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Decimal Fractions Class 5 Problem Set 42 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 42 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Question 1.
Subtract the following :

(1) 25.74 – 13.42
Solution:

(2) 206.35 – 168.22
Solution:

(3) 63.4 – 31.8
Solution:

(4) 63.43 – 31.8
Solution:

(5) 63.4 – 31.83
Solution:

(6) 8.23 – 5.45
Solution:

(7) 18.23 – 9.45
Solution:

(8) 78.03 – 41.65
Solution:

Question 2.
Vrinda was 1.48 m tall. After a year, her height became 1.53 m. How many centimeters did her height increase in a year?
Answer:

∴ 5 cm height has increased in a year.

Something more

Decimals used for measurement

We need to measure distance, mass (weight) and volume every day. We use suitable units for these measurements. Kilometre, metre and centimeter for distance; litre, millilitre for volume and kilogram and gram for mass are the units that are used most of the time.

All these units are decimal units. In this method, gram, metre and litre are taken as the basic units for mass, distance and volume respectively. Units larger than these increase 10 times at every step and smaller units become \(\frac{1}{10}\) of the previous unit at each step.

Look at the table of these units given below.

The origin of the terms kilo, hecto… milli is in the Greek or Latin language. Their English equivalents are given in brackets along with the terms.

Decimal Fractions Problem Set 42 Additional Important Questions and Answers

Subtract the following:

(1) 304.17 – 95.28
Solution:

(2) 72.84 – 36.96
Solution:

(3) 9.17 – 5.88
Solution:

(4) 100 – 49.99
Solution:

(5) Atul has 56.25 and Anup has 65. Whose amount is more? How much?
Solution:

∴ Anup’s amount is more by ₹ 8.75

Maharashtra Board Class 5 Maths Solutions

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

Multiples and Factors Class 5 Problem Set 33 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 8 Multiples and Factors Problem Set 33 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 8 Multiples and Factors

Question 1.
(1) Write five three-digit numbers that are multiples of 2.
Answer:
100, 102, 104, 106, 108.

(2) Write five three-digit numbers that are multiples of 5.
Answer:
100, 105, 110, 115, 120.

(3) Write five three-digit numbers that are multiples of 10.
Answer:
100, 110, 120, 130, 140.

Question 2.
Write 5 numbers that are multiples of 2 as well as of 3.
Answer:
2 as well as of 3 means 2 and 3 that is multiples of 6.
They are 6, 12, 18, 24, 30.

Question 3.
A ribbon is 3 metres long. Can we cut it into 50 cm pieces and have nothing left over? Write the reason why or why not.
Answer:
3 metres = 300 cm.
We can cut it into 50 cm pieces.
Since 300 is exactly divisible 50.
That is 300 is multiples of 50.
300 ÷ 50 = 6
We will get 6 pieces, nothing is left over.

Question 4.
A ribbon is 3 metres long. I need 8 pieces of ribbon each 40 cm long. How many centimetres shorter is the ribbon than the length I need?
Answer:
1 piece of 40 cm, so for 8 pieces ribbon needed is 40 x 8 = 320 cm.
But ribbon is 3 metre = 300 cm long.
So ribbon is shorter by 320 – 300 = 20 cm.

Question 5.
If the number given in the table is divisible by the given divisor, put ✓ in the box. If it is not divisible by the divisor, put ✗ in the box.

Answer:

Prime and composite numbers

Some numbers are given in the tables below. Write all of their factors.

Dada : What do you notice on studying the table?

Ajay : The number 1 is a factor of every number. Some numbers have only 1 and the number itself as factors. For example, the only factors of 3 are 1 and 3. Similarly, the factors of 2 are only 1 and 2 and the factors of 19 are only 1 and 19. Some numbers have more than two factors.

Dada : Numbers like 2, 3, 19 which have only two factors are called prime numbers.

A number which has only two factors, 1 and the number itself, is called a prime number.

Ajay : What do we call numbers like 4, 6 and 16 which have more than two factors?

Dada : Numbers like 4, 6 and 16 are called composite numbers.

A number which has more than two factors is called a composite number.

Dada : Think carefully and tell me whether 1 is a prime or composite number.

Ajay : The number 1 has only one factor, 1 itself, so I can’t answer your question.

Dada : You’re right. 1 is considered neither a prime number nor a composite number.

1 is a number which is neither prime nor composite.

Multiples and Factors Problem Set 33 Additional Important Questions and Answers

Question 1.
Write five three-digit numbers that are multiples of 3.
Answer:
102, 105, 108, 111, 114.

Question 2.
Write five two-digit numbers that are multiples of 7.
Answer:
14, 21, 28, 35, 42

Question 3.
Write five three-digit numbers that are multiples of 4.
Answer:
112, 116, 120, 124, 128.

Question 4.
Write 5 numbers that are multiples of 3 as well as 5.
Answer:
3 as well as 5 means 3 and 5. i.e. multiples of 15.
They are 15, 30, 45, 60, 75.

Question 5.
A string is 4 metres long. Can we cut it into 50 cm pieces and have nothing left over?
Answer:
4 metres = 400 cm.
We can cut it into 50 cm pieces.
Since 400 is exactly divisible by 50.
That is 400 is multiple of 50 400 + 50 = 8
We will get 8 pieces. Nothing is left over.

Question 6.
A paper Is 2 metres long. I need 8 pieces of paper each 30 cm long. How many centimetres shorter is the paper than the length I need?
Answer:
A piece of 30 cm, so for 8 pieces paper needed is 30 x 8 = 240 cm.
But paper is 2 metre = 2 x 100 = 200 cm long.
So paper is shorter by 240 – 200 = 40 cm

Question 7.
If the number given in the table is divisible by the given divisor, put P in the box. If it is not divisible by the divisor, put in the box.
Answer:

Maharashtra Board Class 5 Maths Solutions

• Multiples and Factors Problem Set 32 Class 5 Maths Solutions
• Multiples and Factors Problem Set 33 Class 5 Maths Solutions
• Multiples and Factors Problem Set 34 Class 5 Maths Solutions
• Multiples and Factors Problem Set 35 Class 5 Maths Solutions

Circles Class 5 Problem Set 29 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 29 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 7 Circles

Question 1.
If the radius of a circle is 5 cm, what will its diameter be?
Solution :
Diameter
= 2 x radius
= 2 x 5 = 10 cm

Question 2.
If the diameter of a circle is 6 cm, what will its radius be?
Solution :
Radius
= diameter ÷ 2
= 6 ÷ 2
= 3 cm

Question 3.
Complete the following table by filling in the blanks.

Answer:

The interior and the exterior of a circle

We play ‘Land and Sea’ inside a circle on the playground. In this game, the children inside the circle are in the ‘sea’, while the children outside the circle are on ‘land’.

In the picture alongside, K, L, M and N are points on a circle with centre T.

The coloured area inside the circle in the picture is the interior of the circle. P, Q, R and T are points in the interior of the circle.

A, B, C and D are points in the exterior of the circle.

Circles Problem Set 29 Additional Important Questions and Answers

Question 1.
If the radius of a circle is 3.5 cm, what will its diameter be?
Solution :
Diameter
= 2 x radius
= 2 x 3.5
= 7 cm

Question 2.
If the diameter of a circle is 5 cm, what will its radius be?
Solution :
Radius
= Diameter ÷ 2
= 5 ÷ 2
= 2.5 cm

Maharashtra Board Class 5 Maths Solutions

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

Preparation for Algebra Class 5 Problem Set 54 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 54 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 16 Preparation for Algebra

Question 1.
Using brackets, write three pairs of numbers whose sum is 13. Use them to write three equalities.
Answer:
(7 + 6), (8 + 5), (9 + 4). since 7 + 6
= 13,8 + 5
= 13, 9 + 4
= 13.

(7 + 6)
= (8 + 5), (7 + 6)
= (9 + 4) or (8 + 5)
= (9 + 4).

Question 2.
Find four pairs of numbers, one for each of addition, subtraction, multiplication and division that make the number 18. Write the equalities for each of them.
Answer:
(9 + 9), (20 – 2), (9 x 2), (36 ÷ 2).
since 9 + 9
= 18, 20 – 2
= 18, 9 x 2
= 18 and 36 + 2
= 18, so (9 + 9)
= (20 – 2)
= (9 x 2)
= (36 ÷ 2).

Inequality
The values of 7 + 5 and 7 × 5 are 12 and 35 respectively. It means that they are not equal. To represent ‘not equal’, the symbol ‘≠’ is used.

To show that (7 + 5) and (7 × 5) are not equal, we write (7 + 5) ≠ (7 × 5) in short.

This kind of representation is called an ‘inequality’.

(9 – 5) ≠ (15 ÷ 3) means that the expressions (9 – 5) and (15 ÷ 3) are not equal.

If two expressions are not equal, one of them is greater or smaller than the other.

To show greater or lesser values, we use the symbols ‘<’ and ‘>’. Therefore, these symbols can also be used to show inequalities.

The value of (9 – 5) is 4 and the value of (15 ÷ 3) is 5. 4 < 5, so the relation between (9 – 5) and (15 ÷ 3) can be shown as (9 – 5) < (15 ÷ 3) or (15 ÷ 3) > (9 – 5).

Fill in the boxes between the expressions with <, = or > as required.

(1) (9 + 8) [ ] (30 ÷ 2)
9 + 8 = 17,
30 ÷ 2 = 15
17 > 15
Therefore (9 + 8) [ > ] (30 ÷ 2)

(2) (16 × 3) (4 × 12)
16 × 3 = 48,
4 × 12 = 48,
48 = 48
Therefore (16 × 3) [ = ] (4 × 12)

(3) (16 – 5) [ ] (2 × 7)
16 – 5 = 11,
2 × 7 = 14,
11 < 14
Therefore (16 – 5) [ < ] (2 × 7)

Write a number in the box that will make this statement correct.
(1) (7 × 2) = ( [ ] – 6)

The value of the expression 7 × 2 is 14, so the number in the box has to be one that gives 14 when 6 is subtracted from it. Subtracting 6 from 20 gives us 14.

Therefore (7 × 2) = ( [ 20 ] – 6 )
(2) (24 ÷ 3) < (5 + [ ] )
The value of the expression 24 ÷ 3 is 8, so the number in the box has to be such that when it is added to 5, the sum is greater than 8.

Now, 5 + 1 = 6, 5 + 2 = 7, 5 + 3 = 8. So the number in the box has to be greater than 3.

Therefore, writing any number like 4, 5, 6 … onwards will do. It means that this problem has several answers. (24 ÷ 3) < (5 + [ 4 ] ) is one among many answers. Even if that is true, writing only one answer will be enough to complete this statement.

Preparation for Algebra Problem Set 54 Additional Important Questions and Answers

Question 1.
Fill in the blanks.
(1) 7 + 3 = …………….. – ……………..
(2) 7 + 3 = …………….. x ……………..
(3) 7 + 3 = …………….. + ……………..
Answer:
(1) 7 + 3 = 10 and 12 – 2 = 10 or 15 – 5 = 10
(2) 7 + 3 = 10 and 10 x 1 = 10 or 5 x 2 = 10
(3) 7 + 3 = 10 and 20 + 2 = 10 or 30 + 3 = 10

Question 2.
Write the proper number in the box.
(1) 7 + 8 = 10 + [ ]
(2) 7 + 8 = 20 – [ ]
(3) 7 + 8 = 30 + [ ]
(4) 7 + 8 = 5 x [ ]
Answer:
(1) 7 + 8 = 15 so, 10 + [ ] = 15.
∴ [ ] = 15 – 10 = 5

(2) 7 + 8 = 15 s0, 20 – [ ] = 15.
∴[ ] = 20 – 15 = 5

(3) 7 + 8 = 15 so, 30 + [ ] = 15.
∴ [ ] = 30 + 15 = 2

(4) 7 + 8 = 15 so, 5 x [ ] = 15.
∴[ ] = 15 + 5 = 3

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

Operations on Fractions Class 6 Maths Chapter 4 Practice Set 12 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 12 Answers Solutions.

Std 6 Maths Practice Set 12 Solutions Answers

Question 1.
Multiply:
i. \(\frac{7}{5} \times \frac{1}{4}\)
ii. \(\frac{6}{7} \times \frac{2}{5}\)
iii. \(\frac{5}{9} \times \frac{4}{9}\)
iv. \(\frac{4}{11} \times \frac{2}{7}\)
v. \(\frac{1}{5} \times \frac{7}{2}\)
vi. \(\frac{9}{7} \times \frac{7}{8}\)
vii. \(\frac{5}{6} \times \frac{6}{5}\)
viii. \(\frac{6}{17} \times \frac{3}{2}\)
Solution:
i. \(\frac{7}{5} \times \frac{1}{4}\)

ii. \(\frac{6}{7} \times \frac{2}{5}\)

iii. \(\frac{5}{9} \times \frac{4}{9}\)

iv. \(\frac{4}{11} \times \frac{2}{7}\)

v. \(\frac{1}{5} \times \frac{7}{2}\)

vi. \(\frac{9}{7} \times \frac{7}{8}\)

vii. \(\frac{5}{6} \times \frac{6}{5}\)

viii. \(\frac{6}{17} \times \frac{3}{2}\)

Question 2.
Ashokrao planted bananas on \(\frac { 2 }{ 7 }\) of his field of 21 acres. What is the area of the banana plantation?
Solution:
Area of banana plantation is \(\frac { 2 }{ 7 }\) of 21
∴ Area of banana plantation

∴ Area of banana plantation is 6 acres

Question 3.
Of the total number of soldiers in our army, \(\frac { 4 }{ 9 }\) are posted on the northern border and one-third of them on the north-eastern border. If the number of soldiers in the north is 5,40,000, how many are posted in the north-east?
Solution:
Number of soldiers posted on northern border = 5,40,000
Since, number of soldiers in north-east = one third of the soldiers on northern border
∴ Number of soldiers in the north-east

∴ The number of soldiers in the north-east is 1,80,000.

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

Pictographs Class 5 Problem Set 52 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 14 Pictographs Problem Set 52 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 14 Pictographs

Question 1.
Stocks of various types of grains stored in a warehouse are as given below. Make a pictograph based on the information given.

Answer:
Scale :1 picture = 8 sacks

Question 2.
Information about the various types of vehicles in Wadgaon is given below. Make a pictograph for this data.

Answer:

Question 3.
The numbers of the various books kept in a cupboard in the school library are given below. Make a pictograph showing the information given.

Answer:

Activity
Collect information based on the points given below and make a pictograph for each.

• Which crops are grown on the farms owned by students in your class? (Vegetables, grains, pulses, fruits, etc.)
• Which storybooks do your classmates like? (fairytales, stories about kings and queens, historical stories, stories about saints, picture stories, etc.)
• What do your classmates want to be when they grow up ? (doctor, teacher, farmer, engineer, officer, etc.)

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Solve the following

Question 1.
Information regarding the number of pages of novel book read in different days by Rosi are as follows. Make a pictograph showing the information given.

Question 2.
Different types of currency notes had with Shamin are as follows. Make a pictograph showing the information given.

Question 3.
Different types of colour of scooters sold by a merchant are as follows. Make a pictograph showing the data given.

Question 4.
Ajhount of sales of goods in rupees for the first four days of a week are as follows. Make a pictograph from the information given below.

Answer:
(1) All the given numbers can be divided by 2, 5, and 10. 1 picture of 10 pages will be convenient scale so 6 pictures for 60 pages. 4 pictures for 40, 3 for 30 and 2 for 20.
(2) All the given numbers can divided by 2 only, so 1 picture for 2 notes will be the scale. So, 4 pictures for notes of ? 500, 5 for ? 100 notes, 3 for ? 50 and 2 for ? 10.
(3) All given numbers are divisible by 3, so 1 picture for 3 scooters will be the scale. So, 2 pictures for white, 3 for Red, 4 for Black and 1 pictures for Yellow.
(4) All the given numbers can be divided by 2, 5,10, 25 and 50. So, 1 picture for 50 rupees will be convenient scale. So, draw 3 pictures for Monday, 4 for Tuesday, 5 for Wednesday and 2 for Thursday.
(5) Number of books are multiples of 50. Therefore Take number of pictures = 5,4,2, 1, 3 respectively.

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

HCF-LCM Class 6 Maths Chapter 9 Practice Set 25 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 25 Answers Solutions.

Std 6 Maths Practice Set 25 Solutions Answers

Question 1.
Find out the LCM of the following numbers.
i. 9,15
ii. 2,3,5
iii. 12,28
iv. 15,20
v. 8,11
Solution:
i. Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Multiples of 15 = 15, 30, 45
∴ LCM of 9 and 15 = 45

ii. Multiples of 2 = 2, 4,6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 5 = 5, 10, 15, 20, 25, 30
∴ LCM of 2,3 and 5 = 30

iii. Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Multiples of 28 = 28, 56, 84
∴ LCM of 12 and 28 = 84

iv. Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120
Multiples of 20 = 20, 40, 60
∴ LCM of 15 and 20 = 60

v. Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88
∴ LCM of 8 and 11 = 88

Question 2.
Solve the following problems:
i. On the playground, if the children are made to stand for drill either 20 to a row or 25 to a row, all rows are complete and no child is left out. What is the lowest possible number of children in that school?

ii. Veena has some beads. She wants to make necklaces with an equal number of beads in each. If She makes necklaces of 16 or 24 or 40 beads, there is no bead left over. What is the least number of beads with her?

iii. An equal number of laddoos have been placed in 3 different boxes. The laddoos in the first box were distributed among 20 children equally, the laddoos in the second box among 24 children and those in the third box among 12 children. Not a single laddoo was left over. What was the minimum number of laddoos in the three boxes altogether?

iv. We observed the traffic lights at three different squares on the same big road. They turn green every 60 seconds, 120 seconds and 24 seconds. When the signals are switched on at 8 o’clock in the morning, all the lights were green. How long after that will all three signals turn green simultaneously again?

v. Given the fractions \(\frac { 13 }{ 45 }\) and \(\frac { 22 }{ 75 }\). Write their equivalent fractions with same denominators and add the fractions.
Solution:
i. The lowest possible number of children is equal to the lowest common multiple of 20 and 25.
Multiples of 20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
Multiples of 25 = 25, 50, 75, 100
∴ LCM of 20 and 25 = 100
∴ The least number of students in the school is 100.

ii. The least number of beads with Veena is equal to the lowest common multiple of 16,24 and 40.
Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240, 256, 272, 288
Multiples of 24 = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240
Multiples of 40 = 40, 80, 120, 160, 200, 240
∴ LCM of 16, 24 and 40 = 240
∴ The least number of beads with Veena are 240.

iii. The lowest common multiple of 20,24 and 12 gives the minimum number of laddoos in one box.
Multiples of 20 = 20, 40, 60, 80, 100,120, 140, 160, 180, 200
Multiples of 24 = 24, 48, 72, 96, 120
Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
∴ LCM of 20, 24 and 12 = 120
∴ Minimum number of ladoos in 1 boxes =120
∴ Minimum number of ladoos in 3 boxes = 3 x 120 = 360
∴ The minimum number of ladoos in 3 boxes are 360.

iv. All three signals will turn green for lowest common multiple of 60 seconds, 120 seconds and 24 seconds.
Multiples of 60 = 60, 120, 180, 240, 300, 360, 420, 480
Multiples of 120 = 120, 240, 360
Multiples of 24 = 24, 48, 72, 96, 120
LCM of 60, 120 and 24 = 120
Since, 60 seconds = 1 minute
∴ 120 seconds = 2 minutes
∴ The signals will turn green simultaneously again after 120 seconds i.e. 2 minutes.

v. The lowest common multiple of 45 and 75 gives the same denominator.
Multiples of 45 = 45, 90, 135, 180, 225, 270, 315, 360, 405, 450
Multiples of 75 = 75, 150, 336
∴ LCM of 45 and 75 = 225

Maharashtra Board Class 6 Maths Chapter 9 HCF-LCM Practice Set 25 Intext Questions and Activities

Question 1.
Pravin, Bageshri and Yash are cousins who live in the same house. Pravin is an Army Officer. Bageshri is studying in a Medical College in another city. Yash lives in a nearby town in a hostel. Pravin can come home every 120 days.
Bageshri comes home every 45 days and Yash, every 30 days. All three of them left home at the same time on the 15th of June 2016. Their parents said, “We shall celebrate like a festival the day you all come home together.” Mother asked Yash, “What day will that be?”
Yash said, “The number of days after which we come back together must be divisible by 30, 120. That means we shall be back together on the 10th of June next year. That will certainly be a for us!”
How did Yash find the answer? (Textbook pg. no. 49)

Solution:
The day when Pravin, Bageshri and Yash come back together is lowest common multiple of 30, 45 and 120.
Multiples of 30: 30, 60, 90, 120, 150, 180,210, 240, 270, 300, 330, 360
Multiples of 45: 45, 90, 135, 180, 225, 270, 315, 360
Multiples of 120: 120, 240, 360
∴ They will come together after 360 days
Day when they left home = 15th June
∴ Day when they come back together = 15th June + 360 days
= 10th June next year
∴ Pravin, Bageshri and Yash will come back together on 10th June next year.

Question 2.
A Maths Riddle! (Textbook pg. no. 50)
We have four papers. On each of them there is a number on one side and some information on the other. The numbers on the papers are 7, 2, 15, 5. The information on the papers is given below in random order.
i. A number divisible by 7
ii. A prime number
iii. An odd number
iv. A number greater than 100
If the number on every paper is mismatched with the information on its other side, what is the number on the paper which says ‘A number greater than 100?
Solution:

6th Std Maths Digest Pdf Download

• Bar Graphs Practice Set 18 Class 6 Maths Solution
• Bar Graphs Practice Set 19 Class 6 Maths Solution
• Symmetry Practice Set 20 Class 6 Maths Solution
• Symmetry Practice Set 21 Class 6 Maths Solution
• Divisibility Practice Set 22 Class 6 Maths Solution
• HCF-LCM Practice Set 23 Class 6 Maths Solution
• HCF-LCM Practice Set 24 Class 6 Maths Solution
• HCF-LCM Practice Set 25 Class 6 Maths Solution

Perimeter and Area Class 5 Problem Set 50 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 12 Perimeter and Area

Question 1.
The length of the side of each square is given below. Find its area.

(1) 12 metres
Solution:
Area of a square
= side x side
= 12 x 12

(2) 6 cm
Solution:
Area of a square
= side x side
= 6 x 6
= 36 sq.cm.

(3) 25 metres
Solution:
Area of a square
= side x side
= 25 x 25
= 625 sq.m.

(4) 18 cm
Solution:
Area of a square
= side x side
= 18 x 18
= 324 sq.cm.

Question 2.
If the cost of 1 sq m of a plot of land is 900 rupees, find the total cost of a plot of land that is 25 m long and 20 m broad.
Solution:
Area of the rectangular plot
= length x breadth
= 25 x 20
= 500 sq.m.

Cost of the plot of land
= Area of the plot x rate
= 500 x 900
= 4,50,000 rupees

Question 3.
The side of a square is 4 cm. The length of a rectangle is 8 cm and its width is 2 cm. Find the perimeter and area of both figures.
Solution:
Perimeter of a square = 4 x side
= 4 x 4
= 16 cm

Area of a square = side x side
= 4 x 4
= 16 sq.cm.

Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 8 + 2 x 2
= 16 + 4
= 20 cm

Area of a rectangle
= length x breadth
= 8 x 2
= 16 sq.cm.

Question 4.
What will be the labour cost of laying the floor of an assembly hall that is 16 m long and 12 m wide if the cost of laying 1 sq m is 80 rupees?
Solution:
Area of rectangular floor
= length x breadth
= 16 x 12
= 192 sq.cm.

The cost of laying 1 sq.m, is 80 rupees.
Hence, the cost of laying 192 sq.m.
= 192 x 80
= 15,360 rupees.
∴ ₹ 15,360

Question 5.
The picture alongside shows some squares. Find out how many squares with the same measures will fit in the empty space in the figure.

Solution:
length of the empty space = 4 – 1 = 3 cm
breadth of the empty space = 3 – 1 = 2 cm
square in empty space
= length x breadth
= 3 x 2 = 6 sq.cm.

∴ 6 squares will fit in the empty space in the figure

Question 6.
Divide the figure given alongside into four parts in such a way that the area and shape of each part is the same. Colour the parts with different colours.

Fair and square

As shown in the figure alongside, a square plot of land owned by the government contains four houses and a well right in the centre. The government has to divide the houses and the land between four poor persons according to the following conditions.
(1) Each person must get only one house.
(2) The shape and area of the land must be the same.
(3) Each person must be able to use the well without trespassing on any one else’s land.

Show the appropriate divisions in four different colours.

Activity
Using a graph paper, find out the area of different rectangles and squares.

Perimeter and Area Problem Set 50 Additional Important Questions and Answers

Question 1.
15 cm
Solution:
Area of a square
= side x side
= 15 x 15
= 225 sq.cm.

Question 2.
21 cm
Solution:
Area of a square = side x side
= 21 x 21
= 441 sq.cm.

Solve the following:

Question 1.
The side of a square hall is of length 8 m. If it is tiled with a tile of length 4 m and breadth 2 m., how many tiles will be required?
Solution:
Area of the square hall
= side x side
= 8 x 8
= 64 sq.m.

Area of 1 rectangular tile
= length x breadth
= 4 x 2

∴ 8 sq.m.

For 8 sq.m., 1 tile is required,
but for 64 sq.m = \(\frac{64}{8}\) = 8 tiles required

∴ 8 tiles required

Question 2.
Perimeter of a square is 16 cm. What is the length of each side? What is the area of the square?
Solution:
Perimeter of square = 4 x side
16 = 4 x side

∴ side of a square = \(\frac{16}{4}\) = 4 cm
Area of the square
= side x side
= 4 x 4
= 16 sq.cm

∴ Side of square is 4 cm and area of the sqaure is 16 sq.cm

Question 3.
Write the perimeter of each figure in the box given below it.

Answer:
24 cm

Answer:
18 cm

Answer:
21 cm

Answer:
20 cm

Question 4.
Two squares of side 2 cm is cut out of two corners of a larger square with side 5 cm (see the figure). What will be the perimeter of the remaining shape?

Answer:
20 cm

Question 5.
Match the columns ‘A’ and ‘B’

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

Question 6.
Solve the following word problems:
(1) What is the perimeter of a rectangle having length 9 cm and its breadth 6 cm?
(2) The sides of a rectangular field are having length 150 m and breadth 100 m. Find the perimeter of field.
(3) If each side of a square is 8 cm then what is the perimeter of the square?
(4) A rectangular garden of length 650 m and breadth 350 m. Mohan makes four rounds daily. How many kilometres does he walk everyday?
(5) One square field is having one side of it is of 225 m, Soham makes 6 rounds of the square field daily. How much distance is covered by him? Write it in km and m.
(6) A rectangular field whose length is 58 m and breadth is 32 m. Fencing the field by 4 rounds with a wire, what length of wire is required? If the cost of 1 m wire is ? 75, then what is the expenditure of fencing the field?
(7) A length of a rectangular classroom is 8 m and its breadth is 5 m. A wooden strip is to be fitted along the four walls to hang charts and pictures. What is the length of the wooden strip required?
(8) The side of a square table is 1.5 m. To fit a strip of tin sheet around the table, how many metres of strip is required?
(9) What is the perimeter of the triangle whose sides are 13.8 cm, 17.6 cm and 10.6 cm?
(10) The sides of some squares are given below. Find their areas.
(i) 11 cm (ii) 23 cm (iii) 9 cm
(11) Find the perimeter and area of the following:
(i) square of side 6 cm
(ii) Rectangle: length 12 cm and breadth 6 cm
(12) A rectangular land is having its length 25 m and breadth 16 m. If the cost of 1 sq.m, land is ? 1,500, then what will be cost of land?
(13) The side of a square plot is 10 metre. It is tiled at the rate of 50 rupees per sq.m. What will be cost of tiling the floor?
(14) A wall of length 25 m and breadth 12 m. It is painted at the rate of 60 rupees per sq.m. What will be cost of painting the wall?
Answer:
(1) 30 cm
(2) 500 m
(3) 32 cm
(4) 8 km
(5) 5 km 400 m
(6) ₹ 54,000
(7) 26 m
(8) 6 m
(9) 42 cm

(10) (i) 121 sq. cm
(ii) 529 sq.cm
(iii) 81sq.cm.

(11) (i) Perimeter = 24 cm, Area = 36 sq.cm,
(ii) Perimeter = 36 cm, Area = 72 sq. cm

(12) 6,00,000 rupees
(13) 5,000 rupees
(14) t 18,000

Question 7.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

Perimeter of Rectangle
(1) XYZW = [ ] cm
(2) CDEF = [ ] cm
(3) JKLM = [ ] cm
(4) NOPQ = [ ] cm
Answer:
(1) 12 cm
(2) 10 cm
(3) 10 cm
(4) 8 cm

Question 8.
Find the area of the. following figures. (All small squares are having side 1 cm)

Answer:
5 sq.cm

Answer:
4 sq.cm

Answer:
9 sq.cm

Answer:
16 sq.cm.

Question 9.
The pictures below shows some squares. Find out how many squares with the same measures will fit in the empty space in the figures.

Answer:
(1) 9
(2) 9

Question 10.
Fill in the blanks:

Answer:
(1) Area = 24 sq.cm., Perimeter = 22 cm
(2) Breadth = 4 cm, Perimeter = 20 cm
(3) Length = 2 cm, Perimeter = 8 cm

Maharashtra Board Class 5 Maths Solutions

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

Measuring Time Class 5 Problem Set 43 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 10 Measuring Time Problem Set 43 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 10 Measuring Time

Question 1.
Write the time shown in each clock in the box given below it.
(1)
Answer:
25 minutes past 2

(2)
Answer:
50 minutes past 7

(3)
Answer:
5 minutes past 8

(4)
Answer:
40 minutes past 4

Question 2.
Draw the hands of the clock to show the time given in the box.
(1)
Four-thirty
Answer:

(2)
Quarter past nine
Answer:

(3)
Quarter to five
Answer:

(4)
20 minutes past 11
Answer:

Question 3.
If a bus that leaves Nashik at 5 o’clock in the morning reaches Pune that same day at ten-thirty in the morning, how long does the journey take?
Solution:

∴ Bus took 5 hrs 30 min

Question 4.
A play that was to start at nine fifteen at night was delayed by half an hour because of a power outage. What time did the play start?
Solution:

∴ Play started 9:45 at night

Question 5.
If a train leaves Mumbai at ten-fifteen at night and reaches Nagpur at one forty the next afternoon, how long does the journey take?
Solution:
Ten fifteen at night to 12 mid night is

12 mid night to next 1:40 afternoon = 13 hours 40 minutes
Total times

∴ Total time of the journey is 15 hrs. 25 min:

Learning about seconds

This clock is showing 5 minutes past 3. We know this because of the position of the hour and minute hands. There is another hand in the picture called the second hand. This hand moves swiftly. The second is a very small unit used to measure time less than a minute.

The face of a clock is a circle divided into 60 equal parts. When the second-hand moves one part, it takes one second. When it completes one round of the clock face, it moves across all 60 parts. This takes 60 seconds. In the same time, the minute hand moves one place, which means that one minute is over.

It means that, 1 minute is equal to 60 seconds.

1 minute = 60 seconds

The clock in the picture above shows 5 minutes and 50 seconds past 3.

20 minutes and 10 seconds past 7
15 minutes and 40 seconds past 10

Seconds are used on various occasions such as measuring temperature with a thermometer, measuring heartbeats or timing a race.

Ante meridiem and post meridiem

Shripati was sitting at home at night, tired. There were guests at home. They asked, “You must have worked very hard in the fields today. How long were you working?”

Shripati said, “I was in the field from six o’clock to eight o’clock.” Someone asked, “You’re this tired even though you were in the field for only two hours?”

Shripati said, “No, no, I was in the field from 6 in the morning till 8 at night! Now tell me how many hours I spent in the field.”

The guests had not understood what Shripati said at first. To avoid such mistakes, it has been internationally agreed that as the clock strikes 12 midnight, one day ends and the next day begins. From that moment on, the clock shows the time for the next day. When one hour passes after 12 midnight, it is 1’o’clock. After that, it is 2, 3, 4, …, 12 o’clock in serial order. After 12 noon, again it is 1, 2, 3, …, 12 o’clock in serial order. The time before 12 noon is stated as ante meridiem or am. The time after 12 noon is stated as post meridiem or pm.

This method of measuring time is called the 12 hour clock.

Shripati was in the field from 6 am to 8 pm or for 14 hours.

The 24 hour clock
The 24 hour clock is used to avoid this division of the day into ante meridiem and post meridiem. This method is used in timetables for trains, planes, buses and long distance boat journeys. In this method, instead of starting again from 1, 2, 3 after 12 noon, we continue with 13, 14, 15,…,24. In a 24 hour digital watch, time is shown only in the form of numbers. It does not have hands. In such a clock, 20 minutes past 6 in the morning is shown as ‘6:20’ and 20 minutes past 6 in the evening is shown as ‘18:20’.

23:59 means 59 minutes after 23 and one minute later, 24 hours will be complete. The digital clock will show this time as 00:00 at midnight and the day will change. At that time, a 12 hour clock shows 12 midnight.

Study the following table to see how different times of the day are shown in the 12 hour and 24 hour clocks.

The timetables of some trains going from Badnera to Nagpur are given below. Observe the use of the 24 hour clock in the timetable.

Measuring Time Problem Set 43 Additional Important Questions and Answers

Write the time shown in each clock in the box given below it.
(1)
Answer:
15 minutes past 6

(2)
Answer:
30 minutes past 9

Draw the hands of the clock to sho the time given in the box.
(1) 5 minutes to four
Answer:

(2) 35 minutes past to 2
Answer:

(3) Sujata left home 6:30 and returned at 11. How much time did she spend away from home?
Solution:

∴ 4 hrs. 30min. spent away from home.

(4) A speech that started at 4:20 in the afternoon ended at 5:45. How long was the speech?
Solution:

∴ Speech was for 1 hr. 25 min.

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

Operations on Fractions Class 6 Maths Chapter 4 Practice Set 11 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 11 Answers Solutions.

Std 6 Maths Practice Set 11 Solutions Answers

Question 11.
What fractions do the points A and B show on the number lines below?

Solution:
(1) Each unit is divided in 6 parts
A is 5th division from 0
∴ \(A=\frac { 5 }{ 6 }\)

B is 10th division from 0
∴ \(B=\frac { 10 }{ 6 }\)

(2) Each unit is divided in 5 parts
A is 3rd division from 0
∴ \(A=\frac { 3 }{ 5 }\)

B is 7th division from 0
∴ \(B=\frac { 7 }{ 5 }\)

(3) Each unit is divided in 7 parts
A is 10th division from 0
∴ \(A=\frac { 10 }{ 7 }\)

B is 3rd division from 0
∴ \(B=\frac { 3 }{ 7 }\)

Question 2.
Show the following fractions on the number line:
i. \(\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5}\)
ii. \(\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4}\)
Solution:
i. \(\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5}\)

ii. \(\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4}\)

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 11 Intext Questions and Activities

Question 1.
If we want to show the fractions \(\frac{3}{10}, \frac{9}{20}, \frac{19}{40}\) on the number line, how big should the unit be? (Textbook pg. no. 24)
Solution:
The denominators of the given fractions are not equal.
The numbers in the denominators 10, 20 and 40 have common multiple 40.
∴ Making the denominators equal, we get

∴ To represent these fractions on the numbers line, each main unit should be divided into 40 equal sub-units.

Therefore,
\(\frac{3}{10}=\frac{12}{40}\) is represented on 12th mark from 0.
\(\frac{9}{20}=\frac{18}{40}\) is represented on 18th mark from 0 and
\(\frac { 19 }{ 40 }\) is represented on 19 mark from 0.

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