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Fractions Class 5 Problem Set 22 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 22 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

Question 1.
Add the following:

\(\text { (1) } \frac{1}{8}+\frac{3}{4}\)
Solution:
The smallest common multiple of 4 and 8 is 8. So making 8 is the common denominator of the given fractions.

Answer:
\(\frac{7}{8}\)

\(\text { (2) } \frac{2}{21}+\frac{3}{7}\)
Solution:
21 is the multiple of 7. So making 21 as denominator of both the fractions.

Answer:
\(\frac{11}{21}\)

\(\text { (3) } \frac{2}{5}+\frac{1}{3}\)
Solution:
Least common multiple of 5 and 3 is 15. So making common denominator of both the fractions 15.

Answer:
\(\frac{11}{15}\)

\(\text { (4) } \frac{2}{7}+\frac{1}{2}\)
Solution:
Smallest common multiple of 2 and 7 is 14. So, making denominator of both the fractions 14.

Answer:
\(\frac{11}{14}\)

\(\text { (5) } \frac{3}{9}+\frac{3}{5}\)
Solution:
Smallest common multiple of 9 and 5 is 45.

Answer:
\(\frac{42}{45}\)

Question 2.
Subtract the following:

\(\text { (1) } \frac{3}{10}-\frac{1}{20}\)
Solution:
20 is the multiples of 10. So,


Answer:
\(\frac{5}{20}\)

\(\text { (2) } \frac{3}{4}-\frac{1}{2}\)
Solution:
4 is the multiple of 2. So,

Answer:
\(\frac{1}{4}\)

\(\text { (3) } \frac{6}{14}-\frac{2}{7}\)
Solution:
14 is the multiples of 7. So,

Answer:
\(\frac{2}{14}\)

\(\text { (4) } \frac{4}{6}-\frac{3}{5}\)
Solution:
Smallest common multiple of 6 and 5 is 30. So,

Answer:
\(\frac{2}{30}\)

\(\text { (5) } \frac{2}{7}-\frac{1}{4}\)
Solution:
Smallest common multiple of 7 and 4 is 28.

Answer:
\(\frac{1}{28}\)

A fraction of a collection and a multiple of a fraction

\(\frac{1}{4}\) of a collection of 20 dots – \(\frac{1}{2}\) of a collection of 20 dots

\(\frac{3}{4}\) of a collection of 20 dots

Twice 5 is 10 – \(\frac{1}{2}\) times 10

Thrice 5 – \(\frac{1}{3}\) times 15

\(\frac{1}{3}\) times 15

Meena has 5 rupees. Tina has twice as many rupees. That is, Tina has 5 × 2 = 10 rupees. Meena has half as many rupees as Tina, that is, \(\frac{1}{2}\) of 10, or, 5 rupees.

Ramu has to travel a distance of 20 km. If he has travelled \(\frac{4}{5}\) of the distance by car, how many kilometres did he travel by car?
\(\frac{4}{5}\) of 20 km is 20 × \(\frac{4}{5}\). So, we take \(\frac{1}{5}\) of 20, 4 times.
\(\frac{1}{5}\) of 20 = 4. 4 times 4 is 4 × 4 = 16.
It means that 20 × \(\frac{4}{5}\) = 16.
Ramu travelled a distance of 16 kilometres by car.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

\(\text { (1) } \frac{5}{6}+\frac{1}{12}\)
Solution:
12 is the multiple of 6

Answer:
\(\frac{11}{12}\)

\(\text { (2) } \frac{1}{9}+\frac{2}{3}\)
Solution:
Here 9 is the multiples of 3. So, making like fractions of denominator 9, we get

Answer:
\(\frac{7}{9}\)

Subtract the following:

\(\text { (1) } \frac{4}{9}-\frac{2}{5}\)
Solution:
Common multiple of 9 and 5 is 45

Answer:
\(\frac{2}{45}\)

Answer:
\(\frac{3}{8}\)

Maharashtra Board Class 5 Maths Solutions

• Fractions Problem Set 17 Class 5 Maths Solutions
• Fractions Problem Set 18 Class 5 Maths Solutions
• Fractions Problem Set 19 Class 5 Maths Solutions
• Fractions Problem Set 20 Class 5 Maths Solutions
• Fractions Problem Set 21 Class 5 Maths Solutions
• Fractions Problem Set 22 Class 5 Maths Solutions
• Fractions Problem Set 23 Class 5 Maths Solutions

Ratio-Proportion Class 6 Maths Chapter 11 Practice Set 29 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 11 Ratio-Proportion Class 6 Practice Set 29 Answers Solutions.

Std 6 Maths Practice Set 29 Solutions Answers

Question 1.
If 20 metres of cloth costs Rs 3600, find the cost of 16 m of cloth.
Solution:
Cost of 20 metres of cloth = Rs 3600
∴ Cost of 1 metre of cloth = \(\frac{\text { cost of } 20 \text { metres of cloth }}{20}=\frac{3600}{20}\)
= Rs 180
∴ Cost of 16 metres of cloth = Cost of 1 metre of a cloth × 16
= 180 x 16 = Rs 2880
∴ The cost of 16 metres of cloth is Rs 2880.

Question 2.
Find the cost of 8 kg of rice, if the cost of 10 kg is Rs 325.
Solution:
Cost of 10 kg rice = Rs 325
∴ Cost of 10 kg rice

Cost of 8 kg rice = Cost of 1 kg rice x 8

∴ The cost of 8 kg rice is Rs 260.

Question 3.
If 14 chairs cost Rs 5992, how much will have to be paid for 12 chairs?
Solution:
Cost of 14 chairs = Rs 5992
∴ Cost of 1 chairs

= Rs 428
∴ Cost of 12 chairs = Cost of 1 chair x 12
= 428 x 12 = Rs 5136
∴ The amount to be paid for 12 chairs is Rs 5136.

Question 4.
The weight of 30 boxes is 6 kg. What is the weight of 1080 such boxes?
Solution:
Weight of 30 boxes = 6 kg
∴ Weight of 1 box

∴ Weight of 1080 boxes = Weight of 1 box x 1080

∴ The weight of 1080 boxes is 216 kg.

Question 5.
A car travelling at a uniform speed covers a distance of 165 km in 3 hours. At that same speed,
a. How long will it take to cover a distance of 330 km?
b. How far will it travel in 8 hours?
Solution:
Distance covered in 3 hours = 165 km
Distance covered in 1 hour

= 55 km
a. Time required to covered a distance of 330 km

= 6 hours
∴ The time required to cover a distance of 330 km is 6 hours.

b. Distance traveled in 8 hours = Distance covered in 1 hour x 8
= 55 x 8 = 440 km
∴ The distance traveled in 8 hours is 440 km.

Question 6.
A tractor uses up 12 litres of diesel while ploughing 3 acres of land. How much diesel will be needed to plough 19 acres of land?
Solution:
Diesel required to plough 3 acres of land =12 litres
∴ Diesel required to plough 1 acre of land

= 4 liters
∴ Diesel required to plough 19 acres of land = Diesel required to plough 1 acre of land x 19
= 4 x 19 = 76 litres
∴ Diesel needed to plough 19 acres of land is 76 litres.

Question 7.
At a sugar factory, 5376 kg of sugar can be obtained from 48 tonnes of sugarcane. If Savitatai has grown 50 tonnes of sugarcanes, how much sugar will it yield?
Solution:
Sugar obtained from 48 tonnes of sugarcane = 5376 kg
∴ Sugar obtained from 48 tonnes of sugarcane

∴ Sugar obtained from 50 tonnes of sugarcane = Sugar obtained from 1 tonne of sugarcane x 50
= 112 x 50 = 5600 kg
∴ 50 tonnes of sugarcane will yield 5600 kg of sugar.

Question 8.
In an orchard, there are 128 mango trees in 8 rows. If all the rows have an equal number of trees, how many trees would there be in 13 rows?
Solution:
Number of mango trees in 8 rows =128
Number of mango trees in 1 row

∴ Number of mango trees in 13 rows = Number of mango trees in 1 row x 13
= 16 x 13 = 208
∴ The number of mango trees in 13 rows are 208.

Question 9.
A pond in a field holds 120000 litres of water. It costs Rs 18000 to make such a pond. How many ponds will be required to store 480000 litres of water, and what would be the expense?
Solution:
Capacity of 1 pond = 1,20,000 litres
Total quantity of water = 4,80,000 litres
∴ Number of ponds required

Amount required to make 1 pond = Rs 18,000
∴ Amount required to make 4 ponds = Amount required to make 1 pond x 4
= 18,000 x 4 = Rs 72,000
∴ The number of ponds required to store 4,80,000 litres of water is 4, and the expense incurred in making the ponds is Rs 72,000.

Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 29 Intext Questions and Activities

Question 1.
Vijaya wanted to gift pens to seven of her friends on her birthday. When she went to a shop to buy them, the shopkeeper told her the rate for a dozen pens.
i. Can you help Vijaya to find the cost of 7 pens?
ii. If you find the cost of one pen, you can also find the cost of 7, right? (Textbook pg. no. 59)

Solution:
Cost of 12 pens = Rs 84.
∴ Cost of 12 pens

∴ Cost of 7 pens = Cost of one pen x Number of pens = 7 × 7
∴ Cost of 7 pens = Rs 49
∴ The cost of 7 pens (Rs 49) can be found by unitary method.

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

Patterns Class 5 Problem Set 53 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 15 Patterns Problem Set 53 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 15 Patterns

Question 1.
Find the square numbers from the list given below.
5, 9, 12, 16, 50, 60, 64, 72, 80, 81
Answer:
9,16, 64, 81, 4, 25, 49 are square numbers.

Question 2.
Which are the triangular numbers in the given list?
3, 6, 8, 9, 12, 15, 16, 20, 21, 42
Answer:
3, 6, 15, 21, 28, 10, 45, 55 are triangular numbers.

Question 3.
Name a number which is square as well as triangular.
Answer:
36 is square as well as triangular number.

Question 4.
If 4 is the first square number, which is the tenth one?
Answer:
121 is the tenth square number.

Question 5.
If 3 is the first triangular number, which is the tenth one?
Answer:
66 is the tenth triangular number.

Think about it.

• How will you decide if a given number is a square number?
• How will you decide if a given number is a triangular number?
• How many square numbers do you think there are?
• How many triangular numbers do you think there are?

Activity

Make a collection of pictures in which you can see square or triangular numbers.

Patterns in floor tiles

The tiles in each picture below form a specific pattern. Observe that there is no gap or open ground between two tiles.

On a large piece of card sheet, draw several shapes like the one shown alongside. Colour half of them. Cut them all out and separate them.

One pattern made of these shapes is shown alongside. Make some other patterns of your own.

Cut out many pieces of each of the shapes shown alongside. Join them in a pattern like floor tiles.

Note the pattern and complete the design.

Make your own shapes and use them to make patterns for sari and shawl borders, etc.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Solve the following :

Question 1.
If 4 is the first square number which is the eighth one?
Answer:
81 is the eighth square number.

Question 2.
If 3 is the first triangular number which is the eighth one?
Answer:
45 is the eighth triangular number.

Question 3.
Classify the following into square numbers and triangular numbers.
3, 4, 9,10,15,16; 45, 49, 64, 66, 81, 91
Answer:
Square Numbers : 4, 9,16, 49, 64, 81
Triangular Numbers : 3, 10, 15, 45, 66, 91

Question 4.
Find out the numbers which are neither square nor triangular numbers from the following.
4, 5, 6, 8, 9, 10, 14, 15, 16, 25, 26, 27, 28.
Answer:
5, 8 14, 26 and 27

Question 5.
(1) If 4 is the first square number, which is the fifth one?
(2) If 3 is the first triangular number, which is the sixth one?
(3) Write all the square numbers between 20 and 80.
(4) Write all the triangular numbers between 20 and 80.
(5) Write the greatest two-digit square numbers as well as triangular numbers.
(6) Write the next three square numbers, 36, 49, 64,…….,
(7) Write the next three triangular numbers 36, 45, 55,
Answer:
(1) 36
(2) 28
(3) 25, 36, 49, 64
(4) 21, 28, 36, 45, 55, 66, 78
(5) 81, 91
(6) 81, 100, 121
(7) 66, 78, 91

Question 6.
Match the columns

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

Maharashtra Board Class 5 Maths Solutions

• Three Dimensional Objects and Nets Problem Set 51  Class 5 Maths Solutions
• Pictographs Problem Set 52 Class 5 Maths Solutions
• Patterns Problem Set 53 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 54 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 55 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 56 Class 5 Maths Solutions

Equations Class 6 Maths Chapter 10 Practice Set 26 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 26 Answers Solutions.

Std 6 Maths Practice Set 26 Solutions Answers

Question 1.
Different mathematical operations are given in the two rows below. Find out the number you get in each operation and make equations.

1. 1. 16 ÷ 2,
   2. 5 × 2,
   3. 9 + 4,
   4. 72 ÷ 3,
   5. 4 + 5,

1. 8 × 3,
2. 19 – 10,
3. 10 – 2,
4. 37 – 27,
5. 6 + 7

Solution:

1. 16 ÷ 2 = 8
2. 5 × 2 = 10
3. 9 + 4 = 13
4. 72 ÷ 3 = 24
5. 4 + 5 = 9
6. 8 × 3 = 24
7. 19 – 10 = 9
8. 10 – 2 = 8
9. 37 – 27 = 10
10. 6 + 7 = 13

∴ The equations are

1. 16 ÷ 2 = 10 – 2
2. 5 × 2 = 37 – 27
3. 9 + 4 = 6 + 7
4. 72 ÷ 3 = 8 x 3
5. 4 + 5 = 19 – 10

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

Basic Concepts in Geometry Class 6 Maths Chapter 1 Practice Set 1 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 1 Basic Concepts in Geometry Class 6 Practice Set 1 Answers Solutions.

Std 6 Maths Practice Set 1 Solutions Answers

Question 1.
Look at the figure alongside and name the following:
i. Collinear points
ii. Rays
iii. Line segments
iv. Lines

Solution:

Question 2.
Write the different names of the line.

Solution:
The different names of the given line are line l, line AB, line AC, line AD, line BC, line BD and line CD.

Question 3.

Solution:
(i – Line),
(ii – Line Segment),
(iii – Plane),
(iv – Ray)

Question 4.
Observe the given figure. Name the parallel lines, the concurrent lines and the points of concurrence in the figure.

Solution:

Maharashtra Board Class 6 Maths Chapter 1 Basic Concepts in Geometry Intext Questions and Activities

Question 1.
Complete the rangoli. Then, have a class discussion with the help of the following questions:

1. What kind of surface do you need for making a rangoli?
2. How do you start making a rangoli?
3. What did you do in order to complete the rangoli?
4. Name the different shapes you see in the rangoli.
5. Would it be possible to make a rangoli on a scooter or on an elephant’s back?
6. When making a rangoli on paper, what do you use to make the dots?

Solution:

1. For making a rangoli, I need a flat surface.
2. I can start making a rangoli by drawing equally spaced dots on the flat surface using a chalk.
3. In order to complete the rangoli, I joined the dots by straight lines to make a design.
4. In the rangoli, I find various shapes such as square, rectangle and triangles of two different size.
5. No. It won’t be possible to make a rangoli on a scooter or on an elephant’s back as they do not have a flat surface.
6. When making a rangoli on paper, I made use of scale and pencil to make equally spaced dots.

Question 2.
Write the proper term, ‘intersecting lines’ or ‘parallel lines’ in each of the empty boxes. (Textbook pg. no. 4)

Solution:
i. Intersecting Lines
ii. Parallel Lines
iii. Intersecting Lines

Question 3.
Draw a point on the blackboard. Every student now draws a line that passes through that point. How many such lines can be drawn? (Textbook pg. no. 2)
Solution:
An infinite number of lines can be drawn through one point.

Question 4.
Draw a point on a paper and use your ruler to draw lines that pass through it. How many such lines can you draw? (Textbook pg. no. 2)
Solution:
An infinite number of lines can be drawn through one point.

Question 5.
There are 9 points in the figure. Name them. (Textbook pg. no. 3)
i. If you choose any two points, how many lines can pass through the pair?
ii. Which three or more of these nine points lie on a straight line?
iii. Of these nine points, name any three or more points which do not lie on the same line.

Solution:
i. One and only one line can be drawn through two distinct , points.
ii. Points A, B, C and D lie on the same line. Points F, G and C lie on the same line.
iii. Points E, F, G, H and I do not lie on the same line, points A, B, E, H and I do not lie on the same line.

Question 6.
Observe the picture of the game being played. Identify the collinear players, non-collinear players, parallel lines and the plane. (Textbook pg. no. 4)

Solution:

Question 7.
In January, we can see the constellation of Orion in the eastern sky after seven in the evening. Then it moves up slowly in the sky. Can you see the three collinear stars in this constellation? Do you also see a bright star on the same line some distance away? (Textbook pg. no. 4)

Solution:

1. The three stars shown by points C, D and E are collinear.
2. The star shown by point H lies on the same line as the stars C, D and E.

Question 8.
Maths is fun! (Textbook pg. no. 5)
Take a flat piece of thermocol or cardboard, a needle and thread. Tie a big knot or button or bead at one end of the thread. Thread the needle with the other end. Pass the needle up through any convenient point P. Pull the thread up, leaving the knot or the button below. Remove the needle and put it aside. Now hold the free end of the thread and gently pull it straight. Which figure do you see? Now, holding the thread straight, turn it in different directions. See how a countless number of lines can pass through a single point P.

Solution:

1. The pulled thread forms a straight line.
2. An infinite number of lines can be drawn through one point.

Question 9.
Choose the correct option for each of the following questions:
i. ______ is used to name a point.
(A) Capital letter
(B) Small letter
(C) Number
(D) Roman numeral
Solution :
(A) Capital letter

ii. A line segment has two points showing its limits. They are called_____
(A) origin
(B) end points
(C) arrow heads
(D) infinite points
Solution :
(B) end points

iii. An arrow head is drawn at one end of the ray to show that it is _____ on that side.
(A) finite
(B) ending
(C) infinite
(D) broken
Solution :
(C) infinite

iv. Lines which lie in the same plane but do not intersect are said to be ____ to each other.
(A) intersecting
(B) collinear
(C) parallel
(D) non-collinear
Solution :
(C) parallel

Question 10.
Determine the collinear and non-collinear points in the figure alongside:

Solution:
Collinear points:

1. Points A, E, H and C.
2. Points B, E, I and D.

Non-collinear points:
Points B, G, F and I

Question 11.
Look at the figure alongside and answer the questions given below:
i. Name the parallel lines.
ii. Name the concurrent lines and the point of concurrence.
iii. Write the different names of line PV.

Solution:
i. Parallel lines:
a. line l and line n
b. line p, line q, line r and line s

ii. Concurrent Lines: line q, line m, line n
Point of Concurrence: point S

iii. line l, line PT, line PR, line PV, line RT, line RV and line TV.

Question 12.
Name the different line segments and rays in the given figure:

Solution:
Line Segments:
seg UV, seg OY, seg OX, seg OV and seg OU

Rays:
ray OV, ray OX, ray OY and ray UV.

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

Number Work Class 5 Problem Set 5 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 2 Number Work

Question 1.
Write the place value of the underlined digit.

(1) 78, 95,210
(2) 14, 95,210
(3) 3,52,749
(4) 50,000
(5) 89, 99,988
Answer:
(1) Here, the underlined digit 7 is in ten lakhs place.
So, its place value is 70,00,000 (70 lakhs)

(2) Here, the underlined digit 4 is in lakhs place.
So, its place value is 4,00,000 (4 lakhs)

(3) Here, the underlined digit 5 in ten thousands place.
So, its place value is 50,000 (50 thousands)

(4) Here, the underlined digit ‘0’ is in the unit place.
Hence, its place value is 0 (zero)

(5) Here, the underlined digit 9 is in ten thousands place
So, its place value is 90,000 (90 thousands)

Question 2.
Write the numbers in their expanded form.
(1) 56, 43, 215
(2) 70, 815
(3) 8, 35, 999
(4) 8, 88, 889
(5) 92, 32, 992
Answer:
(1) 56,43,215: 50,00,000 + 6,00,000 + 40,000 + 3,000 + 200 + 10 + 5
(2) 70,815 : 70,000 + 800 + 10 + 5
(3) 8,35,999 : 8,00,000 + 30,000 + 5,000 + 900 + 90 + 9
(4) 8,88,889 : 8,00,000 + 80,000 + 8,000 + 800 + 80 + 9
(5) 92,32,992: 90,00,000 + 2,00,000 + 30,000 + 2,000 + 900 + 90 + 2

Question 3.
Write the place name and place value of each digit in the following numbers.
(1) 35, 705
Answer:
Digit 3 is in ten thousands place, its place value is 30,000
Digit 5 is in thousands place, its place value is 5,000
Digit 7 is in hundreds place, its place value is 700
Digit 0 is in ten place, its place value is 0
Digit 5 is in units place, its place value is 5

(2) 7, 82, 899
Answer:
Digit 7 is in lakhs place, its place value is 7.0. 000
Digit 8 is in ten thousands place, its place value is 80,000
Digit 2 is in thousands place, its place value is 2,000
Digit 8 is in hundreds place, its place value is 800
Digit 9 is- in ten place, its place value is 90 Digit 9 is in units place, its place value is 9

(3) 82, 74, 508
Answer:
Digit 8 is in ten lakhs place, its place value is 80,00,000
Digit 2 is in lakhs place, its place value is 2.0. 000
Digit 7 is in ten thousands place, its place value is 70,000
Digit 4 is in thousands place, its place value is 4,000
Digit 5 is in hundreds place, its place value is 500
Digit 0 is in ten place, its place value is 0
Digit 8 is in units place, its place value is 8

Question 4.
The expanded form of the number is given. Write the number.
(1) 60, 000 + 4000 + 600 + 70 + 9
(2) 9, 00, 000 + 20,000 + 7000 + 800 + 5
(3) 20,00,000 + 3,00,000 + 60,000 + 9000 + 500 + 10 + 7
(4) 7,00,000 + 80,000 + 4000 + 500
(5) 80,00,000 + 50,000 + 1000 + 600 + 9
Answer:
(1) The number is 64,679
(2) The number is 9,27,805
(3) The number is 23,69,517
(4) The number is 7,84,500
(5) The number is 80,51,609

An interesting dice game

Prepare a table with the name of each player, as shown below.
In front of each name, there are boxes to make seven-digit numbers.

Game 1 :
The first player throws the dice and writes that number in any one of the boxes in front of his/her name. You can write only one number in each box and once it is written, you cannot change its place. The other players do the same till all the boxes are filled and each one gets a seven-digit number. The one with the largest number is the winner.

Game 2 :
Use the same table, but you may write the number (you get on throwing the dice) in any box in front of anyone’s name. The one with the largest number is the winner.

Game 3 :
The rules are the same as for game 2, but the one with the smallest number is the winner.

Bigger and smaller numbers

Hamid : How do we determine the smaller or bigger number when we are dealing with six- or seven-digit numbers ?

Teacher : You have learnt how to do that with five-digit numbers. The number with the bigger ten thousands digit is the bigger number. If they are the same, we look at the thousands digits to determine the smaller or bigger number.

Now, can you tell how to compare six- or seven-digit numbers ?

Hamid : Yes, I can. First, we’ll look at the ten lakhs digits. If they are the same, we’ll look at the digits in the lakhs place. If those are equal, we look at the ten thousands place to tell the smaller or bigger number and so on. Besides, we might be able to tell which of the numbers is bigger, just by looking at the number of digits in each number. Right ?

Teacher : Absolutely ! The number with more digits is the bigger number.

Roman Numerals Problem Set 5 Additional Important Questions and Answers

Question 1.
Write the place value of the underlined digit.

(1) 81,67,303
Answer:
Here, the underlined digit 7 is in thousands place.
So, its place value is 7,000 (7 thousands)

(2) 41,35,062
Answer:
Here, the underlined digit 6 is in ten’s place.
So, its place value is 60 (sixty)

(3) 90,31,265
Answer:
Here, the underlined digit 3 is in ten thousands place.
So, its place value is 30,000 (30 thousands)

Question 2.
Write the numbers in their expanded form.
Answer:
(1) 51,03,640: .50,00,000 + 1,00,000 + 3,000 + 600 + 40
(2) 60,60,600: 60,00,000 + 60,000 + 600
(3) 71,45,042 : 70,00,000 + 1,00,000 + 40,000 + 5,000 + 40 + 2

Question 3.
Write the place name and place value of each digit in the following numbers.

(1) 1,88,919
Answer:
Digit 1 is in lakhs place, its place value is 1,00,000
Digit 8 is in ten thousands place, its place value is 80,000
Digit 8 is in thousands place, its place value is 8,000
Digit 9 is in hundreds place, its place value is 900
Digit 1 is in ten place, its place value is 10
Digit 9 is in units place, its place value is 9

Question 4.
The expanded form write the number.

(1) 40,00,000 + 5,00,000 + 10,000 + 3,000 + 200 + 70+8
Answer:
The number is 45,13,278

(2) 80,000 + 300 + 40 + 1
Answer:
The number is 80,341

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

Operations on Fractions Class 6 Maths Chapter 4 Practice Set 13 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 13 Answers Solutions.

Std 6 Maths Practice Set 13 Solutions Answers

Question 1.
Write the reciprocals of the following numbers:

1. 7
2. \(\frac { 11 }{ 3 }\)
3. \(\frac { 5 }{ 13 }\)
4. 2
5. \(\frac { 6 }{ 7 }\)

Solution:

1. \(\frac { 1 }{ 7 }\)
2. \(\frac { 3 }{ 11 }\)
3. \(\frac { 13 }{ 5 }\)
4. \(\frac { 1 }{ 2 }\)
5. \(\frac { 7 }{ 6 }\)

Question 2.
Carry out the following Divisions:
i. \(\frac{2}{3} \div \frac{1}{4}\)
ii. \(\frac{5}{9} \div \frac{3}{2}\)
iii. \(\frac{3}{7} \div \frac{5}{11}\)
iv. \(\frac{11}{12} \div \frac{4}{7}\)
Solution:
i. \(\frac{2}{3} \div \frac{1}{4}\)

ii. \(\frac{5}{9} \div \frac{3}{2}\)

iii. \(\frac{3}{7} \div \frac{5}{11}\)

iv. \(\frac{11}{12} \div \frac{4}{7}\)

Question 3.
There were 420 students participating in the Swachh Bharat Campaign. They cleaned \(\frac { 42 }{ 75 }\) part of the town, Sevagram. What part of Sevagram did each student clean if the work was equally shared by all?
Solution:
Total number of students = 420
Part of town cleaned by all the students = \(\frac { 42 }{ 75 }\)
∴ Part of town cleaned by one student

∴ Part of town cleaned bv one student is \(\frac { 1 }{ 750 }\)

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 13 Intext Questions and Activities

Question 1.
Ramanujan’s Magic square. (Textbook pg. no. 28)

• Add the four numbers in the rows, the columns and along the diagonals of this square.
• What is the sum?
• Is it the same every time?
• What is the peculiarity?
• Look at the numbers in the first row, 22 – 12 – 1887. Find out why this date is special.

Obtain and read a biography of the great Indian mathematician Srinivasa Ramanujan.
Solution:
Sum of the numbers in each row:
i. 22 + 12 + 18 + 87 = 139
ii. 88 + 17 + 9 + 25 = 139
iii. 10 + 24 + 89 + 16 = 139
iv. 19 + 86 + 23 + 11 = 139

Sum of the numbers along the diagonals:
i. 22 + 17 + 89 + 11 = 139
ii. 87 + 9 + 24 + 19 = 139

Sum of the numbers in each column:
i. 22 + 88 + 10 + 19 = 139
ii. 12 + 17 + 24 + 86 = 139
iii. 18 + 9 + 89 + 23 = 139
iv. 87 + 25 + 16 + 11 = 139

∴ We observe that the sum of the numbers in each of the rows, the columns and along each diagonal remains the same every time. The numbers in the first row 22 – 12 – 1887 is the birth date of Srinivasa Ramanujan.

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

Banks and Simple Interest Class 6 Maths Chapter 14 Practice Set 35 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 14 Banks and Simple Interest Class 6 Practice Set 35 Answers Solutions.

Std 6 Maths Practice Set 35 Solutions Answers

Question 1.
At a rate of 10 p.c.p.a., what would be the interest for one year on Rs 6000?
Solution:
Principal Amount = Rs 6000
Rate of Interest = 10 p.c.p.a.
Let interest on principal Rs 6000 be Rs x.
By taking ratio of the interest to the principal for both, we obtain an equation x 10

∴ The interest for one year is Rs 600.

Question 2.
Mahesh deposited Rs 8650 in a bank at a rate of 6 p.c.p.a. How much money will he get at the end of the year in all?
Solution:
Principal Amount = Rs 8650
Rate of interest = 6 p.c.p.a.
Let interest on principal Rs 8650 be Rs x
By taking ratio of the interest to the principal for both, we obtain an equation

∴ Amount received at the end of the year = Principal amount + Interest
= Rs 8650 + Rs 519
= Rs 9169
∴ Mahesh will get Rs 9169 at the end of the year.

Question 3.
Ahmad Chacha borrowed Rs 25,000 at 12 p.c.p.a. for a year. What amount will he have to return to the bank at the end of the year?
Solution:
Principal Amount = Rs 25,000
Rate of interest = 12 p.c.p.a.
Let interest on principal Rs 25,000 be Rs x
By taking ratio of the interest to the principal for both, we obtain an equation

Amount to be returned to the bank at the end of the year = Principal Amount + Interest
= Rs 25,000 + Rs 3,000
= Rs 28,000
Ahmad Chacha has to return Rs 28,000 to the bank at the end of the year.

Question 4.
Kisanrao wanted to make a pond in his field. He borrowed Rs 35,250 from a bank at an interest rate of 6 p.c.p.a. How much interest will he have to pay to the bank at the end of the year?
Solution:
Principal Amount = Rs 35,250
Rate of interest = 6 p.c.p.a.
Let interest on principal Rs 35,250 be Rs x
By taking ratio of the interest to the principal for both, we obtain an equation

∴ Kisanrao will have to pay an interest of Rs 2115 to the bank at the end of the year.

Maharashtra Board Class 6 Maths Chapter 14 Banks and Simple Interest Practice Set 35 Intext Questions and Activities

Question 1.
Study the figure given below and answer the following questions (Textbook pg. no. 74)

1. In the above picture, who are the people shown to be using bank services?
2. What does the symbol on the bag in the centre stand for?
3. What do the arrows in the given picture tell you?

Solution:

1. Students, farmers, Women’s savings groups, industrialists / professionals and traders / businessmen are shown to be using bank services.
2. The symbol on the bag in the center stands for rupees.
3. The arrows tell us about the monetary transactions taking place. In simple words, it explains the give and take relationship.

Question 2.
Visit the Bank (Textbook pg. no. 74)

• Teachers should organise a visit to a bank. Encourage the children to obtain some preliminary information about banks.
• Help them to fill some bank forms and slips for withdrawals and deposits.
• If there is no bank nearby, teachers could obtain specimen forms and get the children to fill them in class.
• Give a demonstration of banking transactions by setting up a mock bank in the school.
• Invite participation of parents who work in banks or other bank employees to give the children more detailed information about banking.

Solution:
(Students should attempt this activity with the help of their teacher/parents.)

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

Measuring Time Class 5 Problem Set 44 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 44 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 10 Measuring Time

Question 1.
The time below is given by the 12 hour clock. Write the same by the 24 hour clock.
(1) 30 minutes past 10 in the morning –
(2) 10 minutes past 8 in the morning –
(3) 20 minutes past 1 in the afternoon –
(4) 40 minutes past 5 in the evening –
Answer:
(1) [10:30]
(2) [8:10]
(3) [13:20]
(4) [17:40]

Question 2.
Match the following.
12 hour clock – 24 hour clock
(1) 9:10 am – 23:10
(2) 2:10 pm – 7:25
(3) 5:25 pm – 14:10
(4) 11:10 pm – 9:10
(5) 7:25 am – 17:25
Answer:
(1) – d
(2) – c
(3) – e
(4) – a
(5) – b

Examples of time measurement

Example (1) If Abdul started working on the computer at 11 in the morning and finished his work at 3:30 in the afternoon, how long did he work?

Method 1 :
From 11 in the morning to 12 noon, it is 1 hour. From 12 noon to 3:30 in the afternoon, it is 3 hours and 30 minutes. Therefore, the total time is 4 hours and 30 minutes.

Method 2 :
According to the 24 hour clock, 11’o’clock in the morning is 11:00 and 3:30 in the afternoon is 15:30.

Abdul worked for a total of 4 hours and thirty minutes, or four and a half hours.

Example (2) Add : 4 hours 30 min + 2 hours 45 min

Example (3) Subtract : 5 hr 30 min – 2 hr 45 min

45 minutes cannot be subtracted from 30 minutes. Therefore, we borrow 1 hour and convert it into 60 minutes for the subtraction.

Example (4) Amruta travelled by bus for 3 hours 40 minutes and by motorcycle for 1 hour 45 minutes. How long did she spend travelling?

(60 + 25) minutes are 85 minutes, that is, 1 hour and 25 minutes.
Let us add this 1 hour to 4 hours.

Therefore, Amruta travelled for a total of 5 hours and 25 minutes.

Measuring Time Problem Set 44 Additional Important Questions and Answers

The time is given by the 12-hour clock. Write the same by the 24-hour clock.

(1) 15 minutes past 9 in the evening –
(2) 12 midnight –
Answer:
[21:15]
[0o:00]

The time below is given by the 24-hour clock. Write the same by the 12-hour clock.

(1) 20:20 =
(2) 9:30 =
(3) 23:00 =
(4) 4:00 =
(5) 12:00 =
(6) 00:00 =
Answer:
[8:20 pm]
[9:30 am]
[11 pm]
[4 am]
[12 noon] or [12:00]
[12 midnight]

Maharashtra Board Class 5 Maths Solutions

• Measuring Time Problem Set 43 Class 5 Maths Solutions
• Measuring Time Problem Set 44 Class 5 Maths Solutions
• Measuring Time Problem Set 45 Class 5 Maths Solutions

Profit-Loss Class 6 Maths Chapter 13 Practice Set 34 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 34 Answers Solutions.

Std 6 Maths Practice Set 34 Solutions Answers

Question 1.
Cost price Rs 1600, selling price Rs 2800.
Solution:
Sanju bought goods worth Rs 1600 and sold them for Rs 2800. What was his profit in percentage?
Cost price = Rs 1600, Selling price = Rs 2800
Profit = Selling price – Cost price
= 2800 – 1600
= Rs 1200
Let Sanju make profit of x%.

∴ x = 75%
∴ Sanju made a profit of 75%.

Question 2.
Cost price Rs 2000, selling price Rs 1900.
Solution:
Rakhi bought books worth Rs 2000 and sold them for Rs 1900. What was her loss in percentage?
Cost price = Rs 2000, Selling price = Rs 1900
Loss = Cost price – Selling price .
= 2000 – 1900
= Rs 100
Let Rakhi incur a loss of x%.

∴ x = 5%
∴ Rakhi suffered a loss of 5%.

Question 3.
Cost price of 8 articles is Rs 1200 each, selling price Rs 1400 each.
Solution:
Pallavi bought 8 tables for Rs 1200 each and sold them for Rs 1400 each. What was the percentage of her profit or loss?
Cost price of 1 table = Rs 1200
∴ Cost price of 8 tables = 1200 x 8 = Rs 9600
Selling price of 1 table = Rs 1400
∴ Selling price of 8 tables = 1400 x8 = Rs 11200
Selling price is greater than the cost price.
∴ Pallavi made a profit.
∴ Profit = Selling price – Cost price
= 11200 – 9600
= Rs 1600
Let Pallavi make a profit of x%.

∴ Pallavi made a profit of \(16\frac { 2 }{ 3 }\) %.

Question 4.
Cost price of 50 kg grain Rs 2000, Selling price Rs 43 per kg.
Solution:
Ramesh bought 50 kg grains for Rs 2000 and sold it at the rate of Rs 43 per kg. Find the percentage of profit or loss.
Cost price of 50 kg grains = Rs 2000
Selling price of 1 kg grains = Rs 43
∴ Selling price of 50 kg grains= 43 x 50
= Rs 2150
Selling price is greater than the cost price.
∴ Ramesh made a profit.
∴ Profit = Selling price – Cost price
= 2150 – 2000
= Rs 150
Let Ramesh make profit of x%.

∴ Ramesh made a profit of \(7\frac { 1 }{ 2 }\) %.

Question 5.
Cost price Rs 8600, Transport charges Rs 250, Portage Rs 150, Selling price Rs 10000.
Solution:
Faruk bought a fridge for Rs 8600. He spent Rs 250 on transport and Rs 150 on portage.
If he sold the fridge for Rs 10,000, what was his percent profit or loss?
Total cost price of a fridge = Cost of fridge + Transportation cost + Portage
= 8600 + 250 + 150
= Rs 9000
∴ Selling price = Rs 10,000
Selling price is greater than the total cost price.
∴ Faruk made a profit.
Profit = Selling price – Total cost price
= 10000 – 9000
= Rs 1000
Let Faruk make a profit of x% on cost price.

∴ Faruk made a profit of \(11\frac { 1 }{ 9 }\) %.

Question 6.
Seeds worth Rs 20500, Labour Rs 9700, Chemicals and fertilizers Rs 5600, selling price Rs 28640.
Solution:
Ramchandra bought sunflower seeds worth Rs 20500. He spent Rs 9700 on labour and Rs 5600 on chemicals and fertilizers. He sold it for Rs 28640. What is the percentage of profit or loss?
Total cost price = Cost of seeds + Labour cost + Cost of chemicals and Fertilizers
= 20500 + 9700 + 5600
= Rs 35800
Selling price = Rs 28,640
The total cost price is greater than selling price.
∴ Ramchandra suffered a loss.
Loss = Total cost price – Selling price
= 35800- 28640
= Rs 7160
Let Ramchandra incur a loss of x%.

∴ x = 20%
∴ Ramchandra incurred a loss of 20%.

Maharashtra Board Class 6 Maths Chapter 13 Profit-Loss Practice Set 34 Intext Questions and Activities

Question 1.
Maths is fun! (Textbook pg. no. 72)

Arpita used 4 matchsticks to make a square. Then she took 3 more sticks and arranged them to make 2 squares. Another 3 sticks helped her to make 3 squares. How many sticks are needed to make 7 such squares in the same way? How many sticks are needed to make 50 squares?
Solution:
Matchsticks needed to make 7 squares = 4 + (6 × 3)
= 22
Matchsticks needed to make 50 squares= 4 + (49 × 3)
= 151

Question 2.
Project (Textbook pg. no. 72)
i. Relate instances of profit and loss that you have experienced. Express them as problems and solve the problems.
ii. Organize a fair. Gain the experience of selling things/trading. What was the expenditure on preparing or obtaining the good to be sold? How much were the sales worth? Write a composition about it or enact this entire transaction.
Solution:
(Students should attempt the activities on their own.)

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