Class 6 Math Solutions WBBSE Chapter 21 Fundamental Concept Of Ratio And Proportion
We take lengths of our pen caps and lengths of pens without cap and try to take their ratio and write them in a table like below.
Let us see the picture count the number ratio in its lowest form and write the ratio.
Class 6 WBBSE Math Solutions Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21
Question 1. I have taken measurement of the floor of my living room. I see that the rectangular floor is of 8 m length and 5 m in breadth. Let’s find the ratio of its length to its breadth. Let’s also write if the ratio is ratio of greater or lesser inequality.
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Solution:
⇒ Length of the floor = 8 m.
⇒ Breadth of the floor = 5 m.
∴ Ratio of length and breadth = 8:5
It is a ratio of greater inequality.
Question 2. Sabita is making garlands of chinarose and marigold, if she made 12 garlands of chinarose and 15 of marigold, then let’s find what is the ratio of the number of garlands of chinarose and marigold. Let’s also write if the ratio is a ratio of greater or lesser inequality.
Solution:
Given
Sabita is making garlands of chinarose and marigold, if she made 12 garlands of chinarose and 15 of marigold,
⇒ Number of garlands of china rose = 12 and
⇒ Number of garlands of marigold = 15
∴ Ratio of garlands of qhinarose and marigold = 12:15 = 4:5
It is a ratio of lesser inequality.
Question 3. The ratio of my age to Sutapa’s age is 5: 6. If my age is 10 years, let’s find the age of Sutapa.
Solution:
⇒ The ratio of my age to Sutapa’s age = 5:6
⇒ or, \(\frac{\text { My age }}{\text { Sutapa’s age }}=\frac{5}{6}\)
⇒ or, \(\frac{10 \text { years }}{\text { Sutapa’s age }}=\frac{5}{6}\)
∴ 5 x Sutapa’s age = 6 x 10 years 6×10
∴ Sutapa’s age = \(\frac{6 \times 10}{5}\) years = 12 years.
Class 6 WBBSE Math Solutions
Question 4. My mother gave sweets to me and Raju. Me and Raju ate sweets in the ratio of 1:3. Let’s find how many sweets my mother gave us [Let’s try any 4 values].
⇒ \(\frac{\text { No. of sweets I ate }}{\text { No. of sweets Raju ate }}=\frac{1}{3}=\frac{2}{6}\) i.e., mother gave (2 + 6) = 8 sweets also she can give (1 + 3) = 4 sweets
Solution:
\(\frac{\text { No. of sweets I ate }}{\text { No. of sweets Raju ate }}=\frac{1}{3}=\frac{2}{6}\) = \(\frac{3}{9}=\frac{4}{12}\)
⇒ In 1st case, mother gave us = 1 +3 = 4 sweets
⇒ In 2nd case, mother gave us = 2 + 6 = 8 sweets
⇒ In 3rd case, mother gave us = 3 + 9 = 12 sweets
⇒ In 4th case, mother gave us = 4 +12 = 16 sweets
Question 5. Today 10 of us have come to the park to play. If the ratio of number of girls and boys is 2:3 in our group, let’s find how many girls and boys have come to play.
Solution:
Given
Today 10 of us have come to the park to play. If the ratio of number of girls and boys is 2:3 in our group,
⇒ Total number of students = 10
⇒ Ratio of number of girls and boys = 2:3
∴ Number of girl students = 2/5 parts
∴ Number of boy students = 3/5 parts
⇒ Number of girl students = 2/5 x 10 = 4
⇒ Number of boy students = 3/5 x 10 = 6
Question 6. Father bought 4 pairs of bananas from market. If brother and sister ate those bananas in the ratio of 1 : 3, let’s find how many bananas each of them ate.
Solution:
Given
Father bought 4 pairs of bananas from market. If brother and sister ate those bananas in the ratio of 1 : 3,
⇒ Total number of bananas = 4 pairs = 4 x 2 = 8
⇒ Ratio of bananas brother and sister ate = 1:3
∴ Number of bananas brother ate = \(\frac{1}{1+3} \times 8=\frac{1}{4} \times 8=2\)
Number of bananas sister ate = \(\frac{3}{1+3} \times 8=\frac{3}{4} \times 8=6\)
Class 6 Maths Solutions WBBSE Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.1
Question 1. Let’s find if the following numbers are in proportion or not:
1. 13, 52,30,120
Solution:
⇒ 1st term x 4th term = 13 x 120 = 1560
⇒ 2nd term x 3rd term = 52 x 30 = 1560
∴ 13: 52 :: 30:120 (In proportion)
2. 22,11,72, 36
Solution:
⇒ 1 st term x 4th term = 22 x 36 = 792
⇒ 2nd term x 3rd term = 11 x 72 = 792
∴ 22: 11:: 72: 36 (In proportion)
3. 45, 27,15, 25
Solution: 45,27, 15,25
⇒ 1st term x 4th term = 45 x 25 = 1125
⇒ 2nd term x 3rd term = 27 x 15 = 405
∴ 1st term x 4th term 2nd term x 3rd term
∴ 45, 27,15, 25 are not in proportion.
Class 6 Maths Solutions WBBSE
4. 18, 20, 27, 30
Solution:
⇒ 1st term x 4th term = 18 x 30 = 540
⇒ 2nd term x 3rd term = 20 x 27 = 540
∴ 18: 20:: 27: 30 (In proportion)
5. 11,22, 36, 72
Solution:
⇒ 1st term x 4th term = 11 x 72 = 792
⇒ 2nd term x 3rd term = 22 x 36 = 792 (In proportion) .
Question 2. Let’s write if the folowing relations are True/False.
1. 4.5 litre: 13.5 litre:: 4 kg: 12 kg
Solution:
⇒ 4.5 litre: 13.5 litre:: 4kg: 12 kg
⇒ 4.5: 13.5:: 4: 12
⇒ 1st term x 4th term = 4.5 x 12 = 54
⇒ 2nd term x 3rd term = 13.4 x 4 = 54
∴ The relation is True.
Class 6 Maths Solutions WBBSE
2. 12 km: 8 km:: 61 km:: 40 km
Solution:
⇒ 12 km: 8 km:: 60 km: 40 km
⇒ or, 12: 8:: 60: 40
⇒ 1st term x 4th term = 12 x 40 = 480
⇒ 2nd term x 3rd term = 8 x 60 = 480
∴ The relation is True.
3. 20 men: 45 men:: 180 rupee: 270 rupee
Solution:
⇒ 20 men: 45 men:: 180 rupees: 270 rupees
∴ 1 st term x 4th term = 20 x 270 = 5400
⇒ 2nd term x 3rd term = 45 x 180 = 8100
∴ 20:45 ≠ 180:270
∴ The relation is False.
4. 15m: 9m :: 35m: 21m.
Solution:
⇒ 15m: 9 m :: 35 m: 21 m
⇒ or 15:9:: 35:21
⇒ 1st term x 4th term = 15 x 21 = 315
⇒ 2nd term x 3rd term = 9 x 35 = 315
∴ The relation is True.
WBBSE Math Solutions Class 6 Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.2
Question 1. Let’s find if the following numbers are in proportion or not. If they are in proportion, let’s write all the possible proportions with them.
1. 3,15, 4, 20
Solution:
1st case: 3, 15, 4, 20
⇒ 1st term x 4th term = 3 x 20 = 60
⇒ 2nd term x 3rd term = 15 x 4 = 60
∴ 3: 15:: 4.20
2nd case: 3, 4,15, 20
⇒ 1st term x 4th term = 3 x 20 = 60
⇒ 2nd term x 3rd term = 4 x 15 = 60
∴ 3:4:: 15: 20
3rd case: 15, 3, 20, 4
⇒ 1st term x 4th term = 15 x 4 = 60
⇒ 2nd term x 3rd term = 3 x 20 = 60
∴ 15: 3:: 20: 4
4th case: 15, 20, 3, 4
⇒ 1st term x 4th term = 15×4 = 60
⇒ 2nd term x 3rd term = 20 x 3 = 60
∴ 15: 20:: 3: 4
WBBSE Math Solutions Class 6
2. 6,18,7, 21
Solution:
1st case: 6,18, 7, 21
⇒ 1st term x 4th term = 6×21 = 126
⇒ 2nd term x 3rd term = 18 x 7 = 126
∴ 6:18:: 7: 21
2nd case: 6, 7,18, 21
⇒ 1st term x 4th term = 6 x 21 =126
⇒ 2nd term x 3rd term = 7 x 18 = 126
∴ 6: 7:: 18: 21
3rd case: 18, 6, 21: 7
⇒ 1st term x 4th term = 18 x 7 = 126
⇒ 2nd term x 3rd term = 6 x 21 =126
∴ 18: 6:: 21: 7
4th case: 18, 21,6, 7
⇒ 1st term x 4th term = 18 x 7 = 126
⇒ 2nd term x 3rd term = 21 x 6 = 126
∴ 18: 21:: 6: 7
WBBSE Math Solutions Class 6
3. 5, 15, 7, 21
Solution:
1st case: 5,15, 7, 21
⇒ 1st term x 4th term = 5 x 21 =105
⇒ 2nd term x 3rd term = 15×7 = 105
∴ 5: 15:: 7: 21
2nd case: 15, 5, 21,7
⇒ 1st term x 4th term = 15 x 7 = 105
⇒ 2nd term x 3rd term = 5×21 =105
∴ 15: 5 :: 21: 7
3rd case: 5, 7, 15, 21
⇒ 1st term x 4th term = 5 x 21 = 105
⇒ 2nd term x 3rd term = 7 x 15 = 105
∴ 5:7:: 15: 21
4th case: 15, 21,5, 7
⇒ 1st term x 4th term = 15 x 7 = 105
⇒ 2nd term x 3rd term = 21 x 5 = 105
∴ 15:21:: 5: 7
4. 7, 21,4, 12
Solution:
1st case: 7, 21, 4,12
⇒ 1st term x 4th term = 7 x 12 = 84
⇒ 2nd term x 3rd term =21 x 4 = 84
∴ 7:21:: 4: 12
2nd Case: 21, 7,12, 4
⇒ 1st x 4th term = 21 x 4 = 84
⇒ 2nd term 3rd term = 7 x 12 = 84
∴ 21: 7:: 12: 4
3rd case: 7, 4, 21, 12
⇒ 1st term x 4th term = 7 x 12 = 84
⇒ 2nd term x 3rd term = 4×21 =84
∴ 7:4:: 21: 12
4th case 21,12, 7, 4
⇒ 1st term x 4th term = 21 x 4 = 84
⇒ 2nd term x 3rd term = 12 x 7 = 84
∴ 21: 12:: 7: 4
WBBSE Class 6 Maths Solutions
5. 3,15,10,50
Solution:
1st case: 3, 15,10, 50
⇒ 1st term x 4th term = 3 x 50 = 150
⇒ 2nd term x 3rd term = 15 x 10 = 150
∴ 3: 15:: 10:50
2nd case: 15, 3, 50,10
⇒ 1st term x 4th term = 15 x 10 = 150
⇒ 2nd term x 3rd term = 3 x 50 = 150
∴ 15:3 :: 50:15
3rd case: 3, 10,15, 50
⇒ 1st term x 4th term = 3 x 50 = 150
⇒ 2nd term x 3rd term = 10×15=150
∴ 3: 10:: 15: 50
4th case: 15, 50, 3,10
⇒ 1st term x 4th term = 15 x 10 = 150
⇒ 2nd term x 3rd term = 50 x 3 = 150
∴ 15: 50:: 3: 10
6. 2, 6, 7, 21.
Solution:
1st case: 2, 6, 7, 21
⇒ 1 st term x 4th term = 2 x 21 = 42
⇒ 2nd term x 3rd term = 6 x 7 = 42
∴ 2:6:: 7: 21
WBBSE Class 6 Maths Solutions
2nd case: 6, 2, 21,7
⇒ 1 st term x 4th term = 6 x 7 = 42
⇒ 2nd term x 3rd term = 2 x 21 =42
∴ 6: 2:: 21: 7
3rd case: 2, 7, 6, 21
⇒ 1st term x 4th term = 2 x 21 = 42
⇒ 2nd term x 3rd term = 7 x 6 = 42
∴ 2: 7:: 6: 21
4th case: 6, 21,2, 7
⇒ 1st term x 4th term = 6 x 7 = 42
⇒ 2nd term x 3rd term = 21 x 2 = 42
∴ 6: 21:: 2: 7
Question 2. Let’s take 4 positive whole numbers:
1. 4,6,8,12
Solution:
1st case: 4, 6, 8, 12
⇒ 1st term x 4th term = 4 x 12 = 48
⇒ 2nd term x 3rd term = 6 x 8 = 48
∴ 4: 6:: 8: 12
2nd case: 6, 4,12, 8
⇒ 1st term x 4th term = 6 x 8 = 48
⇒ 2nd term x 3rd term = 4 x 12 = 48
∴ 6:4 :: 12:8
3rd case: 6,12, 4, 8
⇒ 1st term x 4th term = 6 x 8 = 48
⇒ 2nd term x 3rd term = 12 x 4 = 48
∴ 6:12 :: 4:8
4th case: 12, 6, 8, 4
⇒ 1st term x 4th term = 12 x 4 = 48
⇒ 2nd term x 3rd term = 6 x 8 = 48
∴12: 6:: 8:4
WBBSE Class 6 Maths Solutions
Question 3.
1. 3.5, 7, 2, 4
Solution: 3.5, 7, 2, 4
1st case: 3.5, 7, 2, 4
⇒ 1st term x 4th term = 3.5 x 4 = 14
⇒ 2nd term x 3rd term = 7×2 = 14
∴ 3.5: 7:: 2: 4
2nd case: 7, 3.5, 4, 2
⇒ 1st term x 4th term = 7×2 = 14
⇒ 2nd term x 3rd term = 3.5 x 4 = 14
∴ 7: 3.5:: 4:2
3rd case: 3.5, 2, 7, 4
⇒ 1st term x 4th term = 3.5 x 4 = 14
⇒ 2nd term x 3rd term = 2×7 = 14
∴ 3.5: 2:: 7: 4
4th case: 7,4, 3.5, 2
⇒ 1st term x 4th term = 7×2 = 14
⇒ 2nd term x 3rd term = 4 x 3.5 = 15
∴ 7: 4:: 3.5:2
2. 1.5, 4.5, 2.5; 7.5
Solution: 1.5, 4.5, 2.5, 7.5
1st case: 1.5, 4.5, 2.5, 7.5
⇒ 1st term x 4th term = 1.5 x 7.5 = 11.25
⇒ 2nd term x 3rd term = 4.5 x 2.5 = 11.25
∴ 1.5: 4.5:: 2.5: 7.5
2nd case: 4.5, 7.5,1.5, 2.5
⇒ 1st term x 4th term = 4.5 x 2.5 = 11.25
⇒ 2nd term x 3rd term = 7.5 x 1.5 = 11.25
∴ 4.5: 7.5:: 1.5:2.5
WBBSE Class 6 Maths Solutions
3rd case: 1.5, 2.5, 4.5, 7.5
⇒ 1st term x 4th term = 1.5 x 7.5 = 11.25
⇒ 2nd term x 3rd term = 2.5 x 4.5 = 11.25
∴1.5: 2.5:: 4.5 4.5: 4.5
4th case: 4.5,1.5, 7.5, 2.5
⇒ 1st term x 4th term = 4.5 x 2.5 = 11.25
⇒ 2nd term x 3rd term = 1.5 x 7.5 = 11.25
∴ 4.5: 1.5:: 7.5:2.5
3. 0.35,1.05, 0.09, 0.27
Solution: 0.35,1.05, 0.09, 0.27
1st case: 0.35,1.05,0.09, 0.27
⇒ 1st term x 4th term = 0.35 x 0.27 = 0.0945
⇒ 2nd term x 3rd term = 1.05 x 0.09 = 0.0945
∴ 0.35:1.05 :: 0.09:0.27
2nd case: 1.05, 0.35, 0.27, 0.09
⇒ 1st term x 4th term = 1.05 x 0.09 = 0.0945
⇒ 2nd term x 3rd term = 0.35 x 0.27 = 0.0945
∴ 1.05: 0.35:: 0.27: 0.09
3rd case: 0.35,1.05,0.09, 0.27
⇒ 1st term x 4th term = 0.35 x 0.27 = 0.0945
⇒ 2nd term x 3rd term = 1.05 x 0.09 = 0.0945
∴ 0.35:1.05 :: 0.09:0.27
WBBSE Class 6 Maths Solutions
4th case: 1.05,0.27,0.35, 0.09
⇒ 1st term x 4th term = 1.05 x 0.09 = 0.0945;
⇒ 2nd term x 3rd term = 0.27 x 0.35 = 0.0945
∴ 1.05: 0.27 ::0.35:0.09
4. Let’s take 4 positive decimal numbers:
Solution:
1st case: 2.4,4.8,7.5,15
⇒ Here, 1st term x 4th term = 2.4 x 15 = 36
⇒ 2nd term x 3rd term F 4.8 x 7.5 = 36
= 2.4: 4.8:: 7.5:15
2nd case: 4.8, 2.4,15, 7.5
⇒ Here, 1 st term x 4th term = 4.8 x 7.5 = 36
⇒ 2nd term x 3rd term = 2.4 x 15 = 36
∴ 4.8: 2.4:: 15:7.5
3rd case: 15, 7.5, 4.8, 2.4
⇒ Here, 1 st term x 4th term = 15 x 2.4 = 36
⇒ 2nd term x 3rd term = 7.5 x 4.8 = 36
∴ 15:7.5:: 4.8 : 2.4
4th case: 2.4, 7.5, 4.8,15
⇒ Here, 1 st term x 4th term = 2.4 x 15 = 36
⇒ 2nd term x 3rd term = 7.5 x 4.8 = 36
∴ 2.4: 7.5:: 4.8:15
Question 4.
1. \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\)
Solution:
1st case: \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\)
⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)
⇒ 2nd term x 3rd term = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)
∴ \(\frac{1}{2}: \frac{1}{3}:: \frac{1}{4}: \frac{1}{6}\)
2nd case: \(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}, \frac{1}{4}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)
⇒ 2nd term x 3 rd term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)
∴ \(\frac{1}{3}: \frac{1}{2}:: \frac{1}{6}: \frac{1}{4}\)
Class 6 Math WBBSE Solutions
3rd case: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{3}, \frac{1}{6}\)
⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)
⇒ 2nd term x 3rd term = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)
∴ \(\frac{1}{2}: \frac{1}{4}:: \frac{1}{3}: \frac{1}{6}\)
4th case: \(\frac{1}{3}, \frac{1}{6}, \frac{1}{2}, \frac{1}{4}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{4} \times=\frac{1}{12}\)
⇒ 2nd term x 3rd term = \(\frac{1}{6} \times \frac{1}{2}=\frac{1}{12}\)
∴ \(\frac{1}{3}: \frac{1}{6}:: \frac{1}{2}: \frac{1}{4}\)
2. \(\frac{1}{5}, \frac{1}{10}, \frac{1}{16}, \frac{1}{32}\)
1st case: \(\frac{1}{5}, \frac{1}{10}, \frac{1}{16}, \frac{1}{32}\)
⇒ 1st term x 4th term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)
⇒ 2nd term x 3rd term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)
∴ \(\frac{1}{5}: \frac{1}{10}:: \frac{1}{16}: \frac{1}{32}\)
2nd case: \(\frac{1}{10}, \frac{1}{5}, \frac{1}{32}, \frac{1}{16}\)
⇒ 1st term x 4th term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)
⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)
∴ \(\frac{1}{10}: \frac{1}{5}:: \frac{1}{32}: \frac{1}{16}\)
3rd case: \(\frac{1}{5}, \frac{1}{16}, \frac{1}{10}, \frac{1}{32}\)
⇒ 1st term x 4th term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)
⇒ 2nd term x 3rd term = \(\frac{1}{16} \times \frac{1}{10}=\frac{1}{160}\)
∴ \(\frac{1}{5}: \frac{1}{16}:: \frac{1}{10}: \frac{1}{32}\)
Class 6 Math WBBSE Solutions
4th case: \(\frac{1}{10}, \frac{1}{32}, \frac{1}{5}, \frac{1}{16}\)
⇒ 1st term x 4th term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)
⇒ 2nd term x 3rd term = \(\frac{1}{32} \times \frac{1}{5}=\frac{1}{160}\)
∴ \(\frac{1}{10}: \frac{1}{32}:: \frac{1}{5}: \frac{1}{16}\)
3. \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{18}\)
1st case: \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{18}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)
⇒ 2nd term x 3rd term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)
∴ \(\frac{1}{3}: \frac{1}{6}:: \frac{1}{9}: \frac{1}{18}\)
2nd case: \(\frac{1}{6}, \frac{1}{3}, \frac{1}{18}, \frac{1}{9}\)
⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)
⇒ 2nd term x 3rd term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)
∴ \(\frac{1}{6}: \frac{1}{3}:: \frac{1}{18}: \frac{1}{9}\)
3rd case: \(\frac{1}{3}, \frac{1}{9}, \frac{1}{6}, \frac{1}{18}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)
⇒ 2nd term x 3rd term = \(\frac{1}{9} \times \frac{1}{6}=\frac{1}{54}\)
∴ \(\frac{1}{3}: \frac{1}{9}:: \frac{1}{6}: \frac{1}{18}\)
4th case: \(\frac{1}{6}, \frac{1}{18}, \frac{1}{3}, \frac{1}{9}\)
⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)
⇒ 2nd term x 3rd term = \(\frac{1}{18} \times \frac{1}{3}=\frac{1}{54}\)
∴ \(\frac{1}{6}: \frac{1}{18}:: \frac{1}{3}: \frac{1}{9}\)
4. \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)
Solution:
1st case: \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)
⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
2nd case: \(\frac{1}{24}, \frac{1}{8}, \frac{1}{15}, \frac{1}{5}\)
⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
∴ \(\frac{1}{24}: \frac{1}{8}:: \frac{1}{15}: \frac{1}{5}\)
3rd case: \(\frac{1}{8}, \frac{1}{5}, \frac{1}{24}, \frac{1}{15}\)
⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{24}=\frac{1}{120}\)
∴ \(\frac{1}{8}: \frac{1}{5}:: \frac{1}{24}: \frac{1}{15}\)
4th case: \(\frac{1}{24}, \frac{1}{15}, \frac{1}{8}, \frac{1}{5}\)
⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{15} \times \frac{1}{8}=\frac{1}{120}\)
∴ \(\frac{1}{24}: \frac{1}{15}:: \frac{1}{8}: \frac{1}{5}\)
5. \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)
Solution:
1st case: \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{3}{10}=\frac{1}{20}\)
⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{4}=\frac{1}{20}\)
∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{1}{4}: \frac{3}{10}\)
∴ \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{20}\)
4th case: \(\frac{1}{6}, \frac{1}{18}, \frac{1}{3}, \frac{1}{9}\)
⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)
⇒ 2nd term x 3rd term = \(\frac{1}{18} \times \frac{1}{3}=\frac{1}{54}\)
∴ \(\frac{1}{6}: \frac{1}{18}:: \frac{1}{3}: \frac{1}{9}\)
Class 6 Math WBBSE Solutions
6. \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)
1st case: \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)
⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
2nd case: \(\frac{1}{24}, \frac{1}{8}, \frac{1}{15}, \frac{1}{5}\)
⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
∴ \(\frac{1}{24}: \frac{1}{8}:: \frac{1}{15}: \frac{1}{5}\)
3rd case: \(\frac{1}{8}, \frac{1}{5}, \frac{1}{24}, \frac{1}{15}\)
⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{24}=\frac{1}{120}\)
∴ \(\frac{1}{8}: \frac{1}{5}:: \frac{1}{24}: \frac{1}{15}\)
4th case: \(\frac{1}{24}\), \(\frac{1}{15}\), \(\frac{1}{8}\), \(\frac{1}{5}\)
⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)
⇒ 2nd term x 3rd term = \(\frac{1}{15} \times \frac{1}{8}=\frac{1}{120}\)
∴ \(\frac{1}{24}: \frac{1}{15}:: \frac{1}{8}: \frac{1}{5}\)
7. \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)
1st case: \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)
⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{3}{10}=\frac{1}{20}\)
⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{4}=\frac{1}{20}\)
∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{1}{4}: \frac{3}{10}\)
∴ \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{20}\)
2nd case: \(\frac{1}{3}, \frac{1}{5}, \frac{3}{10}, \frac{1}{4}\)
= \(\frac{1}{5} \times \frac{3}{10}=\frac{3}{50}\)
∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{3}{10}: \frac{1}{4}\)
Again, \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}, \frac{1}{5} \times \frac{3}{10}=\frac{3}{50}\)
∴ \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\) are not in proportion.
8. \(\frac{8}{3}, 8, \frac{5}{3}, 8\)
1st case: \(\frac{8}{3} \times 8=\frac{64}{3}, 8 \times \frac{5}{3}=\frac{40}{3}\)
2nd case: \(\frac{8}{3} \times \frac{5}{3}=\frac{40}{.3}, 8 \times 8=64\)
∴ \(\frac{8}{3}, 8, \frac{5}{3}, 8\) are not in proportion.
Class 6 Math WBBSE Solutions Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.3
Question 1. Which of the following cases can be represented in ratio? Let’s write.
1. My friend Jatita’s weight and her height.
Solution: My friend Jatita’s weight and her height are not possible.
2. In this month, number of days I went to school and number of days my friend Jahir went to school.
Solution: In this month, the number of days I went to school and number of days my friend Jahir went to school — Possible.
3. The money I had and the money I spent.
Solution: The money I had and the money I spent — Possible.
4. Litres of water in my water bottle and temperature of water.
Solution: Litres of water in my water bottle and temperature of water — Not possible.
5. How long I played today and how long my brother played.
Solution: How long I play today and how long my brother played — Possible.
Question 2. Let’s express the following quantities in ratio and identify the ratio of greater inequality and the ratio of lesser inequality.
1. 10 kg, 15 kg.
Solution: 10 kg: 15 kg
= 10:15
= 2:3
(Ratio of lesser inequality).
2. 27 things and 18 things
Solution: 27 things: 18 things
= 27:18 = 3:2
(Ratio of greater inequality).
3. Rs. 30 and Rs. 22.50
Solution: Rs. 30: Rs. 22.50
= 30: 22.5
= 6 : 4.5 = 4:3
(Ratio of greater inequality).
4. 4.9 litre and 8.4 litre
Solution: 4.9 litre: 8.4 litre
= 4.9: 84
= 49 : 84
= 7:12
(Ratio of lesser inequality).
5. 52 m and 78 m
Solution: 52 m: 78 m = 52: 78
=4:6
= 2:3
(Ratio of lesser inequality).
6. 1 hr 24 mins and 6 hrs 18 min
Solution: 1 hr 24 min: 6 hr 18 min
= (60 + 24) min : (360 + 10) min = 84 min : 378 m
= 12:54
= 2:9
(Ratio of lesser inequality).
Question 3. A bamboo stick is 2 m long. 0.75 m of it is coloured red and rest part is colored white.
Solution:
Length of the bamboo stick = 2 m.
Portion of red colour = 0.75 m.
∴ Portion of white colour = (2 – 0.75) m = 1.25 m.
1. Let us find the ratio between the total length of bamboo and the portion painted red.
Solution: Length of bamboo: Length of red portion
= 2 : 0.75 = 2 : 75/100
= 2 : 3/4 = 8 : 3
2. Let’s find the ratio between the total length of bamboo and the portion painted white.
Solution: Length of bamboo: Length of white portion
= 2: 1.25 = 2: 125/100
= 2 : 5/4 = 8.5
3. Let’s find the ratio between the part coloured red and that coloured white.
Solution: Length of red portion: Length of white portion = 0.75:1.25
= 75:125
= 3:5
Class 6 Math Solutions WBBSE
Question 4. The ratio of length and breadth of my room is 7: 5. Let’s write 4 possible perimeters of the room in the above ratio.
Solution: Length: Breadth = 7:5
∴ \(\frac{\text { Length }}{\text { Breadth }}=\frac{7}{5}=\frac{14}{10}=\frac{28}{20}=\frac{56}{40}\)
- Perimeter = 2(7 + 5) = 2 x 12 = 24 units
- Perimeter = 2(14 + 10)= 2 x 24 = 48 units
- Perimeter = 2(28 + 20) = 2 x 48 = 96 units
- Perimeter = 2(56 + 40) = 2 x 96 = 192 units.
Question 5. I have 26 stamps. Myself and Mita shall divide the stamps in the ratio of 8: 5. Let’s find how many stamps each of us will have.
Solution:
Total number of stamps = 26
∴ \(\frac{\text { No. of Stamps of me }}{\text { No. of Stamps of Mita }}=\frac{8}{5}=8: 5\)
No. of Stamps of mine = 8/13 x 26 = 16
No. of stamps of Mita = 5/13 x 26 = 10
Question 6. The ratio of my textbooks and story books is 4:3. If number of textbooks are 28, let’s find the number of story books and a total number of books.
Solution:
The ratio of my textbooks and story books = 4:3
⇒ Total number of textbooks = 28.
∴ \(\frac{\text { No. of my text books }}{\text { No. of story books }}=\frac{4}{3}\)
⇒ or, \(\frac{28}{\text { No. of story books }}=\frac{4}{3}\)
⇒ 4 x No. of story books = 3 x 28
⇒ No. of story books = \(\frac{3 \times 28}{4}\) = 21.
∴ Total number of books = No. of text books + No. of storybooks
= 28 + 21 = 49
Question 7. In a particular type of jewellery the ratio of gold and silver is 4: 7. In such type of jewelry ornament, how many milligrams of gold has been mixed with 357 milligrams of silver, it is to be calculated.
Solution:
In a particular type of jewellery the ratio of gold and silver is 4: 7. In such type of jewelry ornament,
Ratio of Gold and Silver = 4:7
∴ \(\frac{\text { Gold }}{\text { Silver }}=\frac{4}{7}\)
⇒ \(\frac{\text { Weight of Gold }}{\text { Weight of Silver }}=\frac{4}{7}\)
⇒ \(\frac{\text { Weight of Gold }}{357 \mathrm{mg}}=\frac{4}{7}\)
7 x weight of Gold = \(4 \times 357 \mathrm{mg}\)
Weight of Gold = \(\frac{4 \times 357}{7} \mathrm{mg} . \)
∴ Weight of Gold = 204 mg
Question 8. Let’s write the ratio of the three angles of an equilateral triangle.
Solution:
Each angle of an equilateral triangle = 60°
∴ 1st angle : 2nd angle : 3rd angle = 60°: 60°: 60°
= 1:1:1.
Class 6 Math Solutions WBBSE
Question 9. Let’s write the ratio of the angles of a right-angled isosceles triangle.
Solution:
The three angles of the right-angled isosceles triangle are 90°, 45°, and 45°.
∴ 1st angle : 2nd angle : 3rd angle = 90° : 45°, 45°
= 2:1:1 .
Question 10. Let me divide Rs. 210 among Fatema and Sakira in the ratio of 3:4. Let me find how much each would get.
Solution:
Total Amount = Rs. 210.
Amount of Fatema: Amount of Sakira = 3:4
∴ Amount of money Fatema received
Amount of money Sakira received = Rs. (210 – 90) = Rs. 120.
Question 11. Mohit bought 6 bananas for Rs. 18 from a vendor and Raju bought 2 dozen of bananas for Rs. 72 from another vendor. Let’s express in ratio to find who paid more for bananas.
Solution:
⇒ Mohit bought 6 bananas for Rs. 18.
∴ Mohit bought 1 banana for Rs. 18/6 = Rs. 3.
⇒ Raju bought 2 dozen (24) bananas for Rs. 72.
∴ Raju bought 1 banana for Rs. 72/24 = Rs. 3.
∴ Ratio of prices = 3:3 = 1 :1 (Equal amount)
Question 12. The distances of Ayesha and Kamal’s house from our school are 1 km and 600 m respectively. Today Ayesha and Kamal came to school in 20 min and 12 min respectively. Let’s find, expressed in ratios, if they reached school at the same time or one of them has reached earlier.
Solution:
Given
The distances of Ayesha and Kamal’s house from our school are 1 km and 600 m respectively. Today Ayesha and Kamal came to school in 20 min and 12 min respectively.
Distance of Ayesha’s house from school: Distance of Kamal’s house from school
= 1 km : 600 m
= 1000 m: 600 m = 5:3
⇒ Again, time taken by Ayesha: time taken by Kamal = 20 min: 12 min
= 5:3
∴ Their speeds are equal.
Question 13. Let’s find which of the following ratios are equal ratios
1. 3:3 and 5:5
Solution: = 1: 1 and 1: 1
∴ Equal ratio
2. 20 : 24 and 25 : 30
Solution: 20: 24 and 25: 30
= 5: 6 and 5: 6.
∴ Equal ratio.
3. 1: 9 and 9:18
Solution: 1:9 and 9: 18
= 1: 9 and 1: 2
∴ Not equal ratio
4. 28 : 21 and 20:15
Solution: 28:21 and 20:15
= 4 : 3 and 4 : 3
∴ Equal ratio
5. 1.4: 0.6 and 6.3 : 2.7
Solution: 1.4 : 0.6 and 6.3 : 2.7
= 14 : 6 and 63 : 27
= 7 : 3 and 7 : 3
∴ Equal ratio
6. 52 : 39 and 44 : 33
Solution: 52: 39 and 44: 33
= 4 : 3 and 4 : 3
∴ Equal ratio
Question 14. Let’s verify which of the following numbers are in proportion:
1. 9, 7, 36, 28
Solution: 9, 7, 36, 28
1st term x 4th term = 9 x 28 = 252
2nd term x 3rd term = 7 x 36 = 252
∴ 9, 7, 36, and 28 are in proportion.
2. 12, 30,14, 70
Solution: 12, 30, 14, 70
1st term 4th term = 1 2 x 70 = 840
2nd term x 3rd term = 30 x 1 4 = 420
∴ 12, 30, 14, 70 are not in proportion.
3. 24, 6, 108, 27
Solution: 24, 6, 108, 27
⇒ 1st term x 4th term = 24 x 27 = 648
⇒ 2nd term x 3rd term = 6 x 1 08 = 648
∴ 24, 6, 108, 27 are in proportion.
4. \(\frac{1}{2}, 1, \frac{3}{5}, 1 \frac{1}{5}\)
Solution: \(\frac{1}{2}, 1, \frac{3}{5}, 1 \frac{1}{5}\)
= \(\frac{1}{2}, 1, \frac{3}{5}, \frac{6}{5}\)
⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{6}{5}=\frac{3}{5}\)
⇒ 2nd term x 3rd term = \(1 \times \frac{3}{5}=\frac{3}{5}\)
∴ \(\frac{1}{2}, 1, \frac{3}{5}, 1 \frac{1}{5}\) are in proportion.
5. 72, 45, 70, 25
Solution: 72, 45, 70, 25
⇒ 1st term x 4th term = 72 x 25 = 1800
⇒ 2nd term x 3rd term = 45 x 70 = 31 50
∴ 72, 45, 70, 25 are not in proportion
6. _______ Let’s take + positive whole numbers.
Solution: 7, 8, 35, 40
1 st term x 4th term = 7 x 40 = 280
2nd term x 3rd term = 8 x 35 = 280
∴ 7, 8, 35, 40 are in proportion.
Class 6 Math Solutions WBBSE
Question 15. Let’s first try to find, if the following numbers are in proportion, let’s find all other possible proportions for each of them.
1. 60, 2,10,12
Solution: 60, 2,10,12
Case 1: Here, 1st term x 4th term = 60 x 12 = 720
⇒ 2nd term x 3rd term = 2 x 10 = 20
∴ 60: 2 ≠10: 12
Case 2: 60, 12, 10,2
⇒ Here, 1st term x 4th term = 60 x 2 = 120
⇒ 2nd term x 3rd term = 12 x 10 = 120
∴ 60: 12 :: 10: 2
Case 3: 60, 12,12, 2
⇒ Here, 1 st term x 4th term = 60 x 2 = 120
⇒ 2nd term x 3rd term = 10 x 12 = 120
∴ 60 : 10 :: 12 : 2
Case 4: 10,2,60, 12
⇒ Here, 1 st term x 4th term = 10 x 12 = 120
⇒ 2nd term x 3rd term = 2 x 60 = 120
∴ 10: 2:: 60: 12
2. 4,10, 6,15
Solution: 4,10, 6,15
⇒ Case 1: 4, 10, 6,15
⇒ Here, 1 st term x 4th term = 4 x 15 = 60
⇒ 2nd term x 3rd term = 6 x 10 = 60
∴ 4 : 10 :: 6: 15
Case 2: 4, 6,10, 15
⇒ Here, 1 st term x 4th term = 4 x 15 = 60
⇒ 2nd term x 3rd term = 6 x 10 = 60
∴ 4 : 6:: 10 : 15
Case 3: 60, 10, 12,2
⇒ Here, 1st term x 4th term = 10 x 6
⇒ 60 2nd term x 3rd term = 4 x 15 = 60
∴ 10 : 4:: 15 : 6
Case 4: 10, 15, 4,6
⇒ Here, 1st term x 4th term = 10 x6 = 60
⇒ 2nd term x 3rd term = 15 x 4 = 60
∴ 10: 15:: 4: 6
3. 8, 9, 24, 2
Solution: 8, 9, 24, 2
Case 1: 8, 24, 2, 9
⇒ Here, 1 st term X 4th term = 8 x 9 = 72
⇒ 2nd term x 3rd term = 24 x 2 = 48
∴ 8: 24 ≠ 2: 9
Case 2: 8, 9, 2, 24
⇒ Here, 1st term x 4th term = 8 x 24 = 192
⇒ 2nd term x 3rd term = 9×2 = 18
∴ 8: 9 ≠ 2: 24
Case 3: 8, 24, 9, 2
⇒ Here, 1st term x 4th term = 8×2 = 16
⇒ 2nd term x 3rd term = 24 x 9 = 216
∴ 8 : 24 ≠ 9:2
∴ 8, 9, 24, 2 are not in proportion
4. 3, 5,15, 25
Solution: 3, 5, 15, 25
Case 1: 5, 3, 25, 15
⇒ Here, 1 st term x 4th term = 5 x 15 = 75
⇒ 2nd term x 3rd term = 3 x 25 = 75
∴ 5 : 3:: 25 :15
Case 2: 5, 25, 3, 15
⇒ Here, 1 st term x 4th term = 5 x 15 = 75
⇒ 2nd term x 3rd term = 25 x 3 = 75
∴ 5: 25 :: 3:15
Case 3: 25, 5, 15, 3
⇒ Here, 1st term x 4th term = 25 x 3 = 75
⇒ 2nd term x 3rd term = 5×15 = 15
∴ 10: 4:: 15:6
Case 4: 25,15, 5, 3
⇒ Here, 1st term x 4th term = 25 x 3 = 75
⇒ 2nd term x 3rd term = 15 x 5 = 75
∴ 25 : 15 :: 5 : 3
5. 45, 5,75,5
Solution: 45, 5, 75: 5
⇒Here, 45 x 5 = 225 75 X 5 = 375
∴ 45:5 ≠ 75.5
⇒ Again, 45 x 75 5 x 5 = 25
∴ 45 : 5 ≠ 5: 75
∴ 45, 5, 75, and 5 are not in proportion.
6. 24, 4, 36, 6
Solution: 24, 4, 36, 6
Case 1: 24, 4, 36, 6
⇒ Here, 1 st term x 4th term = 24 x 6 = 144
⇒ 2nd term x 3rd term = 4 x 36 = 144
∴ 24 : 4:: 36 : 6
Case 2: 24, 36, 4, 6
⇒ Here, 1st term x 4th term = 24 x 6 = 144
⇒ 2nd term x 3rd term = 36 x 4 = 144
∴ 24 : 36 :: 4: 6
Case 3: 4, 24, 6, 36
⇒ Here,. 1 st term x 4th term = 4 x 36 = 144
⇒ 2nd term x 3rd term = 24 x 6 = 144
∴ 4: 24 6: 36
Case 4: 4, 6, 24, 36
⇒ Here, 1 st term x 4th term = 4 x 36 = 144
⇒ 2nd term x 3rd term = 6 x 24 = 144
∴ 4:6:: 24 : 36
Class 6 Math Solutions WBBSE
Question 16. The height of my friend Priya is 160 cm and the height of her mother is 170 cm. Again, Priya’s weight is 40 kg and her mother’s weight is 42.5 kg. Let’s find if their weights are in proportion to their heights.
Solution:
Given
The height of my friend Priya is 160 cm and the height of her mother is 170 cm. Again, Priya’s weight is 40 kg and her mother’s weight is 42.5 kg.
⇒ Height of Priya = 160 m.
⇒ Height of Priya’s mother = 170 m.
∴ Ratio of their heights = 160 m: 170 m.
= 16:17
⇒ Again, weight of Priya = 40 kg.
⇒ Weight of Priya’s mother = 42.5 kg.
∴ Ratio of their weights = 40 kg: 42.5 kg.
= 40: 425/10
= 16:17
∴ Their weights are in proportion to their heights.