frac{6000 m}{60}\) = 100 m
In 20 minutes, distance covered = 20 × 100 m = 2000 m = 2 km

Total points scored = 84 Ratio = 10 : 11
Sum of the ratio = 10 + 11 = 21
21 parts = 84 points
1 part = \(\frac{84}{21}\) = 4 points
Kavitha = 11 parts
Kaarmugilan = 10 parts
Foints scored by Kavitha = 11 parts = 11 × 4 points = 44 points

Karmegam Runs made in 9 overs = 54
Runs made in 1 over = \(\frac{54}{9}\) = 6 runs
Asif Runs made in 11 overs = 77
Runs made in 1 over = \(\frac{77}{11}\) = 7 runs
∴ Asif’s run rate is better than Karmegam.

Myself
Cost of 6 apples = Rs 90
Cost of 1 apple = \(\frac{Rs 90}{6}\) = Rs 15
Friend’s purchase
Cost of 5 apples = Rs 70
Cost of 1 apple = \(\frac{70}{5}\) = Rs 14
∴ Friend’s purchase is better than mine.Objective Type Questions

(i) The Elephant: the Lion
= 20 : 80 = \(\frac{20}{80}\) = \(\frac{1}{4}\) = 1 : 4
(ii) the Lion : the Cheetah
= 80 : 100 = \(\frac{80}{100}\) = \(\frac{4}{5}\) = 4 : 5
(iii) the Elephant: the Cheetah
= 20 : 100 = \(\frac{20}{100}\) = \(\frac{1}{5}\) = 1 : 5
The ratio of Elephant to Cheetah is the least.

Administrators : teachers : students = 5 : 50 : 1500 = 1 : 10 : 300
If the school grows to 1800 students then 10 parts = teachers
1 part = administrators
300 parts = 1800
1 part = \(\frac{1800}{300}\) = 6
10 parts = 6 × 10 = 60
So, if the school grows to 1800 students the new ratio is administrators : teachers: students
6 : 60 : 1800

Let x be the part owned by C then A : B : C = 2(2x) : 2x : x = 4x : 2x : x
A : B = 4x : 2x = 2 : 1
B : C = 2x : x = 2 : 1
A : B : : B : C. i.e, They are in proportion.

Number of times Antony brushes a day = 2
Number of times Antony brushes a week = 2 × 7 = 14
Number of times Shabeen brushes a day = 1
Number of times Shabeen brushes a week = 1 × 7 = 7
Number of times Antony brushes : Number of times Antony brushes = 14 : 7 = 2 : 1
The required ratio = 2 : 1

Red : blue = 35 : 30 = 7 : 6
Different ways (i) 7 : 6
(ii) 14 : 12;
(iii) 21 : 18;
(iv) 28 : 24

Std VI – teachers: students = 6 : 12 = 1 : 2
Std VII – teachers : students = 9 : 27 = 1 : 3
Std VIII – teachers : students = 4 : 16 = 1 : 4
Std VIII has the least ratio.

6 : ……. = …….. : 15
Product of the extremes = 6 × 15 = 90
Set of suitable numbers
1 and 90, 2 and 45, 3 and 30, 5 and 18, 6 and 15