Bhagya

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 6th Maths Solutions Chapter 7 Fractions Ex 7.6

Question 1.
Solve:

Solution:

Question 2.
Sarita bought \(\frac{2}{5}\) metre of ribbon and Lalita \(\frac{3}{4}\) metre of ribbon. What is the total length of the
ribbon they bought?
Solution:
Ribbon bought by Sarita = \(\frac{2}{5}\) m
And Ribbon bought by Lalita = \(\frac{3}{4}\) m

Therefore, they bought 1\(\frac{3}{20}\) m of ribbon.

Question 3.
Naina was given 1\(\frac{1}{22}\) piece of cake and Najma was given 1\(\frac{1}{3}\) piece of cake. Find the total amount of cake was given to both of them.
Solution:
Cake taken by Naina = 1\(\frac{1}{2}\) piece
And cake taken by Najma = 1\(\frac{1}{3}\) piece
Total cake taken = \(1 \frac{1}{2}+1 \frac{1}{3}=\frac{3}{2}+\frac{4}{3}\)

Question 4.
Fill in the boxes:

Solution:

Question 5.
Complete the addition-subtraction box.

Solution:

Question 6.
A piece of wire \(\frac{7}{8}\) metre long broke into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Solution:
Total length of wire = \(\frac{7}{8}\) metre
Length of first part = \(\frac{1}{4}\) metre

Therefore, the length of remaining part is \(\frac{5}{8}\) metre

Question 7.
Nandini’s house is \(\frac{9}{10}\) metre km from her school. She 10
walked some distance and then took a bus for \(\frac{1}{2}\) metre km to reach the school. How far did she walk?
Solution:
Total distance between school and house = \(\frac{9}{10}\) km
Distance covered bv bus = \(\frac{1}{2}\) km

Therefore, distance covered by walking is \(\frac{2}{5}\) km.

Question 8.
Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is \(\frac{5}{6}\) th full and Samuel’s shelf is \(\frac{2}{5}\)th full. Whose bookshelf is more full? By what fraction?
Solution:

∴ Asha’s bookshelf is more covered than Samuel.
Difference \(=\frac{25}{30}-\frac{12}{30}=\frac{13}{30}\)

Question 9.
Jaidev takes 2\(\frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Solution:

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable
(a) 17 = x + 7
(b) (t – 7) > 5
(c) \(\frac{4}{2}\) = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 ×2) + p
(k) 20 = 5y
(l) \(\frac{3q}{2}\) < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3× 5
(o) 7 – x = 5
Solution:
(a) 17 = x + 7 is an equation with variable as both the sides are equal. The variable is x.
(b) (t – 7) > 5 is not an equation as L.H.S. is greater than R.H.S.
(c) \(\frac{4}{2}\) = 2 is not an equation with variable.
(d) (7 × 3) – 19 = 8 is not an equation with variable.
(e) 5 × 4 – 8 = 2x is an equation with variable as both the sides are equal. The variable is x.
(f) x – 2 = 0 is an equation with variable as both the sides are equal. The variable is
(g) 2m < 30 is not an equation as L.H.S. is less than R.H.S.
(h) 2n + 1 = 11 is an equation with variable as both the sides are equal. The variable is n.
(i) 7 = (11 × 5) – (12 × 4) is not an equation with variable.
(j) 7 = (11 × 2) + p is an equation with variable as both the sides are equal. The variable is p.
(k) 20 = 5y is an equation with variable as both the sides are equal. The variable is
(l) \(\frac{3q}{2}\) < 5 is not an equation as L.H.S. is less than R.H.S.
(m) z + 12 > 24 is not an equation as L.H.S. is greater than R.H.S.
(n) 20 – (10 – 5) = 3 × 5 is not an equation with variable.
(o) 7 – x = 5 is an equation with variable as both the sides are equal. The variable is x.

Question 2.
Complete the entries in the third column of the table.


Solution:

Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60        (10, 5, 12, 15)
(b) n + 12 = 20      (12, 8, 20, 0)
(c) p – 5 = 5       (0, 10, 5, -5)
(d) \(\frac{q}{2}\) = 7   (7, 2, 10, 14)
(e) r – 4 = 0     (4, -4, 8, 0)
(f) x + 4 = 2      (-2, 0, 2, 4)
Solution:
(a) 5m = 60
Putting the given values in L.H.S., we get
5 × 10 = 50
∵ L.H.S. ≠ R.H.S.
∴ m = 10 is not the solution.
5 × 5 = 25
∵ L.H.S. ≠ R.H.S.
∴ m = 5 is not the solution.
5 × 12 = 60
∵ L.H.S. = R.H.S.
∴ m = 12 is a solution.
5 x 15 = 75
∵ L.H.S. ≠ R.H.S.
∴ m = 15 is not the solution.

(b) n + 12 = 20
Putting the given values in L.H.S., we get
12 + 12 = 24
∵ L.H.S. ≠ R.H.S.
∴ n = 12 is not the solution.
8 + 12 = 20
∵ L.H.S. = R.H.S.
∴ n = 8 is a solution.
20 + 12 = 32
∵ L.H.S. ≠ R.H.S.
∴ n = 20 is not the solution.
0 + 12 = 12
∵ L.H.S. ≠ R.H.S.
∴ n = 0 is not the solution.

(c) p – 5 = 5
Putting the given values in L.H.S., we get
0 – 5 = -5
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not the solution.
10 – 5 = 5
∵ L.H.S. = R.H.S.
∴ p = 10 is a solution.
5 – 5 = 0
∵ L.H.S. ≠ R.H.S.
∴ p = 5 is not the solution.
-5 – 5 = -10
∵ L.H.S. ≠ R.H.S.
∴ p = -5 is not the solution.

(d) \(\frac{q}{2}\) = 7
Putting the given values in L.H.S., we get
\(\frac{7}{2}\)
∵ L.H.S. ≠ R.H.S.
∴ q = 7 is not the solution.
\(\frac{2}{2}\) = 1
∵ L.H.S. ≠ R.H.S.
∴ q = 2 is not the solution.
\(\frac{10}{2}\) = 5
∵ L.H.S. ≠ R.H.S.
∴ q = 10 is not the solution.
\(\frac{14}{2}\) = 7
∵ L.H.S. = R.H.S.
∴ q = 14 is a solution.

(e) r – 4 = 0
Putting the given values in L.H.S., we get
4 – 4 = 0
∵ L.H.S. = R.H.S.
∴ r = 4 is a solution.
-4 – 4 = -8
∵ L.H.S. ≠ R.H.S.
∴ r = -4 is not the solution.
8 – 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ r = 8 is not the solution.
0 – 4 = -4
∵ L.H.S. ≠ R.H.S.
∴ r = 0 is not the solution.

(f) x + 4 = 2
Putting the given values in L.H.S., we get
-2 + 4 = 2
∵ L.H.S. = R.H.S.
∴ x = -2 is a solution.
0 + 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ x = 0 is not the solution.
2 + 4 = 6
∵ L.H.S. ≠ R.H.S.
∴ x = 2 is not the solution.
4 + 4 = 8
∵ L.H.S. ≠ R.H.S.
∴ x = 4 is not the solution.

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

(c) Complete the table and find the solution of the equation z/3 = 4 using the table

(d) Complete the table and find the solution to the equation m – 7 = 3.

Solution:

Question 5.
Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Solution:
(i) According to given information, we have
3(4) + x = 34 ⇒ 12 + x = 34
⇒ x = 34 – 12 ⇒ x = 22

(ii) According to given information, we have
x + 7 = 23
⇒ x = 23 – 7 ⇒ x = 16

(iii) According to given information, we have
x – 6 = 11
⇒ x = 11 + 6 ⇒ x = 17 .

(iv) According to given information, we have
x = 22-x ⇒ x + x = 22 ⇒ 2x = 22
⇒ x = \(\frac{22}{2}\) ⇒ x = 11

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 7 Fractions Ex 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:

Solution:

Question 2.
Solve:

Solution:

Question 3.
Shubham painted \(\frac{2}{3}\) of the wall space in his room. His sister Madhavi helped and painted \(\frac{1}{3}\) of the wall space. How much did they paint together?
Solution:
Fraction of wall painted by Shubham = \(\frac{2}{3}\)
Fraction of wall painted by Madhavi = \(\frac{1}{3}\)
Total painting by both of them \(=\frac{2}{3}+\frac{1}{3}=\frac{2+1}{3}=\frac{3}{3}=1\)
Therefore, they painted complete wall.

Question 4.
Fill in the missing fractions.

Solution:

Question 5.
Javed was given \(\frac{5}{7}\) of a basket of oranges. What fraction of oranges was left in the basket?
Solution:
Consider the total number of oranges to be the whole portion or 1.
Fraction of oranges left = \(1-\frac{5}{7}\)

Thus, \(\frac{2}{7}\) oranges was left in the basket.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.4

Question 1.
Answer the following:
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 metres less than 3 times the breadth of the hall. What is the length, if the breadth is b metres?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Solution:
(a) Sarita’s present age = y years
(i) After 5 years, her age will be (y + 5) years.
(ii) 3 years ago, her age was (y – 3) years.
(iii) Age of her grandfather = 6 × (Sarita’s age) = (6 × y) years = 6y years
(iv) Age of her grandmother = age of her grandfather – 2 years = (6y – 2) years
(v) Age of her father = 3 (Sarita’s age) + 5 years = (3y + 5) years

(b) Breadth of the hall = b m
Length of the hall = 3 (breadth) – 4m = 3b m – 4 m = (3b – 4) m

(c) Height of the box = h cm
Length of the box = 5 × height = 5 h cm
Breadth of the box = Length – 10 cm = (5h – 10) cm

(d) Meena’s position = s
Beena’s position = 8 steps ahead = s + 8
Leena’s position = 7 steps behind = s – 7
∴ Total number of steps = 4s – 10

(e) Speed of the bus = v km/h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Therefore, total distance = (5v + 20) km

Question 2.
Change the following statements using expressions into statements in ordinary language.
(For example. Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs ₹ p. A book costs ₹ 3 p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution:
(a) A book costs 3 times the cost of a notebook.
(b) The number of marbles in Tony’s box is 8 times the marbles on the table.
(c) Total number of students in the school is 20 times the number of students in our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.

Question 3.
(a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?
{Hint: Think of Munnu’s younger brother.)
Can you guess what (x + 4) may show? What (3x+ 7) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past.
What will the following expression indicate? y + 7, y – 3, y + \(4 \frac{1}{2}\), y – \(2 \frac{1}{2}\).
(c) Given n students in the class like football, what may 2n show? What may \(\frac{n}{2}\) show?
(Hint: Think of games other than football).
Solution:
(a) Munnu’s age = x years Age of his. younger brother who is 2 years younger than him = (x – 2) years
Age of his elder brother who is 4 years elder than him = (x + 4) years
Age of his father whose age is 7 years more than thrice of his age = (3x + 7) years

(b) Sara’s present age = y years
Her age in past = (y – 3), (y – \(2 \frac{1}{2}\))
Her age in future = (y + 7), (y + \(4 \frac{1}{2}\))

(c) Number of students like football = n
Number of students like hockey is twice the number of students liking football,
i. e., 2n
Number of students like tennis is half the number of students liking football,
i.e., \(\frac{n}{2}\)

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.3

Question 1.
Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint : Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 × 8) + 7; make the other expressions.)
Solution:
(a) (8 × 5) – 7
(b) (8 + 5) – 7
(c) (8 × 7) – 5
(d) (8 + 7) – 5
(e) 5 × (7 + 8)
(f) 5 + (7 × 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)

Question 2.
Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3 x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p
Solution:
(c) and (d) have expressions with numbers only.

Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
(a) z + 1, z – 1, y + 17, y – 17
(b) 17y, \(\frac{y}{17}\), 5z
(c) 2y + 17, 2y – 17
(d) 7m, – 7m + 3, – 7m – 3
Solution:
(a) z + 1 = 1 added to z
z – 1 = 1 subtracted from z
y + 17 = 17 added to y
y – 17 = 17 subtracted from y

(b) 17y = y multiplied by 17
\(\frac{y}{17}\) = y divided by 17
5z = z multiplied by 5 .

(c) 2y + 17 = First y multiplied by 2, then 17 added to the product
2y – 17 = First y multiplied by 2, then 17 subtracted from the product

(d) 7m = m multiplied by 7
-7m + 3 = First m multiplied by -7, then 3 added to the product
-7m – 3 = First m multiplied by -7, then 3 subtracted from the product

Question 4.
Give expressions for the following cases.
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) -p multiplied by 5
(g) -p divided by 5
(h) p multiplied by – 5
Solution:
(a) p + 7
(b) p – 7
(c) 7p
(d) \(\frac{p}{7}\)
(e) -m – 7
(f) -5p
(g) \(\frac{-p}{5}\)
(h) -5p

Question 5.
Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to 16.
Solution:
(a) 2m + 11
(b) 2m – 11
(c) 5y + 3
(d) 5y – 3
(e) -8y
(f) -8y + 5
(g) 16 – 5 y
(h) -5y + 16

Question 6.
(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution:
(a) t + 4, t – 4, 4 – t, 4t, \(\frac{t}{4}\), \(\frac{4}{t}\)
(b) 2 y + 7, 2y – 7, 7y + 2,7y – 2 and so on.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 7 Fractions Ex 7.4

Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘ ‘=’ ‘>’ between the fractions:

(c) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}\) and \(\frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.

Solution:

Question 2.
Compare the fractions and put an appropriate sign.

Solution:
(a)
\(\frac{3}{6}\) and \(\frac{5}{6}\) are like fractions.
Also, denominator of \(\frac{5}{6}\) is greater than denominator of \(\frac{3}{6}\)

(b)
\(\frac{1}{7}\) and \(\frac{1}{4}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{7}\) is greater than denominator of \(\frac{1}{4}\)

(c)
\(\frac{4}{5}\) and \(\frac{5}{5}\) are like fractions.
Also, numerator of \(\frac{5}{5}\) is greater than numerator of \(\frac{4}{5}\)

(d) \(\frac{3}{5}\) and \(\frac{3}{7}\) are unlike fractions with same numerator. Also, denominator of \(\frac{3}{7}\) is greater than denominator of \(\frac{3}{5}\).

Question 3.
Make five more such pairs and put appropriate signs.
Solution:

Question 4.
Look at the figures and write ‘<‘ or ‘>’, ‘=’ between the given pairs of fractions.

make five more such problems and solve them with your friends.
Solution:

Question 5.
How quickly can you do this ? Fill appropriate sign (‘<‘, ‘=’, ‘>’)

Solution:
(a) \(\frac{1}{2}\) and \(\frac{1}{5}\) are unlike fractions with same numerator. Also denominator of \(\frac{1}{5}\) is greater than denominator of \(\frac{1}{2}\).

(b) \(\frac{2}{3}\) and \(\frac{3}{6}\) are unlike fractions with different numerator.

(c) \(\frac{3}{5}\) and \(\frac{2}{3}\) are unlike fractions with different numerator.

(d) \(\frac{3}{4}\) and \(\frac{2}{8}\) are unlike fractions with different numerators.

(e) \(\frac{3}{5}\) and \(\frac{6}{5}\) are like fractions. Also, numerator of \(\frac{6}{5}\) is greater than numerator of \(\frac{3}{9}\).

(f) \(\frac{7}{9}\) and \(\frac{3}{9}\) are like fractions. Also, numerator of \(\frac{7}{9}\) is greater than numerator of \(\frac{3}{9}\).

(g) \(\frac{1}{4}\) and \(\frac{1}{2}\) are unlike fractions with different numerators.

Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.


Solution:

Question 7.
Find answers to the following. Write and indicate how you solved them.


Solution:

Question 8.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{2}{5}\) of the same book. Who read less ?
Solution:
Ila read 25 pages out of 100 pages.
Fraction of reading the pages

Question 9.
Rafiq \(\frac{3}{6}\) of an hour, while Rohit exercised for \(\frac{3}{4}\) of an hour. Who exercised for a longer time?
Solution:
Rafiq exercised \(\frac{3}{6}\) of an hour.
Rafiq exercised \(\frac{3}{4}\) of an hour.
Since, \(\frac{3}{4}>\frac{3}{6}\)
Therefore, Rohit exercised for a longer time.

Question 10.
In a Class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
In class A, 20 passed in first class out of 25.
∴ Fraction of first class passed students \(=\frac{20}{25}=\frac{4}{5}\)
In class B, 24 passed in first class out of 30.
∴ Fraction of first class passed students \(=\frac{24}{30}=\frac{4}{5}\)
Hence, both classes have same fraction of student getting first class.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Solution:
Side of equilateral triangle = l
Therefore, perimeter of equilateral triangle = 3 × side = 3l

Question 2.
The side of a regular hexagon (see fig.) is denoted by l.

Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)
Solution:
Side of regular hexagon = l
Therefore, perimeter of regular hexagon = 6 × side = 6l

Question 3.
A cube is a three-dimensional figure as shown in the given figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Solution:
Length of one edge of a cube = l
Number of edges in a cube = 12
Therefore, total length of the edges of a cube = 12 × l = 12l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure, AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).

Solution:
Since, diameter of the circle is always twice the radius of the circle.
Therefore, d = 2r

Question 5.
To find sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54.
Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution:
(a + b) + c = a + (b + c), where a, b and c are any three numbers.

MP Board Class 6th Maths Solutions