Category: Class 5

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 27

Question 1.
Give two examples of parallel lines you can see in your environment.
Answer:
(i) Bars on the window.
(ii) Horizontal lines in the notebook are the examples of parallel lines.

Question 2.
Give two examples of perpendicular lines you can see in your environment.
Answer:
(i) The angles formed by a pole and its shadow on the ground.
(ii) The adjacent sides of a notebook.

Question 3.
Look at the pictures given below. Decide whether the lines given in each picture are parallel or perpendicular to each other and write the answer in the box.

Answer:

Angles Problem Set 27 Additional Important Questions and Answers

Question 1.
Give some examples of perpendicular lines from capital letters of the English alphabet.
Answer:

Question 2.
Complete the following:

Answer:
(1) ZDEF or ZFED, Vertex E, arms are ED

Question 3.
Which of the following figures show angle?

Answer:
Figure 3

Question 4.
(A) Measure the angles given below and write the measure in the given boxes:


Answer:
(1) 30°
(2) 135°
(3) 90°
(4) 50°
(5) 90°
(6) 150°

B) Classify the above figures according to types of the angles.
Answer:
Acute angles: (1) and (4),
Right angles: (3) and (5),
Obtuse angles: (2) and (6)

Question 5.
Draw and name the following angles with the help of a protractor:








Answer:
Students to draw angles.

Question 6.
Look at the pictures given below. Decide whether the lines given in each picture are parallel or perpendicular to each other and write the answer in the box:

Answer:
(1) Parallel lines
(2) Perpendicular lines
(3) Perpendicular lines
(4) Parallel lines

Question 7.
Say, true or false of the following:
(1) Parallel lines do not intersect each other.
(2) Pole and its shadow on the ground makes a cute angle.
(3) Angle between two parallel lines is 90°.
(4) Angle between two intersecting lines may or may not be 90°.
Answer:
(1) True
(2) False
(3) False
(4) True

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 32 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 32

Write the factors of the following numbers.

(1) 8
Answer:
8 is exactly divisible by 1, 2, 4, 8.
So, 1, 2, 4, 8 are factors of 8.

(2) 5
Answer:
5 is exactly divisible by 1, 5.
So, 1, 5 are factors of 5.

(3) 14
Answer:
14 is exactly divisible by 1, 2, 7, 14.
So, 1, 2, 7, 14 are the factors 14.

(4) 10
Answer:
10 is exactly divisible by 1, 2, 5, 10.
So, 1, 2, 5, 10 are the factors of 10.

(5) 7
Answer:
7 is exactly divisible by 1, 7.
So, 1, 7 are factors of 7.

(6) 22
Answer:
22 is exactly divisible by 1, 2, 11, 22.
So, 1, 2, 11, 22 are the factors of 22.

(7) 25
Answer:
25 is exactly divisible by 1, 5, 25.
So, 1, 5, 25 are the factors of 25.

(8) 32
Answer:
32 is exactly divisible by 1, 2, 4, 8,16, 32.
So, 1, 2, 4, 8, 16, 32 are the factors of 32.

(9) 33
Answer:
33 is exactly divisible by 1, 3, 11, 33.
So, 1, 3, 11, 33 are the factors of 33.

Multiples

Dada : You know what a divisor and a dividend is. Do you know what a multiple is?

Anju : I don’t know what a multiple is, but I think it must be related to multiplication.

Dada : Right ! Let me give you an example. You can solve 20 ÷ 5, can’t you?

Anju : Yes. When we divide the dividend 20 by the divisor 5, the quotient is 4 and the remainder is 0.

Dada : When the division of a dividend leaves no remainder, the dividend is said to be a multiple of the divisor. In such a case, the dividend is the product of the divisor and the quotient. Here, 20 is a multiple of 5, but 21 is not.

Now tell me, can we divide 84 chalksticks into groups of six?

Suraj : Let me divide by 6. 84 can be divided exactly by 6 and the quotient is 14. Thus, we can make 14 groups of 6. So, 84 is the multiple of 6 and 6 is a factor of 84.

Dada : If the number of chalksticks is 6, 12, 18, 36 or 84, then we can make exact groups of 6 with none left over. It means that 6, 12, 18, 36 and 84 are multiples of 6, or that they are exactly divisible by 6. To see whether the number of chalksticks is a multiple of 6, divide that number by 6. If the remainder is 0, the number is a multiple of 6.

Each number in the 3 times table is exactly divisible by 3 or is a multiple of 3. Similarly, the numbers in the 7 times table are multiples of 7. Numbers in the 9 times table are multiples of 9.

We use this idea all the time. Let me ask you a few questions so as to make it clear. I have a 200 ml measure. Will I be able to measure out 1 litre of milk with it?

Suraj : I litre is 1000 ml. 1000 = 200 × 5, which means that 1000 is a multiple of 200. So we can measure out 1 litre of milk with the 200 millilitre measure. 5 measures of 200 ml make 1 litre.

Dada : Can we measure out one and a half litres of milk with the 200 ml measure?

Anju : One and a half litres is 1500 ml. 1500 is not divisible by 200. So, it is not a multiple of 200. So the 200 ml measure cannot be used to measure out one and a half litres of milk.

Dada : I have 400 grams of chana. I have to make pouches of 60 grams each. Is that possible, if I don’t want any left overs?

Anju : No. 400 is not a multiple of 60.

Dada : How much more chana will I need to make those pouches of 60 grams each?

Anju : We will have to find the multiple of 60 that comes directly after 400. 60 × 6 = 360, 60 × 7 = 420. So, we need 20 grams more of chana.

Tests for divisibility

Study the 2 times table and see which numbers appear in the units place. Similarly, divide 52, 74, 80, 96 and 98 by 2 to see if they are exactly divisible by 2. What rule do we get for determining whether a number is a multiple of 2?

Now study the 5 and 10 times tables.

See what rules you get for finding multiples of 5 and 10, that is, numbers divisible by 5 and 10.

Test for divisibility by 2 : If there is 0, 2, 4, 6 or 8 in the units place, the number is a multiple of 2, or is exactly divisible by 2.

Test for divisibility by 5 : Any number with 5 or 0 in the units place is a multiple of 5 or, is divisible by 5.

Test for divisibility by 10 : Any number that has 0 in the units place is a multiple of 10.

Multiples and Factors Problem Set 32 Additional Important Questions and Answers

Question 1.
Write the factors of the following numbers.

(1) 45
Answer:
45 is exactly divisible by 1, 3, 5, 9,15, 45.
So, 1, 3, 5, 9, 15, 45 are the factors of 45.

(2) 48
Answer:
48 is exactly divisible by 1, 2, 3, 4, 6, 8,12,16, 24, 48.
So, 1, 2, 3, 4, 6, 8,12,16, 24, 48 are the factors of 48.

(3) 60
Answer:
60 is exactly divisible by 1, 2, 3, 4, 5, 6,10,12, 15, 20, 30, 60
So, 1, 2, 3, 4, 5, 6,10, 12, 15, 20, 30, 60 are the factors of 60.

Question 2.
Is 8 a factor of 60?
Answer:
No, since 60 is not exactly divisible by 8.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 26 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 26

Question 1.
Draw and name the following angles with the help of a protractor.

(1) 60°
Answer:

(2) 120°
Answer:

(3) 90°
Answer:

(4) 150°
Answer:

(5) 30°
Answer:

(6) 165°
Answer:

(7) 45°
Answer:

Types of angles

∠ABC is a right angle.
Angles of measure 90° are called right angles.

∠RST measures less than 90°, that is, less than a right angle.
An angle which measures less than a right angle is called an acute angle.
∠RST is an acute angle.

∠LMN measures more than 90°, that is, more than a right angle.
An angle which measures more than a right angle is called an obtuse angle.
∠LMN is an obtuse angle.

Activity :

Making a right angle by folding
(1) Fold a sheet of paper roughly in half.
(2) Make another fold in the paper at any point on the first fold, as shown in the picture.
(3) Now unfold the paper. You will find two lines. The angle between those two lines will be a right angle.

With the help of a protractor, verify that the measure of this angle is 90°.

Parallel and perpendicular lines
Parallel lines

The bars on the window in the picture are parallel to each other.
The steps on the ladder in the picture are parallel to each other.
The vertical legs of the ladder are parallel to each other.

1. Take a rectangular piece of paper.

2. Fold it in such a way that one edge falls exactly on the opposite edge.

3. Make another fold in the same way.

4. Unfold the paper and trace the lines made by the folds, with a pencil.

___________ The lines traced with the pencil are parallel to each other.
______________________ The lines shown alongside are not of equal length, yet they are parallel to each other.

Parallel lines do not intersect, that is, they do not cut each other, no matter how far they are extended on either side.

Take a ruler as shown in the picture.

Using a pencil, draw lines along both sides of the ruler. Put the ruler aside. The two lines are parallel to each other.
In this way, we can use several rectangular objects to draw parallel lines.

Perpendicular lines

We have seen many objects standing straight on the ground. These objects form a right angle with their shadows.

For example, the angle formed by a pole and its shadow on the ground is 90° or a right angle. Similarly, adjacent sides of wooden planks or books also form angles of 90°.

When two lines form an angle of 90° with each other, they are said to be perpendicular to each other. To show that two lines are perpendicular, a symbol as shown the figure is drawn between them.

Measure the angle between any two adjacent sides of your notebook. Since it is a right angle, the two sides are perpendicular to each other.

Look at this picture of a page of a notebook.

The horizontal lines on the paper are parallel to each other. However, the vertical margin line on the side forms a right angle with the horizontal lines, therefore, it is perpendicular to the horizontal lines.

Angles Problem Set 26 Additional Important Questions and Answers

(1) 80°
Answer:

(2) 55°
Answer:

(3) 55°
Answer:

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

Maharashtra State Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 2

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:

Two-digit numbers	Reading a number
37	Thirty-seven
80	Eighty
49	Forty-nine
65	Sixty-five
28	Twenty-eight
54	Fifty-four
92	Ninety-two
71	Seventy-one
16	Sixteen
22	Twenty-two
Three-digit numbers	Reading a number
504	Five hundred and four
386	Three hundred eighty-six
430	Four hundred thirty
891	Eight hundred ninety-one
615	Six hundred fifteen
267	Two hundred sixty-seven
900	Nine hundred
173	One hundred seventy-three
766	Seven hundred sixty-six
258	Two hundred and fifty-eight
Four-digit numbers	Reading a number
3,817	Three thousand eight hundred and seventeen
4,059	Four thousand fifty-nine
9,611	Nine thousand six hundred and eleven
7,413	Seven thousand four hundred thirteen
5,608	Five thousand six hundred and eight
Four-digit numbers	Reading a number
2,009	Two thousand and nine
6,420	Six thousand four hundred and twenty
1,357	One thousand three hundred and fifty-seven
8,172	Eight thousand one hundred and seventy-two
6,156     –	Six thousand one hundred and fifty-six
Five-digit numbers	Reading a number
41,309	Forty-one thousand, three hundred and nine
68,527	Sixty-eight thousand five hundred and twenty seven
50,348	Fifty thousand three hundred and forty eight
76,052	Seventy-six thousand and fifty-two
21,546	Twenty-one thousand five hundred and forty-six
10,358	Ten thousand three hundred and fifty-eight
94,215	Ninety-four thousand two hundred and fifteen
36,104	Thirty-six thousand one hundred and four
89,157	Eighty-nine thousand one hundred and fifty-seven
72,560	Seventy-two thousand five hundred and sixty

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

Answer:

	Devnagari numerals	International numerals	Number written in words
(1)	२,३५९	2,359	Two thousand three hundred and fifty nine
(2)	३२,७५६	32,756	Thirty two thousand seven hundred and fifty Six
(3)	६७,८५९	67,859	Sixty seven thousand eight hundred and fifty Nine
(4)	१,०३४	1,034	One thousand and thirty four
(5)	२७,८९५	27,895	Twenty seven thousand eight hundred and ninety five

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:

Talukas	No. of Bags	Numbers in words
Kopargaon	12,740	Twelve thousand seven hundred and forty
Shevgaon	28,095	Twenty-eight thousand and ninety-five
Karjat	31,608	Thirty-one thousand six hundred and eight
Sangamner	10,792	Ten thousand seven hundred and ninety-two

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:

	Devnagari numerals	International numerals	The number written in words
(1)	५,५१८	5,518	Five thousand five hundred and eighteen
(2)	४९,८०९	49,809	Forty-nine thousand eight hundred and nine
(3)	७,२५६	7,256	Seven thousand two hundred and fifty-six

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

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

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 22

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

\(\text { (2) } \frac{1}{2}+\frac{3}{4}-\frac{7}{8}\)

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

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 20

Question 1.
Add the following
\(\text { (1) } \frac{1}{5}+\frac{3}{5}\)
Answer:
\(\frac{1}{5}+\frac{3}{5}=\frac{1+3}{5}=\frac{4}{5}\)

\(\text { (2) } \frac{2}{7}+\frac{4}{7}\)
Answer:
\(\frac{2}{7}+\frac{4}{7}=\frac{2+4}{7}=\frac{6}{7}\)

\(\text { (3) } \frac{7}{12}+\frac{2}{12}\)
Answer:
\(\frac{7}{12}+\frac{2}{12}=\frac{7+2}{12}=\frac{9}{12}\)

\(\text { (4) } \frac{2}{9}+\frac{7}{9}\)
Answer:
\(\frac{2}{9}+\frac{7}{9}=\frac{2+7}{9}=\frac{9}{9}=1\)

\(\text { (5) } \frac{3}{15}+\frac{4}{15}\)
Answer:
\(\frac{3}{15}+\frac{4}{15}=\frac{3+4}{15}=\frac{7}{15}\)

\(\text { (6) } \frac{2}{7}+\frac{1}{7}+\frac{3}{7}\)
Answer:
\(\frac{2}{7}+\frac{1}{7}+\frac{3}{7}=\frac{2+1+3}{7}=\frac{6}{7}\)

\(\text { (7) } \frac{2}{10}+\frac{4}{10}+\frac{3}{10}\)
Answer:
\(\frac{2}{10}+\frac{4}{10}+\frac{3}{10}=\frac{2+4+3}{10}=\frac{9}{10}\)

\(\text { (8) } \frac{4}{9}+\frac{1}{9}\)
Answer:
\(\frac{4}{9}+\frac{1}{9}=\frac{4+1}{9}=\frac{5}{9}\)

\(\text { (9) } \frac{5}{8}+\frac{3}{8}\)
Answer:
\(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}=1\)

Question 2.
Mother gave \(\frac{3}{8}\) of one guava to Meena and \(\frac{2}{8}\) of the guava to Geeta. What part of the guava did she give them altogether?
Solution:
\(\frac{3}{8}+\frac{2}{8}=\frac{3+2}{8}=\frac{5}{8}\) given altogether
Answer:
\(\frac{5}{8}\) part of guava given altogether

Question 3.
The girls of Std V cleaned \(\frac{3}{4}\) of a field while the boys cleaned \(\frac{1}{4}\). What part of the field was cleaned altogether?
Solution:
Girls cleaned + Boys cleaned
\(\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)
Answer:
Full whole field cleaned altogether.

Subtraction of like fractions

A figure is divided into 5 equal parts and 4 of them are colored. That is, \(\frac{4}{5}\) part of the figure is coloured.

Now, we remove the colour from one of the coloured parts. That is, we subtract \(\frac{1}{5}\) from \(\frac{4}{5}\). The remaining coloured part is \(\frac{3}{5}\). Therefore, \(\frac{4}{5}\) – \(\frac{1}{5}\) = \(\frac{4-1}{5}\) = \(\frac{3}{5}\).

When subtracting a fraction from another like fraction, we write the difference between the numerators in the numerator and the common denominator in the denominator.

Example (1) Subtract : \(\frac{7}{13}\) – \(\frac{5}{13}\)

These two fractions have a common denominator. So, we shall subtract the second numerator from the first and write the denominator as it is.
\(\frac{7}{13}-\frac{5}{13}=\frac{7-5}{13}=\frac{2}{13}\)

Example (2) If Raju got \(\frac{5}{12}\) part of a sugarcane and Sanju got \(\frac{3}{12}\) part, how much was the extra part that Raju got?

To find out the difference, we must subtract.
\(\frac{5}{12}-\frac{3}{12}=\frac{5-3}{12}=\frac{2}{12}\). Thus, Raju got \(\frac{2}{12}\) extra.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

\(\text { (1) } \frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)
Answer:
\(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1\)

\(\text { (2) } \frac{4}{10}+\frac{1}{10}+\frac{3}{10}+\frac{2}{10}\)
Answer:
\(\frac{4}{10}+\frac{1}{10}+\frac{3}{10}+\frac{2}{10}=\frac{4+1+3+2}{10}=\frac{10}{10}=1\)

\(\text { (3) } \frac{1}{2}+\frac{1}{2}\)
Answer:
\(\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)

Solve the following word problems:

Question 1.
of journey travelled by A and of journey travelled by B. What part of the journey travelled by both field was cleaned altogether?
Solution:
Travelled by A + Travelled by B
\(\frac{3}{5}+\frac{2}{5}=\frac{3+2}{5}=\frac{5}{5}=1\)
Answer:
Full (whole) journey travelled by both.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 19

Write the proper symbol from < , > , or = in the box.

Answer:
=

Answer:
>

Answer:
<

Answer:
=

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
>

Answer:
=

Answer:
=

Answer:
>

Answer:
>

Answer:
<

Answer:
>

Addition of like fractions
Example (1) 3/7 + 2/7 = ?
Let us divide a strip into 7 equal parts. We shall colour 3 parts with one colour and 2 parts with another.
The part with one colour is 3/7, and that with the other colour is 2/7.
The total coloured part is shown by the fraction 5/7.
It means that, \(\frac{3}{7}+\frac{2}{7}=\frac{3+2}{7}=\frac{5}{7}\)

Example (2) Add : \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}\)
The total coloured part is \(\frac{3}{8}+\frac{2}{8}+\frac{1}{8}=\frac{3+2+1}{8}=\frac{6}{8}\)

When adding like fractions, we add the numerators of the two fractions and write the denominator as it is.
Example (3) Add : 2/6 + 4/6 \(\frac{2}{6}+\frac{4}{6}=\frac{2+4}{6}=\frac{6}{6}\)
However, we know that 6/6 means that all 6 of the 6 equal parts are taken. That is, 1 whole figure is taken. Therefore, 6/6 = 1.

Note that:
If the numerator and denominator of a fraction are equal, the fraction is equal to one.
That is why, \(\frac{7}{7}=1 ; \frac{10}{10}=1 ; \frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=\frac{5}{5}=1\)
Remember that, if we do not divide a figure into parts, but keep it whole, it can also be written as 1.
This tells us that \(1=\frac{1}{1}=\frac{2}{2}=\frac{3}{3}\) and so on.
You also know that if the numerator and denominator of a fraction have a common divisor, then the fraction obtained by dividing them by that divisor is equivalent to the given fraction.
\(\frac{5}{5}=\frac{5 \div 5}{5 \div 5}=\frac{1}{1}=1\)

Fractions Problem Set 19 Additional Important Questions and Answers

Question 1.

Answer:
>

Question 2.

Answer:
=

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 18 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 18

Convert the given fractions into like fractions.

Solution :
8 is the multiple of 4 So, make 8, the common denominator \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\).Thus 6/8 and 5/8are the required like fractions.

Solution :
The number 35 is a multiple of both 5 and 7 So, making 35 as the common denominater \(\frac{3}{5}=\frac{3 \times 7}{5 \times 7}=\frac{21}{35}, \frac{3}{7}=\frac{3 \times 5}{7 \times 5}=\frac{15}{35}\) Therefore, 21/35 and 15/35 are required like fractions.

Solution :
Here 10 is the multiples of 5. So make 10 as the common denominator \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\). Thus 8/10 and 3/10 are required like fractions.

Solution :
Least common multiple of 9 and 6 is 18. So, make, 18 as the common denominator. \(\frac{2}{9}=\frac{2 \times 2}{9 \times 2}=\frac{4}{18}, \frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}\). Thus, 4/18 and 3/18 are the required like fractions.

Solution :
Least common multiple of 4 and 3 is 12 So, make 12 as common denominator \(\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}, \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}\). so, \(\frac{3}{12}, \frac{8}{12}\) are required like fractions.

Solution :
Least common multiple of 6 and 5 is 30 So, make 30 as common denominator \(\frac{5}{6}=\frac{5 \times 5}{6 \times 5}=\frac{25}{30}, \frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\) So, \(\frac{25}{30}, \frac{24}{30}\) are required like fractions.

Solution :
Least common multiple of 8 and 6 is 24 So, make 24 as common denominator \(\frac{3}{8}=\frac{3 \times 3}{8 \times 3}=\frac{9}{24}, \frac{1}{6}=\frac{1 \times 4}{6 \times 4}=\frac{4}{24}\) So, \(\frac{9}{24}, \frac{4}{24}\) are required like fractions.

Solution :
Least common multiple of 6 and 9 is 18 So, make 18 as common denominator \(\frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}, \frac{4}{9}=\frac{4 \times 2}{9 \times 2}=\frac{8}{18}\) So, 3/18 and 8/18 are the required like fractions.

Comparing like fractions
Example (1) A strip was divided into 5 equal parts. It means that each part is 1/5 . The coloured part is \(\frac{3}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\).

The white part is \(\frac{2}{5}=\frac{1}{5}+\frac{1}{5}\). The coloured part is bigger than the white part. This tells us that 3/5 is greater than 2/5. This is written as 3/5 > 2/5.

Example (2) This strip is divided into 8 equal parts. 3 of the parts have one colour and 4 have another colour. Here, 3/8 < 8/4.

In like fractions, the fraction with the greater numerator is the greater fraction.

Comparing fractions with equal numerators
You have learnt that the value of fractions with numerator 1 decreases as the denominator increases.

Even if the numerator is not 1, the same rule applies so long as all the fractions have a common numerator. For example, look at the figures below. All the strips in the figure are alike.
2 of the 3 equal parts of the strip
2 of the 4 equal parts of the strip
2 of the 5 equal parts of the strip
The figure shows that 2/3 > 2/4 > 5/2.

Of two fractions with equal numerators, the fraction with the greater denominator is the smaller fraction.

Comparing unlike fractions
Teacher : Suppose we have to compare the unlike fractions 3/5 and 4/7. Let us take an example to see how this is done. These two boys are standing on two blocks. How do we decide who is taller?

Sonu : But the height of the blocks is not the same. If both blocks are of the same height, it is easy to tell who is taller.

Nandu : Now that they are on blocks of equal height, we see that the boy on the right is taller.

Teacher : The height of the boys can be compared when they stand at the same height. Similarly, if fractions have the same denominators, their numerators decide which fraction is bigger.

Nandu : Got it! Let’s obtain the same denominators for both fractions.

Sonu : 5 × 7 can be divided by both 5 and 7. So, 35 can be the common denominator.

To compare unlike fractions, we convert them into their equivalent fractions so that their denominators are the same.

Fractions Problem Set 18 Additional Important Questions and Answers

Question 1.
\(\frac{5}{9}, \frac{17}{36}\)
Solution :
36 is the multiple of 9 So, make 36 the common denominator \(\frac{5}{9}=\frac{5 \times 4}{9 \times 4}=\frac{20}{36}\), Thus 20/36 and 17/36 are the required like fractions.

Question 2.
\(\frac{5}{6}, \frac{7}{9}\)
Solution:
Least common multiple of 6 and 9 is 18 So, make 18 as the common denominator \(\frac{5}{6}=\frac{5 \times 3}{6 \times 3}=\frac{15}{18}, \quad \frac{7}{9}=\frac{7 \times 2}{9 \times 2}=\frac{14}{18}\) So, 15/18 and 14/18 are the required like fractions.

Question 3.
\(\frac{7}{11}, \frac{3}{5}\)
Solution:
Least common multiple of 11 and 5 is 55 So, make 55 as the common denominator. \(\frac{7}{11}=\frac{7 \times 5}{11 \times 5}=\frac{35}{55}, \frac{3}{5}=\frac{3 \times 11}{5 \times 11}=\frac{33}{55}\). Thus 35/55 and 33/55 are required like fractions.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17

Question 1.
Write the proper number in the box.

Answer:
Here 20 = 2 x 10

Answer:
Here 15 = 3 x 5

Answer:
Here 18 = 9 x 2
hehce,

Answer:
Here 40 + 5 = 8,
hence,

Answer:
Here 26 ÷ 2 = 13,
hence,

Answer:
Here 6 ÷ 2 = 3,
hence,

Answer:
Here 4 ÷ 4 = 1,
hence,

Answer:
Here 25 ÷ 5 = 5,
hence,

Question 2.
Find an equivalent fraction with denominator 18, for each of the following fractions.

Solution:

Question 3.
Find an equivalent fraction with denominator 5, for each of the following fractions.

Solution:

Question 4.
From the fractions given below, pair off the equivalent fractions.

Solution:

Question 5.
Find two equivalent fractions for each of the following fractions.

Solution:

Like fractions and unlike fractions

Fractions such as \(\frac{1}{7}, \frac{4}{7}, \frac{6}{7}\) whose denominators are equal, are called ‘like fractions’.
Fractions such as \(\frac{1}{3}, \frac{4}{8}, \frac{9}{11}\) which have different denominators are called unlike fractions’.

Converting unlike fractions into like fractions

Example (1) Convert 5/6 and 7/9 into like fractions.
Here, we must find a common multiple for the numbers 6 and 9.
Multiples of 6 : 6, 12, 18, 24, 30, 36, ……..
Multiples of 9 : 9, 18, 27, 36, 45 ……..
Here, the number 18 is a multiple of both 6 and 9. So, let us make 18 the denominator of both fractions.

Thus, 15/18 and 1418 are like fractions, respectively equivalent to 5/6 and 7/9.
Here, 18 is a multiple of both 6 and 9. We could also choose numbers like 36 and 54 as the common denominators.

Example (2) Convert 4/8 and 5/16 into like fractions.
As 16 is twice 8, it is easy to make 16 the common denominator.
\(\frac{4}{8}=\frac{4 \times 2}{8 \times 2}=\frac{8}{16}\) Thus, 8/16 and 5/16 are the required like fractions.

Example (3) Find a common denominator for 4/7 and 3/4.
The number 28 is a multiple of both 7 and 4. So, make 28 the common denominator. . Therefore, 16/28 and 21/28 are the required like fractions.

Fractions Problem Set 17 Additional Important Questions and Answers

Question 1.
Find two equivalent fractions for each of the following fraction:

Solution:

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 13 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 13

Solve the following mixed word problems:

Question 1.
The Forest Department planted 23,078 trees of khair, 19,476 of behada besides trees of several other kinds. If the Department planted 50,000 trees altogether, how many trees were neither of khair nor of behada?
Solution:
2 3 0 7 8 Trees of khair
+
1 9 4 7 6 Trees of behada
4 2 5 5 4 Trees of khair and behada

5 0 0 0 0 Total trees planted
–
4 2 5 5 4 Khair and behada trees planted
7 4 4 6 Other kind of trees planted

Answer:
7,446 trees planted other than khair and behada trees.

Question 2.
A city has a population of 37,04,926. If this includes 11,24,069 men and 10,96,478 women, what is the number of children in the city?
Solution:
1 1 2 4 0 6 9 Men
+
1 0 9 6 4 7 8 Women
2 2 2 0 5 4 7 Total of men and women

3 7 0 4 9 2 6 Total population
–
2 2 2 0 5 4 7 Men and women
1 4 8 4 3 7 9 No. of children

Answer:
Number of children in the city is 14,84,379.

Question 3.
The management of a certain factory had 25,40,600 rupees in the labour welfare fund. From this fund, 12,37,865 rupees were used for medical expenses, 8,42,317 rupees were spent on the education of the workers’ children and the remaining was put aside for a canteen. How much money was put aside for the canteen?
Solution:
₹ 1 2 3 7 8 6 5 Medical expenses
₹ 8 4 2 3 1 7 Education for workers children
₹ 2080182 Spent for medical and education.
₹ 2 5 4 0 6 0 0 Labour welfare fund
₹ 2 0 8 0 1 8 2 Medical & education
₹ 4 6 0 4 1 8 Kept a side for canteen


Answer:
₹ 4,60,418 put aside for the canteen,

Question 4.
For a three-day cricket match, 13,608 tickets were sold on the first day and 8,955 on the second day. If, altogether, 36,563 tickets were sold in three days, how many were sold on the third day?
Solution:
1 3 6 0 8 Tickets sold on 1st day
+
8 9 5 5 Ticket sold on 2nd day
2 2 5 6 3 Tickets sold on 1st and 2nd day
3 6 5 6 3 Tickets sold in 3 days
–
2 2 5 6 3 Tickets sold in 2 days
1 4 0 0 0 Tickets sold on 3rd day


Answer:
₹ 14,000 tickets sold on the third day.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

Solve the following mixed word problems:

Question 1.
A man had ₹ 1,65,346 in the bank. He deposited ₹ 2,47,190 in the bank, then he gave a cheque of ₹ 3,18,649 to Ashutosh. How much’is the balance in the bank now?
Solution:
₹ 1 6 5 3 4 6 Had in the bank
+
₹ 2 4 7 1 9 0 Deposited in the bank
₹ 4 1 2 5 3 6 Total balance
₹ 4 1 2 5 3 6 Total
–
₹ 3 1 8 6 4 9 Gave to Ashutosh
₹ 9 3 8 8 7 Balance in the bank

Answer:
Balance in the bank ₹ 93,887.

Question 2.
Vighanesh had ₹ 36,28,500 from this amount he gave ₹ 15,04,930 to his wife and ₹ 10,13,825to his son. How much amount left with him?
Solution:
₹ 3 6 2 8 5 0 0 Vighanesh had
–
₹ 1 5 0 4 9 3 0 given to wife
₹ 2 1 2 3 5 7 0 Total

₹ 2 1 2 3 5 7 0 Balance
–
₹ 1 0 1 3 8 2 5 Gave to son
₹ 1 1 0 9 7 4 5 Left with him

Answer:
₹ 11,09,745 left with Vighanesh.

Question 2.
Add the following:

(1) 3 0 5 8 3
+
1 2 3 2 9
_____________
_____________
Answer:
42912

(2) 4 5 3 7 8
+
4 4 6 2 2
_____________
_____________
Answer:
90000

(3) 7 5 0 3 8
+
1 7 4 1 8
_____________
_____________
Answer:
92456

(4) 2 2 1 0 5
+
3 9 6 5 1
_____________
_____________
Answer:
61756

Question 3.
Add the following:
(1) 63,348 + 74,35,631
(2) 9,65,247 + 3,28,925
(3) 7,61,856 + 1,45,437
(4) 33,23,057 + 35,28,436
(5) 3,451 + 62,507 + 3,40,678
(6) 48 + 38,41,705 + 98,314
(7) 25,38,781 + 328 + 16,508
(8) 29,145 + 40,37,615 + 8,70,469
Answer:
(1) 74,98,979
(2) 12,94,172
(3) 9,07,293
(4) 68,51,493
(5) 4,06,636
(6) 39,40,065
(7) 25,55,617
(8) 49,37,229

Question 4.
Match the equal numbers in three columns

Column (A)	Column (B)	Column (C)
(1) Thirteen thousand plus two hundred	(a) 304 + 500	(i) 80,704
(2) Eight thousand plus seventy	(b) 13,000 + 200	(ii) 804
(3) Three hundred and four plus five hundred	(c) 80,000 + 704	(iii) 8070
(4) Eighty thousand plus seven hundred and four	(d) 8,000 + 70	(iv) 13,200

Answer:
(1) b – iv
(2) d – iii
(3) a – ii
(4) c – i

Question 5.
Subtract the following:

(1) 7 6 3 8 5
–
5 7 6 3 7
____________
____________
Answer:
18,748

(2) 5 6 0 4 7
–
3 2 3 7 8
____________
____________
Answer:
23,669

(3) 8 2 3 5 6
–
4 1 5 6 3 9
____________
____________
Answer:
36,927

(4) 4 5 4 2 9
–
3 5 9 6 8
____________
____________
Answer:
04,788

(5) 7 4 3 5 0 8
–
4 1 5 6 3 9
____________
____________
Answer:
3,27,869

(6) 2 4 8 1 3 6 7
–
1 7 8 4 2 7 8
____________
____________
Answer:
6,97,089

(7) 5 9 3 1 6 6 5
–
4 3 6 5 7 4 9
____________
____________
Answer:
19,79,109

(8) 8 0 5 1 4 3 6
–
4 3 6 5 7 4 9
____________
____________
Answer:
36,85,687

Question 6.
Solve the following examples :
(1) (a) 64,83,217 – 23,94,128 + 16,84,579
(b) 36,94,523 + 28,17,689 – 50,49,876
(c) 83,47,215 – 38,58,386 – 25,74,978
(d) 3,72,190 + 2,18,310 – 1,56,900
(e) 36,00,800 – 27,91,978 – 3,01,005
(f) 51,51,515 – 5,55,555 + 6,66,006
Answer:
(a) 57,73,668
(b) 14,62,336
(c) 19,13,851
(d) 4,33,600
(e) 5,07,817
(f) 52,61,966