ML Aggarwal Solutions for Class 7 Maths Chapter 6 Ratio and Proportion help students to understand the concepts precisely and also develop a strong base in the subject. The solutions improve the problem-solving skills of students, which is very important in achieving high marks in academics. This also helps them pursue their desired course in higher studies. Here, students can download ML Aggarwal Solutions for Class 7 Maths Chapter 6 Ratio and Proportion PDF from the links provided below.
This chapter contains questions on determining the ratio and proportion of various word problems. Students are suggested to practise these solutions regularly to ace the exam with ease. The solutions provide a thorough explanation of each and every concept covered based on the latest syllabus of the prescribed board.
ML Aggarwal Solutions for Class 7 Maths Chapter 6 Ratio and Proportion
Access ML Aggarwal Solutions for Class 7 Maths Chapter 6 Ratio and Proportion
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 is 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, and 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 and 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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