Class 5

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 5 Fractions Problem Set 23 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23

Question 1.
What is \(\frac{1}{3}\) of each of the collections given below?

(1) 15 pencils
(2) 21 balloons
(3) 9 children
(4) 18 books
Answer:
(1) 15 pencils → \(\frac{1}{3}\) of 15 = 5, 15 ÷ 3 = 5 pencils.
(2) 21 baloons → \(\frac{1}{3}\) of 21 = 7,21 ÷ 3 = 7 baloons.
(3) 9 children → \(\frac{1}{3}\) of 9 = 3, 9 ÷ 3 = 3 chi1dren.
(4) 18 books → \(\frac{1}{3}\) of 18 = 6, 18 ÷ 3 = 6 books.

Question 2.
What is \(\frac{1}{5}\) of each of the following?
(1) 20 rupees
(2) 30 km
(3) 15 litres
(4) 25 cm
Answer:
(1) 20 rupees → \(\frac{1}{5}\) of 20 = 4, 20 ÷ 5 = 4 rupees.
(2) 30 km → \(\frac{1}{5}\) of 30 = 6, 30 ÷ 5 = 6km.
(3) 15 litres → \(\frac{1}{5}\) of 15 = 3, 15 ÷ 5 = 3 litres.
(4) 25 cm → \(\frac{1}{5}\) of 25 = 5, 25 ÷ 5 = 5cm.

Question 3.
Find the part of each of the following numbers equal to the given fraction.

(1) \(\frac{2}{3}\) of 30
Solution:
\(\frac{2}{3}\) x 30 So, we take \(\frac{1}{3}\) of 30, twice
\(\frac{1}{3}\) x 30 = 10, twice of 10 is 2 x 10 = 20
It means that \(\frac{2}{3}\) x 30 = 20

(2) \(\frac{7}{11}\) of 22
Solution:
\(\frac{7}{11}\) x 22 So, we take of 22, 7 times
\(\frac{1}{11}\) x 22 = 2, seven times of 2 is 2 x 7 = 14

(3) \(\frac{3}{8}\) of 64
Solution:
\(\frac{3}{8}\) x 64 So, we take \(\frac{1}{8}\) of 64, thrice
\(\frac{1}{8}\) x 64 = 8, 3 times 8 is 3 x 8 = 24

(4) \(\frac{5}{13}\) of 65
Solution:
\(\frac{5}{13}\) x 65 So, we take \(\frac{1}{13}\) of 65, 5 times
\(\frac{1}{13}\) x 65 = 55 times of 5 is 5 x 5 = 25

Mixed fractions

Half of each of the three circles is coloured. That is, 3 parts, each equal to \(\frac{1}{2}\) of the circle, are coloured.

The coloured part is \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\), that is, \(\frac{3}{2}\) or 1 + \(\frac{1}{2}\).

1 + \(\frac{1}{2}\) is written as 1 \(\frac{1}{2}\). 1 \(\frac{1}{2}\) is read as ‘one and one upon two’.

In the fraction 1 \(\frac{1}{2}\), 1 is the integer part and \(\frac{1}{2}\) is the fraction part. Hence, such fractions are called mixed fractions or mixed numbers. 2 \(\frac{1}{4}\), 3 \(\frac{2}{5}\), 7 \(\frac{4}{9}\) are all mixed fractions.

Fractions in which the numerator is greater than the denominator are called improper fractions.

\(\frac{3}{2}\), \(\frac{5}{3}\) are improper fractions. We can convert improper fractions into mixed fractions.

For example,

Activities
1. Colour the Hats.

In the picture alongside :
Colour \(\frac{1}{3}\) of the hats red.
Colour \(\frac{3}{5}\) of the hats blue.
How many hats have you coloured red?
How many hats have you coloured blue?
How many are still not coloured?

2. Make a Magic Spinner.

Take a white cardboard disc. As shown in the figure, divide it into six equal parts.

Colour the parts red, orange, yellow, green, blue and violet.

Make a small hole at the centre of the disc and fix a pointed stick in the hole.

Your magic spinner is ready.

What fraction of the disc is each of the coloured parts?
Give the disc a strong tug to make it turn fast. What colour does it appear to be now?

The Clever Poet

There was a king who had a great love for literature. A certain poet knew that if the king read a good poem it made him very happy. Then the king would give the poet an award. Once, the poet composed a good poem. He thought if he showed it to the king, he would win a prize. So, he went to the king’s palace. But, it was not easy to meet the king. You had to pass a number of gates and guards. The first guard asked the poet why he wanted to meet the king. So, the poet told him the reason. Seeing the chance of getting a share of the award, the guard demanded, ‘You must

give me \(\frac{1}{10}\) of your prize. Only then will I let you go in.’ The poet could do nothing but agree. The second guard stopped him and said, ‘I will let you go in only if you promise me \(\frac{2}{5}\) of your prize.’ The third guard, too, was a greedy man. He said, ‘I will not let you go, unless you promise me \(\frac{1}{4}\) of your prize.’ The king’s palace was just a little distance away. Now, the poet told the guard, ‘Why only \(\frac{1}{4}\), I shall give you half the prize!’ The guard was pleased and let him in.

The king liked the poem. He asked the poet, ‘What is the prize you want?’ ‘I shall be happy if Your Majesty awards me 100 lashes of the whip.’ The king was surprised. ‘Are you out of your mind!’ he exclaimed. ‘I have never met anyone so crazy as to ask for a whipping !’

‘Your Majesty, if you wish to know the reason, the three palace guards must be called here.’ When the guards came, the poet explained, ‘Your Majesty, all of them have a share in the 100 lashes that you have awarded to me. Each of them has fixed his own share of the prize I get. The first guard

must get \(\frac{1}{10}\) of the award, that is, [ ] lashes. The second must get \(\frac{2}{5}\), which is [ ], and the third must get half the award, that is, [ ] lashes !’ The king could now see how greedy the guards were and how clever the poet was. He saw to it that each guard got the punishment he deserved. He gave the poet a prize for his poem. He also gave him an extra 100 gold coins for exposing the greed of the guards.

What was the clever idea of the poet which the king appreciated so much?

Fractions Problem Set 23 Additional Important Questions and Answers

Question 1.
What is \(\frac{1}{3}\) of each of the collections given below?

(1) 24 marbles →
(2) 6 erasers →
Answer:
(1) 24 marbles → \(\frac{1}{3}\) of 24 = 8, 24 ÷ 3 = 8 marbles.
(1) 6 erasers → \(\frac{1}{3}\) of 6 = 2, 6 ÷ 3 = 2 erasers.

Question 2.
What is \(\frac{1}{5}\) of each of the following?

(1) 35 gm →
(2) 40m →
Answer:
(1) 35 gm → \(\frac{1}{5}\) of 35 = 7, 35 ÷ 5 = 7 gm.
(2) 40m → \(\frac{1}{5}\) of 40 = 8, 40 ÷ 5 = 8m.

Question 3.
Find the part of each of the following numbers equal to the given fraction:

(1) \(\frac{7}{9}\) of 45
Solution:
\(\frac{7}{9}\) x 45 So, we take \(\frac{1}{9}\) of 45, 7 times
\(\frac{1}{9}\) x 45 = 5, 7 times of 5 is 7 x 5 = 35

(2) \(\frac{3}{7}\) of 28
Solution:
\(\frac{3}{7}\) x 28 So, we take \(\frac{1}{7}\) of 28, thrice
\(\frac{1}{7}\) x 28 = 4, 3 times of 4 is 4 x 3 = 12

Question 4.
Find the proper number in the box:








Answer:
(1) 3
(2) 36
(3) 3
(4) 7
(5) 8, 18
(6) 12, 6
(7) 9, 16, 20, 24
(8) 15, 20, 35, 36, 55

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

Answer:

Question 6.
Find an equivalent fraction with numerator 30 for each of the following fractions.

Answer:

Question 7.
Find two equivalent fractions for each of the following fraction.
\(\text { (1) } \frac{5}{7}\)
\(\text { (2) } \frac{8}{9}\)
\(\text { (3) } \frac{7}{13}\)
Answer:
(1)
(2)
(3)

Question 8.
Match the columns (A) and (B) for having equivalent fractions:

	(A)	(B)
(1)	\(\frac{3}{4}\)	(a) \(\frac{15}{27}\)
(2)	\(\frac{5}{9}\)	(b) \(\frac{2}{3}\)
(3)	\(\frac{7}{11}\)	(c) \(\frac{27}{36}\)
(4)	\(\frac{8}{12}\)	(d) \(\frac{28}{44}\)

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

Question 9.
Convert the given fractions into like fractions:
\(\text { (1) } \frac{1}{10}, \frac{2}{3}\)
\(\text { (2) } \frac{3}{7}, \frac{4}{5}\)
\(\text { (3) } \frac{1}{3}, \frac{3}{5}\)
\(\text { (3) } \frac{1}{4}, \frac{2}{5}\)
Answer:
\(\text { (1) } \frac{3}{30}, \frac{20}{30}\)
\(\text { (2) } \frac{15}{35}, \frac{28}{35}\)
\(\text { (3) } \frac{5}{15}, \frac{9}{15}\)
\(\text { (3) } \frac{5}{20}, \frac{8}{20}\)

Question 10.
Write the proper symbol from <, > or = in the box:





Answer:
(1) >
(2) >
(3) >
(4) >
(5) >

Question 11.
Add the following:
\(\text { (1) } \frac{1}{6}+\frac{2}{6}\)
\(\text { (2) } \frac{1}{4}+\frac{3}{4}\)
\(\text { (3) } \frac{5}{13}+\frac{2}{13}+\frac{3}{13}\)
\(\text { (4) } \frac{2}{9}+\frac{3}{7}\)
\(\text { (5) } \frac{3}{11}+\frac{2}{3}\)
\(\text { (6) } \frac{1}{10}+\frac{4}{5}\)
Answer:
\(\text { (1) } \frac{3}{6}\)
\(\text { (2) } \frac{4}{4}\)
\(\text { (3) } \frac{10}{13}\)
\(\text { (4) } \frac{41}{63}\)
\(\text { (5) } \frac{31}{33}\)
\(\text { (6) } \frac{9}{10}\)

Question 12.
Subtract the following:
\(\text { (1) } \frac{5}{6}-\frac{1}{6}\)
\(\text { (2) } \frac{3}{5}-\frac{2}{5}\)
\(\text { (3) } \frac{7}{16}-\frac{3}{16}-\frac{1}{16}\)
\(\text { (4) } \frac{5}{6}-\frac{7}{12}\)
\(\text { (5) } \frac{13}{16}-\frac{5}{8}\)
\(\text { (6) } \frac{4}{9}-\frac{3}{10}\)
Answer:
\(\text { (1) } \frac{4}{6}\)
\(\text { (2) } \frac{1}{5}\)
\(\text { (3) } \frac{3}{16}\)
\(\text { (4) } \frac{3}{12}\)
\(\text { (5) } \frac{3}{13}\)
\(\text { (6) } \frac{13}{90}\)

Question 13.
What is \(\frac{1}{4}\) of each of the collections given below:
(1) 20 marbles
(2) 12 pens
(3) 24 notebooks
(4) 8 ladoos
Answer:
(1) 5 marbles
(2) 3 pens
(3) 6 notebooks
(4) 2 ladoos

Question 14.
What is \(\frac{1}{6}\) of each of the following:
(1) 18 bananas
(2) 12 gms
(3) 30 metres
(4) 24 ₹
Answer:
(1) 3 bananas
(2) 2 gms
(3) 5 metres
(4) 4 ₹

Question 15.
Find the part of each of the following numbers equal to the given fraction.
(1) \(\frac{2}{5}\) of 25
(2) \(\frac{3}{7}\) of 21
(3) \(\frac{4}{9}\) of 36
(4) \(\frac{4}{17}\) of 34
Answer:
(1) 10
(2) 9
(3) 16
(4) 8

Question 16.
Printed price of. the book was 80. Vikram purchased the book by paying of the printed price of the book. How much he paid for the book?
Answer:
64 ₹

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 6 Angles Problem Set 25 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 25

Measure the angles given below and write the measure in the given boxes.

Answer:
(1) 40°
(2) 120°
(3) 90°
(4) 85°

Drawing an angle of the given measure
Example Draw ∠ABC of measure 70°.
B is the vertex of∠ABC and BA and BC are its arms.

1. First draw arm BC with a ruler.

2. Since B is the vertex, we must draw a 70° angle at that point.

Put the centre of the protractor on B. Place the protractor so that the baseline lies on arm BC. Count the divisions starting from the 0 near point C. Mark a point with your pencil at the division that shows 70°. Lift the protractor.

Draw a line from vertex B through the point marking the 70° angle. Name the other end of the line A.

∠ABC is an angle of measure 700.
Rahul and Sayali drew ∠PQR of measure 800 as shown below.

Teacher : Have Rahul and Sayali drawn the angles correctly?
Shalaka : Sir, Rahul’s angle is wrong. Sayali’s angle is correct.
Teacher : Why is Rahul’s angle wrong?
Rahul : I counted 10, 20, 30…from the left and drew the angle at 80.
Teacher : Rahul measured the angle from the left. Under the baseline on the left of Q, there is nothing. The arm of the angle is on the right of Q. Therefore, the point should have been marked 80° counting from the right side, that is, on the side on which point R lies.

Angles Problem Set 25 Additional Important Questions and Answers

Answer:
60°

Answer:
110°

Answer:
100°

Answer:
90°

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 24 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 24

Complete the following table.

Answer:

The protractor

A protractor is used to measure an angle and also to draw an angle according to a given measure.

The picture opposite shows a protractor.

A protractor is semi-circular in shape.

The semi-circular edge of a protractor is divided into 180 equal parts. Each part is ‘one degree’. ‘One degree’ is written as ‘1°’.

The divisions on a protractor, i.e., the degrees can be marked in two ways. The divisions 0, 10, 20, 30,…180 are marked anticlockwise or from right to left; the divisions 0, 10, 20, 30,…180 are also marked clockwise, or serially from left to right.

The centre of the circle of which the protractor is a half part, is called the centre of the protractor. A diameter of that circle is the baseline or line of reference of the protractor.

Measuring angles

Observe how to measure ∠ABC given alongside, using a protractor.

1. First, put the centre of the protractor on the vertex B of the angle. Place the baseline of the protractor exactly on arm BC. The arms of the angle do not reach the divisions on the protractor.

2. At such times, set the protractor aside and extend the arms of the angle. Extending the arms of the angle does not change the measure of the angle.

3. The angle is measured starting from the zero on that side of the vertex on which the arm of the angle lies. Here, the arm BC is on the right of the vertex B. Therefore, count the divisions starting from the 0 on the right. See which mark falls on arm BA. Read the number on that mark. This number is the measure of the angle.

The measure of∠ABC is 40°.

We can measure the same ∠ABC by positioning the protractor differently.

1. First put the centre point of the protractor on vertex B of the angle. Align the baseline of the protractor with arm BA.
2. Find the 0 mark on the side of BA. Count the marks starting from the 0 on the side of point A. See which mark falls on arm BC. Read the number at that point.

Observe that here, too, the measure of ∠ABC is 40°.

See how the angles given below have been measured with the help of a protractor.

Chapter 6 Angles Problem Set 24 Additional Important Questions and Answers

Question 1.
Observing the adjacent figure, answer the following questions. Write the name of (1) all angles (2) Vertex of the angles (3) aims of the angles.
Solution:
(1) ZXOY, ZYOZ, ¿XOZ
(2) Vertex of the angles is ‘O’
(3) arms of Z)OY are OX and UY
arms of ZYOZ are UY and OZ
arms of ZXOZ are OX and OZ

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 21 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 21

Question 1.
Subtract the following.

\(\text { (1) } \frac{5}{7}-\frac{1}{7}\)
Answer:

\(\text { (2) } \frac{5}{8}-\frac{3}{8}\)
Answer:

\(\text { (3) } \frac{7}{9}-\frac{2}{9}\)
Answer:

\(\text { (4) } \frac{8}{11}-\frac{5}{11}\)
Answer:

\(\text { (5) } \frac{9}{13}-\frac{4}{13}\)
Answer:

\(\text { (6) } \frac{7}{10}-\frac{3}{10}\)
Answer:

\(\text { (7) } \frac{9}{12}-\frac{2}{12}\)
Answer:

\(\text { (8) } \frac{10}{15}-\frac{3}{15}\)
Answer:

Question 2.
\(\frac{7}{10}\) of a wall is to be painted. Ramu has painted 410 of it. How much more needs to be painted?
Solution:
To be painted – painted

Answer:
\(\frac{3}{10}\) more needs to be painted.

Addition and subtraction of unlike fractions

Example (1) Add : \(\frac{2}{3}+\frac{1}{6}\)

First let us show the fraction \(\frac{2}{3}\) by coloring two of the three equal parts on a strip.

You have learnt to add and to subtract fractions with common denominators. Here, we have to add the fraction \(\frac{1}{6}\) to the fraction \(\frac{2}{3}\).

So let us divide each part on this strip into two equal parts. \(\frac{4}{6}\) is a fraction equivalent to \(\frac{2}{3}\). Now, as 16 is to be added to \(\frac{2}{3}\) i.e. to \(\frac{4}{6}\), we shall colour one more of the six parts on the strip. Now, the total coloured part is \(\frac{5}{6}\).

Therefore,
That is,

Example (2) Add : \(\frac{1}{2}+\frac{2}{5}\)
Here, the smallest common multiple of the two denominators is 10. So, we shall change the denominator of both fractions to 10.

Example (3) Add : \(\frac{3}{8}+\frac{1}{16}\)
Here, 16 is twice 8. So, we shall change the denominator of both fractions to 16.

Example (4) Subtract : \(\frac{3}{4}-\frac{5}{8}\)
Let us make 8 the common denominator of the given fractions.

Example (5) Subtract : \(\frac{4}{5}-\frac{2}{3}\)
The smallest common multiple of the denominators is 15. So, we shall change the denominator of both fractions to 15.

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

\(\text { (1) } \frac{9}{14}-\frac{3}{14}\)
Answer:

\(\text { (2) } \frac{5}{6}-\frac{3}{6}\)
Answer:

\(\text { (3) } \frac{9}{16}-\frac{5}{16}\)
Answer:

\(\text { (4) } \frac{7}{8}-\frac{3}{8}-\frac{1}{8}\)
Answer:

(5) \(\frac{7}{9}\) part of the work done by Neha and Supriya together. \(\frac{5}{9}\) part of this work was done by Neha. How much work done by Supriya?
Solution:
Total work done – work done by Neha = work done by Supriya

Answer:
\(\frac{2}{9}\) work done by Supriya

(6) Mr. Sharma is \(\frac{14}{9}\) m tall. Mrs. Sharma is \(\frac{4}{9}\) shorter than him. What is Mrs. Sharma’s height?
Solution:
Mrs. Sharma’s height

Answer:
Mrs. Sharma’s height = \(\frac{10}{9}\)

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 15

Question 1.
Solve the following and write the quotient and remainder.
(1) 1284 ÷ 32
Solution :

Quotient = 40
Remainder = 4

(2) 5586 ÷ 87
Solution :

Quotient = 64
Remainder =18

(3) 1207 ÷ 27
Solution:

Quotient = 44
Remainder =19

(4) 8543 ÷ 41
Solution :

Quotient = 208
Remainder =15

(5) 2304 ÷ 43
Solution:

Quotient = 53
Remainder = 25

(6) 56,741 ÷ 26
Solution:

Quotient =2182
Remainder = 9

Question 2.
How many hours will it take to travel 336 km at a speed of 48 km per hour?
Solution:
Time = Distance ÷ Speed

Answer:
It will take 7 hours.

Question 3.
Girija needed 35 cartons to pack 1400 books. There are an equal number of books in every carton. How many books did she pack into each carton?
Solution:
No. of cartons x No. of books in each carton = Total no. of books 35 x No. of books in each carton = 1400 No. of books in each carton = 1400 35

Answer:
She packs 40 books in each carton.

Question 4.
The contribution for a picnic was 65 rupees each. Altogether, 2925 rupees were collected. How many had paid for the picnic?
Solution:

Answer:
45 persons paid for the picnic.

Question 5.
Which number, on being multiplied by 56, gives a product of 9688?
Solution:

Answer:
173

Question 6.
If 48 sheets are required for making one notebook, how many notebooks at the most will 5880 sheets make and how many sheets will be left over?
Solution:

Answer:
122 notebooks can be made and 24 sheets left over.

Question 7.
What will the quotient be when the smallest five-digit number is divided by the smallest four-digit number?
Solution:
Smallest five-digit number is 10,000 and smallest four-digit number is 1,000.
So, 10000 ÷ 1000 = 10

Answer:
Quotient = 10

Mixed examples

A farmer brought 140 trays of chilli seedlings. Each tray had 24 seedlings. He planted all the seedlings in his field, putting 32 in a row. How many rows of chillies did he plant?

Let us find out the total number of seedlings when there were 24 seedlings in each of the 140 trays. We shall multiply 140 and 24.

Total number of seedlings 3,360.
To find out how many rows were planted with 32 seedlings in each row, we shall divide 3,360 by 32.
The quotient is 105.
Therefore, the number of rows is 105.
Carry out the multiplication of 105 × 32 and verify your answer.

Multiplication and Division Problem Set 15 Additional Important Questions and Answers

Solve the following and write the quotient and remainder.

(1) 9148 ÷ 37
Solution

Quotient = 247
Remainder = 9

(2) 1175 ÷ 15
Solution :

Quotient =78
Remainder = 5

Solve the following word problems:

(1) If 45 kg of sugar cost 1305 rupees, what is the rate of sugar per kg?
Solution:

Answer:
The rate per kg of sugar is 29 rupees.

(2) 17 people spent ₹ 83,475. How much did each person spend and what is the amount left?
Solution:

Answer:
Each person spent ₹ 4,910 and the amount left is ₹ 5