Bhagya

MP Board Class 6th Maths Solutions Chapter 8 Decimals Ex 8.2

Question 1.
Complete the table with the help of these boxes and use decimals to write the number,


Solution:

Question 2.
Write the numbers given in the following place value table in

Solution:

Question 3.
Write the following decimals in the place value table.
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Solution:

Question 4.
Write each of the following as decimals.

Solution:
(a) 20 + 9 + 0.4 + 0.01 = 29.41
(b) 137 + 0.05 = 137.05
(c) 23 + 0.2 + 0.006 = 23.206
(d) 700 + 20 + 5 + 0.09 = 725.09

Question 5.
Write each of the following decimals in words.
(a) 0.03
(b) 1.20
(c) 108.56
(d) 10.07
(e) 0.032
(f) 5.008
Solution:
(a) Zero point zero three
(b) One point two zero
(c) One hundred and eight point five six
(d) Ten point zero seven
(e) Zero point zero three two
(f) Five point zero zero eight

Question 6.
Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.06
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
Solution:
(a) 0.06 lies between 0 and 0.1.
(b) 0.45 lies between 0.4 and 0.5.
(c) 0.19 lies between 0.1 and 0.2.
(d) 0.66 lies between 0.6 and 0.7.
(e) 0.92 lies between 0.9 and 1.0.
(f) 0.57 lies between 0.5 and 0.6.

Question 7.
Write as fractions in lowest terms.
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.18
(e) 0.25
(f) 0.125
(g) 0.066
Solution:

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.4

Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students. Which activity is preferred by most of the students other than playing?
Solution:

Reading story books is preferred by most of the students other than playing.

Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choice.
Solution:

Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

(a) In which year were the maximum number of bicycles manufactured?
(b) In which year were the minimum number of bicycles manufactured?
Solution:

(a) In 2002, the maximum number of bicycles were manufactured.
(b) In 1999, the minimum number of bicycles were manufactured.

Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent the above information and answer the following questions, (take 1 unit length = 20 thousands)
(a) Which two age groups have same Population?
(b) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Solution:

(a) Age groups 30 – 44 and 45 – 59 have same population.
(b) 80,000 + 40,000 = 1,20,000 senior citizens are there in the town.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.3

Question 1.
The bar graph given alongside shows the amount of wheat purchased by the government during the year 1998 – 2002.

Read the bar graph and write down your observations. In which year was
(a) the wheat production maximum?
(b) the wheat production minimum?
Solution:
(a) In 2002, production of wheat was maximum.
(b) In 1998, production of wheat was minimum.

Question 2.
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.

Now answer the following questions :
(a) What information does the above bar graph give?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(c) On which day were the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day were the minimum number of shirts sold?
(e) How many shirts were sold on Thursday?
Solution:
(a) The bar graph shows the sale of shirts in a ready made shop from Monday to Saturday.
(b) 1 unit length = 5 shirts
(c) On Saturday, maximum number of shirts, i.e., 60 shirts were sold.
(d) On Tuesday, minimum number of shirts were sold.
(e) On Thursday, 35 shirts were sold.

Question 3.
Observe this bar graph which shows the marks obtained by Aziz in half-yearly examination in different subjects.

Answer the given questions.
(a) What information does the bar graph give?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the name of the subjects and marks obtained in each of them.
Solution:
(a) The bar graph shows the marks obtained by Aziz in half-yearly examination in different subjects.
(b) Hindi
(c) Social Studies
(d) Hindi – 80, English – 60, Mathematics – 70, Science – 50, Social Studies – 40.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.3

Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solution:
(a) Area of rectangle = length × breadth
= 3 cm × 4 cm = 12 cm²

(b) Area of rectangle = length × breadth
= 12 m × 21 m = 252 m²

(c) Area of rectangle = length × breadth
= 2 km × 3 km = 6 km²

(d) Area of rectangle = length × breadth
= 2 m × 70 cm = 2 m × 0.7 m = 1.4 m²

Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution:
(a) Area of square = side × side
= 10 cm × 10 cm = 100 cm²

(b) Area of square = side × side
= 14 cm × 14 cm = 196 cm²

(c) Area of square = side × side
= 5 m × 5 m = 25 m²

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) Area of rectangle = length × breadth
= 9m × 6m = 54m²

(b) Area of rectangle = length × breadth
= 17 m × 3 m = 51 m²

(c) Area of rectangle = length × breadth
= 4 m × 14 m = 56 m²
Thus, rectangle (c) has the largest area, i.e. 56 m² and rectangle (b) has the smallest area, i.e., 51 m².

Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
SolutionL
Length of rectangle = 50 m
Area of rectangle = 300 m²
Since, area of rectangle = length × breadth
Therefore, breadth = \(\frac{\text { area of rectangle }}{\text { length }}\)
= \(\frac{300}{50}\) m = 6 m
Thus, the breadth of the garden is 6 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 Rs. 8 per hundred sq m?
Solution:
Length of land = 500 m
Breadth of land = 200 m
Area of land = length × breadth
= 500 m × 200 m = 1,00,000 sq m
Cost of tiling 100 sq m of land = Rs. 8
∴ Cost of tiling 1,00,000 sq m of land
= Rs. \(\frac{8 \times 100000}{100}\) = Rs. 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:
Length of table-top = 2 m
Breadth of table-top = 1 m 50 cm = 1.50 m
∴ Area of table-top = length × breadth
= 2 m × 1.50 m = 3 m²

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 room = 4 m
And breadth of room = 3 m 50 cm = 3.50 m
∴ Area of carpet = length × breadth
= 4 m × 3.50 m = 14 m²

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 floor = 5 m
And breadth of floor = 4 m
Area of floor = length × breadth
= 5m × 4m = 20m²
Now, side of square carpet = 3 m
Area of square carpet = side × side
= 3m × 3m = 9m²
∴ Area of floor that is not carpeted
= 20 m² – 9 m² = 11 m²

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 square flower bed = 1 m Area of square flower bed = side × side
= 1m × 1m = 1m²
∴ Area of 5 square flower beds = (1 × 5) m²
= 5 m²
Now, length of land = 5 m
And breadth of land = 4 m
∴ Area of land = length × breadth = 5m × 4m
= 20 m²
∴ Area of remaining part
= Area of land – Area of 5 flower beds
= 20 m² – 5 m² = 15 m²

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

Solution:
(a) We have,

Area of square HKLM = 3 × 3 cm² = 9 cm²
Area of rectangle I]CH = 1 × 2 cm² = 2 cm²
Area of square FEDG = 3 × 3 cm² = 9 cm²
Area of rectangle ABCD = 2 × 4 cm² = 8 cm²
∴ Total area of the figure = (9 + 2 + 9 + 8) cm² = 28 cm²

(b) We have,

Area of rectangle ABCD = 3 × 1 cm² = 3 cm²
Area of rectangle BJEF = 3 × 1 cm² = 3 cm²
Area of rectangle FGHI = 3 × 1 cm² = 3 cm²
∴ Total area of the figure = (3 + 3 + 3) cm² = 9 cm²

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

Solution:
(a) We have,

Area of rectangle ABCD = 2 × 10 cm² = 20 cm²
Area of rectangle DEFG = 10 × 2 cm² = 20 cm²
∴ Total area of the figure = (20 + 20) cm²
= 40 cm²

(b) We have,

There are 5 squares each of side 7 cm.
Area of one square = 7 × 7 cm² = 49 cm²
∴ Area of 5 squares = 5 × 49 cm² = 245 cm²

(c) We have,

Area of rectangle ABCD = 5 × 1 cm² = 5 cm²
Area of rectangle EFGH = 4 × 1 cm² = 4 cm²
∴ Total area of the figure = (5 + 4) cm²
= 9 cm²

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:
(a) Area of rectangular region
= length × breadth = 100 cm × 144 cm = 14400 cm²
Area of one tile = 12 cm × 5 cm = 60 cm²

Thus, 240 tiles are required.

(b) Area of rectangular region
= length × breadth = 70 cm × 36 cm = 2520 cm²
Area of one tile = 12 cm × 5 cm = 60 cm²
∴ Number of tiles
= \(\frac{\text { Area of rectangular region }}{\text { Area of one tile }}=\frac{2520}{60}\) = 40
Thus, 42 tiles are required.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.2

Question 1.
Total number of animals in five villages are as follows:
Village A : 80 Village B : 120
Village C : 90 Village D : 40
Village E : 60
Prepare a pictograph of these animals using one symbol to represent 10 animals and answer the following questions :
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals : village A or village C?
Solution:

(a) 6 symbols represent animals of village E.
(b) Village B has the maximum number of animals.
(c) Village C has more animals than village A.

Question 2.
Total number of students of a school in different years is shown in the following table.

A. Prepare a pictograph of students using one symbol img 2 to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Solution:
A.

(a) 6 symbols represent total number of students in the year 2002.
(b) Five completed and one incomplete symbols represent total number of students for the year 1998.
B.

Pictograph B is more informative than pictograph A.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.2

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

Solution:
(a) Number of filled squares = 9
∴ Area covered by filled squares
= (9 × 1) sq units = 9 sq units

(b) Number of filled squares = 5
∴ Area covered by filled squares
= (5 × 1) sq units = 5 sq units

(c) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(d) Number of filled squares = 8
∴ Area covered by filled squares
= (8 × 1) sq units = 8 sq units

(e) Number of filled squares = 10
∴ Area covered by filled squares
= (10 × 1) sq units = 10 sq units

(f) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(g) Number of fully-filled squares = 4
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (4 × 1) sq units = 4 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (4 + 2) sq units = 6 sq units

(h) Number of filled squares = 5 .
∴ Area covered by filled squares
= (5 × 1) sq units = 5 sq units

(i) Number of filled squares = 9
∴ Area covered by filled squares
= (9 × 1) sq units = 9 sq units

(j) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(k) Number of fully-filled squares = 4
Number of half-filled squares = 2
∴ Area covered by fully-filled squares
= (4 × 1) sq units = 4 sq units
Area covered by half-filled squares
= (2 × \(\frac{1}{2}\)) sq units = 1 sq units
∴ Total area = (4 + 1) sq units = 5 sq units

(l) Number of fully-filled squares = 3,
Number of half-filled squares = 2,
Number of more than half-filled squares = 4
and number of less than half-filled squares = 4.
Now, estimated area covered by
fully-filled squares = 3 sq units,
half-filled squares = (2 × \(\frac{1}{2}\)) sq units
= 1 sq unit,
more than half-filled squares = 4 sq units
and less than half-filled squares
= 0 sq unit
∴ Total area = (3 + 1 + 4 + 0) sq units
= 8 sq units.

(m) Number of fully-filled squares = 7,
Number of more than half-filled squares = 7
and number of less than half-filled squares = 5
Estimated area covered by
fully-filled squares = 7 sq units,
more than half-filled squares = 7 sq units
and less than half-filled squares = 0 sq unit
∴ Total area = (7 + 7 + 0) sq units = 14 sq units

(n) Number of fully-filled squares = 10,
Number of more than half-filled squares = 8
and number of less than half-filled squares = 5
Estimated area covered by
fully-filled squares = 10 sq units,
more than half-filled squares = 8 sq units
less than half-filled squares = 0 sq unit
∴ Total area = (10 + 8 + 0) sq units
= 18 sq units.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.1

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
Solution:

(a) 5 + 4 + 3 = 12 students obtained marks equal to or more than 7.
(b) 2 + 3 + 3 = 8 students obtained marks below 4.

Question 2.
Following is the choice of sweets of 30 students of Class VI.
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solution:

(b) Ladoo 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
Solution:

(a) 4 appeared minimum number of times.
(b) 5 appeared maximum number of times.
(c) 1 and 6 appeared equal number of 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) Flow 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:
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village C has 8 – 5 = 3 more tractors than village B.
(iv) Total number of tractors = 6 + 5 + 8 + 3 + 6 = 28

Question 5.
The number of girl students in each class of a co-educational 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) Flow many girls are there in Class VII?
Solution:

(a) Class VIII has the minimum number of girl students.
(b) No, the number of girls in Class VI is greater than the number of girls in Class V.
(c) There are 12 girls in Class VII.

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 is 14.
(b) Maximum number of bulbs were sold on Sunday.
(c) Same number of bulbs were sold on Wednesday and Saturday.
(d) Minimum number of bulbs were sold on Wednesday and Saturday.
(e) The total number of bulbs were sold in the given week = 86
Number of cartons required for 9 bulbs = 1
∴ Number of cartons required for 86 bulbs = 86 ÷ 9 = 9.55 = 10
Therefore, 10 cartons were needed in the given week.

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 baskets.
(b) 700 fruit baskets were sold by Anwar.
(c) Anwar, Martin and Ranjit Singh have sold more than 600 baskets.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.1

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

Solution:
(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm

(b) Perimeter = Sum of all the sides
= 23 cm + 35cm + 40 cm + 35cm = 133 cm

(c) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e) Perimeter = Sum of all the sides
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm

(f) Perimeter = Sum of all the sides
= 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:
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 +10) cm = 2 × 50 cm = 100 cm = 1 m

Thus, the total length of tape required is 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 table-top = 2 m 25 cm = 2.25 m
Breadth of table-top = 1 m 50 cm = 1.50 m
Perimeter of table-top = 2 × (length + breadth)
= 2 × (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
Thus, perimeter of table-top is 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 wooden strip
= Perimeter of photograph
= 2 × (length + breadth)
= 2 (32 + 21) cm = 2 × 53 cm = 106 cm
Thus, the length of the wooden strip required is 106 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:
Since 4 rows of wires are needed. Therefore, the total length of wire is equal to 4 times the perimeter of land.
Perimeter of land = 2 × (length + breadth)
= 2 × (0.7 + 0.5) km = 2 × 1.2 km = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire
= 4 × 2400 m = 9600 m = 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) Perimeter of ∆ABC
= AB + BC + CA
= 3 cm+ 5 cm+ 4 cm
= 12 cm

(b) Perimeter of equilateral ∆ABC
= 3 × side
= 3 × 9 cm
= 27 cm

(c) Perimeter of ∆ABC
= AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, perimeter of triangle is 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
Perimeter of a regular hexagon
= 6 × length of one side
= 6 × 8m
= 48m
Thus, the perimeter of the regular hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of square = 4 × side
⇒ 20 m = 4 × side ⇒ side = \(\frac{20}{4}\) m = 5m
Thus, the side of square is 5 m.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of regular pentagon = 5 × side
⇒ 100 cm = 5 × side ⇒ side = \(\frac{100}{5}\) cm = 20 cm
Thus, the side of regular pentagon is 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:
Length of string = Perimeter of each shape
(a) Perimeter of square = 4 × side
⇒ 30 cm = 4 × side ⇒ side = \(\frac{30}{4}\) cm = 7.5 cm
Thus, the length of each side of square will be 7.5 cm.

(b) Perimeter of equilateral triangle = 3 × side
⇒ 30 cm = 3 × side ⇒ side = \(\frac{30}{3}\) cm = 10 cm
Thus, the length of each side of equilateral triangle will be 10 cm.

(c) Perimeter of regular hexagon = 6 × side
⇒ 30 cm = 6 × side ⇒ side = \(\frac{30}{6}\) cm = 5 cm
Thus, the length of each side of regular hexagon will be 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:
Let the length of third side be x cm. Length of other two sides are 12 cm and 14 cm.
Now, perimeter of triangle = 36 cm
⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36
⇒ x = 36 – 26 ⇒ x = 10
Thus, the length of third side is 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per metre.
Solution:
Side of square park = 250 m
Perimeter of square park = 4 × side
= 4 × 250 m = 1000 m
Since, cost of fencing for 1 metre = Rs. 20
Therefore, cost of fencing for 1000 metres
= Rs. 20 × 1000 = Rs. 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per metre.
Solution:
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park = 2 × (length + breadth)
= 2 × (175 + 125) m = 2 × 300 m = 600 m
Since, cost of fencing park for 1 metre = Rs. 12
Therefore, cost of fencing park for 600 m = Rs. 12 × 600 = Rs. 7,200

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:
Distance covered by Sweety
= Perimeter of square park = 4 × side
= 4 × 75 m = 300 m
Thus, distance covered by Sweety is 300 m.
Now, distance covered by Bulbul
= Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) m = 2 × 105 m = 210 m
Thus, Bulbul covers a distance of 210 m.
So, Bulbul covers less distance.

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

Solution:
(a) Perimeter of square = 4 × side
= 4 × 25 cm = 100 cm

(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) cm = 2 × 50 cm = 100 cm

(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) cm = 2 × 50 cm = 100 cm

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm
= 100 cm
Thus, all the figures have same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with side of \(\frac{1}{2}\) m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [see fig. (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 [see fig. (ii)]?
(c) Which has a 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) Side of one small square = \(\frac{1}{2}\) m
∴ Side of given square = \(\frac{3}{2}\) m
Perimeter of square = 4 × side
= 4 × \(\frac{3}{2}\) m = 6 m

(b) Perimeter of given figure
= Sum of all sides = 20 × \(\frac{1}{2}\) m = 10 m

(c) The arrangement cross has greater perimeter.
(d) It is not possible to determine the arrangement with perimeter greater than 10 m.

MP Board Class 6th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Find the values of the letters in each of the following and give reasons for the steps involved.
Question 1.

Solution:
We have to find the values of A and B. We add the digits column-wise.
In C₂ : A + 5 gives 2, that is, a number whose ones digit is 2 i.e., 12.
∴ A + 5 = 12 ⇒ A = 12 – 5 = 7
In C₁ : 1 (Carry of C₂) + 3 + 2 = B
⇒ 6 = B
∴ We get A = 7, B = 6.

Question 2.

Solution:
We have to find the values of A, B and C.
We add the digits column-wise.
In C₃ : A + 8 gives 3, that is, a number whose ones digit is 3 i.e., 13.
Therefore, A + 8 = 13 ⇒ A = 13 – 8 = 5
In C₂ : 4 + 9 + 1 (Carry of C₃) = 14 gives B = 4
In C₁ : 1 (Carry of C₂) = 1
Therefore, A = 5, B = 4, C = 1.

Question 3.

Solution:
We have to find the value of A.
Since, the ones digit of A × A = A
⇒ A can be 0, 1, 5, 6
If A = 0, then 10 × 0 = 0 ≠ 90
If A = 1, then 11 × 1 = 11 ≠ 91
If A = 5, then 15 × 5 = 75 ≠ 95
If A = 6, then 16 × 6 = 96 = 96
Hence, A = 6.

Question 4.

Solution:
We have to find the values of A and B.
We add the digits column-wise.
In C₂ : B + 7 – A …………… (i)
In C₁ : A + 3 + carry = 6 …………. (ii)
Where, carry is the carry from (i)
Let carry = 0
∴ A + 3 = 6 or A = 3
Putting value of A in (i),
6 + 7 = 3 or 6 = – 4
But B can not be a negative number.
⇒ Carry = 1
In (ii), A + 3 + 1 = 6
⇒ A = 2
Putting value of A in (i), B + 7 = 2
But this is not possible
∴ 6 + 7 = 12 ⇒ 6 = 5.

Question 5.

Solution:
We have to find the values of A, 6 and C.
Now, looking at ones place
B × 3 = 6
This is possible only when B = 0, 5
Now, writing the equation for multiplication as
3 × (10A + B) = 100C + 10A + B
⇒ 30A + 3B = 100C + 10A + B
⇒ 20A + 2B = 100C
⇒ 50C = 10A + B ………….. (i)
But R.H.S. is the number we multiplied by 3 to obtain CAB.
∴ AB is a multiple of 50.
⇒ B = 0
Equation (i) becomes,
50C = 10A ⇒ 5C = A
Since, A is a single digit number.
∴ C = 1 & A = 5
Hence, A = 5, B = 0, C= 1.

Question 6.

Solution:
We have to find the values of A, B and C.
Now, looking at ones place
B × 5 = B
This is possible only when B = 0, 5
Also, writing the equation for multiplication
5 × (10A + B) = 100C + 10A + B
⇒ 50A + 5B = 100C + 10A + B
⇒ 40A + 4B = 100C
⇒ 25C = 10A + B ……………. (i)
But, RHS is the number we multiplied by 5 to obtain result CAB
AB is a multiple of 25.
⇒ B = 0 or 5
(1) Taking B = 0
EQ. (i) becomes
25C = 10A ⇒ 5C = 2A
The only integral one digit solution for this is A = 5, C = 2
(2) Taking B = 5
EQ. (i) becomes
25C = 10A + 5
or 5C = 2A +1
The integral one digit solutions are
A = 2, C = 1
or A = 7, C = 3
Hence, we have three solutions
A = 5, B = 0, C = 2
A = 2, B = 5, C = 1
A = 7, B = 5, C = 3

Question 7.

Solution:
We have to find the values of A and B.
Writing the multiplication equation
6 × (10A + B) = 100B + 10B + B
⇒ 60A + 6B = 111B
⇒ 60A = 105B
⇒ 4A = 7B
The only integral one digit solution is
A = 7, B = 4

Question 8.

Solution:
We have to find the values of A and B. We add the digits column-wise.
In C₂ : 1 + B = 10
⇒ B = 9
In C₁ : A + 1 + 1 (Carry of C₂) = B
⇒ A + 2 = 9
⇒ A = 7
∴ A = 7, B = 9

Question 9.

Solution:
We have to find the values of A and B.
We add the digits column-wise.
In C₃ : B + 1 = 8
⇒ B = 7
In C₂ : A + B = 11 {∵ B is 7 and A can’t be negative}
⇒ A + 7 = 11 ⇒ A = 4
∴ A = 4, B = 7

Question 10.

Solution:
We have to find the values of A and B.
We add the digits column-wise.
In C₃: A + B = 9
In C₂ : 2 + A = 10 (∵ A can’t be negative}
⇒ A = 8
∴ 8 + B = 9
⇒ B = 1
∴ A = 8, B = 1

MP Board Class 8th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 8 Decimals Ex 8.4

Question 1.
Express as rupees using decimals.
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solution:

Question 2.
Express as metres using decimals,
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Solution:

Question 3.
Express as cm using decimals.
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solution:

Question 4.
Express as km using decimals
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solution:

Question 5.
Express as kg using decimals.
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solution:

MP Board Class 6th Maths Solutions