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Decimal Fractions Class 6 Maths Chapter 5 Practice Set 15 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 15 Answers Solutions.

Std 6 Maths Practice Set 15 Solutions Answers

Question 1.
Write the proper number in the empty boxes.

Solution:

Question 2.
Convert the common fractions into decimal fractions:
i. \(\frac { 3 }{ 4 }\)
ii. \(\frac { 4 }{ 5 }\)
iii. \(\frac { 9 }{ 8 }\)
iv. \(\frac { 17 }{ 20 }\)
v. \(\frac { 36 }{ 40 }\)
vi. \(\frac { 7 }{ 25 }\)
vii. \(\frac { 19 }{ 200 }\)
Solution:
i. \(\frac { 3 }{ 4 }\)

ii. \(\frac { 4 }{ 5 }\)

iii. \(\frac { 9 }{ 8 }\)

iv. \(\frac { 17 }{ 20 }\)

v. \(\frac { 36 }{ 40 }\)

vi. \(\frac { 7 }{ 25 }\)

vii. \(\frac { 19 }{ 200 }\)

Question 3.
Convert the decimal fractions into common fractions:
i. 27.5
ii. 0.007
iii. 90.8
iv. 39.15
v. 3.12
vi. 70.400
Solution:
i. 27.5
= \(\frac { 275 }{ 10 }\)

ii. 0.007
= \(\frac { 7 }{ 1000 }\)

iii. 90.8
= \(\frac { 908 }{ 10 }\)

iv. 39.15
= \(\frac { 3915 }{ 100 }\)

v. 3.12
= \(\frac { 312 }{ 100 }\)

vi. 70.400
= 70.4
= \(\frac { 704 }{ 10 }\)

6th Std Maths Digest Pdf Download

• Operations on Fractions Practice Set 9 Class 6 Maths Solution
• Operations on Fractions Practice Set 10 Class 6 Maths Solution
• Operations on Fractions Practice Set 11 Class 6 Maths Solution
• Operations on Fractions Practice Set 12 Class 6 Maths Solution
• Operations on Fractions Practice Set 13 Class 6 Maths Solution
• Decimal Fractions Practice Set 14 Class 6 Maths Solution
• Decimal Fractions Practice Set 15 Class 6 Maths Solution
• Decimal Fractions Practice Set 16 Class 6 Maths Solution
• Decimal Fractions Practice Set 17 Class 6 Maths Solution

Equations Class 6 Maths Chapter 10 Practice Set 27 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 27 Answers Solutions.

Std 6 Maths Practice Set 27 Solutions Answers

Question 1.
Rewrite the following using a letter:
i. The sum of a certain number and 3.
ii. The difference is obtained by subtracting 11 from another number.
iii. The product of 15 and another number.
iv. Four times a number is 24.
Solution:
i. Let the number be x.
∴ x + 3 represents the sum of a certain number x and 3.

ii. Let the number be x.
∴ x – 11 represents the number obtained by subtracting 11 from another number x.

iii. Let the number be x.
∴ 15x represents the product of 15 and another number x.

iv. Let the number be x.
∴ 4x = 24 represents four the product of a number x four times.

Question 2.
Find out which operation must be done on both sides of these equations in order to solve them:

1. x + 9 = 11
2. x – 4 = 9
3. 8x = 24
4. \(\frac { x }{ 6 }\) = 3

Solution:

1. Subtract 9 from both sides.
2. Add 4 to both sides.
3. Divide both sides by 8.
4. Multiply both sides by 6.

Question 3.
Given below are some equations and the values of the variables. Are these values the solutions to those equations?

Solution:

i. y – 3 = 11
∴ y – 3 + 3 = 11 + 3
…. (Adding 3 to both sides)
∴ y + 0 = 14
∴ y = 14

ii. 17 = n + 7
∴ 17 – 7 = n + 7 – 7
…. (Subtracting 7 from both sides)
∴ 17 + (-7) = n + 7 – 7
∴ 10 = n
∴  n = 10

iii. 30 = 5x
∴ \(\frac{30}{5}=\frac{5x}{5}\)
…. (Dividing both sides by 5)
∴  6 = 1x
∴ 6 = x
∴  x = 6

iv. \(\frac { m }{ 2 }\) = 14
∴ \(\frac { m }{ 2 }\) × 2 = 14 × 2
…. (Multiplying both sides by 2)
\(\frac { m\times2 }{ 2\times1 }\) = 28
∴ m = 28

Question 4.
Solve the following equations:
i. y – 5 = 1
ii. 8 = t + 5
iii. 4x = 52
iv. 19 = m – 4
v. \(\frac { p }{ 4 }=9\)
vi. x + 10 = 5
vi. m – 5 = -12
vii. p + 4 = -1
Solution:
i. y – 5 = 1
∴y – 5 + 5 = 1 + 5
…. (Adding 5 to both sides)
∴y + 0 = 6
∴y = 6

ii. 8 = t + 5
∴8 – 5 = t + 5 – 5
……(Subtracting 5 from both sides)
∴8 + (-5) = t + 0
∴ 3 = t
∴t = 3

iii. 4x = 52
∴\(\frac{4x}{4}=\frac{52}{4}\)
…. (Dividing both sides by 4)
∴ 1x = 13
∴ x = 13

iv. 19 = m -4
∴ 19 + 4 = m – 4 + 4
…. (Adding 4 to both sides)
∴ 23 = m + 0
∴ m = 23

v. \(\frac { p }{ 4 }\) = 9
∴ \(\frac { p }{ 4 }\) × 4 = 9 × 4 …. (Multiplying both sides by 4)
∴ \(\frac { p\times4 }{ 4\times1 }=36\)
∴ 1p = 36
∴ p = 36

vi. x + 10 = 5
∴ x + 10 – 10 = 5 – 10
…. (Subtracting 10 from both sides)
∴ x + 0 = 5 + (-10)
∴ x = -5

vii. m – 5 = -12
∴m – 5 + 5 = – 12 + 5
…. (Adding 5 to both sides)
∴m + 0 = -7
∴m = -7

viii. p + 4 = – 1
∴p + 4 – 4 = -1 – 4
…. (Subtracting 4 from both sides)
∴p + 0 = (-1) + (-4)
∴P = -5

Question 5.
Write the given information as an equation and find its solution:
i. Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?

ii. Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?

iii. Archana bought some kilograms of wheat. She requires 12 kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?
Solution:
i. Let the number of sheep before selling be x.
∴ x – 34 = 176
∴ x – 34 + 34 = 176 + 34 ….(Adding 34 to both sides)
∴ x + 0 = 210
∴ x = 210
The number of sheep with Haraba before selling is 210.

ii. Let the total number of bottles be x.
∴ x – 7 = 12
∴ x – 7 + 7 = 12 + 7 ….(Adding 7 to both sides)
∴ x + 0 = 19
∴ x = 19
Weight of jam in each bottle = 250g
∴ Total weight of jam = 19 × 250g = 4750 g = \(\frac { 4750 }{ 1000 }\)kg = 4.75 kg
∴ The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

iii. Let the total wheat bought by Archana be x kg.
Wheat used in 1 month = 12 kg
∴ Wheat used in 3 months = 3 × 12 = 36 kg
∴ x – 36 = 14
∴ x – 36 + 36 = 14 + 36 ….(Adding 36 to both sides)
∴ x + 0 = 50
∴ x = 50
∴ The total amount of wheat bought by Archana was 50 kg.

6th Std Maths Digest Pdf Download

• Equations Practice Set 26 Class 6 Maths Solution
• Equations Practice Set 27 Class 6 Maths Solution
• Ratio-Proportion Practice Set 28 Class 6 Maths Solution
• Ratio-Proportion Practice Set 29 Class 6 Maths Solution
• Percentage Practice Set 30 Class 6 Maths Solution
• Profit-Loss Practice Set 31 Class 6 Maths Solution
• Profit-Loss Practice Set 32 Class 6 Maths Solution
• Profit-Loss Practice Set 33 Class 6 Maths Solution
• Profit-Loss Practice Set 34 Class 6 Maths Solution
• Banks and Simple Interest Practice Set 35 Class 6 Maths Solution

Circles Class 5 Problem Set 30 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 30 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.

Answer:

The circumference of a circle

Take a bowl with a circular edge.

Wind a string once around the bowl and make a full circle around it.
Unwind this circle and straighten it out as shown.
Measure the straightened part with a ruler. The length of that part is the circumference of the circle or of the bowl.

An arc of a circle
Shown alongside is a plastic bangle. If the bangle breaks at points A and B, it will split into two parts as shown in the picture.

Each of these parts is an arc of a circle.

On the given circle, there are two points P and Q. These two points have divided the circle into two parts. Each of these parts is an arc of the circle.

This means that P and Q have created two arcs. P and Q are the end points of both arcs.

From the name ‘arc PQ’, we cannot say which of the two arcs we are speaking of. So, an additional point is taken on each arc. This point is used to give each arc a three-letter name. In the figure, there are two arcs, arc PSQ and arc PRQ.

Circles Problem Set 30 Additional Important Questions and Answers

Question 1.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.
Answer:

Question 2.
Draw a circle and take points A, B, C on the circle. L, M, N in the interior of the circle, P, Q, R in the exterior of the circle.
Answer:

Maharashtra Board Class 5 Maths Solutions

• Circles Problem Set 28 Class 5 Maths Solutions
• Circles Problem Set 29 Class 5 Maths Solutions
• Circles Problem Set 30 Class 5 Maths Solutions
• Circles Problem Set 31 Class 5 Maths Solutions

Perimeter and Area Class 5 Problem Set 49 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 12 Perimeter and Area

Question 1.
How much wire will be needed to make a rectangle 7 cm long and 4 cm wide?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 7 + 2 x 4
= 14 + 8
= 22 cm

∴ 22 cm wire will be needed to make a rectangle

Question 2.
If the length of a rectangle is 20 m and its width is 12m, what is its perimeter?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2×20 + 2×12
= 40 + 24
= 64 m
∴ Perimeter is 64 m

Question 3.
Each side of a square is 9 m long. Find its perimeter.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 9
= 36 m
∴ Perimeter is 36 m

Question 4.
If we take 4 rounds around a field that is 160 m long and 90 m wide, what is the distance we walk in kilometres?
Solution:
Perimeter of a rectangular field
= 2 x length + 2 x breadth
= 2 x 160 + 2 x 90
= 320 + 180
= 500 m

In one round distance walked is 500 m, hence, distance walked in 4 rounds
= 500 x 4
= 2000 m
= 2km
∴ The distance walked in 4 rounds is 2 km

Question 5.
Sanju completes 12 rounds around a square park every day. If one side of the park is 120 m, find out in kilometres and metres the distance that Sanju covers daily.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 120
= 480 m

So, in one round the distance can be covered is 480 m, hence in 12 rounds the distance can be covered is
= 480 x 12
= 5760 m
= 5000 m + 760 m

∴ Sanju covers 5 km 760 m daily

Question 6.
The length of a rectangular plot of land is 50 m and its width is 30 m. A triple fence has to be put along its edges. If the wire costs 60 rupees permetre, what will be the total cost of the wire needed for the fence?
Solution:
Perimeter of a rectangular plot
= 2 x length + 2 x breadth
= 2 x 50 + 2 x 30
= 100 + 60 – 160 m
For a triple fence, wire needed
= 3 x 160 = 480 m

Cost of the wire needed
= wire needed x rate
= 480 x 60
= 28800 rupees
∴ The total cost of the wire needed for the fence is ₹ 28,800

Question 7.
A game requires its players to run around a square playground. Each side of the playground is 20 m long. One player took 5 rounds around the playground. How many metres did he run altogether?
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 20
= 80 m

In one round 80 m.
So in 5 round
= 80 x 5
= 400
= 400 m

∴ He runs altogether = 400 m

Question 8.
Four rounds of wire fence have to be put around a field. If the field is 60 m long and 40 m wide, how much wire will be needed?
Solution:
Perimeter of rectangular field
= 2 x length + 2 x breadth
= 2 x 60 + 2 x 40
= 120 + 80
= 200 m
Hence, wire required for 4 rounds
= 200 x 4
= 800 m

∴ Wire required for 4 rounds
= 800 m

Question 9.
The sides of a triangle are 24.7cm, 20.4 cm and 10.5 cm respectively. What is the perimeter of the triangle?
Solution:
Perimeter of triangle
= 24.7 + 20.4 + 10.5
= 55.6

∴ The perimeter of a triangle
= 55.6 cm

Question 10.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

(1) Perimeter of
rectangle ABCD
= [ ] cm
(2) Perimeter of
rectangle EFGH
= [ ] cm
(3) Perimeter of
square PQRS
= [ ] cm
(4) Perimeter of
rectangle STUV
= [ ] cm
Solution:

(1) Perimeter of a rectangle ABCD
= 2 x length + 2 x breadth
= 2 x 3.5 + 2 x 2.5
= 7 + 5
= 12 cm

∴ 12 cm

(2) Perimeter of a rectangle EFGH
= 2 x length + 2 x breadth
= 2 x 3.8 + 2 x 1.3
= 7.6 + 2.6
= 10.2 cm

∴ 10.2 cm

(3) Perimeter of a rectangle PQRS
= 2 x length + 2 x breadth
= 2 x 2.4 + 2 x 2.4
= 4.8+ 4.8
= 9.6 cm

∴ 9.6 cm

(4) Perimeter of a rectangle STUV
= 2 x length + 2 x breadth
= 2 x 3 + 2 x 2
= 6 + 4
= 10 cm

∴ 10 cm

(5) Perimeter of a triangle LMN
= 1.5 + 2.5 + 2
= 6 cm

∴ 6 cm

Area : Revision

Of the figures given above, figure ABCD has six squares of 1 cm each inside it. It means that its area is 6 sq cm.

In the same way, count the squares in each figure and write its area.
(1) Area of MNRS = [ ] sq cm
(2) Area of EFGH = [ ] sq cm
(3) Area of PQRS = [ ] sq cm
(4) Area of IJKL = [ ] sq cm

Atul : Sir, why is the unit for area written as sq cm? We measure the sides in centimetres.

Teacher : Centimetre is a standard unit of length. In order to measure area, we need a standard unit of area. For this, a square with a side 1 cm is taken as the standard unit. The area of this square is 1 square centimetre. That is why this unit is written as sq cm, in short.

To measure large areas like fields, parks and playgrounds, a square with side 1 m, that is, an area of 1 sq m, is taken as the standard unit.

To measure the areas oftalukas or districts, a square with side 1km, or 1sq km is the standard unit used.

Formula for the area of a rectangle

(1) In the rectangle ABCD given alongside, 1 cm divisions were marked off on each side. The points on opposite sides were joined as shown in the figure. The length of the sides of each square thus created is 1cm. Therefore, the area of each square is 1 sq cm.

In ABCD, 3 rows with 5 squares each have been created.
The number of squares in rectangle ABCD is 3 × 5 = 15.
Therefore, the area of rectangle ABCD is 15 sq cm.
Here, the length of the figure is 5 cm and its breadth is 3 cm.
Note that the product of 3 and 5 is 15.

(2) In the rectangle with sides 4 cm and 2 cm, make squares of 1 sq cm each as shown above. Count the number of squares.

Note that here too, the number of squares formed are the same as the product of the length and width of the rectangle.

Therefore, The area of a rectangle = length × breadth

Formula for the area of a square

(1) Look at the square given alongside. The side of the square is 3 cm long. 9 squares of 1 cm each are formed within this square.

Therefore, the area of this square is 9 sq cm.

Here, there are 3 rows with 3 squares each, i.e., there are 3 × 3 = 9 squares.
The length of each side of the square is 3 cm.
The product of two sides of the square is 3 × 3 = 9.

(2) Measure the area of a square with side 5 cm, in the same way.
The answer will be 25 sq cm.
Note that 5 × 5 = 25

Therefore, The area of a square = length of a side × length of a side

It is not necessary to divide a square or rectangle into small squares every time you calculate their area. The advantage of a formula is that you can calculate the area simply by substituting the appropriate values.

Word problems
Example (1) What is the area of a rectangle of length 20 cm and width 15 cm?
Area of a rectangle = length × breadth
= 20 × 15 = 300.
Therefore, the area of the rectangle is 300 sq cm.

Example (2) A wall that is 4 m long and 3 m wide has to be painted. If the labour charges are ₹ 25 per sq m, what is the cost of labour for painting this wall?

First let us calculate the area of the wall to be painted.
Area of the wall = length of the wall × breadth of the wall = 4 × 3 = 12
Thus, the area of the wall is 12 sq m.
Labour cost of 1 sq m is 25 rupees.
So the labour cost for 12 sq m will be = 12 × 25 = 300
The cost of labour for painting the wall will be 300 rupees.

Example (3) What will be the area of a square with sides 15 cm?
Area of a square = length of side × length of side
= 15 × 15 = 225
The area of the square is 225 sq cm.

Example (4) One side of a square room is 4 m. If the cost of labour for laying 1 sq m of the floor is 35 rupees, what will be the total cost of labour?
First we must find the area of the square room.
Area of the square room = length of side × length of side = 4 × 4 = 16
Therefore, the area of the square room is 16 sq m.
The labour cost of laying 1 sq m of flooring is 35 rupees.
Therefore, the cost of laying 16 sq m of flooring is 16 × 35 = 560 rupees.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Devendra walks five rounds of a square garden everyday. If the side of the garden is 150 m, how many kilometres does Devendra walk every morning?
Solution:
Perimeter of a square garden
= 4 x one side of the garden
= 4 x 150
= 600 m

In 5 rounds walking
= 5 x 600
= 3000 m
= 3 km
3 km

Question 2.
The length of a rectangular play ground is 75 m and its breadth is 50 m. Rupali walks four rounds. How many kilometres did she walk?
Solution:
Perimeter of rectangle
= 2 x length + 2 x breadth
= 2 x 75 + 2 x 50
= 150 + 100
= 250 m

In 4 rounds walking
= 4 x 250
= 1000 m
= 1 km

∴ 1 km

Question 3.
Length of the rectangle is 10 cm and its breadth is 8 cm and one square is side 9 cm. Whose perimetre is more? By how much?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 10 + 2 x 8
= 20 + 16
= 36 cm ……………….. (i)

Perimeter of a square
= 4 x length of side
= 4 x 9
36 cm ……………….. (ii)
From (i) and (ii) perimeter of both is equal.

∴ perimeter of both is equal

Maharashtra Board Class 5 Maths Solutions

• Problems on Measurement Problem Set 46 Class 5 Maths Solutions
• Problems on Measurement Problem Set 47 Class 5 Maths Solutions
• Perimeter and Area Problem Set 48 Class 5 Maths Solutions
• Perimeter and Area Problem Set 49 Class 5 Maths Solutions
• Perimeter and Area Problem Set 50 Class 5 Maths Solutions

Decimal Fractions Class 5 Problem Set 39 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Question 1.
Write how many rupees and how many paise.

(1) ₹ 58.43
Answer:
58 rupees 43 paise.

(2) ₹ 9.30
Answer:
9 rupees 30 paise.

(3) ₹ 2.30
Answer:
2 rupees 30 paise.

(4) ₹ 2.3
Answer:
2 rupees 30 paise.

Question 2.
Write how many rupees in decimal form.

(1) 6 rupees 25 paise
Answer:
₹ 6.25

(2) 15 rupees 70 paise
Answer:
₹ 15.70

(3) 8 rupees 5 paise
Answer:
₹ 8:05

(4) 22 rupees 4 paise
Answer:
₹ 22.04

(5) 720 paise
Answer:
₹ 7.20

Question 3.
Write how many metres and how many centimetres.

(1) 58.75 m
Answer:
58 m 75 cm

(2) 9.30 m
Answer:
9 m 30 cm

(3) 0.30 m
Answer:
30 cm

(4) 0.3 m
Answer:
30 cm

(5) 1.62 m
Answer:
1 m 62 cm

(6) 91.4 m
Answer:
91 cm 40 cm

(7) 7.02 m
Answer:
7 m 2 cm

(8) 0.09 m
Answer:
9 cm

Question 4.
Write how many metres in decimal form.

(1) 1 m 50 cm
Answer:
1.5 m

(2) 50 m 40 cm
Answer:
50.40 m

(3) 50 m 4 cm
Answer:
50.04 m

(4) 734 cm
Answer:
7.34 m

(5) 10 cm
Answer:
0.1 m

(6) 2 cm
Answer:
0.02 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 6.9 cm
Answer:
6 cm 9 mm

(2) 20.4 cm
Answer:
20 cm 4 mm

(3) 0.8 cm
Answer:
8 mm

(4) 0.5 cm
Answer:
5 mm

Question 6.
Write how many centimetres in decimal form.
(1) 7 cm 1 mm
Answer:
7.1 cm

(2) 16 mm
Answer:
1.6 cm

(3) 144 mm
Answer:
14.4 cm

(4) 8 mm
Answer:
0.8 cm

Writing half, quarter, three-quarters and one and a quarter in decimal form

‘Half’ is usually written as \(\frac{1}{2}\). To convert this fraction into decimal form, the denominator of \(\frac{1}{2}\) must be converted into an equivalent fraction with denominator 10.

\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) so the decimal form of \(\frac{1}{2}\) will be \(\frac{5}{10}\) or 0.5 Just as \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) = 0.5, note that \(\frac{1}{2}=\frac{1 \times 50}{2 \times 50}=\frac{50}{100}\) = 0.50

Therefore, ‘half’ is written as ‘0.5’ or 0.50’. ‘Quarter’ and ‘three quarters’ are written in fractions as \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively. Let us convert them into decimal fractions. 10 is not divisible by 4. Therefore, the denominators of \(\frac{1}{4}\) and \(\frac{3}{4}\) cannot be made into fractions with multiples of 10. However, 4 × 25 = 100, so the denominator can be 100.

A quarter \(=\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}=0.25\)
and Three quarters \(=\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}=0.75\)
One and a quarter = 1 \(\frac{1}{4}\) = 1.25
One and a half = 1 \(\frac{1}{2}\) = 1.50 = 1.5
One and three quarters = 1 \(\frac{3}{4}\) = 1.75
Seventeen and a half = 17 \(\frac{1}{2}\) = 17.50 = 17.5

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

Question 1.
Write how many rupees and how many paise.

(1) ₹ 147.5
Answer:
1 hundred and 47 rupees 50 paise.

(2) ₹ 40.4
Answer:
40 rupees and 40 paise.

Question 2.
Write how many rupees in decimal form.

(1) 105 paise
Answer:
₹ 1.05

(2) 6 rupees 6 paise
Answer:
₹ 6.06

(3) 20 rupees 20 paise
Answer:
₹ 20.2

Question 3.
Write how many metres and how many centimetres.

(1) 1.1m =
Answer:
1 m 10 cm

(2) 120 cm =
Answer:
1 m 20 cm

(3) 24.8 m =
Answer:
24 m 80 cm

(4) 0.5 m =
Answer:
50 cm

Question 4.
Write how many metres in decimal form.

(1) 110 cm =
Answer:
1.1m

(2) 60 cm =
Answer:
0.6 m

Question 5.
Write how many centimetres and how many millimetres.

(1) 0.1 cm =
Answer:
1 mm

(2) 10.5 cm =
Answer:
10 cm 5 mm

Question 6.
Write how form. many centimetres in decimal
(1) 1 mm =
Answer:
0.1 cm

(2) 100 mm =
Answer:
10.0 cm

Maharashtra Board Class 5 Maths Solutions

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions

Number Work Class 5 Problem Set 2 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 2 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 write ten each of two-, three-, four- and five-digit numbers. Read the numbers.
Answer:

Question 2.
Fill in the blanks in the table below.

Answer:

Question 3.
As a part of the ‘Avoid Plastic Project’, Zilla Parishad schools made and provided paper bags to provision stores and greengrocers. Read the talukawise numbers of the bags and write the numbers in words.

Answer:

Question 4.
How many rupees do they make?
(1) 20 notes of 1000 rupees, 5 notes of 100 rupees and 14 notes of 10 rupees.
(2) 15 notes of 1000 rupees, 12 notes of 100 rupees, 8 notes of 10 rupees and 5 coins of 1 rupee.
Answer:

Question 5.
Write the biggest and the smallest five-digit numbers that can be made using the digits 4, 5, 0, 3, 7 only once.
Answer:
Biggest five digit number is 75,430 Smallest five digit number is 30,457

Question 6.
The names of some places and their populations are given below. Use this information to answer the questions that follow.

Tala : 40,642
Gaganbawada : 35,777
Bodhwad : 91,256
Moregaon : 87,012
Bhamragad : 35,950
Velhe : 54,497
Ashti : 76,201
Washi : 92,173
Morwada : 85,890

(1) Which place has the greatest population? What is its population?
(2) Which place, Morwada or Moregaon, has the greater population?
(3) Which place has the smallest population? How much is it?
Answer:
(1) Washi has the greatest population. Population of Washi is 92,173
(2) Moregaon has the greater population.
(3) Gaganbawada has the smallest population. Its population is 35,777

Introducing six-digit numbers
Teacher : How much, do you think, is the price of a four-wheeler?
Ajay : Maybe about six or seven lakh rupees.
Teacher : Do you know exactly how much one lakh is?
Ajay : It’s a lot, isn’t it? More than even ten thousand, right?
Teacher : Yes, indeed ! Let’s find out just how much. What is 999 + 1?
Ajay : One thousand.
Teacher : You have learnt to write 99000, too. Now, if you add 1000 to that, you will get one hundred thousand. That’s what we call one lakh.
Vijay : 9999+1 is 10,000 (ten thousand). We had made the ten thousands place for it. Can we make a place for one lakh too in the same way?
Teacher : Yes, of course. Carry out the addition 99,999 + 1 and see what you get.

Here we keep carrying over till we have to make a place for the ‘lakh’ on the left of the ten thousands place. And we write the last carried over one in that place. The sum we get is read as ‘one lakh’.
Vijay : Kishakaka bought a second-hand car for two and a half lakh rupees.
Ajay : How much is two and a half lakh?
Teacher : One lakh is 100 thousand. So, half a lakh is 50 thousand. Because, half of 100 is 50.

Vijay : That means two and a half lakh is 2 lakh 50 thousand.
Teacher : Now write this number in figures.
Vijay : 2,50,000.
Teacher : We have seen that a hundred thousand is 1 lakh. If we have 1000 notes of 100 rupees, how many rupees would they make?
Vijay : 1000 notes of 100 rupees would make 1 lakh rupees.

Reading six-digit numbers
(1) 2,35,705 : two lakh thirty-five thousand seven hundred and five
(2) 8,00,363 : eight lakh three hundred and sixty-three
(3) 3,07,899 : three lakh seven thousand eight hundred and ninety-nine
(4) 9,00,049 : nine lakh forty-nine
(5) 5,30,735 : five lakh thirty thousand seven hundred and thirty-five

Writing six-digit numbers in figures
(1) Eight lakh, nine thousand and forty-three : There are 8 lakhs in this number. There are no ten thousands, so we write 0 in that place. As there are 9 thousands, we write 9 in the thousands place. We write 0 in the hundreds place as there are no hundreds. Forty-three is equal to 4 tens and 3 units, so in the tens and units places we write 4 and 3 respectively. In figures : 8,09,043.

When writing numbers in figures, write the digit in the highest place first and then, in each of the next smaller places, write the proper digit from 1 to 9. Write 0, if there is no digit in that place. For example, if the number eight lakh, nine thousand and forty-three is written as ‘89043’, it is wrong. It should be written as 8,09,043. Here, we have to write zero in the ten thousands place.

(2) Four lakh, twenty thousand, five hundred : In this figure, there aren’t any thousands in the thousands place, so we write 0 in it. Since there are five hundreds, we write 5 in the hundreds place. There are no tens and units, hence, we write 0 in those places. In figures : 4,20,500.

Roman Numerals Problem Set 2 Additional Important Questions and Answers

Question 1.
Fill in the blanks in the table below:
Answer:

Question 2.
Solve the following:

(1) In an election, the First candidate received 58,735 votes, the Second candidate received 65,500, the Third candidate received 85,450 and the Fourth candidate got 09,689 votes. Read the numbers of the votes and write the numbers in words.
Answer:
First candidate – 58,735 – Fifty-eight thousand seven hundred and thirty-five
Second candidate – 65,500 – Sixty-five thousand five hundred
Third candidate – 85,450 – Eighty-five thousand four hundred and fifty
Fourth candidate – 09,689 – Nine thousand six hundred and eighty-nine

Question 3.
How many rupees do they make?
*(1) 10 notes of 2,000 rupees, 5 notes of 100 rupees and 14 notes of 10 rupees.
Solution:
10 notes of 2,000 rupees = 10 x 2,000 ₹ 20,000
5 notes of 100 rupees = 5 x 100 = ₹ 500
14 notes of 10 rupees = 14 x 10 = ₹ 140
Total = ₹ 20,640

∴ They make, twenty thousand, six hundred and forty.

*(2) 7 notes of 2,000 rupees, 12 notes of loo rupees, 8 notes of 10 rupees and 5 coins of 1 rupee
Solution:
7 notes of 2,000 rupees = 7 x 2,000 = ₹ 14,000
12 notes of 100 rupees = 12 x 100 = ₹ 1,200
8 notes of 10 rupees = 8 x 10 = ₹ 80
5 coins of 1 rupee = 5 x 1 = ₹ 5
Total = ₹ 15,285

∴ They make, fifteen thousand, two hundred and eighty five.

(3) 4 notes of 2,000 rupees, 6 notes of 100 rupees and 12 notes of 10 rupees
Solution:
4 notes of 2,000 rupees = 4 x 2,000 = ₹ 8,000
6 notes of 100 rupees = 6 x 100 = ₹ 600
12 notes of 10 rupees = 12 x 10 = ₹ 120
Total = ₹ 8,720

∴ They make, eight thousand, seven hundred and twenty.

(4) 5 notes of 2,000 rupees, 9 notes of 500 rupees, 8 notes of 100 rupees, 7 notes of 50 rupees, 6 notes of 20 rupees and 5 note of 10 rupees
Solution:
5 notes of 2,000 rupees 5 x 2,000 = ₹ 10,000
9 notes of 500 rupees = 9 x 500 = ₹ 4,500
8 notes of 100 rupees = 8 x 100 = ₹ 800
7 notes of 50 rupees = 7 x 50 = ₹ 350
6 notes of 20 rupees = 6 x 20 = ₹ 120
5 notes of 10 rupees = 5 x 10 = ₹ 50
Total = ₹ 15,820

∴ They make, fifteen thousand, eight hundred and twenty.

*Question 4.
Write the biggest and the smallest numbers using all the given digits in every number. Use each digit only once.
(1) 4, 8, 0, 2, 6, 5;
(2) 2, 6, 7, 1, 4;
(3) 5, 9, 6, 1, 4, 3;
(4) 9, 4, 1, 3, 6;
(5) 5, 3, 0, 0, 2
Answer:
(1) Biggest six digit number is 8,65,420 Smallest six digit number is 2,04,568
(2) Biggest five digit number is 76,421 Smallest five digit number is 12,467
(3) Biggest six digit number is 9,65,431 Smallest six digit number is 1,34,569
(4) Biggest five digit number is 96,431 Smallest five digit number is 13,469
(5) Biggest five digit number is 53,200 Smallest five digit number is 20,035

Class 5 Maths Solution Maharashtra Board

• Roman Numerals Problem Set 1 Class 5 Maths Solutions
• Number Work Problem Set 2 Class 5 Maths Solutions
• Number Work Problem Set 3 Class 5 Maths Solutions
• Number Work Problem Set 4 Class 5 Maths Solutions
• Number Work Problem Set 5 Class 5 Maths Solutions
• Number Work Problem Set 6 Class 5 Maths Solutions

Circles Class 5 Problem Set 31 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 31 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
In the figure given alongside, points S, L, M, and N are on the circle. Answer the questions with the help of the diagram.

(1) Write the names of the arcs with end-points S and M.
Answer:
Arcs with the end-points S and M are, arc SLM and arc SNM.

(2) Write the names of the arcs with the end-points L and N.
Answer:
Arcs with the end-points L and N are, arc LMN and arc LSN.

Question 2.
Write the names of arcs that points A, B, C, and D in the given circle give rise to.
Answer:
Arcs with end-points A and C are, arc ABC and arc ADC.
Arcs with end-points B and D are, arc BAD and arc BCD.

Question 3.
Give the names of the arcs that are made by points P, Q, R, S, and T in the figure.
Answer:
Taking end-points : P and R, arc PQR, arc PTR.
Taking end-points: Q and S, arc QRS, arc QTS
Taking end-points : R and T, arc RST, arc RPT
Taking end-points : S and P, arc STP, arc SRP
Taking end-points: Q and T, arc QPT, arc QST

Question 4.
Measure and note down the circumference of different circular objects. (It is convenient to use a measuring tape for this purpose.)

Chapter 7 Circles Problem Set 31 Additional Important Questions and Answers

Question 1.
Draw circles with the radii given below:
(1) 1.2 cm
Answer:

(2) 2.5 cm
Answer:

(3) 3.3 cm
Answer:

Question 2.
Write true or false for the following statements:
(1) Longest chord is a diameter.
(2) Centre is not lying on the diameter.
(3) All chords are of equal length.
(4) All chords passes through the centre.
Answer:
(1) True
(2) False
(3) False
(4) False

Question 3.
Match the cplumns (A) and (B):

Answer:
(1) ↔ (c),
(2) ↔ (a),
(3) ↔ (d),
(4) ↔ (b)

Question 4.
Complete the following table by filling in the blanks:

Answer:
(1) 6 cm
(2) 10 cm
(3) 34 cm
(4) 9 cm

Question 5.
From the following figure, fill in the blanks:

(1) If OP = 4cm then AB = _________ cm, OA = _________ cm, OB = _________ cm.
(2) If AB = 10 cm then OA = _________ cm, OB = _________ cm, OP = _________ cm.
Answer:
(1) AB = 8 cm, OA = 4 cm, OB = 4 cm
(2) OA = 5 cm, OB = 5 cm, OP = 5 cm

Question 6.
In the table below, write the names of the points in the interior and exterior of the circle and those on the circle.

Answer:
Points in the exterior of the circle are A, F and G.
Points in the interior of the circle are O, E and B, and Points on the circle are C, D and H.

Question 7.
Radius of a circle with centre P is 4 cm. Fill in the blanks.
(1) The, point A is at a distance of 5 cm from the centre P. Flence the point A lies in the ________ of the circle.
(2) Point B is at a distance of 4 cm. from the centre P. Hence the point B lies ________ circle.
(3) Point C lies at a distance 3 cm from the centre P. Hence it lies in the ________ of the circle.
Answer:
(1) Exterior
(2) on the circle
(3) interior

Question 8.
Solve the following:
(1) What is the length of the diameter of a circle of radius 6 cm?
(2) What is the length of the radius of a circle of diameter 14 cm?
(3) Give the names of the arcs that are made by points X, Y, Z and W in this picture.

(4) Give the names of the arc that are made by points E, F, G and H, taking end points E and G.

(5) If the diameter of a circle is 7 cm. what is the length of the circumference? (Use measure tap)
Answer:
(1) 12 cm
(2) 7 cm
(3) Having end-points X and Z, arc XYZ and arc XWZ
Having end-points Y and W, arc YZW and arc YXW
(4) By end-points E and G, arc EFG and arc EHG
(5) 22 cm

Maharashtra Board Class 5 Maths Solutions

• Circles Problem Set 28 Class 5 Maths Solutions
• Circles Problem Set 29 Class 5 Maths Solutions
• Circles Problem Set 30 Class 5 Maths Solutions
• Circles Problem Set 31 Class 5 Maths Solutions

Triangles and their Properties Class 6 Maths Chapter 15 Practice Set 36 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 15 Triangles and their Properties Class 6 Practice Set 36 Answers Solutions.

Std 6 Maths Practice Set 36 Solutions Answers

Question 1.
Observe the figures below and write the type of the triangle based on its angles:

Solution:
i. right angled
ii. Obtuse angled
iii. acute angled

Question 2.
Observe the figures below and write the type of the triangle based on its sides:

Solution:
i. equilateral
ii. scalene
iii. isosceles

Question 3.
As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.

Solution:
The two roads which Avinash can choose to go to school are

1. Road AB + Road BC
2. Road AC

The three roads together form ∆ABC.
Road AC is shorter because the sum of the lengths of any two sides (side AB + side BC) of a triangle is always greater than the third side (side AC).

Question 4.
The lengths of the sides of some triangles are given. Say what types of triangles they are.

1. 3 cm, 4 cm, 5 cm
2. 3.4 cm, 3.4 cm, 5 cm
3. 4.3 cm, 4.3 cm, 4.3 cm
4. 3.7 cm, 3.4 cm, 4 cm

Solution:

1. Since, no two sides have equal lengths, the given triangle is a scalene triangle.
2. Since, two sides have equal length, the given triangle is an isosceles triangle.
3. Since, all the three sides have equal lengths, the given triangle is an equilateral triangle.
4. Since, no two sides have equal lengths, the given triangle is a scalene triangle.

Question 5.
The lengths of the three segments are given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer.
i. 17 cm, 7 cm, 8 cm
ii. 7 cm, 24 cm, 25 cm
iii. 9 cm, 6 cm, 16 cm
iv. 8.4 cm, 16.4 cm, 4.9 cm
v. 15 cm, 20 cm, 25 cm
vi. 12 cm, 12 cm, 16 cm
Solution:
i. The lengths of the three sides are 17 cm, 7 cm, 8 cm.
a. 7 cm + 17 cm = 24 cm, greater than 8 cm
b. 8 cm +17 cm = 25 cm, greater than 7 cm
c. 7 cm + 8 cm =15 cm, not greater than 17 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 17 cm, 7 cm, 8 cm.

ii. The lengths of the three sides are 7 cm, 24 cm, 25 cm.
a. 7 cm + 24 cm = 31 cm, greater than 25 cm
b. 25 cm + 7 cm = 32 cm, greater than 24 cm
c. 24 cm + 25 cm = 49 cm, greater than 7 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 7 cm, 24 cm, 25 cm.

iii. The lengths of the three sides are 9 cm, 6 cm, 16 cm.
a. 9 cm + 16 cm = 25 cm, greater than 6 cm
b. 6 cm + 16 cm = 22 cm, greater than 9 cm
c. 9 cm+ 6 cm =15 cm, not greater than 16 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 9 cm, 6 cm, 16 cm.

iv. The lengths of the three sides are 8.4 cm, 16.4 cm, 4.9 cm.
a. 8.4 cm + 16.4 cm = 24.8 cm, greater than 4.9 cm
b. 4.9 cm + 16.4 cm = 21.3 cm, greater than 8.4 cm
c. 8.4 cm + 4.9 cm = 13.3 cm, not greater than 16.4 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 8.4 cm, 16.4 cm, 4.9 cm.

v. The lengths of the three sides are 15 cm, 20 cm, 25 cm.
a. 15 cm + 20 cm = 35 cm, greater than 25 cm
b. 25 cm + 20 cm = 45 cm, greater than 15 cm
c. 15 cm + 25 cm = 40 cm, greater than 20 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 15 cm, 20 cm, 25 cm.

vi. The lengths of the three sides are 12 cm, 12 cm, 16 cm.
a. 12 cm + 12 cm = 24 cm, greater than 16 cm
b. 12 cm + 16 cm = 28 cm, greater than 12 cm
c. 12 cm + 16 cm = 28 cm, greater than 12 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 12 cm, 12 cm, 16 cm.

Maharashtra Board Class 6 Maths Chapter 15 Triangles and their Properties Practice Set 36 Intext Questions and Activities

Question 1.
In the given figure, some points and some line segments joining them have been drawn. Which of these figures is a triangle? Which figure is not a triangle? Why not? (Textbook pg. no. 77)

Solution:
ABC it is a closed figure with three sides. Hence, ABC is a triangle.
PQRS has three sides but it is not a closed figure. Hence, PQRS is not a triangle.

Question 2.
As seen above, ∆ABC has three sides. Line segment AB is one side. Write the names of the other two sides. ∆ABC has three angles. ∠ABC is one among them. Write the names of the other angles. (Textbook pg. no. 77)
Solution:
The names of other two sides are: seg BC and seg AC
The names of other angles are: ∠BCA and ∠CAB

Question 3.
Measure the sides of the following triangles in centimeters, using a divider and ruler. Enter the lengths in the table below. What do you observe? (Textbook pg. no. 77)

Solution:

We observe that,

1. ∆ABC is an equilateral triangle,
2. ∆PQR is an isosceles triangle, and
3. ∆XYZ is a scalene triangle.

Question 4.
Measure all the angles of the triangles given below. Enter them in the following table. (Textbook pg. no. 78)

Solution:

1. ADEF is an acute angled triangle,
2. APQR is a right angled triangle,
3. ALMN is an obtuse angled triangle.

Question 5.
Observe the set squares in your compass box. What kind of triangles are they? (Textbook pg. no. 78)

Solution:
The first set square is a scalene triangle and also a right angled triangle.
The second set square is an isosceles triangle and also a right angled triangle.

Question 6.
Properties of a triangle. (Textbook pg. no. 79)
Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?

Solution:
The three angles of the triangle form a straight angle.
∴ m∠A + m∠B + m∠C = 180°
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 7.
Properties of a triangle (Textbook pg. no. 79)
Take a triangular piece of paper and make three different types of marks near the three angles. Take a point approximately at the center of the triangle. From this point, draw three lines that meet the three sides. Cut the paper along those lines. Place the three angles side by side as shown. See how the three angles of a triangle together form a straight angle, or, an angle that measures 180°.

Solution:
The three angles of the triangle form a straight angle.
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 8.
Draw any triangle on a paper. Name its vertices A, B, C. Measure the lengths of its three sides using a divider and scale and enter them in the table. (Textbook pg. no. 79)

Solution:

6th Std Maths Digest Pdf Download

• Triangles and their Properties Practice Set 36 Class 6 Maths Solution
• Quadrilaterals Practice Set 37 Class 6 Maths Solution
• Quadrilaterals Practice Set 38 Class 6 Maths Solution
• Geometrical Constructions Practice Set 39 Class 6 Maths Solution
• Geometrical Constructions Practice Set 40 Class 6 Maths Solution
• Three Dimensional Shapes Practice Set 41 Class 6 Maths Solution

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 14 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 14 Answers Solutions.

Std 6 Maths Practice Set 14 Solutions Answers

Question 1.
In the table below, write the place value of each of the digits in the number 378.025.

Solution:

Question 2.
Solve :
i. 905.5 + 27.197
ii. 39 + 700.65
iii. 40 + 27.7 + 2.451
Solution:
i. 905.5 + 27.197

ii. 39 + 700.65

iii. 40 + 27.7 + 2.451

Question 3.
Subtract:
i. 85.96 – 2.345
ii. 632.24 – 97.45
iii. 200.005 – 17.186
Solution:
i. 85.96 – 2.345

ii. 632.24 – 97.45

iii. 200.005 – 17.186

Question 4.
Avinash traveled 42 km 365 m by bus, 12 km 460 in by car and walked 640 m. How many kilometers did he travel altogether? (Write your answer in decimal fractions)
Solution:
Distance traveled in bus = 42 km 365 m
= 42 km + \(\frac { 365 }{ 1000 }\) km
= 42 km + 0.365 km

= 42.365 km
Distance travelled in car = 12 km 460 m
= 12 km + \(\frac { 460 }{ 1000 }\) km
= 12 km + 0.460 km

= 12.460 km
Distance walked = 640 m
= \(\frac { 640 }{ 1000 }\) = 0.640 km
∴ Total distance travelled = Distance travelled in bus + Distance travelled in car + Distance walked
= 42.365 + 12.460 + 0.640

= 55.465 km
∴ Distance travelled altogether by Avinash is 55.465 km.

Question 5.
Ayesha bought 1.80 m of cloth for her salwaar and 2.25 for her kurta. If the cloth costs Rs 120 per metre, how much must she pay the shopkeeper?
Solution:
Total length of cloth bought = 1.80 m + 2.25 m
= 4.05 m

Cost of 1 m of cloth = Rs 120
∴ Cost of 4.05 m of cloth = 4.05 x 120

∴ Amount to be paid to the shopkeeper is Rs 486.

Question 6.
Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750 g to the children in her neighbourhood. How much of it does she have left?
Solution:
Total weight of watermelon = 4.25 kg
Weight of watermelon given to children = 1 kg 750 g
= 1 kg + \(\frac { 750 }{ 1000 }\) kg
= 1 kg + 0.75 kg

= 1.75 kg
∴ Weight of watermelon left = Total weight of watermelon – Weight of watermelon given to children
= 4.25 kg – 1.75 kg

= 2.5 kg
∴ Weight of watermelon left with Sujata is 2.5 kg.

Question 7.
Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?
Solution:
Speed at which Anita is driving = 85.6 km per hr.
Speed limit = 55 km per hr.
∴ Anita should reduce her speed by 85.6 km per hr – 55 km per hr.

= 30.6 km per hr.
∴ Anita should reduce her speed by 30.6 km per hour to be within the speed limit.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 14 Intext Questions and Activities

Question 1.
Nandu went to a shop to buy a pen, notebook, eraser and a paint box. The shopkeeper told him the prices. A pen costs four and a half rupees, an eraser one and a half, a notebook six and a half and a paintbox twenty-five rupees and fifty paise. Nandu bought one of each article. Prepare his bill.
If Nandu gave a 100 rupee note, how much money does he get back? (Textbook pg. no. 29)

Nandu will get __ rupees back.
Solution:
100 – 38 = 62.00
Nandu will get Rs 62 rupees back.

Question 2.
Take a pen and notebook with you when you go to the market with your parent. Note the weight of every vegetable your mother buys. Find out the total weight of those vegetables. (Textbook pg. no. 30)
Solution:
(Students should attempt this activity on their own.)

6th Std Maths Digest Pdf Download

• Operations on Fractions Practice Set 9 Class 6 Maths Solution
• Operations on Fractions Practice Set 10 Class 6 Maths Solution
• Operations on Fractions Practice Set 11 Class 6 Maths Solution
• Operations on Fractions Practice Set 12 Class 6 Maths Solution
• Operations on Fractions Practice Set 13 Class 6 Maths Solution
• Decimal Fractions Practice Set 14 Class 6 Maths Solution
• Decimal Fractions Practice Set 15 Class 6 Maths Solution
• Decimal Fractions Practice Set 16 Class 6 Maths Solution
• Decimal Fractions Practice Set 17 Class 6 Maths Solution

Angles Class 6 Maths Chapter 2 Practice Set 2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 2 Answers Solutions.

Std 6 Maths Practice Set 2 Solutions Answers

Question 1.
Match the following:

Solution:
(i – Straight Angle),
(ii – Reflex Angle),
(iii – Complete Angle),
(iv – Zero Angle).

Question 2.
The measures of some angles are given below. Write the type of each angle:

1. 75°
2. 0°
3. 215°
4. 360°
5. 180°
6. 120°
7. 148°
8. 90°

Solution:

1. Acute angle
2. Zero angle
3. Reflex angle
4. Complete angle
5. Straight angle
6. Obtuse angle
7. Obtuse angle
8. Right angle

Question 3.
Look at the figures below and write the type of each of the angles:

Solution:
a. Acute angle
b. Right angle
c. Reflex angle
d. Straight angle
e. Zero angle
f. Complete angle

Question 4.
Use a protractor to draw an acute angle, a right angle and an obtuse angle:
Solution:

[Note: Students may draw acute and obtuse angles of measure other than the ones given.]

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 2 Intext Questions and Activities

Question 1.
Look at the angles shown in the pictures below. Identify the type of angle and write its name below the picture: (Textbook pg. no. 6)

Solution:

Question 2.
Complete the following table: (Textbook pg. no. 6)

Solution:

6th Std Maths Digest Pdf Download

• Basic Concepts in Geometry Practice Set 1 Class 6 Maths Solution
• Angles Practice Set 2 Class 6 Maths Solution
• Angles Practice Set 3 Class 6 Maths Solution
• Integers Practice Set 4 Class 6 Maths Solution
• Integers Practice Set 5 Class 6 Maths Solution
• Integers Practice Set 6 Class 6 Maths Solution
• Integers Practice Set 7 Class 6 Maths Solution
• Integers Practice Set 8 Class 6 Maths Solution