Srinivas

Well-designed AP 6th Class Maths Guide Chapter 12 Ratio and Proportion InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 12th Chapter Mensuration InText Questions

Try these (Page No: 152)

Question 1.
In a class, there are 20 boys and 40 girls. What is the ratio of the number of boys to the number of girls?
Solution:
Number of boys in the class = 20
Number of girls in the class = 40
∴ Ratio of number of boys to the number of girls = \(\frac { 20 }{ 40 }\) = \(\frac { 1 }{ 2 }\)
Ratio = 1 : 2

Question 2.
Ravi walks 6 km in an hour while Roshan walks 4 km in an hour. What is the ratio of the distance covered by Ravi to the distance covered by Roshan?
Solution:
Distance covered by Ravi = 6 km in an hour
Distance covered by Roshan = 4 km in an hour
Ratio of the distance covered by Ravi to the distance covered by Roshan

Try these (Page No: 154)

Question 1.
Saurabh takes 15 minutes to reach school from his house and Sachin takes one hour to reach school from his house. Find the ratio of the time taken by Saurabh to the time taken by Sachin.
Solution:
Time taken by Saurabh = 15 minutes
Time taken by Sachin = 1 hour = 60 minutes
(converting the time into same units)
∴ Ratio of the time taken by Saurabh to the time taken by Sachin

Ratio = 1 : 4

Question 2.
Cost of a toffee is 50 paise and cost of a chocolate is ₹ 10. Find the ratio of the cost of a toffee to the cost of a chocolate.
Solution:
Cost of a toffee = 50 paise
Cost of a chocolate = ₹ 10
= 10 × 100 = 1000 paise
∴ The ratio of the cost of a toffee to the cost of chocolate
Ratio = 1 : 20

Question 3.
In a school, there were 73 holidays in one year. What is the ratio of the number of holidays to the number of days in one year ?
Solution:
Number of holidays in one year = 73
Number of days in one year = 365
∴ Ratio of the number of holidays to days in one year =

Ratio = 1 : 5.

Try these (Page No: 158)

Question 1.
Find the ratio of number of notebooks to the number of books in your bag.
Solution:
Student’s Activity.

Question 2.
Find the ratio of number of desks and chairs in your classroom.
Solution:
Student’s Activity.

Question 3.
Find the number of students above twelve years of age in your class. Then, find the ratio of number of students with age above twelve years and the remaining students.
Solution:
Student’s Activity.

Question 4.
Find the ratio of number of doors and the number of windows in your classroom.
Solution:
Student’s Activity.

Question 5.
Draw any rectangle and find the ratio of its length to its breadth.
Solution:
Student’s Activity.

Try these (Page No: 170)

Question 1.
Check whether the given ratios are equal, i.e. they are in proportion. If yes, then write them in the proper form.
1. 1 : 5 and 3 : 15
Solution:
1 : 5 and 3 : 15
3 : 15 = \(\frac { 3 }{ 15 }\) = \(\frac{3 /div 3}{15 /div 3}\) = \(\frac { 1 }{ 5 }\) = 1 : 5
So, 1 : 5 and 3 : 15 are equal ratios.
∴ Thus 1 : 5 : : 3 : 15

Question 2.
2 : 9 and 18 : 81
Solution:
2 : 9 and 18 : 81
18 : 81 = \(\frac { 18 }{ 81 }\) = \(\frac{18 /div 9}{81 /div 9}\) = \(\frac { 2 }{ 9 }\) = 2 : 9
So, 2 : 9 and 18 : 81 are equal ratios.
∴ Thus 2 : 9 :: 18 : 81

Question 3.
15 : 45 and 5 : 25
Solution:
15 : 45 and 5 : 25
15 : 45 = \(\frac { 15 }{ 45 }\) = \(\frac{15 /div 15}{45 /div 15}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
5 : 25 = \(\frac { 5 }{ 25 }\) = \(\frac{5 /div 5}{25 /div 5}\) = \(\frac { 1 }{ 5 }\) = 1 : 5
Since 15 : 45 and 5 : 25 are not equal ratios.
15 : 45 and 5 : 25 are not in proportion.

Question 4.
4 : 12 and 9 : 27
Solution:
4 : 12 and 9 : 27
4 : 12 = \(\frac { 4 }{ 12 }\) = \(\frac{4 /div 4}{12 /div 4}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
9 : 27 = \(\frac { 9 }{ 27 }\) = \(\frac{9 /div 9}{27 /div 9}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
So, 4 : 12 and 9 : 27 are equal ratio
Thus 4 : 12 :: 9 : 27.

Question 5.
₹ 10 to ₹ 15 and 4 to 6
Solution:
₹ 10 to ₹ 15 and 4 to 6
Ratio of ₹ 10 to ₹ 15
= \(\frac{₹ 10}{₹ 15}\) = \(\frac{10}{15}\) = \(\frac{10 \div 5}{15 /div 5}\) = \(\frac { 2 }{ 3}\) = 2 : 3
Ratio of 4 to 6 = \(\frac{4}{6}\) = \(\frac{4 \div 2}{6 /div 2}\) = \(\frac { 2 }{ 3}\) = 2 : 3
So, ₹ 10 to ₹ 15 and 4 to 6 ratios are equal ratios.
₹ 10 : ₹ 15 :: 4 : 6

Try these (Page No: 176)

Question 1.
Prepare five similar problems and ask your friends to solve them,
Solution:
Student’s Activity.

Question 2.
Read the table and fill in the boxes.

Solution:
Distance travelled by Karan in 2 hours = 8 km
Distance travelled by Karan in 1 hour = \(\frac { 8 km }{ 2 }\) = 4 km
∴ Distance travelled by Karan in 4 hours = 4 km × 4 = 16 km
∴ _____ = 16 km
Distance travelled by Kriti in 2 hours = 6 km
Distance travelled by Kriti in 1 hour = \(\frac { 6 km }{ 2 }\) = 3 km
∴ _____ = 3 km
Distance travelled by Kriti in 1 hour = 3 km
∴ Distance travelled by Kriti in 4 hours = 3 km × 4 = 12 km
∴ _____ = 12 km

Well-designed AP 6th Class Maths Guide Chapter 12 Ratio and Proportion Exercise 12.3 offers step-by-step explanations to help students understand problem-solving strategies.

Algebra Class 6 Exercise 12.3 Solutions – 6th Class Maths 12.3 Exercise Solutions

Question 1.
If the cost of 7 m of cloth is ₹ 147, find the cost of 5 m of cloth.
Solution:
Using unitary method, we have cost of 7 m of cloth = ₹ 1470
Cost of 1 m cloth = \(\frac{₹ 1470}{7}\)
Cost of 5 m cloth =
Thus, the required cost = ₹ 1050

Question 2.
Ekta earns ₹ 3000 in 10 days. How much will she earn in 30 days ?
Solution:
Money earned by Ekta in 10 days = ₹ 3000
Money earned by her in 1 day = \(₹\left[\frac{3000}{10}\right]\)
Money earned by Ekta in 30 days =
Thus, money earned by Ekta in 30 days = ₹ 9000

Question 3.
If it has rained 276 mm in the last 3 days, how mahy cm of rain will fall in one full week (7 days) ? Assume that the rain continues to fall at the same rate.
Solution:
In last 3 days the rain falls = 276 mm
In 1 day the rain falls = \(\frac{276 mm}{3}\)
In 7 days the rain will fall

Question 4.
Cost of 5 kg of wheat is ₹ 91 : 50.
a) What will be the cost of 8 kg of wheat?
Solution:
Cost of 5 kg of wheat = ₹ 91.50
Cost of 1 kg of wheat = \(\frac{₹ 91.50}{5}\)
Cost of 8 kg of wheat = \(₹\left(\frac{91.50}{5} \times 8\right)\) = ₹ 146.40
Thus, the required cost = ₹ 146.40

b) What quantity of wheat can be purchased in ₹ 183 ?
Solution:
The quantity of wheat purchased in ₹ 91.50 = 5 kg
The quantity of wheat purchased in ₹ 1 = \(\frac{5}{91.50}\)
The quantity of wheat purchased in ₹ 183 = \(₹\left(\frac{5}{91.50} \times 183\right)\) = 10 kg
Thus, the required quantity of wheat = 10 kg

Question 5.
The temperature dropped 15 degree celsius in the last 30 days. If the rate of temperature drop. remains the same, how many degrees will the temperature drop in the next ten days?
Solution:
In last 30 days the quantity of drop in temperature = 15 degree celcius
In last 1 day the quantity of drop in temperature = \(\frac{15}{30}\) degree celcius
In last 10 days the quantity of drop in temperature =
= 5 degree celcius
Thus the required drop in temperature in last 10 days = 5 degree celcius.

Question 6.
Shaina pays ₹ 15000 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same ?
Solution:
Amount of rent paid in 3 months = ₹ 15000
Amount of rent paid in 1 month = \(₹\left(\frac{15000}{3}\right)\)
Amount of rent paid in whole year

Thus the amount of rent paid by Shaina whole year = ₹ 60,000

Question 7.
Cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹ 90 ?
Solution:
1 dozen = 12 units
4 dones = (12 × 4) units = 48 units
4 dozen bananas = 48 bananas
₹ 180 is cost of 4 dozen = 12 × 4 = 48 bananas
₹ 1 is cost of = \(\frac { 48 }{ 180 }\)

Thus, the required number of bananas = 24 bananas.

Question 8.
The weight of 72 books is 97 kg. What is the weight of 40 such books?
Solution:
Weight of 72 books = 9 kg
Weight of 1 book = \(\frac { 9 }{ 72 }\) kg
∴ Weight of 40 books =

Hence, the required weight of 40 such books = 5 kg

Question 9.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km ?
Solution:
To cover 594 km. the amount of diesel required – 108 litres.
To cover 1 km, the amiount of diesel will be required = \(\frac { 108 }{ 594 }\) litres.
So, to cover 1650 km , the amount of diesel required = \(\frac { 108 }{ 594 }\) × 1650 = 300 litres.

Question 10.
Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pens cheaper ?
Solution:

Thus, Manish got the pens cheaper than Raju.

Question 11.
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Solution:
Anish:
Number of runs made by Anish in 6 overs = 42
Number of runs made by him in 1 over = \(\frac { 42 }{ 6 }\) = 7 runs
Anup:
Number of runs made by Anup in 7 overs = 63
Number of runs made by him in 1 over = \(\frac { 63 }{ 7 }\) = 9 runs > 7 runs
9 runs > 7 runs
Thus, Anup has made more runs per over.

Well-designed AP 6th Class Maths Guide Chapter 12 Ratio and Proportion Exercise 12.2 offers step-by-step explanations to help students understand problem-solving strategies.

Algebra Class 6 Exercise 12.2 Solutions – 6th Class Maths 12.2 Exercise Solutions

Question 1.
Determine if the following are in proportion.
a) 15, 45, 40, 120
Solution:
15 and 45 = \(\frac { 15 }{ 45 }\) = \(\frac{15 \div 15}{45\div15}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
40 and 120 = \(\frac { 40 }{ 45 }\) = \(\frac{40 \div 40}{120\div 40}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
∴ Since 1 : 3 = 1 : 3
Yes, 15, 45, 40 and 120 are in proportion.

b) 33,121,9,96
Solution:
33 and 121 = \(\frac{33}{121}\) = \(\frac{33 \div 11}{121\div 11}\) = \(\frac { 3 }{ 11 }\) = 3 : 11
9 and 96 = \(\frac{9}{96}\) = \(\frac{9 \div 3}{96\div 3}\) = \(\frac { 3 }{ 32 }\) = 3 : 32
∴ Since 3 : 11 ≠ 3 : 32
No, 33, 121, 9 and 96 are not in proportion.

c) 24,28,36,48
Solution:
24 and 28 = \(\frac{24}{28}\) = \(\frac{24\div 4}{28\div 4}\) = \(\frac { 6 }{ 7 }\) = 6 : 7
36 and 48 = \(\frac{36}{48}\) = \(\frac{36\div 12}{48\div 12}\) = \(\frac { 3 }{ 4 }\) = 3 : 4
∴ Since 6 : 7 ≠ 3 : 4
No, 24, 28, 36 and 48 are not in proportion.

d) 32,48,70,210
Solution:
32 and 48 = \(\frac{36}{48}\) = \(\frac{32\div 16}{48\div 16}\) = \(\frac { 2 }{ 3 }\) = 2 : 3
70 and 210 = \(\frac{70}{210}\) = \(\frac{70\div 70}{210\div 70}\) = \(\frac { 1 }{ 3 }\) = 1 : 3
∴ Since 2 : 3 ≠ 1 : 3
No, 32, 48, 70 and 210 are not in proportion.

e) 4,6,8,12
Solution:
4 and 6 = \(\frac{4}{6}\) = \(\frac{4\div 2}{6\div 2}\) = \(\frac { 2 }{ 3 }\) = 2 : 3
8 and 12 = \(\frac{8}{12}\) = \(\frac{8\div 4}{12\div 4}\) = \(\frac { 2 }{ 3 }\) = 2 : 3
∴ Since 2 : 3 = 2 : 3
Yes, 4,6,8 and 12 are in proportion.

f) 33,44,75,100
Solution:
33 and 44 = \(\frac{33}{44}\) = \(\frac{33\div 11}{44\div 11}\) = \(\frac { 3 }{ 4 }\) = 3 : 4
75 and 100 = \(\frac{75}{100}\) = \(\frac{75\div 25}{100\div 25}\) = \(\frac { 3 }{ 4 }\) = 3 : 4
∴ Since 3 : 4 = 3 : 4
Yes, 33,44,75 and 100 are in proportion.

Question 2.
Write True (T) or False (F) against each of the following statements :
a) 16 : 24 :: 20 : 30
Solution:
16 : 24 :: 20 : 30
16 : 24 = \(\frac{16}{24}\) = 2 : 3
20 : 30 = \(\frac{20}{30}\) = 2 : 3
2 : 3 = 2 : 3
∴ The given statement (a) is True

b) 21 : 6 :: 35 : 10
Solution:
21 : 6 :: 35 : 10
21 : 6 = \(\frac { 1 }{ 2 }\) ⇒ \(\frac { 1 }{ 2 }\) = 7 : 2
35 : 10 = \(\frac { 35 }{ 10 }\) ⇒ \(\frac { 7 }{ 2 }\) = 7 : 2
7 : 2 = 7 : 2
∴ The given statement (b) is True

c) 12 : 8 :: 28 : 12
Solution:
12 : 18 = \(\frac { 12 }{ 18 }\) = 2 : 3
28 : 12 = \(\frac { 28 }{ 12 }\) ⇒ \(\frac { 7 }{ 3 }\) = 7 : 3
2 : 3 = 7 : 3
∴ The given statement (c) is False
∴ The given statement (c) is False

d) 8 : 9 :: 24 : 27
Solution:
24 : 27 = \(\frac { 24 }{ 27 }\) = 8 : 9
8 : 9 = 8 : 9
∴ The given statement (d) is True

e) 5.2 : 3.9 :: 3 : 4
Solution:
5.2 : 3.9 = \(\frac { 5.2 }{ 3.9 }\) = \(\frac{5.2 \times 10}{3.9 \times 10}\) = \(\frac { 52 }{ 39 }\) ⇒ \(\frac { 4 }{ 3 }\) = 4 : 3
4 : 3 ≠ 3 : 4
∴ The given statement (e) is false
∴ The given statement (e) is False

f) 0.9 : 0.36 :: 10 : 4
Solution:
0.9 : 0.36 :: 10 : 4

5 : 2 = 5 : 2.
∴ The given statement (f) is True

Question 3.
Are the following statements true ?
a) 40 persons : 200 persons = ₹ 15 : ₹ 75
Solution:
40 persons : 200 persons
= \(\frac { 40 }{ 200 }\) = \(\frac{40 \div 40}{200 \div 40}\) = \(\frac { 1 }{ 5 }\) = 1 : 5
₹ 15 : ₹ 75 = \(\frac { 15 }{ 75 }\) = \(\frac{15 \div 15}{75 \times 15}\) = \(\frac { 1 }{ 5 }\) = 1 : 5
∴ Statement (a) is True

b) 7.5 litres : 15 litres = 5 kg : 10 kg
Solution:
7.5 litres : 15 litres
= \(\frac{7.5}{15}\) = \(\frac{75}{150}\) = \(\frac{75 \div 75}{150 \div 75}\) = \(\frac { 1 }{ 2 }\) = 1 : 2
5 kg : 10 kg = \(\frac { 5 }{ 10 }\) = \(\frac{5 \div 5}{10 \div 5}\) = \(\frac { 1 }{ 2 }\) = 1 : 2
∴ Statement (b) is True

c) 99 kg : 45 kg = ₹ 44 : ₹ 20
Solution:
99 kg : 45 kg = \(\frac { 99 }{ 45 }\) = \(\frac{99 \div 9}{45 \div 9}\) = \(\frac { 11 }{ 5 }\) = 11 : 5
₹ 44 : ₹ 20 = \(\frac { 44 }{ 20 }\) = \(\frac{44 \div 4}{20 \div 4}\) = \(\frac { 11 }{ 5 }\) = 11 : 5
∴ Statement (c) is True

d) 32 m : 64 m = 6 sec : 12 sec
Solution:
32 m : 64 m = \(\frac { 32 }{ 64 }\) = \(\frac{32 \div 32}{64 \div 32}\) = \(\frac { 1 }{2 }\) = 1 : 2
6 sec : 12 sec = \(\frac { 6 }{ 12 }\) = \(\frac{6 \div 6}{12 \div 6}\) = \(\frac { 1 }{2 }\) = 1 : 2
∴ Statement (d) is True

e) 45 km 60 km = 12 hours : 15 hours
Solution:
45 km : 60 km = \(\frac { 45 }{ 60 }\) = \(\frac { 45 /div 15 }{60 /div 15 }\) = \(\frac { 3 }{ 4 }\) = 3 : 4
12 hours : 15 hours = \(\frac { 12 }{ 15 }\) = \(\frac { 12 /div 3 }{15 /div 3 }\) = \(\frac { 4 }{ 5 }\) = 4 : 5
Since 3 : 4 ≠ 4 : 5
∴ Statement (e) is not True.

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 : 1 m and ₹ 40 : ₹ 160
Solution:
25 cm : 1m = 25 cm : 1000 m [∵ 1m = 100 cm]
= \(\frac { 25 }{ 100 }\) = \(\frac { 25 /div 25 }{100 /div 25 }\) = \(\frac { 1 }{ 4 }\) = 1 : 4
₹ 40 : ₹ 160 = \(\frac { 40 }{ 160 }\) = \(\frac { 40 /div 40 }{ 160 /div 40 }\) = \(\frac { 1 }{ 4 }\) = 1 : 4
∴ The given ratios are in proportion.
Extreme terms are 25 cm and ₹ 160
Middle terms are 1 m and ₹ 40

b) 39 litres : 65 litres and 6 bottles : 10 bottles
Solution:
39 litres : 65 litres = \(\frac { 39 }{ 65 }\) = \(\frac { 39 /div 13 }{ 65 /div 13 }\) = \(\frac { 3 }{ 5 }\) = 3 : 5
6 bottles : 10 bottles = \(\frac { 6 }{ 10 }\) = \(\frac { 6/div 2 }{ 10 /div 2 }\) = \(\frac { 3 }{ 5 }\) = 3 : 5
∴ The given ratios are in proportion.
Extreme terms are 39 litres and 10 bottles
Middle terms are 65 litres and 6 bottles

c) 2 kg : 80 kg and 25 g : 625 g
Solution:
2 kg : 80 kg = \(\frac { 2 }{ 80 }\) = \(\frac { 2/div 2 }{ 80 /div 2 }\) = \(\frac { 1 }{ 40 }\) = 1 : 4
25 g : 625 g = \(\frac { 25 }{ 625 }\) = \(\frac { 25/div 25 }{ 625 /div 25 }\) = \(\frac { 1 }{ 25 }\) = 1: 25
Since 1 : 4 ≠ 1 : 25
∴ The given ratios are not in proportion.

d) 200 ml : 2.5 litre and ₹ 4 : ₹ 50
Soluton:
200 ml : 2.5 litre
= \(\frac { 200 }{ 2500 }\) = \(\frac { 200/div 100 }{ 2500 /div 100 }\) = \(\frac { 2 }{ 25 }\) = 2 : 25
₹ 4 : ₹ 50 = \(\frac { 4 }{ 50 }\) = \(\frac { 4/div 2 }{ 50 /div 2 }\) = \(\frac { 2 }{ 25 }\) = 2 : 25
∴ The given ratios are in proportion.
Extreme terms are 200 ml and ₹ 50
Middle terms are 2.5 litre and ₹ 4

Well-designed AP 6th Class Maths Guide Chapter 12 Ratio and Proportion Exercise 12.1 offers step-by-step explanations to help students understand problem-solving strategies.

Algebra Class 6 Exercise 12.1 Solutions – 6th Class Maths 12.1 Exercise Solutions

Question 1.
There are 20 girls and 15 boys in a class.
a) What is the ratio of number of girls to the number of boys?
b) What is the ratio of number of girls to the total number of students in the class?
Solution:
Number of girls in the class = 20
Number of boys in the class = 15

a) The ratio of number of girls to the number of boys = \(\frac { 20 }{ 15 }\) = \(\frac { 4 }{ 3 }\)
Thus the required ratio = 4 : 3

b) Total number of students = 20 + 15 = 35
The ratio of number of girls to the total number of students = \(\frac { 20 }{ 35 }\) = \(\frac { 4 }{ 7 }\)
Thus the required ratio = 4 : 7.

Question 2.
Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of

a) Number of students liking football to number of students liking tennis.
b) Number of students liking cricket to total number of students.
Solution:
Number of students in the class = 30
Number of students liking football = 6
Number of students liking cricket = 12
Number of students liking tennis = 30 – (6 + 12) = 30 – 18 = 12

a) Ratio of number of students liking football to the number of students liking tennis = \(\frac { 6 }{ 12 }\) = \(\frac { 1 }{ 2 }\)
Thus, the required ratio = 1 : 2.

b) The ratio of the number of students liking cricket to the total number of stdudents
= \(\frac { 12 }{ 30 }\) = \(\frac{12 \div 6}{30 \div 6}\) = \(\frac { 2 }{ 5 }\)
Thus, the required ratio = 2 : 5.

Question 3.
See the figure and find the ratio of

a) Number of triangles to the number of circles inside the rectangle.
b) Number of squares to all the figures inside the rectangle.
c) Number of circles to all the figures inside the rectangle.
Solution:
Number of triangles in the given diagram = 3
Number of circles in the given diagram = 2
Number of squares in the given diagram = 2
Total figures = 7
a) The ratio of number of triangles to the number of circles = \(\frac { 3 }{ 2 }\)
Thus, the required ratio = 3 : 2.
b) The ratio of number of squares to number of all figures =\(\frac { 2 }{ 7 }\)
Thus, the required ratio = 2 : 7.
c) The ratio of number of circles to the number of all the figures = \(\frac { 2 }{ 7 }\)
Thus, the required ratio = 2 : 7.

Question 4.
Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.
Solution:
Distance travelled by Hamid = 9 km
Distance travelled by Akhtar = 12 km
Speed of Hamid = 9 km per hour
Speed of Akhtar = 12 km per hour
∴ Ratio of the speed of Hamid to the speed of Akhtar = \(\frac { 9 }{ 12 }\) = \(\frac{9 \div 3}{12 \div 3}\) = \(\frac { 3 }{ 4 }\)
Thus, the required ratio = 3 : 4.

Question 5.
Fill in the following blanks :

Question 6.
Find the ratio of the following :
a) 81 to 108
Solution:
81 to 108 = \(\frac { 81 }{ 108 }\) = \(\frac{81 \div 9}{108 \div 9}\) = \(\frac{9 \div 3}{12 \div 3}\) = \(\frac { 3 }{ 4 }\)
Thus, the ratio = 3 : 4

b) 98 to 63
Solution:
98 to 63 = \(\frac { 98 }{ 63 }\) = \(\frac{98 \div 7}{63 \div 7}\) = \(\frac { 14 }{ 9 }\)
Thus, the ratio = 14 : 9

c) 33 km to 121 km
Solution:
33 km to 121 km = \(\frac { 33 km }{ 121 km }\) = \(\frac { 33 }{ 121 }\) = \(\frac{33 \div 11}{121 \div 11}\) = \(\frac { 3 }{ 11 }\)
Thus, the ratio = 3 : 11

d) 30 minutes to 45 minutes
Solution:
30 minutes to 45 minutes = \(\frac{30 \text { minutes }}{45 \text { minutes }}\) = \(\frac { 30 }{ 45 }\) = \(\frac{30 \div 15}{45 \div 15}\) = \(\frac { 2 }{ 3 }\)
Thus, the ratio = 2 : 3

Question 7.
Find the ratio of the following :
a) 30 minutes to 1.5 hours
Solution:
30 minutes to 1.5 hours
Ratio of 30 minutes to 1.5 hours = Ratio of 30 minutes to 90 minutes.
[ 1.5 hour = 1.5 × 60 minutes = 90 minutes] = \(\frac{30 \text { minutes }}{90 \text { minutes }}\)
= \(\frac { 30 }{ 90 }\) = \(\frac{30 \div 10}{90 \div 10}\) = \(\frac { 3 }{ 9 }\) = \(\frac{3 \div 3}{9\div 3}\) = \(\frac { 1 }{ 3 }\)
Thus, the ratio = 1 : 3.

b) 40 cm to 1.5 m
Solution:
40 cm to 1.5 m
1.5 m = 1.5 × 100 cm = 150 cm
∴ Ratio of 40 cm to 1.5 m
= Ratio of 40 cm to 150 cm
= \(\frac { 40 cm }{ 150 cm }\) = \(\frac { 40 }{ 150 }\) = \(\frac{40 \div 10}{150 \div 10}\) = \(\frac { 4 }{ 15 }\)
Thus, the ratio = 4 : 15.

c) 55 paise to ₹ 1
Solution:
55 paise to ₹ 1 = ₹ 1 = 100 paise
∴ Ratio of 55 paise to ₹ 1
= Ratio of 55 paise to 100 paise
= \(\frac { 55 paise }{ 100 paise }\) = \(\frac { 55 }{ 100 }\) = \(\frac{55 \div 5}{100 \div 5}\) = \(\frac { 11 }{ 20 }\)
Thus, the ratio = 11 : 20

d) 500 ml to 2 litres
Solution:
500 ml to 2 litres = 2 litres = 2 × 1000 ml = 2000 ml.
∴ Ratio of 500 ml to 2 litres
= 500 ml to 2000 ml
= \(\frac { 500 }{ 2000 }\) = \(\frac{500 \div 500}{2000 \div 500}\) = \(\frac { 1 }{ 4 }\)
Thus, the ratio = 1 : 4

Question 8.
In a year, Seema earns ₹ 1,50,000$ and saves ₹ 50,000. Find the ratio of
a) Money that Seema earns to the money she saves.
Solution:
Money earned by Seema = ₹ 1,50,000
Money saved by her = ₹ 50,000
Money spent by her = ₹ 1,50,000 – ₹ 50,000 = ₹ 1,00,000
∴ Ratio of money earned by Seema to the money saved by her

b) Money that she saves to the money she spends.
Sol. Money saved by Seema = ₹ 50,000
Money spent by Seema = ₹ 1,00,000
∴ Ratio of money saved and money spent by Seema

Question 9.
There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Solution:
Number of teachers = 102
Number of students = 3300
∴ Ratio of teachers to the students = \(\frac { 102 }{ 3300 }\) = \(\frac{102 \div 2}{3300 \div 2}\) = \(\frac{51 \div 3}{1650 \div 3}\) = \(\frac { 17 }{ 0550 }\)
Thus the ratio = 17 : 550.

Question 10.
In a college, out of 4320 students, 2300 are girls. Find the ratio of
a) Number of girls to the total number of students.
Solution:
Total number of students = 4320
Number of girls = 2300
∴ Number of boys = 4320 – 2300 = 2020
Ratio of number of girls to the total number of students
\(\frac { 2300 }{ 4320 }\) = \(\frac{2300 \div 10}{4320 \div 10}\) = \(\frac{230 \div 2}{432 \div 2}\) = \(\frac { 115 }{ 216 }\)
Thus, the ratio = 115 : 216.

b) Number of boys to the number of girls.
Solution:
Ratio of number of boys to the number of girls = \(\frac { 2020 }{ 2300 }\) = \(\frac{2020 \div 10}{2300 \div 10}\) = \(\frac{202 \div 2}{230 \div 2}\) = \(\frac { 101 }{ 115 }\)
Thus, the ratio = 101 : 115

c) Number of boys to the total number of students.
Solution:
Ratio of number of boys to the total number of students
= \(\frac { 2020 }{ 4320 }\) = \(\frac{2020 \div 10}{4320 \div 10}\) = \(\frac{202 \div 2}{432 \div 2}\) = \(\frac { 101 }{ 216 }\)
Thus, the ratio = 101 : 216

Question 11.
Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of
a) Number of students who opted basketball to the number of students who opted table tennis.
Solution:
Total number of students = 1800
Number of students opted basket ball = -750
Number of students opted cricket = 800
Number of remaining students who opted table tennis = 1800 – (750 + 800) = 1800 – 1550 = 250

a) Ratio of number of students opted basket ball to the number of students who opted table tennis
= \(\frac{750 \div 10}{250 \div 10}\) = \(\frac{75 \div 25}{25 \div 25}\) = \(\frac { 3 }{ 1 }\)
Thus, the ratio = 3 : 1.

b) Number of students who opted cricket to the number of students opting basketball.
Solution:
Ratio of the students who opted cricket to the number of students opting basket ball
= \(\frac { 800 }{ 750 }\) = \(\frac{800 \div 10}{750 \div 10}\) = \(\frac{80 \div 5}{75 \div 5}\) = \(\frac { 16 }{ 15 }\)
Thus, the ratio = 16 : 15.

c) Number of students who opted basketball to the total number of students.
Solution:
Ratio of number of students who opted basket ball to the total number of students = \(\frac { 750 }{ 1800 }\) = \(\frac{750 \div 10}{1800 \div 10}\) = \(\frac{75 \div 5}{180 \div 5}\) = \(\frac{15 \div 3}{36 \div 3}\) = \(\frac { 5 }{ 12 }\)
Thus, the ratio = 5 : 12.

Question 12.
Cost of a dozen pens is ₹ 180 and cost of 8 ball pens is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Solution:
Cost of 1 dozen pens = ₹ 180 [1 dozen = 12 objects]
So, 12 pens cost = ₹ 180
∴ Cost of 1 pen = \(\frac { ₹ 180 }{ 12 }\) = ₹ 15
Cost of 8 ballpens = ₹ 56
∴ Cost of 1 ballpen = \(\frac { ₹ 56 }{ 8 }\) = ₹ 7
Now, the ratio of cost of 1 pen to cost of 1 ballpen = \(\frac { 15 }{ 7 }\) = 15 : 7
Thus required ratio = 15 : 7

Question 13.
Consider the statement: Ratio of breadth and length of a hall is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall.

Question 14.
Divide 20 pens between Sheela and Sangeeta in the ratio of 3 : 2.
Solution:
Total number of pens = 20
Dividing ratio = 3 : 2
We have 3 + 2 = 5

Thus, Sheela gets 12 pens and Sangeeta gets 8 pens
(OR)
Total number of pens = 20
Dividing ratio = 3 : 2
Let the number of pens Sheela gets = 3x pens
and Sangeeta gets = 2x pens
Total number of pens 3x + 2x = 20 = 5x = 20
x = \(\frac { 20 }{ 5 }\) = 4
∴ Number of pens Sheela gets = 3x = 3(4) = 12
Number of pens Sangeeta gets = 2x = 2(4) = 8
(OR)

So, Sheela gets 12 pens and Sangeeta gets 8 pens.

Question 15.
Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Solution:
Given that total dividing amount = ₹36
Dividing ratio
Money got by Shreya : Money got by Bhoomika = Shreya’s age: Bhoomika age
= 15: 12 = 5 : 4 [\(\frac { 15 }{ 12 }\) = \(\frac{15 \div 3}{12 \div 3}\) = \(\frac { 5 }{ 4 }\)]
Sum = 5 + 4 = 9

(or)
Bhoomika’s share = Total – Shreya’s share = ₹ 36 – ₹ 20 = ₹ 16

Question 16.
Present age of father is 42 years and that of his son is 14 years. Find the ratio of
a) Present age of father to the present age of son.
b) Age of the father to the age of son, when son was 12 years old.
c) Age of father after 10 years to the age of son after 10 years.
d) Age of father to the age of son when father was 30 years old.
Solution:
Present age of father = 42 years
Present age of his son = 14 years
a) Ratio of present age of father to the present age of son
\(\frac { 42 }{ 14 }\) = \(\frac{42 \div 2}{14 \div 2}\) = \(\frac{21 \div 7}{7 \div 7}\) = \(\frac { 3 }{ 1 }\)
Thus, the ratio = 3 : 1

b) When son was 12 years old, Le. 14 – 12 – 2 years ago father’s age = 42 – 2 = 40 years
Now the ratio of the father’s age to son’s age = \(\frac { 40 }{ 12 }\) = \(\frac{40 \div 4}{12 \div 4}\) = \(\frac { 10 }{ 3 }\)
Thus, the ratio = 10 : 3.

c) Age of father after 10 years = 42 + 10 = 52 years
Age of son after 10 years = 14 + 10 = 24 years
Now, ratio of father’s age after 10 years to the age of son after 10 years
= \(\frac { 52 }{ 24 }\) = \(\frac{52 \div 2}{24 \div 2}\) = \(\frac{26 \div 2}{12 \div 2}\) = \(\frac { 13 }{ 6 }\)
Thus, the ratio = 13 : 6.

d) When the father’s age is 30 years, i.e 42 – 30 = 12 ago son age = 14 – 12 = 2 years
Now the ratio of age of father to the age of son when father was 30 years
= \(\frac { 30 }{ 2 }\) = \(\frac{30 \div 2}{2 \div 2}\) = \(\frac { 15 }{ 1 }\)
Thus, the ratio = 15 : 1.

Well-designed AP 6th Class Maths Guide Chapter 11 Algebra Exercise 11.1 offers step-by-step explanations to help students understand problem-solving strategies.

Algebra Class 6 Exercise 11.1 Solutions – 6th Class Maths 11.1 Exercise Solutions

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
a) A pattern of letter T as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 2 × 1 = 2
n = 2, the number of matchsticks required = 2 × 2 = 4
n = 3, the number of matchsticks required = 2 × 3 = 6
∴ The rule = 2 × n = 2 n, where n is number of Ts.

b) A pattern of letter Z as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 3 × 1 = 3
n = 2, the number of matchsticks required = 3 × 2 = 6
n = 3, the number of matchsticks required = 3 × 3 = 9
∴ The rule = 3 × n = 3n, where n is number of Zs .

c) A pattern of letter U as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 3 × 1 = 3
n = 2, the number of matchsticks required = 3 × 2 = 6
n = 3, the number of matchsticks required = 3 × 3 = 9
∴ The rule = 3 × n = 3n, where n is number of Us .

d) A pattern of letter V as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 2 × 1 = 2
n = 2, the number of matchsticks required = 2 × 2 = 4
n = 3, the number of matchsticks required = 2 × 3 = 6
n = 4, the number of matchsticks required = 2 × 4 = 8
∴ The rule = 2 × n = 2n, where n is number of Vs.

e) A pattern of letter E as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 5 × 1 = 5
n = 2, the number of matchsticks required = 5 × 2 = 10
n = 3, the number of matchsticks required = 5 × 3 = 15
∴ The rule = 5 × n = 5n, where n is number of Es.

f) A pattern of letter S as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 5 × 1 = 5
n = 2, the number of matchsticks required = 5 × 2 = 10
n = 3, the number of matchsticks required = 5 × 3 = 15
∴ The rule = 5 × n = 5n, where n is number of Ss.

g) A pattern of letter A as
Solution:

Number of matchsticks required to make the pattern
For n = 1, the number of matchsticks required = 6 × 1 = 6
n = 2, the number of matchsticks required = 6 × 2 = 12
n = 3, the number of matchsticks required = 6 × 3 = 18
∴ The rule = 6 × n = 6n, where n is number of As.

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q. 1 (given above) give us the same rule as that given by L. Which are these ? Why does this happen ?
Solution:
Rule for the following letters.

Letter	Rule
L	2n
V	2n
T	2n

Letter	Rule
C	3n
F	3n
U	3n

From the above tables,
i) We obseerve that the rule is same of L,V and T i.e. 2n as they are required only two matchsticks.
ii) Letters C, F and U have the same rule i.e. 3n, as they required only 3 matchsticks.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:
Number of cadets in a row = 5
Number of rows = n

Number of cadets = n
For n = 1 is 5 × 1 = 5
n = 2 is 5 × 2 = 10
n = 3 is 5 × 3 = 15
∴ Rule is 5n, where n represents the number of rows.

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:
Number of boxes = b
Number of mangoes in a box = 50
Number of unangoes
For n – 1 is 50 × 1 = 50
For n = 2 is 50 × 2 = 100
For n = 3 is 50 × 3 = 150
∴ Rule is 50 b, where b represents the number of boxes.

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution:
Number of students = s
Number of pencils ditributed per student = 5
Number of pencils required
For n = 1 is 5 × 1 = 5
For n =2 is 5 × 2 = 10
For n = 3 is 5 × 3 = 15
∴ Rule is 5s, where s represents the number of students.

Question 6.
A bird files 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution:
Distance covered in 1 minute = 1 km
The flying time = t
Distance covered
For n = 1 is 1 × 1 = 1 km
For n = 2 is 1× 2 = 2 km
For n = 3 is 1 × 3 = 3 km
∴ Rule is 1 × t km, where t represents, the flying time.

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoll have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution:
Number of rows = 9

Number of dots in a row drawn by Radha = 8
∴ The number of dots required
For r = 1 is 9 × 1 = 9
For r = 2 is 9 × 2 = 18
For r = 3 is 9 × 3 = 27
∴ Rule is 9r, where r represents the number of rows.
If r = 8 (i.e. 8 rows), the number of dots = 9 × 8 = 72
r = 10 (i.e. 10 rows), the number of dots = 9 × 10 = 90

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age ? Take Radha’s age to be x years.
Solution:
Radha’s age = x years
Given that Leela’s age = Radha’s age – 4 years
= x years – 4 years
= (x – 4) years

Question 9.
Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution:
Given that the number of laddus given away = l
Number of laddus left = 5
∴ Number of laddus made by mother = l + 5

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box ?
Solution:
Given that, the number of oranges in smaller box = x
∴ Number of oranges in bigger box = 2 (number of oranges in small box) + Number of oranges remain outside)
So, the number of oranges in bigger box = 2x + 10

Question 11.
a) Look at the following matchstick pattern of squares (Figure). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint : If you remove the vertical stick at the end, you will get a pattern of Cs.)

Solution:
Let n be the number of squares.
∴ Number of matchsticks required.
For n = 1 = 3n + 1 = 3 × 1 + 1 = 4
For n = 2 = 3n + 1 = 3 × 2 + 1 = 7
For n = 3 = 3n + 1 = 3 × 3 + 1 = 10
For n = 4 = 3n + 1 = 3 × 4 + 1 = 13
For n, 3 × n + 1 = 3n + 1
∴ Rule is 3n + 1, where n represents the number of squares.

b) Following figure gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Solution:
Let n be the number of triangles.
∴ Number of matchsticks required.
For n = 1 = 2 × 1 = 1
For n = 2 = 2 × 2 = 1
For n = 3 = 2 × 3 + 1
For n = 4 = 2 × 4 + 1
For n, 2 × n + 1 = 2n + 1
∴ Rule is 2n + 1, where n represents the number of matchsticks.

Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 10th Chapter Mensuration InText Questions

Try these (Page No : 104)

Question 1.
Measure and write the length of the four sides of the top of your study table.
AB = _____ cm
BC = _____ cm
CD = _____ cm
DA = _____ cm

Now, the sum of the lengths of the four sides
= AB + BC + CD + DA
= _____ cm + _____ cm + _____ cm + _____ cm = _____ cm
What is the perimeter?
Solution:
Student’s activity.

Question 2.
Measure and write the lengths of the four sides of a page of your notebook. The sum of the lengths of the foursides
= AB + BC + CD + DA
= _____ cm + _____ cm + _____ cm + _____ cm
= _____ cm
What is the perimeter of the page ?
Solution:
Student’s activity.

Question 3.
Meera went to a park 150 m long and 80 m wide. She took one complete round on its boundary. What is the distance covered by her ?
Solution:
Length of the park = 150 m
Breadth of the park = 80 m
Perimeter of the park = 2 (length + breadth)
= 2(150 + 80)
= 2 × 230 = 460 m
So, distance covered by Meena = 460 m.

Question 4.
Find the perimeter of the following figures:

Perimeter = AB + BC + CD + DA
= _____ + _____ + _____ + _____ + _____ + =
Solution:
40 + 10 + 40 + 10 = 100 cm

Perimeter = AB + BC + CD + DA
= _____ + _____ + _____ + _____ = _____
Solution:
5 + 5 + 5 + 5 = 20 cm

Perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= _____ + _____ + _____ + _____ + _____ + _____ + _____+ _____ + _____ = _____
Solution:
1 + 3 + 3 + 1 + 3 + 3 + 1 + 3 + 3 + 1 + 3 + 3 = 28 cm

Perimeter = AB + BC + CD + DE + EF + FA
= _____ – _____ + _____ + _____ + _____ + _____ = _____
Solution:
100 + 120 + 90 + 45 + 60 + 80 = 495 m

Try these (Page No: 108)

Question 1.
Find the perimeter of the following rectangles:

Try these (Page No: 114)

Question 1.
Find various objects from your surroundings which have regular shapes and find their perimeters.
Solution:
Student’s activity.

Try these (Page No : 122)

Question 1.
Draw any circle on a graph sheet. Count the squares and use them to estimate the area of the circular region.
Solution:
Student’s Activity.

Question 2.
Trace shapes of leaves, flower petals and other such objects on the graph paper and find their areas.
Solution:
Student’s Activity.

Try these (Page No : 126)

Question 1.
Find the area of the floor of your classroom.
Solution:
Student’s Activity.

Question 2.
Find the area of any one door in your house.
Solution:
Student’s Activity.

Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration Exercise 10.3 offers step-by-step explanations to help students understand problem-solving strategies.

Mensuration Class 6 Exercise 10.3 Solutions – 6th Class Maths 10.3 Exercise Solutions

Question 1.
Find the areas of the rectangles whose sides are :
a) 3 cm and 4 cm
b) 12 m and 21 m
c) 2 km and 3 km
d) 2 m and 70 cm
Solution:
a) Length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Area of the rectangle
= length × breadth
= 3 cm × 4 cm
= 12sq.cm

b) Length of the rectangle = 12 m
Breadth of the rectangle = 21 m
Area of the rectangle
= length × breadth
= 12 m × 21 m
= 252 sq.m

c) Length of the rectangle = 2 km
Breadth of the rectangle = 3 km
Area of the rectangle
= length × breadth
= 2 km × 3 km
= 6 sq. km.

d) Length of the rectangle = 2 m
Breadth of the rectangle = 70 cm
= 0.70 m
Area of the rectangle
= length × breadth
= 2 m × 0.70 m
= 1.40 sq.m

Question 2.
Find the areas of the squares whose sides are:
a) 10 cm
b) 14 cm
c) 5 m
Solution:
a) Side of the square = 10 cm
∴ Area of the square = side × side
= 10 cm × 10 cm
= 100 sq.cm

b) Side of the square = 14 cm
∴ Area of the square = side × side
= 14 cm × 14 cm
= 196 sq.cm

c) Side of the square = 5 m
∴ Area of the square
= side × side
= 5 m × 5 m
= 25 sq.m

Question 3.
The length and breadth of three rectangles are as given below :
a) 9 m and 6 m
b) 17 m and 3 m
c) 4 m and 14 m
Which one has the largest area and which one has the smallest ?
Solution:
a) Length of the rectangle = 9 m
Breadth of the rectangle = 6 m
∴ Area of the rectangle
= length × breadth
= 9 m × 6 m
= 54 sq.m.

b) Length of the rectangle = 17 m
Breadth of the rectangle = 3 m
∴ Area of the rectangle = length × breadth
= 17 m × 3 m
= 51 sq.m.

c) Length of the rectangle = 4 m
Breadth of the rectangle = 14 m
∴ Area of the rectangle = length × breadth
= 4 m × 14 m
= 56 sq.m.
From the above information
Rectangle (c) has the largest area i.e., 56 sq.m and rectangle
(b) has smallest area i.e. 51 sq.m

Question 4.
The area of a rectangular garden 50 m long is 300 sq m . Find the width of the garden.
Solution:
Length of the rectangular garden = 50 m
Area of the rectangular garden = 300 sq.m.
∴ Width of the rectangular garden = Area + Length
= 300 ÷ 50
= 6 m
∴ Width of the rectangular garden 1 m.

Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m. ?
Solution:
Length of the rectangular plot = 500 m
and breadth = 200 m
Area of rectangular park
= length × breadth
= 500 m × 200 m
= 1,00,000 sq.m
Rate of the tiling the plot = ₹8 per 100 sq.m.
∴ Cost of tiling the garden = 100000 × \(\frac { 8 }{ 106 }\) =₹ 8000
Hence the cost of tiling the rectangular plot = ₹ 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm . What is its area in square metres ?
Solution:
Length of the table – top = 2 m
and its breadth = 1 m 50 cm = 1.50 m
Area of the table top = length × breadth
= 2 m × 1.50 m = 3 m²
Hence, the area of the table top = 3m² (or) 3 sq.m.

Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solution:
Length of the room = 4 m
Breadth of the room =3.5 m
Area of the room = length × breadth
= 4 m × 3.5 m
= 14 m² (or) 14 sq.m
Hence, the area of the carpet needed = 14sq.m

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:
Length of the floor = 5 m
Breadth of the floor = 4m

∴ Area of the floor = length × breadth
= 5 m × 4 m
= 20 sq.m
Side of the square carpet = 3 m
∴ Area of the square carpet = side × side
= 3 m × 3 m
= 9 sq.m
∴ Area of the floor which is not carpeted
= Area of floor – Area of square carpet
= 20 m² – 9m²
= 11 sq.m.

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land ?
Solution:
Side of the square flower bed = 1m
∴ Area of a square flower bed
= side × side
= 1 m × 1 m
= 1 sq.m
So, area of 5 square flower beds = 5 × 1 sq.m = 5

Now, Length of the land = 5 m
Breadth of the land = 4 m
Area of the land = length × breadth = 5 m × 4 m = 20 sq.m.
∴ Area of the remaining part = Area of land – Area of 5 square flower beds.
= 20 sq.m – 5 sq.m = 15 sq.m

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solution:
a) Splitting the given figure into four rectangles, they are I, II, II and IV

Area of the rectangle I
= length × breadth
= 4 cm × 3 cm
= 12 sq.m

Area of the rectangle II
= 3 cm × 2 cm
= 6 sq.m

Area of the rectangle III
= 4 cm × 1 cm
= 4 sq.cm

Area of the rectangle IV
= 3 cm × 2 cm
= 6 sq. cm
Total area of the given figure = 12 sq.cm + 6 sq.cm + 4 sq.cm + 6 sq.cm = 28 sq.cm

b) Splitting the given figure into three rectangles, they are I, II and III

Area of the Rectangle I = length × breadth
= 3 cm × 1 cm
= 3 sq.cm

Area of the rectangle II
= 3 cm × 1 cm
= 3 sq.m

Area of the rectangle III
= 3 cm × 1 cm = 3 sq.cm
∴ Total area of the given figure = 3 sq.cm + 3 sq.cm + 3 sq.cm = 9 sq.cm

Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Solution:
a) Splitting the given figure into two rectangles, they are I and II

Now, area of the rectangle I
= length × breadth
= 10 cm × 2 cm
= 20 sq.cm

Area of the rectangle II
= 10 cm × 2 cm
= 20 sq.cm
∴ Total area of the given figure
= 20 sq.cm + 20 sq.cm
= 40 sq.cm

b) Splitting the given figure into two squares, and one rectangle, they are I and III are squares and il is rectangle.

Now, Area of the square l = side × side = 7 cm × 7 cm = 49 sq.cm

Area of the rectangle II = length × breadth
= 21 cm × 7 cm
= 147 sq.cm

Area of the square III
= 7 cm × 7 cm
= 49 sq.cm
∴ Total area of the given figure = 49 sq.cm + 147 sq.cm + 49 sq.cm
= 245 sq.cm

c) Splitting the given figure into two rectangles, they are I and II

Area of the rectangle I = length × breadth
= 4 cm × 1 cm
= 4 sq.cm

Area of the rectangle II = 5 cm × 1 cm
= 5 sq.cm
∴ Total area of the given figure = 4 sq.cm + 5 sq.cm
= 9 sq.cm

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively :
a) 100 cm and 144 cm
b) 70 cm and 36 cm
Solution:
Length of the tile = 12 cm.
Breadth-of the tile = 5 cm.
∴ Area of the tile = length × breadth = 12 cm × 5 cm = 60 sq.cm.

Now
a) Length of the rectangular region = 100 cm
Breadth of the region = 144 cm
∴ Area of the rectangular region = length × breadth
= 100 cm × 144 cm
= 14400 sq.cm.
∴ Number of tiles needed to cover the whole rectangular region = 14400 sq.cm ÷ 60 sq.cm = 240 tiles

b) Length of the rectangular region = 70 cm
Breadth of the rectangular region = 36 cm
Area of the rectangular region = length × breadth
= 70 cm × 36 cm
= 2520 sq.cm
∴ Number of tiles needed to cover the whole rectangular region
= 2520 sq.cm ÷ 60 sq.cm = 42 tiles.

Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration Exercise 10.2 offers step-by-step explanations to help students understand problem-solving strategies.

Mensuration Class 6 Exercise 10.2 Solutions – 6th Class Maths 10.2 Exercise Solutions

Question 1.
Find the areas of the following figures by counting square :

Solution:
a) Number of full squares = 9
Area of 1 square = 1 sq. unit.
∴ Area of 9 squares = 9 × 1 sq. unit = 9 sq. units.

b) Area of the figure = 5 sq. units

c) Number of full squares = 2
Number of half squares = 4
∴ Area of the figure = 2 + 4 × \(\frac { 1 }{ 2 }\) = 2 + 2 = 4sq. units.

d) Area of the figure = 8 sq. units

e) Area of the figure = 10 sq.units

f) Number of full squares = 2
Number of half squares = 4
∴ Area of the figure = 2 + 4 × \(\frac { 1 }{ 2 }\) = 2 + 2 = 4 sq. units.

g) Number of full squares = 4
Number of half squares = 4
∴ Area of the figure = 4 + 4 × \(\frac { 1 }{ 2 }\) = 4 + 2 = 6 sq. units.

h) Area of the figure = 5 sq.units

i) Area of the figure = 9 sq. units

j) Number of full squares = 2
Number of half square = 4
∴ Area of the figure = 2 + 4 × \(\frac { 1 }{ 2 }\) = 2 + 2 = 4 sq. units.

k) Number of full squares = 4
Number of half squares = 2
∴ Area of the figure = 4 + 2 × \(\frac { 1 }{ 2 }\) = 4 + 1 = 5 sq. units.

l) Number of full squares = 4
Number of squares more than half = 3
Number of half squares = 2
∴ Area of the figure = 4 + 3 + 2× \(\frac { 1 }{ 2 }\) = 4 + 3 + 1 = 8 sq. units.

m) Number of full squares = 6
Number of more than half squares = 8
∴ Area of the figure = 6 + 8 = 14 sq. units.

n) Number of squares = 9
Number of more than half squares = 9
∴ Area of the figure = 9 + 9 = 18 sq. units.

Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration Exercise 10.1 offers step-by-step explanations to help students understand problem-solving strategies.

Mensuration Class 6 Exercise 10.1 Solutions – 6th Class Maths 10.1 Exercise Solutions

Question 1.
Find the perimeter of each of the following figures.

Solution:
a) Perimeter = 2 cm + 1 cm + 5 cm + 4 cm = 12 cm
b) Perimeter = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
c) Perimeter = 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
d) Perimeter = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
e) Perimeter = 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm = 15 cm
f) Perimeter = 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:
Length of the lid = 40 cm
Breadth of the lid = 10 cm
∴ Total length of the tape required
= Perimeter of the rectangular lid
= 2 × [length + breadth]
= 2 × (40 cm + 10 cm)
= 2 × 50 cm = 100 cm or 1 m

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution:
Length of the table top = 2 m 25 cm
Breadth of the table top = 1 m 50 cm
∴ Perimeter of the table top = 2 × [length + breadth]
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (3 m 75 cm)
= 2 × (3.75 cm) = 7.5 m

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively ?
Solution:
Length of the photograph = 32 cm
Breadth of the photograph = 21 cm
∴ Length of the wooden strip required to frame a photograph = Perimeter of photograph
= 2 × [length + breadth]
= 2 × (32 cm + 21 cm)
= 2 × (53 cm)
= 106 cm = 1 m 6 cm

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:
Length of the rectangular piece of land = 0.7 km
Breadth of the rectangular piece of land = 0.5 km
0.7 km + 0.5 km = 1.2 km
∴ Perimeter of the retangular land = 2 × [length + breadth]
= 2 × [0.7 km + 0.5 km]
= 2 × 1.2 km = 2.4 km
∴ Length of the wire needed to 4 rounds of the land = 4 × 2.4 km = 9.6 km.

Question 6.
Find the perimeter of each of the following shapes :
a) A triangle of sides 3 cm, 4 cm and 5 cm.
b) An equilateral triangle of side 9 cm.
c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solution:
a) Side of the triangle = 3 cm, 4 cm, 5 cm

∴ Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm

b) Side of an equilateral triangle = 9 cm

∴ Perimeter of an equilateral triangle = 9 cm + 9 cm + 9 cm = 27 cm

c) Sides of an isosceles triangle = 8 cm, 8 cm, 6 cm

∴ Perimeter of an isosceles triangle = 8 cm + 8 cm + 6 cm = 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Sides of the given triangle = 10 cm, 14 cm, 15 cm.
∴ Perimeter of the triangle = 10 cm + 14 cm + 15 cm = 39 cm

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
Sides of the regular hexagon = 8 cm

∴ Sides of the regular hexagon = 8 cm, 8 cm, 8 cm, 8 cm, 8 cm, 8 cm
[∴ Regular hexagon has 6 equal sides]
∴ Perimeter of the given regular hexagon = 8 cm + 8 cm + 8 cm + 8 cm + 8 cm + 8 cm = 48 cm
(OR)
Perimeter of regular hexagon = 6 × side = 6 × 8 cm = 48 cm.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of the square = 20 cm
Perimeter of the square = 4 × side

4 × side = 20 m
∴ side = 20 ÷ 4 = 5 m

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of the regular pentagon = 100 cm
Number of equal sides of the regular pentagon = 5
Perimeter of the regular pentagon = 5 × side

5 × side = 100 cm
side = 100 ÷ 5 = 20 cm
∴ Length of the each side of the given regular pentagon = 20 cm

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
a) a square ?
b) an equilateral triangle ?
c) a regular hexagon?
Solution:
a) Length of the string = 30 cm
Number of equal sides in a square = 4
∴ Length of each side of the square = 30 cm ÷ 4 = 7.5 cm

b) Length of the string = 30 cm
Number of equal sides in an equilateral triangle = 3
∴ Length of each side of an equilateral triangle = 30 cm ÷ 3 = 10 cm

c) Length of the string = 30 cm
Number of equal sides in an regular hexagon = 6
∴ Length of each side of the regular hexagon = 30 cm ÷ 6 = 5 cm

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side ?
Solution:
Length of two sides of the given triangle
= 12 cm and 14 cm
Perimeter of the triangle = 36 cm
So, length of the third side
= 36 -(12 + 14)
= 36 cm – (12 cm + 14 cm)
= 36 cm – 26 cm = 10 cm

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solution:
Length of the side of the square park = 250 m
∴ Perimeter of the square park = 4 × side
= 4 × 250 m
= 1000 m

Rate of fencing = ₹ 20 m
∴ Cost of fencing = 1000 × ₹ 20 = ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre.
Solution:
Length of the rectangular park = 175 m
Perimeter of the park
= 2 × [length + breadth]
= 2 × [175 m + 125 m]
= 2 × 300 m = 600 m

Rate of fencing = ₹ 12 per meter
∴ Cost of fencing of rectangular park = 600 × ₹ 12 = ₹ 7200.

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m . Who covers less distance ?
Solution:
Side of the square park = 75 m
And its perimeter = 4 × side
= 4 × 75 m = 300 m

Length of the rectangular park = 60 m
Breadth of the rectangular park = 45 m

Perimeter of the rectangular park
= 2 × length + breadth
= 2 × [60 m + 45 m]
= 2 × 105 m
= 210 m
Since 210 m < 300 m
So, Bulbul covers less distance.
So, length of the third side
= 36 – (12 + 14)
= 36 cm – (12 cm + 14 cm)
= 36 cm – 26 cm = 10 cm

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:
a) Perimeter of the given square = 25 cm + 25 cm + 25 cm + 25 cm = 4 × 25 = 100 cm
b) Perimeter of the rectangle = 40 cm + 10 cm + 40 cm + 10 cm = 100 cm
c) Perimeter of the rectangle = 30 cm + 20 cm + 30 cm + 20 cm = 100 cm
d) Perimeter of the given triangle = 30 cm + 30 cm + 40 cm = 100 cm
From the above answers, we conclude that different figures may have equal perimeters.

Question 17.
Avneet buys 9 square paving slabs, each with a side of \(\frac { 1 }{ 2 }\) m. He lays them in the form of a square.
a) What is the perimeter of his arrangement [(Figure (i)] ?
b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Figure (ii)] ?
c) Which has greater perimeter?
d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this ? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Solution:
a) The arrangement is in the form of a square of side
= [\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)]m = 1\(\frac { 1 }{ 2 }\)m = \(\frac { 3 }{ 2 }\)m
Perimeter of the square arrangement = 4 × \(\frac { 3 }{ 2 }\) = 6m

b) Perimeter of cross – arrangement = \(\frac { 1 }{ 2 }\)m + 1 m + 1 m + \(\frac { 1 }{ 2 }\)m + 1m + 1m + 1m + \(\frac { 1 }{ 2 }\)m + 1m + 1m = 10m

c) Since 10 m > 6 m
∴ Cross – arrangement has greater perimeter.

d) Total number of tiles = 9
∴ We have the following arrangement

The above arrangement will also have the greater perimeter.
i.e. \(\frac { 1 }{ 2 }\) + \(\frac { 9 }{ 2 }\) + \(\frac { 1 }{ 2 }\) + \(\frac { 9 }{ 2 }\) = \(\frac { 20 }{ 2 }\) = 10 cm.

Well-designed AP Board Solutions Class 6 Maths Chapter 9 Data Handling InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 9th Chapter Data Handling InText Questions

Do this (Page No: 88)

Question 1.
Collect information regarding the number of family members of your classmates and represent it in the form of a table. Find to which category most students belong.
Table
Make a table and enter the data using tally marks. Find the number that appeared
a) the minimum number of times?
b) the maximum number of times ?
c) same number of times?
Solution:
Student’s Activity.

Do this (Page No: 90)

Question 1.
Pictographs are often used by dailies and magazines to attract readers attention. Collect one or two such.published pictographs and display them in your class. Try to understand what they say. It requires some practice to understand the information given by a pictograph.
Solution:
Student’s Activity.

Well-designed AP Board Solutions Class 6 Maths Chapter 9 Data Handling Exercise 9.1 offers step-by-step explanations to help students understand problem-solving strategies.

Data Handling Class 6 Exercise 9.1 Solutions – 6th Class Maths 9.1 Exercise Solutions

Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

a) Find how many students obtained marks equal to or more than 7.
b) How many students obtained marks below 4 ?
Solution:

a) Number of students who obtained marks equal to or more than 7 = 5 + 4 + 3 = 12
b) Number of-students who obtained marks below 4 – 2 + 3 – 3 – 8

Question 2.
Following is the choice of sweets of 39 students of Class VI.
Ladon, Barf, Ladon, Jalebi, Lido, Rasgulla, Jalebi, Lido, Barf, Rasgulla, Ladon, Jalebi, Jalebi, Rasgulla, Ladon, Rasgalla, Jalebi, Ladon, Rasgulla, Lido, Ladon, Barf, Rangolla, Rasgulla, Jalebi, Ragulla, Lade, Rogulla, Jalebi, Ladon,
a) Arrange the names of sweets in a table using tally marks.
b) Which sweet is preferred by most of the students?
Answer:
a)
b) Ladon is preferred by most of the students.

Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the number that appeared.
a) The minimum number of times
b) The maximum number of times
c) Find those numbers that appear an equal number of times.

From the above table, we get
a) The number 4 appeared 4 times which is the minimum.
b) The number 5 appeared 11 times which is the maximum.
c) The numbers 1 and 6 appear for the same number of times, i.e. 7 times.

Question 4.
Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions.
i) Which village has the minimum number of tractors?
ii) Which village has the maximum number of tractors?
iii) How many more tractors village C has as compared to village B.
iv) What is the total number of tractors in all the five villages?
Solution:
From the given pictograph, we get
i) Village D has the minimum number of tractors, i.e. 3.
ii) Village C has the maximum number of tractors, i.e. 8.
iii) Village C has 3 tractors more than that of the village B. (∵ 8 – 5 = 3)
iv) Total number of tractors in all the villages is 28. (6 + 5 + 8 + 3 + 6 = 28)

Question 5.
The number of girl students in each class of a coeducational middle school is depicted by the pictograph :

Observe this pictograph and answer the following questions :
a) Which class has the minimum number of girl students?
b) Is the number of girls in Class VI less than the number of girls in Class V ?
c) How many girls are there in Class VII?
Solution:

Question 6.
The sale of electric bulbs on different days of a week is shown below:

Observe the pictograph and answer the following questions:
a) How many bulbs were sold on Friday?
b) On which day were the maximum number of bulbs sold?
c) On which of the days same number of bulbs were sold?
d) On which of the days minimum number of bulbs were sold?
e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Solution:
a) Number of bulbs sold on Friday = 7 × 2 = 14
b) On Sunday, the number of bulbs sold = 9 × 2 = 18 which is maximum in number.
c) On Wednesday and Saturday, the same number of bulbs were sold. i.e 4 × 2 = 8
d) The minimum number of bulbs were sold on Wednesday and Saturday i.e. 4 × 2 = 8
e) Total number of bulbs sold in the week = 43 × 2 = 86
Number of bulbs can hold the big carton = 9
∴ Number of cartons need = 86 ÷ 9 = 9\(\frac { 5 }{ 9 }\) = 10
(Next integer of 9\(\frac { 5 }{ 9 }\) is 10 )

Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season :

Observe this pictograph and answer the following questions:
a) Which merchant sold the maximum number of baskets ?
b) How many fruit baskets were sold by Anwar?
c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Solution:
a) Martin sold the maximum number of fruit baskets i.e 9\(\frac { 1 }{ 2 }\) × 100 = 950
b) Number of fruit baskets sold by Anwar = 7 × 100 = 700
c) Anwar, Martin and Ranjit Singh have sold 600 or more number of baskets and planning to buy a godown for the next season.

Well-designed AP Board Solutions Class 6 Maths Chapter 8 Decimals InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 8th Chapter Decimals InText Questions

Try these (Page No : 64)

i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.
Solution:
100 paise = ₹ 1
1 paise = ₹\(\frac { 1 }{ 100 }\); 5 paise = ₹\(\frac { 5 }{ 100 }\)
So, 2 rupees 5 paise = ₹ 2 + 5 paise
=₹2 + ₹\(\frac { 5 }{ 100 }\)
=₹ 2 + ₹ 0.05
= ₹ 2.05
50 paise = ₹\(\frac { 50 }{ 100 }\)
So 2 rupees 50 paise = ₹ 2 + 50 paise
= ₹2 + ₹\(\frac { 50 }{ 100 }\)
= ₹ 2.50

ii) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals ?
Solution:
20 rupees 7 paise = ₹ 20 + 7 paise
= ₹ 20 + ₹\(\frac { 7 }{ 100 }\)
= ₹ 20.07
21 rupees 75 paise = ₹ 21 + 75 paise.
₹ 21 + ₹ 0.75 = ₹ 21.75

Try these (Page No: 66)

Question 1.
Cati you write 4 mm in ‘cm’ using decimals ?
Solution:
Yes, we can write 4 mm in ‘cm’ using decimals.
10 mm = 1 cm; 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 4 mm = \(\frac { 4 }{ 10 }\) cm = 0.4 cm
Hence 4 mm = 0.4 cm.

Question 2.
How will you write 7 cm 5 mm in ‘cm’ using decimals ?
Solution:
We know that 1 cm = 10 mm
Also 1 mm = \(\frac { 1 }{ 10 }\) cm
For converting pm into cm , we have to multiply the given values in mm by \(\frac { 1 }{ 10 }\)
7 cm 5 mm = 7 + 5 × \(\frac { 1 }{ 10 }\) cm
= 7 + \(\frac { 5 }{ 10 }\) cm
= 7 + 0.5 cm = 7.5 cm
∴ 7 cm 5 mm = 7.5 cm

Question 3.
Can you now write 52 m as ‘km’ using decimals? How will you write 340 m as ‘km’ using decimals? How will you write 2008 m in ‘km’ ?
Solution:
1000 m = 1 km
1 m = \(\frac { 1 }{ 1000 }\) km
52 m = \(\frac { 52 }{ 1000 }\) km = 0.052 km.
Hence we can write 52 m as 0.052 km .
340 m = \(\frac { 340 }{ 1000 }\)km = 0.340 km = 0.34 km
2008 m = \(\frac { 2008 }{ 1000 }\)km = 2.008 km.

Try these (Page No: 66)

Question 1.
Can you now write 456 g as ‘kg’ using decimals?
Solution:
Yes, we can write 456 g as kg using decimals.
We know thag 1000 g = 1 kg
1 g = \(\frac { 1 }{ 1000 }\) kg
∴ 456 g = \(\frac { 456 }{ 1000 }\)kg = 0.456 kg

Question 2.
How will you write 2 kg 9g in ‘kg ‘ using decimals ?
Solution:
2 kg + 9 g
= 2kg + 9g
= 2kg + \(\frac { 9 }{ 1000 }\)kg (1g = \(\frac { 1 }{ 1000 }\) kg)
= 2 kg + 0.009 kg
= 2.009 kg
Hence, 2kg 9 g = 2.009 kg.

Try These (Page No: 70)

Question 1.
i) 0.29 + 0.36
Solution:
0.29 + 0.36 = 0.65

ii) 0.7 + 0.08
Solution:
0.7 + 0.08 = 0.78

iii) 1.54 + 1.80
Solution:
1.54 + 1.80 = 3.34

iv) 2.66 + 1.85
Solution:
2.66 + 1.85 = 4.51

Try these (Page No: 74)

Question 1.
Subtract 1.85 from 5.46.
Solution:
5.46 – 1.85 = 3.61

Question 2.
Subtract 5.25 from 8.28.
Solution:
8.28 – 5.25 = 3.03

Question 3.
Subtract 0.95 from 2.29.
Solution:
2.29 – 0.95 = 1.34

Question 4.
Subtract 2.25 from 5.68.
Solution:
5.68 – 2.25 = 3.43