ML Aggarwal Solutions for Class 7, maths Chapter 6

Class 7: Maths Chapter 6 solutions. Complete Class 7 Maths Chapter 6 Notes.

Contents

1 ML Aggarwal Solutions for Class 7 Maths Chapter 6- Ratio and Proportion
2 Download PDF
3 Chapterwise ML Aggarwal Solutions for Class 7 Maths :
4 About ML Aggarwal

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

ML Aggarwal Solutions Class 7 Maths Chapter 6 - 5

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 :

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.