Class 7: Maths Chapter 6 solutions. Complete Class 7 Maths Chapter 6 Notes.
ML Aggarwal Solutions for Class 7 Maths Chapter 6- Ratio and Proportion
ML Aggarwal 7th Maths Chapter 6, Class 7 Maths Chapter 6 solutions
1. Express the following ratios in simplest form:
(i) (1 / 6): (1 / 9)
(ii) :
(iii) (1 / 5): (1 / 10): (1 / 15)
Solution:
(i) (1 / 6): (1 / 9)
Given
Ratio = (1 / 6): (1 / 9)
= (1 / 6) ÷ (1 / 9)
= (1 / 6) × (9 / 1)
We get,
= 3 / 2
= 3: 2
(ii)
:
This can be written as,
(9 / 2): (9 / 8)
Given ratio = (9 / 2): (9 / 8)
(9 / 2) ÷ (9 / 8)
= (9 / 2) × (8 / 9)
We get,
= 4 / 1
= 4: 1
(iii) (1 / 5): (1 / 10): (1 / 15)
L.C.M. of 5, 10, 15 is 30
L.C.M. = 5 × 2 × 3
= 30
= (1 / 5) × 30: (1 / 10) × 30: (1 / 15) × 30
We get,
= 6: 3: 2
2. Find the ratio of each of the following in simplest form:
(i) Rs 5 to 50 paise
(ii) 3 km to 300 m
(iii) 9 m to 27 cm
(iv) 15 kg to 210 g
(v) 25 minutes to 1.5 hours
(vi) 30 days to 36 hours
Solution:
(i) Rs 5 to 50 paise
= 500 paise: 50 paise
= 500 / 50
= 50 / 5
Dividing by 5, we get,
= 10 / 1
= 10: 1
(ii) 3 km to 300 m
WKT
1 km = 1000 m
Hence,
3000 m: 300m
= 3000 / 300
= 30 / 3
Dividing by 3, we get,
= 10 / 1
= 10: 1
(iii) 9 m to 27 cm
WKT
1 m = 100 cm
Hence,
= 9 × 100 cm: 27 cm
= (900 / 27) cm
Dividing by 9, we get,
= 100 / 3
= 100: 3
(iv) 15 kg to 210 g
WKT
1 kg = 1000 g
= 15 × 1000 g: 210 g
= 15000: 210
= (1500 / 21) g
Dividing by 30, we get,
= 500 / 7
= 500: 7
(v) 25 minutes to 1.5 hours
= 25 minutes: (3 / 2) × 60
= 25 / 90
Dividing by 5, we get,
= 5 / 18
= 5: 18
(vi) 30 days to 36 hours
= 30 × 24 hours to 36 hours
= 720: 36
= 720 / 36
Dividing by 36, we get,
= 20 / 1
= 20: 1
3. Which of the following statements are true?
(i) 2.5: 1.5:: 7.0: 4.2
(ii) (1 / 2): (1 / 3) = (1 / 3): (1 / 4)
(iii) 24 men: 16 men = 33 horses: 22 horses.
Solution:
(i) 2.5: 1.5:: 7.0: 4.2
Here,
Product of extremes = 2.5 × 4.2
We get,
= 10.50
Product of means = 1.5 × 7.0
We get,
= 10.50
By cross product rule
Product of extremes = Product of means
Therefore,
2.5: 1.5:: 7.0: 4.2 is a true statement
(ii) (1 / 2): (1 / 3) = (1 / 3): (1 / 4)
Here,
Product of extremes = (1 / 2) × (1 / 4)
We get,
= (1 / 8)
Product of means = (1 / 3) × (1 / 3)
We get,
= (1 / 9)
By cross product rule
Product of extremes ≠ Product of means
Therefore,
(1 / 2): (1 / 3) = (1 / 3): (1 / 4) is not a true statement
(iii) 24 men: 16 men = 33 horses: 22 horses
Product of extremes = 24 × 22
We get,
=528
Product of means = 16 × 33
We get,
= 528
By cross product rule
Product of extremes = Product of means
Therefore,
24 men: 16 men = 33 horses: 22 horses is a true statement
4. Check whether the following numbers are in proportion or not:
(i) 18, 10, 9, 5
(ii) 3, , 4,
(iii) 0.1, 0.2, 0.3, 0.6
Solution:
(i) 18, 10, 9, 5
Here,
Product of extremes = 18 × 5
We get,
= 90
Product of means = 10 × 9
We get,
= 90
By cross product rule
Product of extremes = Product of means
Therefore,
The given numbers, 18, 10, 9, 5 are in proportion
(ii) 3,
, 4,
Here,
Product of extremes = 3 ×
= 3 × (9 / 2)
We get,
= (27 / 2)
Product of means =
× 4
= (7 / 2) × 4
We get,
= 14
By cross product rule
Product of extremes ≠ Product of means
Therefore,
The given numbers, 3,
, 4,
are in proportion
(iii) 0.1, 0.2, 0.3, 0.6
Here,
Product of extremes = 0.1 × 0.6
We get,
= 0.06
Product of means = 0.2 × 0.3
We get,
= 0.06
By cross product rule
Product of extremes = Product of means
Therefore,
The given numbers, 0.1, 0.2, 0.3, 0.6 are in proportion
5. 6 bowls cost Rs 90. What would be the cost of 10 such bowls?
Solution:
Cost of 6 bowls = Rs 90
Let the cost of 10 bowls be Rs x
6: 10 = 90: x
6 × x = 10 × 90 (ad = bc)
6x = 900
x = 900 / 6
We get,
x = Rs 150
Therefore,
Cost of 10 bowls = Rs 150
6. Ten pencils cost Rs 15. How many pencils can be bought with Rs 72?
Solution:
Cost of 10 pencils = Rs 15
Let us assume that the number of pencils bought with Rs 72 = x
10: x = 15: 72
x × 15 = 10 × 72 (bc = ad)
We get,
15x = 720
x = 720 / 15
x = 48
Therefore,
Number of pencils = 48
7. Convert the following speeds into m/sec:
(i) 72 km/h
(ii) 9 km/h
(iii) 1.2 km/minutes
(iv) 600 m/hour
Solution:
(i) 72 km/h
WKT
1 km = 1000 m and
1 hour = 3600 sec
Hence,
1 km / h = (1000) / (3600)
= (5 / 18) m/s
Therefore,
72 km/h = (5 / 18) × 72 m/sec
We get,
= 20 m/ sec
(ii) 9 km/h
1 hour = 3600 sec
1 km = 1000 m
1 km / m = (1000) / (3600)
We get,
= (5 / 18) m/s
Therefore,
9 km/h = (5 / 18) × 9 m/sec
We get,
= (5 / 2)
= 2.5 m/ sec
(iii) 1.2 km/minutes
1 hour = 60 minutes
1.2 km/min = 1.2 × 60 km/h
Now,
1 km/ h = (5 / 18) m/s
1.2 × 60 km/ h = 1.2 × 60 × (5 / 18) m/sec
= 72 × (5 / 18) m/sec
We get,
= 20 m/sec
(iv) 600 m/ hour
= (600) / (1000) km/h
= {(600 × 5) / (1000 × 18)} m/sec
We get,
= (1 / 6) m/sec
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ML Aggarwal Solutions for Class 7 Maths Chapter 6- Ratio and Proportion
Download PDF: ML Aggarwal Solutions for Class 7 Maths Chapter 6- Ratio and Proportion PDF
Chapterwise ML Aggarwal Solutions for Class 7 Maths :
- Chapter 1- Integers
- Chapter 2- Fractions and Decimals
- Chapter 3- Rational Numbers
- Chapter 4- Exponents and Powers
- Chapter 5- Sets
- Chapter 6- Ratio and Proportion
- Chapter 7- Percentage and Its Applications
- Chapter 8- Algebraic Expressions
- Chapter 9- Linear Equations and Inequalitie
- Chapter 10- Lines and Angles
- Chapter 11- Triangles and its Properties
- Chapter 12- Congruence of Triangles
- Chapter 13- Practical Geometry
- Chapter 14- Symmetry
- Chapter 15- Visualising Solid Shapes
- Chapter 16- Perimeter and Area
- Chapter 17- Data Handling
About ML Aggarwal
M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.