Gujarat BoardĀ GSEB Textbook Solutions Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2
Question 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
Solution:
(a) 15, 45, 40, 120
āµ Ratio of 15 and 45 = 15 : 45
\(\frac { 15 }{ 45 }\) = \(\frac{15\div 15}{45 \div 15}\) =\(\frac{1}{3}\) = 1 : 3
[āµ HCF of 15 and 45 is 15]
Ratio of 40 and 120 = 40 : 120
\(\frac { 40 }{ 120 }\) = \(\frac{40\div 40}{120 \div 40}\)
[āµ HCF of 40 and 120 is 40]
= \(\frac { 1 }{ 3 }\) = 1 : 3
ā“ 15 : 45 : : 40 : 120
i.e., 15, 45 40, and 120 are in proportion.
(b) 33, 121, 9, 96
Here, ratio of 33 and 121 = 33 : 121
\(\frac { 33 }{ 121 }\) = \(\frac{33\div 11}{121 \div 11}\)
[āµ HCF of 33 and 121 is 11]
\(\frac { 3 }{ 11 }\) = 3 : 11
Ratio of 9 and 96= 9 : 96
\(\frac { 9 }{ 96 }\) = \(\frac{9\div 3}{96 \div 3}\) =\(\frac{3}{32}\) = 3 : 32
[ āµ HCF of 9 and 96 is 3]
Since, 3 : 11 ā 3 : 32
ā“ 33, 121, 9 and 96 are not in proportion.
(c) 24, 28, 36, 48
āµ Ratio of 24 and 28 = 24 : 28
\(\frac { 24 }{ 48 }\) = \(\frac{24\div 4}{28 \div 4}\) =\(\frac{6}{7}\) = 6 : 7
[āµ HCF of 24 and 28 is 4]
Ratio of 36 and 48 = 36 : 48
\(\frac { 36 }{ 48 }\) = \(\frac{36\div 12}{48 \div 12}\) =\(\frac{3}{4}\) = 3 : 4
[āµ HCF of 36 and 48 is 12]
i.e. 6 : 7 ā 3 : 4
or 24 : 28 ā 36 : 48
ā“ 24, 28, 36 and 48 are not in proportion.
(d) 32, 48, 70, 210
Ratio of 32 and 48 = 32 : 48 = \(\frac { 32 }{ 48 }\) = \(\frac{32\div 16}{48 \div 16}\)
[ āµ HCF of 32 and 48 is 16]
\(\frac { 2 }{ 3 }\) = 2 : 3
Ratio of 70 and 210 = \(\frac { 70 }{ 100 }\) = \(\frac{70\div 70}{210 \div 70}\) =\(\frac{1}{3}\) = 1 : 3
Since 2 : 3 ā 1 : 3
[āµ HCF of 70 and 210 is 70]
i.e. 32 : 48 ā 70 : 210
ā“ 32, 48, 70, 210 are not in proportion.
(e) 4, 6, 8, 12
āµ Ratio of 4 and 6 = 4 : 6
\(\frac { 4 }{ 6 }\) = \(\frac{4\div 2}{6 \div 2}\) =\(\frac{2}{3}\) = 2 : 3
āµ Ratio of 8 and 12 = 8 : 12
[HCF of 4 and 6 is 2]
= \(\frac { 8 }{ 12 }\) = \(\frac{8\div 4}{12 \div 4}\) =\(\frac{2}{3}\) = 2 : 3
ā“4 : 6 = 8 : 12 [HCF of 8 and 12 is 4]
i.e., 4,6,8,12 are in proportion.
(f) 33, 44, 75, 100
āµ Ratio of 33 and 44 = 33 : 44 = \(\frac { 33 }{ 44 }\) = \(\frac{33\div 11}{44 \div 11}\)
[āµ HCF of 33 and 44 is 11]
\(\frac { 3 }{ 4 }\) = 3 : 4
ā“ 33 : 44 = 75 : 100 or 33, 44, 75, 100 are in proportion.
Question 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
Solution:
(a) 16 : 24 : : 20 : 30
We have, 16: 24 = \(\frac { 16 }{ 24 }\) = \(\frac{16\div 8}{24 \div 8}\) =\(\frac{2}{3}\) = 2 : 3
[āµ HCF of 16 and 24 is 8]
and 20 : 30 = \(\frac { 20 }{ 30 }\) = \(\frac{20\div 10}{30 \div 10}\) =\(\frac{2}{3}\) = 2 : 3
[āµ HCF of 20 and 30 is 10]
Since, 16 : 24 = 20 : 30
ā“ 16 : 24 : : 20 : 30 is true.
(b) 21 : 6 : : 35 : 10
We have 21 : 6 = \(\frac { 21 }{ 6 }\) = \(\frac{21\div 3}{6 \div 3}\) =\(\frac{7}{2}\) = 7 : 2
[āµ HCF of 21 and 6 is 3]
and 35 : 10 = \(\frac { 35 }{ 10 }\) = \(\frac{35\div 5}{10 \div 4}\) =\(\frac{7}{2}\) = 7 : 2
[āµ HCF of 35 and 10 is 5]
Since, 21 : 6 = 35 : 10
ā“ 21 : 6 : : 35 : 10 is true.
(c) 12 : 18 : : 28 : 12
We have, 12 : 18 =\(\frac{12}{18}\) = \(\frac{12\div 6}{18 \div 6}\) =\(\frac{2}{3}\) = 2 : 3
[āµ HCF of 12 and 18 is 6]
and 28 : 12 = \(\frac{28}{12}\) = \(\frac{28\div 4}{12 \div 4}\) =\(\frac{7}{3}\) = 7 : 3
[āµ HCF of 28 and 12 is 4]
Since, 2 : 3 ā 7 : 3
ā“ 12 : 18 : : 28 : 12 is not true.
(d) 8 : 9 : : 24 : 27
We have, 24 : 27 = \(\frac{24}{27}\) = \(\frac{24\div 3}{27 \div 3}\) =\(\frac{8}{9}\) = 8 : 9
[āµ HCF of 24 and 27 is 3]
Since, 8 : 9 = 24 : 27
ā“ 8 : 9 : : 24 : 27 is true.
(e) 5.2 : 3.9 : : 3 : 4
We have, 5.2: 3.9 = \(\frac { 5.2 }{ 3.9 }\) = \(\frac { 52 }{ 10 }\) Ć·\(\frac { 39 }{ 10 }\) = \(\frac { 52 }{ 10 }\) Ć \(\frac { 10 }{ 36 }\) = \(\frac { 52 }{ 39 }\)= \(\frac{52\div 13}{39 \div 13}\)
[āµ HCF of 52 and 39 is 13]
\(\frac { 4 }{ 3 }\) = 4 : 3
and it is not equal to 3 : 4
ā“ 5.2: 3.9 ā 3: 4
Thus, 5.2: 3.9: : 3: 4 is not true.
(f) 0.9 : 0.36 : : 10 : 4
We have: 0.9 : 0.36
\(\frac { 9 }{ 10 }\) Ć· \(\frac { 100 }{ 36 }\) = \(\frac { 90 }{ 36 }\) = \(\frac{90\div 18}{36 \div 18}\)
[āµ HCF of 90 and 36 is 18]
\(\frac { 5 }{ 2 }\) = 5 : 2
Thus, 0.9 : 0.36 = 10 : 4
or 10 : 4 =\(\frac{10}{4}\) = \(\frac{10\div 2}{4 \div 2}\) = \(\frac{5}{2}\) = 5 : 2
[āµ HCF of 10 and 4 is 2]
Since, 0.9 : 0.36 = 10 : 4
ā“ 0.9 : 0.36 : : 10 : 4 is true.
Question 3.
Are the following statements true?
(a) 40 persons : 200 persons = ā¹ 15 : ā¹ 75
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
(c) 99 kg : 45 kg = ā¹ 44 : ā¹ 20
(d) 32 m : 64 m = 6 sec : 12 sec
(e) 45 km : 60 km = 12 hours : 15 hours
Solution:
(a) 40 persons : 200 persons = ā¹ 15 : ā¹ 75
āµ 40 persons : 200 persons
\(\frac { 40 persons }{ 200 persons }\) = \(\frac { 40 }{ 200 }\) = \(\frac{40\div 40}{200 \div 40}\) = \(\frac { 1 }{ 5 }\)
[āµ HCF of 40 and 200 is 40]
= 1 : 5
and ā¹ 15 : ā¹ 75
\(\frac { ā¹ 15 }{ ā¹ 75 }\) = \(\frac { 15 }{ 75 }\) = \(\frac{15\div 15}{75 \div 15}\) = \(\frac { 1 }{ 5 }\)
[āµ HCF of 15 and 75 is 15]
= 1 : 5
ā“ The given ratios are equal.
Thus, 40 persons : 200 persons = ā¹ 15 : ā¹ 75 is true.
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
We have, 7.5 litres : 15 litres
\(\frac { 75 }{ 15 }\) = \(\frac { 75 }{ 150 }\) = \(\frac{75\div 75}{150 \div 75}\) = \(\frac { 1 }{ 2 }\)
[āµ HCF of 75 and 150 is 75]
= 1 : 2
and 5 kg : 10 kg = \(\frac { 5}{ 10 }\) = \(\frac { 1 }{ 2 }\) = 1 : 2
āµ The two ratios are equal
ā“ 7.5 litres : 15 litres = 5 kg : 10 kg is true.
(c) 99 kg : 45 kg = ā¹ 44 : ā¹ 20
We have, 99 kg : 45 kg = \(\frac { 99 kg }{ 45 kg }\) =\(\frac{99\div 9}{45 \div 9}\) = \(\frac { 11 }{ 5 }\)
[āµ HCF of 99 and 45 is 9]
= 11 : 5
and ā¹ 44 : ā¹ 20 = \(\frac { ā¹ 44 }{ ā¹ 20 }\) = \(\frac { 44 }{ 20 }\) = \(\frac{44\div 4}{20 \div 4}\) = \(\frac { 11 }{ 5 }\)
[āµ HCF of 44 and 20 is 4]
= 11 : 5
Since, the two ratios are equal
ā“ 99 kg : 45 kg = ā¹ 44 : ā¹ 20 is true.
(d) 32 m : 64 m = 6 sec : 12 sec
We have, 32 m : 64
m = \(\frac { 32 m }{ 64 m }\) = \(\frac{32\div 32}{64 \div 32}\) = \(\frac { 1 }{ 2 }\)
[āµ HCF of 32 and 64 is 32]
= 1 : 2
and 6 sec : 12 sec
\(\frac { 6 sec }{ 12 sec }\) = \(\frac{6\div 6}{12 \div 6}\) = \(\frac { 1 }{ 2 }\)
[āµ HCF of 6 and 12 is 6]
= 1 : 2
Since, the two ratios are equal
ā“ 32 m : 64 m = 6 sec : 12 sec is true.
(e) 45 km : 60 km = 12 hours : 15 hours
Since, 45 km : 60 km
\(\frac { 45 sec }{ 60 sec }\) = \(\frac { 45 }{ 60 }\) = \(\frac{12\div 3}{15 \div 3}\)= \(\frac { 3 }{ 4 }\) = 3 : 4
and 12 hours : 15 hours
\(\frac { 12 hours }{ 15 hours }\) = \(\frac { 12 }{ 15 }\) = \(\frac{12\div 3}{15 \div 3}\)= \(\frac { 4 }{ 5}\) = 4 : 5
Since, 3 : 4 ā 4 : 5
Thus, 45 km : 60 km = 12 hours : 15 hours is false.
Question 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 : l m and ā¹ 40 : ā¹ 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 ā¹ 4 : ā¹ 50
Solution:
(a) 25 cm : 1 m and ā¹ 40 : ā¹ 160
We have, 25 cm : l m
= \(\frac { 25 cm }{ l m }\) = \(\frac { 25 cm }{ 100 cm }\) = \(\frac{25\div 25}{100 \div 25}\) = \(\frac { 1 }{ 4 }\) = 1 : 4
and ā¹ 40 : ā¹ 160
= \(\frac { ā¹ 40 }{ ā¹ 160 }\) = \(\frac { 40 }{ 160 }\)
[āµ HCF of 40 and 160 is 40]
\(\frac{40\div 40}{160 \div 40}\) = \(\frac { 1 }{ 4 }\) = 1 : 4
Since, both the ratios are equal, hence, they form a proportion.
Now, middle terms are 1 m and ā¹ 40 and extreme terms are 25 cm and ā¹ 160.
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
We have, 39 litres : 65 litres
\(\frac { 39 litres }{ 65 litres }\) = \(\frac { 39 }{ 65 }\) = \(\frac{39\div 13}{65 \div 13}\) = \(\frac { 3 }{ 5 }\) = 3 : 5
[āµ HCF of 39 and 65 is 13]
and 6 bottles : 10 bottles
\(\frac { 6 bottles }{ 10 bottles }\) = \(\frac { 6 }{ 10 }\) = \(\frac{6\div 2}{10 \div 2}\) = \(\frac { 3 }{ 5 }\) = 3 : 5
[āµ HCF of 6 and 10 is 2]
Since, both the ratios are equal
ā“ 39 litres : 65 litres :: 6 bottles : 10 bottles form a proportion.
Its middle terms are 65 liters and 6 bottles. Its extreme terms are 39 liters and 10 bottles.
(c) 2 kg : 80 kg and 25 g : 625 g
We have, 2 kg : 80 kg
\(\frac{2}{80}\) = \(\frac{2\div 2}{80 \div 2}\) =\(\frac{1}{4}\) = 1 : 40
[āµ HCF of 2 and 80 is 2]
and 25 g : 625 g
\(\frac{26}{625}\) = \(\frac{25\div 25}{625 \div 25}\) =\(\frac{1}{25}\) = 1 : 25
[āµ HCF of 25 and 625 is 25]
Since, 1 : 40 ā 1 : 25
ā“ The given ratios do not form a proportion.
(d) 200 ml : 2.5 litres and ā¹ 4 : ā¹ 50
We have, 200 ml : 2.5 litres
\(\frac { 200 ml }{ 2.5 litres }\) = \(\frac { 200 ml }{ 2.5 Ć 1000 ml }\)
[āµ 1 litre = 1000 ml]
\(\frac { 200 }{ 2500 }\) = \(\frac{200\div 100}{2500 \div 100}\)
[āµ HCF of 200 and 2500 is 100]
\(\frac { 2 }{ 25 }\) = 2 : 25
and ā¹ 4 : ā¹ 50
\(\frac { ā¹ 4 }{ ā¹ 50 }\) = \(\frac { 4 }{ 50 }\) = \(\frac{4\div 2}{50 \div 2}\) = \(\frac { 2 }{ 25 }\) = 2 : 25
[āµ HCF of 4 and 50 is 2]
Since, the two ratios are equal, i.e., 200 ml : 2.5 litres = ā¹ 4 : ā¹ 50
ā“ They form a proportion.
Its middle terms are 2.5 liters and ā¹ 4. Its extreme terms are 200 ml and ā¹ 50.