## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 22 Simple (Linear) Equations (Including Word Problems)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

IMPORTANT POINTS

1. Simple Equations : A mathematical statement, which states that two expressions are equal, is called simple equation.
2. Properties of Simple Equation :
(i) If same quantity is added to both the sides of simple equation, the sums are equal.
For Example :
x = 6 ⇒ x + a = 6 + a [Adding a on both the sides]
(ii) If same quantity is subtracted from both the sides of simple equation, the remainders are equal.
For Example :
x = 6 ⇒ x-a = 6 – a [Subtracting a on both the sides]
(iii) If both the sides of an equation are multiplied by the same quantity, the products are equal.
For Example :
x = 6 ⇒ a x x = a x 6 i.e. ax = 6a [Multiplying both the sides by a]
(iv) If both the sides of simple equation are divided by the same quantity, the quotients are equal.
For Example :
x = 6 ⇒ $$\frac { x }{ a }$$ = $$\frac { 6 }{ a }$$ [Dividing both the sides by a]

Question 1.
Solve:

Solution:

Question 2.
Solve:

Solution:

Question 3.
Solve:

Solution:

Question 4.
Solve:

Solution:

Question 5.
Solve:

Solution:

Question 1.
Solve:

Solution:

Question 2.
Solve:

Solution:

Question 3.
Solve:

Solution:

### Simple (Linear) Equations Exercise 22C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
5 – x = 3
Solution:

Question 2.
2 – y = 8
Solution:

Question 3.
8.4 – x = -2
Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
1.6z = 8
Solution:

Question 8.
3a = – 2.1
Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.
– 5x = 10
Solution:

Question 12.
2.4z = -4.8
Solution:

Question 13.
2y – 5 = -11
Solution:

Question 14.
2x + 4.6 = 8
Solution:

Question 15.
5y – 3.5 = 10
Solution:

Question 16.
3x + 2 = -2.2
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.
-3y – 2 = 10
Solution:

Question 21.
4z – 5 = 3 – z
Solution:

Question 22.
7x -3x +2 =22
Solution:

Question 23.
6y + 3 = 2y + 11
Solution:

Question 24.
3 (x+5) = 18
Solution:
3 (x+5) = 18

Question 25.
5 (x-2) -2 (x+2) = 3
Solution:

Question 26.
(5x-3) 4=3
Solution:

Question 27.
3(2x+1) -2(x-5) -5 (5-2x) = 16
Solution:

### Simple (Linear) Equations Exercise 22D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
A number increased by 17 becomes 54. Find the number.
Solution:
Let the required number = x
∴ According to the sum :
x+17 = 54
⇒ x = 54-17
⇒ x = 37
Required number = 37

Question 2.
A number decreased by 8 equals 26, find the number.
Solution:
Let required number = A
∴According to the sum :
x – 8 = 26
⇒A = 26 + 8
⇒A = 34
∴Required number = 34

Question 3.
One-fourth of a number added to two- seventh of it gives 135; find the number.
Solution:

Question 4.
Two-fifths of a number subtracted from three-fourths of it gives 56, find the number.
Solution:

Question 5.
A number is increased by 12 and the new number obtained is multiplied by 5. If the resulting number is 95, find the original number.
Solution:

Question 6.
A number is increased by 26 and the new number obtained is divided by 3. If the resulting number is 18; find the original number.
Solution:

Question 7.
The age of a man is 27 years more than the age of his son. If the sum of their ages is 47 years, find the age of the son and his father.
Solution:

Question 8.
The difference between the ages of Gopal and his father is 26 years. If the sum of their ages is 56 years, find the ages of Gopal and his father.
Solution:

Question 9.
When two consecutive natural numbers are added, the sum is 31; find the numbers.
Solution:

Question 10.
When three consecutive natural numbers are added, the sum is 66, find the numbers.
Solution:

Question 11.
A natural number decreased by 7 is 12. Find the number.
Solution:

Question 12.
One fourth of a number added to one- sixth of itself is 15. Find the number.
Solution:

Question 13.
A whole number is increased by 7 and the new number so obtained is multiplied by 5; the result is 45. Find the number.
Solution:

Question 14.
The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Solution:

Question 15.
The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Solution:

Question 16.
Two natrual numbers differ by 6 and sum of them is 36. Find the larger number.
Solution:

Question 17.
The difference between two numbers is 15. Taking the smaller number as x; find:
(i) the expression for larger number.
(if) the larger number, if the sum of these numbers is 71.
Solution:

Question 18.
The difference between two numbers is 23. Taking the larger number as x, find:
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Solution:

Question 19.
Find three consecutive integers such that their sum is 78.
Solution:
Sum of three consecutive numbers = 78
Let first number = x
Then second number = x + 1
and third number = x + 2
Then x + x+1+x + 2 = 78
⇒ 3x + 3 = 78
⇒ 3x = 78 – 3 = 75
⇒ x = $$\frac { 75 }{ 3 }$$ =25
∴First number=25
Second number = 25 + 1 = 26
and third number = 26 + 1 = 27
Then the three required numbers are 25, 26,27

Question 20.
The sum of three consecutive numbers is 54. Taking the middle number as x, find:
(i) expression for the smallest number and the largest number.
(ii) the three numbers.
Solution:
Sum of three consecutive numbers = 54
Middle number = x
(i) The first number = x – 1
and third number = x + 1
(ii) ∴x + x-1+x+1 = 54
⇒ 3x = 54
⇒ x= $$\frac { 54 }{ 3 }$$ = 18
∴First number =18-1 = 17
and third number =18 + 1 = 19
∴Three required numbers are 17, 18,19

### Simple (Linear) Equations Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Solve each of the following equations :
Question i.
2x + 3 = 7
Solution:

Question ii.
2x – 3 = 7
Solution:

Question iii.
2x ÷ 3 = 7
Solution:

Question iv.
3x – 8 = 13
Solution:

Question v.
3y + 8 = 13
Solution:

Question vi.
3y ÷ 8 = 13
Solution:

Question vii.

Solution:

Question viii.

Solution:

Question ix.

Solution:

Question x.
5x – 2.4 = 4.9
Solution:

Question xi.
5y + 4.9 = 2.4
Solution:

Question xii.
48 z + 3.6 = 1.2
Solution:

Question xiii.

Solution:

Question xiv.

Solution:

Question xv.

Solution:

Question xvi.
-3x + 4 = 10
Solution:

Question xvii.
5 = x – 3
Solution:

Question xviii.
8y = 3- 3y
Solution:

Question xix.
4x = 4.9 = 6.5
Solution:

Question xx.
3z + 2 = -4
Solution:

Question xxi.
7y -18 = 17
Solution:

Question xxii.

Solution:

Question xxiii.

Solution:

Question xxiv.

Solution:

Question xxv.
7x -2 = 4x +7
Solution:

Question xxvi.
3y -(y -+2) =4
Solution:

Question xxvii.
3z – 18 = z – (12 -4z)
Solution:

Question xxiii.

Solution:

Question xxix.

Solution:

Question xxx.

Solution:

Question xxxi.
5x – 2x +15 = 27
Solution:

Question xxxii.
5y – 15 = 27 -2y
Solution:

Question xxxiii.
7z + 15 = 3z – 13
Solution:

Question xxxiv.
2 (x -3) – 3 (x-4) =12
Solution:

Question xxxv.
(7y +8) +7= 8
Solution:

Question xxxvi.
2(z-5) +3 (z+2) -(3-5z) =10
Solution:

Question 2.
A natural number decreased by 7 is 12. Find the number.
Solution:

Question 3.
One-fourth of a number added to one-sixth of It is 15. Find the number.
Solution:

Question 4.
A whole number is increased by 7 and the number so obtained is multiplied by 5; the result is 45. Find the whole number.
Solution:

Question 5.
The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Solution:

Question 6.
The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Solution:

Question 7.
Two natural numbers differ by 6 and their sum is 36. Find the larger number.
Solution:

Question 8.
The difference between two numbers is 15. Taking the smaller number as x; find :
(i) the expression for the larger number.
(ii) the larger number, if the sum of these numbers is 71.
Solution:

Question 9.
The difference between two numbers is 23. Taking the larger number as x, find :
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Solution:

Question 10.
Find the three consecutive integers whose sum is 78.
Solution:

Question 11.
The sum of three consecutive numbers is 54. Taking the middle number as x, find :
(i) the expressions for the smallest number and the largest number.
(ii) the three numbers.
Solution:

## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 34 Mean and Median

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 34 Mean and Median

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

### Mean and Median Exercise 34A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the mean of :
(i) 7,10, 4 and 17
(ii) 12, 9, 6,11 and 17
(iii) 3, 1, 5, 4, 4 and 7
(iv) 7, 5, 0, 3, 0, 6, 0, 9, 1 and 4
(v) 2.1, 4.5, 5.2, 7.1 and 9.3
(vi) 5, 2.4, 6.2, 8.9, 4.1 and 3.4

Question 2.
Find the mean of :
(i) first eight natural numbers
(ii) first six even natural numbers
(iii) first five odd natural numbers
(iv) all prime numbers upto 30
(v) all prime numbers between 20 and 40.
(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8
∴Sum of these observations =1+2+3+4+5+6+7+8=36
and, number of their observations = 8
∴Required mean = $$\frac { 36 }{ 8 }$$ = 4.5

(ii) The first six even natural numbers are 1 = 2, 4, 6, 8, 10, 12
∴Sum of these observations = 2, 4, 6, 8, 10, 12 = 42
and, number of their observations = 6
∴ Required mean = $$\frac { 42 }{ 6 }$$ = 7

(iii) The first five odd natural numbers are = 1, 3, 5, 7, 9
∴Sum of these observations =1 + 3 + 5 + 7 + 9 = 25
and, number of their observations = 5
∴Required mean = $$\frac { 25 }{ 5 }$$ = 5

(iv) The all prime numbers upto 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
∴Sum of these observations = 2 + 3 +5 + 7+ 11 + 13 + 17 + 19 + 23 +29 = 129
and, number of their observations = 10
∴Required mean = $$\frac { 129 }{ 10 }$$ = 12-9

(v) All prime numbers between 20 and 40 are 23, 29, 31, 37
Sum of these observations = 23+29 + 31 + 37 = 120 .
and, number of their observations = 4
120
∴ Required mean = $$\frac { 120 }{ 4 }$$ = 30

Question 3.
Height (in cm) of 7 boys of a locality are 144 cm, 155 cm, 168 cm, 163 cm, 167 cm, 151 cm and 158 cm. Find their mean height.
Sum of the values = Sum of heights
= 144 cm + 155 cm + 168 cm + 163 cm + 167 cm + 151 cm + 158 cm = 1106 cm and Number of values = Number of boys = 7

Question 4.
Find the mean of 35, 44, 31, 57, 38, 29, 26,36, 41 and 43.
Sum of the values = 35 + 44 + 31 + 57 + 38 + 29 + 26 + 36 + 41 + 43 = 380
and Number of values = 10

Question 5.
The mean of 18, 28, x, 32, 14 and 36 is 23. Find the value of x. Sum of data

Question 6.
If the mean of x, x + 2, x + 4, x + 6 and x + 8 is 13, find the value of x. Sum of data

### Mean and Median Exercise 34B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the median of
(i) 21, 21, 22, 23, 23, 24, 24, 24, 24, 25 and 25
(ii) 3.2, 4.8, 5.6, 5.6, 7.3, 8.9 and 91
(iii) 17, 23, 36, 12, 18, 23, 40 and 20
(iv) 26, 33, 41, 18, 30, 22, 36, 45 and 24
(v) 80, 48, 66, 61, 75, 52, 45 and 70
Solution:

Question 2.
Find the mean and the median of :
(i) 1,3,4, 5, 9, 9 and 11
(ii) 10,12, 12, 15, 15, 17, 18, 18, 18 and 19
(iii) 2, 4, 5, 8, 10,13 and 14
(iv) 5, 8, 10, 11,13, 16, 19 and 20
(v) 1.2, 1.9, 2.2, 2.6 and 2.9
(vi) 0.5, 5.6, 3.8, 4.9, 2.7 and 4.4.
Solution:

## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling (Including Pictograph and Bar Graph)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 33 Data Handling (Including Pictograph and Bar Graph)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

IMPORTANT POINTS

The word statistics is used for two different meanings.
In the singular sense, it is used as a science or a subject which deals with the collection, classification, tabulation, representation and interpretation of the data.
In the plural sense, it is sometimes used for the numerical facts collected in the form of numbers.
If we have collected information about the heights of class 6 children from ten different schools of Delhi, then this information in the form of numbers is called statistics.
1. Data : Each number, collected for giving a required information, is called the data.

2. Bar Graph (Column Graph) : Bar graph is the simplest form of presenting a data. It consists of bars (usually vertical), all of same widths. The heights of these bars are drawn according to the number they represent.

3. Pie Graph : When the given data is represented by the sectors of a circle, the resulting diagram (graph) obtained is called a pie-graph or a pie-chart.

### Data Handling Exercise 33A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Marks scored by 30 students of class VI are as given below :
38, 46, 33, 45, 63, 53, 40, 85, 52, 75, 60, 73, 62, 22, 69, 43, 45, 33, 47, 41, 29, 43, 37, 49, 83, 44, 55, 22, 35 and 45. State :
(i) the highest marks scored.
(ii) the lowest marks scored.
(iii) the range of marks.
Solution:
(i) Highest marks scored = 85 .
(ii) Lowest marks scored = 22
(iii) Range of marks = 85 – 22 = 63

Question 2.
For the following raw data, form a discrete frequency distribution :
30,32,32, 28,34,34,32,30,30,32,32,34,30,32,32. 28,32,30, 28,30,32,32,30,28 and 30.
Solution:
The required frequency table will be as shown below :

Question 3.
Define :
(i) data
(ii) frequency of an observation.
Solution:
(i) Data : The word data means information in the form of numerical figures.
(ii) Frequency of an observation : The number of times a particular observation occurs is called its frequency.

Question 4.
Rearranage the following raw data in descending order :
5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9 and 5.4. Then write the :
(i) highest value
(ii) lowest value
(iii) range of values
Solution:
Writing these numbers in descending order we get:
6.1, 6.0, 5.9, 5.9, 5.8, 5.8, 5.7, 5.7, 5.6, 5.5, 5.5, 5.4, 5.3, 5.2, 5.1
(i) Highest value = 6.1
(ii) Lowest value = 5.1
(iii) Range of values = Highest value – lowest value = 6.1 -5.1 = 1.0

Question 5.
Represent the following data in the form of a frequency distribution :
52, 56, 72, 68, 52, 68, 52, 68, 52, 60, 56, 72, 56, 60, 64, 56, 48, 48, 64 and 64.
Solution:
The required frequency table wilf be as shown below :

Question 6.
In a study of number of accidents per day, the observations for 30 days were obtained as follows :

Solution:
The required frequency table will be as shown below :

Question 7.
The following data represents the weekly wages (in ₹) of 15 workers in a factory : 900, 850, 800, 850, 800, 750, 950, 900, 950, 800, 750, 900, 750, 800 and 850.
Prepare a frequency distribution table. Now find,
(i) how many workers are getting less than ₹850 per week?
(ii) how many workers are getting more than ₹800 per week?
Solution:
The required frequency table will be as shown below :

(i) Workers getting less than ₹850 per week
No. of workers getting ₹750 = 3 workers
No. of workers getting ₹800 = 4 works
∴ Workers getting less than ₹ 850 = 4 + 3 = 7 workers
(ii) Workers are getting more than ₹800 per week
No. of workers getting ₹850 = 3
No. of workers getting ₹900 = 3
No. of workers getting ₹950 = 2
∴Workers getting more than ₹800 = 3 + 3 + 2 = 8 workers

Question 8.
Using the data, given below, construct a frequency distribution table : 9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9 and 12. Now answer the following :
(i) How many numbers are less than 19?
(ii) How many numbers are more than 20?
(iii) Which of the numbers, given above, is occuring most frequently?
Solution:
The required frequency table will be as shown below :

(i) There are 14 numbers are less than 19.
(ii) There are 2 numbers more than 20.
(iii) 9 is occuring most frequently i.e. 5 times.

Question 9.
Using the following data, construct a frequency distribution table : 46, 44, 42, 54, 52, 60, 50, 58, 56, 62, 50, 56, 54, 58 and 48.
(i) What is the range of the numbers?
(ii) How many numbers are greater than 50?
(iii) How many numbers are between 40 and 50?
Solution:

(i) Range of numbers = Highest number – Lowest number = 62 – 42 = 20
(ii) 9 numbers are greater than 50
(iii) 6 numbers are between 40 and 50 Ans.

### Data Handling Exercise 33B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The sale of vehicles, in a particular city, during the first six months of the year 2016 is shown below :

vehicles sold
Draw a pictograph to represent the above data.
Solution:

Question 2.

Solution:
(i) Cars sold by dealer A = 6 x 50 = 300
Cars sold by dealer D = 4 x 5 = 200 ,
∴ A sold more cars than dealer D by = 300 – 200 = 100
∴A has sold 100 more cars than dealer D.
(ii) No. of cars = 23
Scale = 50 cars
∴Total no. of cars = 23 x 50 = 1150 cars Ans.

Question 3.
The following pictograph shows the number of watches manufactured by a factory, in a particular weeks.

Find
(i) on which day were the least number of w atches manufactured ?
(ii) total number of watches manufatured in the whole week ?
Solution:
(i) On Friday least no. of watches manufactured by = 100 x 5 = 500 watches
(ii) Total no. of watches manufactured in the whole week = 100 x 42.5 = 4250 watches

Question 4.
The number or animals in five villages are as follows :

Prepare a pitctograph of these animals using one symbol to represent 20 animals.
Solution:

Question 5.
The following pictograph shows different subject books which are kept in a school library.

Taking symbol of one book = 50 books, find :
(i) how many History books are there in the library ?
(ii) how many Science books are there in the library ?
(iii) which books are maximum in number ?

Solution:
(i) There are 50 x 4 = 200 History books in the library.
(ii) There are 50 x 5.5 = 275 Science books in the library.
(iii) English books are maximum in number = 500 x 9 = 450 books.

### Data Handling Exercise 33C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The following table gives the number of students in class VI in a school during academic years 2011-2012 to 2015-2016.

Represent the above data by a bar graph.
Solution:

Question 2.
The attendence of a particular class for the six days of a week are as given below :

Draw a suitable graph.
Solution:

Question 3.
The total number of students present in class VI B, for the six days in a week were as given below. Draw a suitable bar graph.

Solution:

Question 4.
The following table shows the population of a particular city at different years :

Represent the above information with the help of a suitable bar graph.
Solution:

Question 5.
In a survey of 300 families of a colony, the number of children in each family was recorded and the data has been represented by the bar graph, given below :

(i) How many families have 2 children each ?
(ii) How many families have no child ?
(iii) What percentage of families have 4 children ?
Solution:
(i) 60 families have 2 children each.
(ii) Zero
(iii) The percentage of families having 4 children = $$\frac { 60 }{ 300 }$$ x 100 = 20%

Question 6.
Use the data, given in the following table, to draw’ a bar graph

Out of A, B, C, D, E and F
(i) Which has the maximum value.
(ii) Which is greater A + D or B + E.
Solution:
(i) D has the maximum value of 350
(ii) A + D = 250 + 350 = 600
B + E = 300 + 275 = 575
Hence A + D is greater.

Question 7.
The bar graph drawn below shows the number of tickets sold during a fair by 6 students A, B, C, D, E and F.

Using the Bar graph, answer the following questions :
(i) Who sold the least number of tickets?
(ii) Who sold the maximum number of tickets ?
(iii) How many tickets were sold by A, B and C taken together ?
(iv) How many tickets were sold by D, E and F taken together ?
(v) What is the average number of tickets sold per student ?
Solution:

Question 8.
The following bar graph shows the number of children, in various classes, in a school in Delhi.

Using the given bar graph, find :
(i) the number of children in each class.
(ii) the total number of children from Class 6 to Class 8.
(iii) how many more children there are in Class 5 compared to Class 6 ?
(iv) the total number of children from Class 1 to Class 8.
(v) the average number of children in a class.

Solution:
(i) In, Class 1 = 100, Class 2 = 90, Class 3 = 100, Class 4 = 80,
Class 5 = 120, Class 6 = 90, Class 7 = 70, Class 8 = 50
(ii) Class 6 = 90, Class 7 = 70, Class 8 = 50, Total number = 210
(iii) Number of student in class 5 = 120, Number of student in class 6 = 90
More children is class 5 = (120 – 90) = 30
(iv) Total number of children class 1 to 8 = 100 + 90+ 100+ 80 + 120 +90 + 70 + 50 = 700

Question 9.
The column graph, given above , shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
(i) On which day were the maximum number of patients examined ?
(ii) On which day were the least number of patients examined ?
(iii) On which days were equal number or patients examined ?
(iv) What is the total number of patients examined in the week ?
Solution:

(i) Tuesday were the maximum number of patients examined.
(ii) Friday were the least number of patients examined.
(iii) Sunday and Thursday were equal number of patient examined.
(iv) Total number of patients examined in the week .
= 50 + 40 + 70 + 60 + 50 + 30 = 300

Question 10.
A student spends his pocket money on various items, as given below :
Books : Rs. 380, Postage : Rs. 30, Cosmetics : Rs. 240, Stationary : Rs. 220 and Entertainment : Rs. 120.
Draw a bar graph to represent his expenses.
Solution:
Amount spent on
Books = Rs. 380
Postage = Rs. 30
Cosmetics = Rs. 240
Stationary = Rs. 220
Entertainment = Rs. 120
The bar graph of the above given data is below.

## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 31 Recognition of Solids

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

### Recognition of Solids Exercise 31 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes :

Solution:
Nets for a cube are (ii), (iii) and (iv).

Question 2.
Draw at least three different nets for making cube.
Solution:

Question 3.
The dimensions of a cuboid are 6 cm, 4 cm and 3 cm. Draw two different types of oblique sketches for this cuboid.
Solution:

Question 4.
Two cubes, each with 3 cm edge, are placed side by side to form a cuboid. For this cuboid, draw :
(i) an oblique sketch
(ii) an isometric sketch.

Solution:

Question 5.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :

In each case, name the object.
Solution:

Question 6.
Look at the solids, drawn below, and fill the given chart.

Solution:

P. Q. Using Euler’s formula, find the values of a, b, c and d.

Solution:

(i) a + 6 — 12 = 2 ⇒a = 2- 6+12 = 14-6 = 8
(ii) b + 5- 9 = 2 ⇒b-2 + 9-5 = 6
(iii) 20 + 12 – c = 2 ⇒ 32 – c = 2 ⇒ c = 32 – 2 ⇒ c = 30
(iv) 6 + d – 12 = 2⇒ d-6 = 2 ⇒ d = 2 + 6 = 8

P.Q. Using an isometric dot paper, draw :
(i) a cube with each edge 3 cm.
(ii) a cuboid measuring 5 cm x 4 cm x 3 cm.

Solution:

Question 7.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.

Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:

Question 8.
The following figure represents mets of some solids. Name the solids.

Solution:
The given nets are of the solid as given below :
(i) Tetrahedron
(ii) Triangular prism
(iii) Cube
(iv) Cuboid

## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 32 Perimeter and Area of Plane Figures

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IMPORTANT POINTS

1. Perimeter: It is the length of the boundry of the given figure.
(i) Perimeter of a triangle = Sum of its three sides.
(ii) Perimeter of rectangle = 2 (length + breadth)
(iii) Perimeter of square = 4 x side.
2. Area: Area is the measure of surface of the plane covered by a closed plane figure. In other words, we can say that area of a closed plane figure is the measure of its interior region.
(i) Area of rectangle = length x breadth
(ii) Area of square = (side)².
3. Units of measurement of perimeter and area :
(i) Perimeter is measured in centimetre (cm) metre (m) or millimetre (mm).
(ii) Area is measured in square mm, square cm or square metre.

### Perimeter and Area of Plane Figures Exercise 32A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
What do you understand by a plane closed figure?
Solution:
Any geometrical plane figure bounded by lines (straight or curved) in a plane is called a plane closed figure.
Each of the following figures is a plane closed figure.

Question 2.
The interior of a figure is called region of the figure. Is this statement true ?
Solution:
Yes. The interior of the figure alongwith its boundary is called region of the figure

Question 3.
Find the perimeter of each of the following closed figures :
Solution:
(i) Required perimeter
= AB + AC + BE + EF + FH + HG + HD
= 15 + 5 + 25 + 10 + 5 + 15 + 25 = 110 cm

(ii) Required perimeter
= AB + AC + CD + DG + BF + EF + EH + GH
= 20 + 4 + 8 + 20 + 4 + 8 + 20 + 4 = 88 cm

Question 4.
Find the perimeter of a rectangle whose:
(i) length = 40 cm and breadth = 35 cm
(ii) length = 10 m and breadth = 8 m
(iii) length = 8 m and breadth = 80 cm
(iv) length = 3.6 m and breadth = 2.4 m
Solution:
(i) length = 40 cm and breadth = 35 cm
∴Perimeter = 2 (length + breadth)
= 2 (40 cm + 35 cm)
= 2 x 75 cm
= 150 cm = $$\frac { 150 }{ 100 }$$
= 1.5 m
(ii) length = 10 m and breadth = 8 m
∴Perimeter = 2 (length + breadth)
= 2 (10 m + 8 m)
= 2 x 18 m = 54 m
(iii) length = 8 m and
Length = 8 m
Breadth = 80 cm= $$\frac { 80 }{ 100 }$$ m = 0.8 m
∴ Perimeter = 2 (length + breadth)
= 2 (8 m + 0.8m)
= 2 x 8.8 m = 17.6 m
(iv) length = 3.6 m and breadth = 2.4 m
∴ Perimeter = 2 (length + breadth)
= 2 (3.6 m + 2.4 m)
= 2 x 6 m = 12 m

Question 5.
If P denotes perimeter of a rectangle, l denotes its length and b denotes its breadth, find :
(i) l, if P = 38cm and b = 7cm
(ii) b, if P = 3.2m and l = 100 cm
(iii) P, if l = 2 m and b = 75cm
Solution:

Question 6.
Find the perimeter of a square whose each side is 1.6 m.
Solution:
∵ Side of the square = 1.6 m
∴ its perimeter = 4 x side
= 4 x 1.6 m
= 6.4 m

Question 7.
Find the side of the square whose pe-rimeter is 5 m.
Solution:

Question 8.
A square field has each side 70 m whereas a rectangular field has length = 50 m and breadth = 40 m. Which of the two fields has greater perimeter and by how much?
Solution:
Perimeter of the square field = 4 x side = 4 x 70m = 280m
Perimeter of rectangular field = 2 (length + breadth)
= 2 (50 m + 40 m)
= 2 x 90 m
= 180 m
∴Square field has greater perimeter by 280 m – 180 m = 100 m

Question 9.
A rectangular field has length = 160m and breadth = 120 m. Find :
(i) the perimeter of the field.
(ii) the length of fence required to enclose the field.
(iii) the cost of fencing the field at the rate of ? 80 per metre.
Solution:
Given = length = 160 m, breadth = 120m
(i) The Perimeter of the field = 2 (l + b)
= 2 (160 m + 120 m)
= 2 x 280
= 560 m
(ii) The length of fence required to enclose the field = The perimeter of the rectan-gular field
= 560 m
(iii) The cost of fencing the field = Length of fence x Rate of fence
= 560 m x ₹80 per metre
= ₹44, 800

Question 10.
Each side of a square plot of land is 55 m. Find the cost of fencing the plot at the rate of ₹32 per metre.
Solution:
∵Perimeter of square field = 4 x its side = 4 x 55 m
∴Length of required fencing = 220 m Now, the cost of fencing = its length x its rate
= 220 m x ₹32 per metre?
= ₹7040

Question 11.
Each side of a square field is 70 cm. How much distance will a boy walk in order to make ?
(i) one complete round of this field ?
(ii) 8 complete rounds of this field ?
Solution:
(i) Distance covered by the boy to make one complete round of the field.
Perimeter of the field : 4 x its side = 4 x 70 = 280 m
(ii) Distance covered by the boy to make 8 complete rounds of this field.
= 280 m x 8 m = 2240 m

Question 12.
A school playground is rectangular in shape with length = 120 m and breadth = 90 m. Some school boys run along the boundary of the play-ground and make 15 complete rounds in 45 minutes. How much distance they run during this period.
Solution:
Length of the rectangular playground = 120 mBreadth of the rectangular playground = 90 m
∴ Perimeter of the rectangular ground = 2(l + b)
= 2(120 + 90) m = 420 m
Thus, in one complete round, boys covers a distance of = 420 m
∴Distance covered in 15 complete rounds = 420 m x 15 = 6300 m

Question 13.
Mohit makes 8 full rounds of a rect-angular field with length = 120 m and breadth = 75 m.
John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?
Solution:
Mohit
Length of the rectangular field = 120
Breadth of the rectangular field = 75 m
∴ Distance covered in one round (perim-eter) = 2(1 + b)
= 2(120 + 75) = 390 m Hence, distance covered in 8 rounds = 390 x 8 m = 3120 m
John
Side of the field = 100 m
∴Distance covered in one round = 4 x a = 4 x 100 = 400 m
Hence, Distance covered in 10 rounds = 400 x 10 m = 400 m
John a covers greater distance then Mohit by = (4000-3120) m = 880 m

Question 14.
The length of a rectangle is twice of its breadth. If its perimeter is 60 cm, find its length.
Solution:
Let the breadth of the field = x cm
∴ its length = 2x
and, its perimeter = 2 x (length + breadth)
= 2 x (2x + x)
= 2(3x)
= 6x cm
Perimeter = 60 cm
⇒ 60 cm = 6x cm
⇒ x = $$\frac { 60 }{ 6 }$$ = 10 cm
∴Breadth = x = 10 cm
Length = 2x = 2 x 10 = 20 cm

Question 15.
Find the perimeter of :
(i) an equilateral triangle of side 9.8 cm.
(ii) an isosceles triangle with each equal side = 13 cm and the third side = 10 cm.
(iii) a regular pentagon of side 8.2 cm.
(iv) a regular hexagon of side 6.5 cm.
Solution:
(i) The perimeter of equilateral triangle = 3 x side
= 3 x 9.8 cm
= 29.4 cm
(ii) Required perimeter = 13 cm + 13 cm + 10 cm
= 36 cm
(iii) Perimeter of given pentagon = 5 x side = 5 x 8.2 cm
= 41 cm
(iv) Perimeter of given hexagon = 6 x side = 6 x 6.5 cm
= 39 cm

Question 16.
An equilateral triangle and d square has equal perimeter. If side of the triangle is 9.6 cm ; what is the length of the side of the square ?
Solution:
Perimeter of equilateral triangle = Perimeter of square Side of triangle = 9.6 cm
∴Perimeter of triangle = 3 x side
= 3 x 9.6 cm = 28.8 cm
> Perimeter of the square = 28.8 cm
4 x the side of square = 28.8 cm
⇒ The side of the square = $$\frac { 28.8 }{ 4 }$$ cm
= 7.2 cm Ans.

Question 17.
A rectangle with length = 18 cm and breadth = 12 cm has same perimeter as that of a regular pentagon. Find the side of the pentagon.
Solution:
Length of rectangle = 18 cm
Breadth of rectangle = 12 cm
∴ Perimeter of rectangle = 2 x (l + b)
= 2 x (18+12)
= 2 x 30 = 60 cm
∵Perimeter, of rectangle = Perimeter of pentagon
60 cm = 5 x side
side = $$\frac { 60 }{ 5 }$$ cm = 12 cm
∴Side of the pentagon = 12 cm Ans.

Question 18.
A regular pentagon of each side 12 cm has same perimeter as that of a regular hexagon. Find the length of each side of the hexagon.
Solution:
Perimeter of regular pentagon = 5 x length of the side
= 5 x 12 cm = 60 cm
Clearly, perimeter of the given pentagon = 60 cm
⇒ 6 x side of hexagon = 60 cm 60
⇒ side of hexagon = $$\frac { 60 }{ 6 }$$cm = 10 cm

Question 19.
Each side of a square is 45 cm and a rectangle has length 50 cm. If the perimeters of both (square and rectangle) are same, find the breadth of the rectangle.
Solution:

Question 20.
A wire is bent in the form of an equilateral triangle of each side 20 cm. If the same wire is bent in the form of a square, find the side of the square.
Solution:
∵Each side of the given equilateral triangle = 20 cm
∴Perimeter of the triangle = 3 x side = 3 x 20 cm = 60 cm ,
∴ Perimeter of the square = Perimeter of equilateral triangle
⇒ 4 x side of square = 60 cm
⇒ The side of the square = $$\frac { 60 }{ 4 }$$
=15 cm

### Perimeter and Area of Plane Figures Exercise 32B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the area of a rectangle whose :
(i) length = 15 cm breadth = 6.4 cm
(ii) Length = 8.5 m breadth = 5 m
(iii) Length = 3.6 m breadth = 90 cm
(iv) Length = 24 cm breadth =180 mm
Solution:

Question 2.
Find the area of a square, whose each side is :
(i) 7.2 cm
(ii) 4.5 m
(iii) 4.1 cm
Solution:
(i) 7.2 cm
Area of the square = (side)² = (7.2 cm)² = 7.2 cm x 7.2 cm = 51.84 cm²
(ii) 4.5 m
Area of the square = (side)² = (4.5 m)² = 4.5 m x 4.5 m = 20.25 m²
(iii) 4.1 cm
Area of the square = (side)² = (4.1 cm)² = 4.1 cm x 4.1 cm = 16.81 cm²

Question 3.
If A denotes area of a rectangle, l represents its length and b represents its breadth, find :
(i) l, if A = 48 cm² and b = 6 cm
(ii) b, if A = 88 m² and l = 8m
Solution:

Question 4.
Each side of a square is 3.6 cm; find its
(i) perimeter
(ii) area.
Solution:
(i) Perimeter = 4 x side
= 4 x 3.6 cm = 14.4 cm
(ii) Area = (side)²
= (3.6 cm)²
= 12.96 cm²

Question 5.
The perimeter of a square is 60 m, find :
(i) its each side its area
(ii) its new area obtained on increasing
(iii) each of its sides by 2 m.
Solution:
Perimeter of a square = 60 m
(i) Perimeter of a square = 4 x side
60 m = 4 x side
$$\frac { 60 }{ 4 }$$ = side 4
∴side = 15 m
(ii) Area of square = (side)² = (15 m)²
= 15 m x 15 m
= 225 m²
(iii) Increased each side = 2 m
Side of square = 15 m
New length of side = (2m + 15m)
= 17m
∴New Area of square = (17m)² = 17m x 17m = 289 m²

Question 6.
Each side of a square is 7 m. If its each side be increased by 3 m, what will be the increase in its area.
Solution:
Each side of square = 7 m
∴Area of square = (side)²= (7 m)²
= 7m x 7m =49m²
∵ Side increased by 3 m
∴Total length of side will be = 3 m + 7 m = 10m
∴Area of square = (10 m)²= 10m x 10 m = 100 m²
∴Increase in area = 100 m² – 49 m² = 51 m²

Question 7.
The perimeter of a square field is numerically equal to its area. Find each side of the square.
Solution:

Question 8.
A rectangular piece of paper has area = 24 cm² and length = 5 cm. Find its perimeter.
Solution:

Question 9.
Find the perimeter of a rectangle whose area = 2600 m² and breadth = 50 m.
Solution:

Question 10.
What will happen to the area of a rectangle, if its length and breadth both are trebled?
Solution:
Let the original length of the rectangle = l and its original breadth = b
∴ its original area = length x breadth i.e A = l – b i. e.
Since,
Increased length -=3l
∴ New area = 3l x 3b = 9 x l x b [∵A = l x b]
⇒ Area of the new rectangle = 9 times than area of original rectangle

Question 11.
Length of a rectangle is 30 m and its breadth is 20 m. Find the increase in its area if its length is increased by 10 m and its breadth is doubled.
Solution:
Length of a rectangle (l) = 30 m,
Breadth of the rectangle (b) = 20 m
Area of rectangle = l x b
= 30 x 20 = 600 m2
Since, the length its increased by 10 m and breadth is doubled
∴New length (l) = (30 + 10) m = 40 m
and new breadth = (20 x 2) m = 40 m
∴New area = l x b = 40 x 40 m2 = 1600 m2
Hence, the increase in the area = (1600 – 600) m2
= 1000 m2

Question 12.
The side of a square field is 16 m. What will be increase in its area, if:
(i) each of its sides is increased by 4 m
(ii) each of its sides is doubled.
Solution:

Question 13.
Each rectangular tile is 40 cm long and 30 cm wide. How many tiles will be required to cover the floor of a room with length = 4.8 m and breadth = 2.4 m.
Solution:

Question 14.
Each side of a square tile is 60 cm. How many tiles will be required to cover the floor of a hall with length = 50 m and breadth = 36 m.
Solution:

Question 15.
The perimeter of a square plot = 360 m. Find :
(i) its area.
(ii) cost of fencing its boundary at the rate of ₹ 40 per metre.
(iii) cost of levelling the plot at ₹60 per square metre.
Solution:
Given, perimeter of square plot = 360 m
∵ Perimeter of the square = 4 x its side
∴4 x side of square = 360 m
⇒ side of the square = $$\frac { 360m }{ 4 }$$ = 90 m
(i) The area of the square field = (side)²
= (90 m)²
= 90 m x 90 m
= 8100 m²
Cost of fencing at ₹ 40 per metre
= 8100 m2 x ₹ 40 per metre
= ₹ 324000
Cost of levelling at₹ 60 per m²
= 8100 m² x ₹60 per m²
= ₹ 486000

Question 16.
The perimeter of a rectangular field is 500 m and its length = 150 m. Find:
(ii) its area.
(iii) cost of ploughing the field at the rate of ₹1.20 per square metre.
Solution:
(i) Perimeter of a rectangle = 2 x (length + breadth)
⇒500 m = 2x(i50m + breadth)
⇒250 m – 150 m = breadth
(ii) Area of rectangular field = length x breadth
= 150 m x 100 m = 15000 m²
(iii) Cost of ploughing the field at the rate of
= ₹1.20 per square m²= area of the field x rate of ploughing = 15000 m² x ₹1.20 per square metre = ₹15000 x 1.20 = ₹18000

Question 17.
The cost of flooring a hall of ₹64 per square metre is ₹2,048. If the breadth of the hall is 5m, find :
(i) its length.
(ii) its perimeter.
(iii) cost of fixing a border of very small width along its boundary at the rate of ₹60 per square metre.
Solution:

Question 18.
The length of a rectangle is three times its breadth. If the area of the rectangle is 1875 sq. cm, find its perimeter.
Solution: