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

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

Std 5 Maths Chapter 12 Perimeter and Area

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

∴ 22 cm wire will be needed to make a rectangle

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

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

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

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

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

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

∴ Sanju covers 5 km 760 m daily

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

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

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

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

∴ He runs altogether = 400 m

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

∴ Wire required for 4 rounds
= 800 m

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

∴ The perimeter of a triangle
= 55.6 cm

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

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

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

∴ 12 cm

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

∴ 10.2 cm

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

∴ 9.6 cm

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

∴ 10 cm

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

∴ 6 cm

Area : Revision

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

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

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

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

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

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

Formula for the area of a rectangle

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

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

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

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

Therefore, The area of a rectangle = length × breadth

Formula for the area of a square

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

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

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

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

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

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

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

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

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

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

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

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

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

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

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

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

∴ 1 km

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

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

∴ perimeter of both is equal

Class 5 Maths Solution Maharashtra Board

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

Decimal Fractions Class 5 Problem Set 39 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 9 Decimal Fractions

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

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

(2) ₹ 9.30
Answer:
9 rupees 30 paise.

(3) ₹ 2.30
Answer:
2 rupees 30 paise.

(4) ₹ 2.3
Answer:
2 rupees 30 paise.

Question 2.
Write how many rupees in decimal form.

(1) 6 rupees 25 paise
Answer:
₹ 6.25

(2) 15 rupees 70 paise
Answer:
₹ 15.70

(3) 8 rupees 5 paise
Answer:
₹ 8:05

(4) 22 rupees 4 paise
Answer:
₹ 22.04

(5) 720 paise
Answer:
₹ 7.20

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

(1) 58.75 m
Answer:
58 m 75 cm

(2) 9.30 m
Answer:
9 m 30 cm

(3) 0.30 m
Answer:
30 cm

(4) 0.3 m
Answer:
30 cm

(5) 1.62 m
Answer:
1 m 62 cm

(6) 91.4 m
Answer:
91 cm 40 cm

(7) 7.02 m
Answer:
7 m 2 cm

(8) 0.09 m
Answer:
9 cm

Question 4.
Write how many metres in decimal form.

(1) 1 m 50 cm
Answer:
1.5 m

(2) 50 m 40 cm
Answer:
50.40 m

(3) 50 m 4 cm
Answer:
50.04 m

(4) 734 cm
Answer:
7.34 m

(5) 10 cm
Answer:
0.1 m

(6) 2 cm
Answer:
0.02 m

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

(1) 6.9 cm
Answer:
6 cm 9 mm

(2) 20.4 cm
Answer:
20 cm 4 mm

(3) 0.8 cm
Answer:
8 mm

(4) 0.5 cm
Answer:
5 mm

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

(2) 16 mm
Answer:
1.6 cm

(3) 144 mm
Answer:
14.4 cm

(4) 8 mm
Answer:
0.8 cm

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

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

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

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

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

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

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

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

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

Question 2.
Write how many rupees in decimal form.

(1) 105 paise
Answer:
₹ 1.05

(2) 6 rupees 6 paise
Answer:
₹ 6.06

(3) 20 rupees 20 paise
Answer:
₹ 20.2

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

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

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

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

(4) 0.5 m =
Answer:
50 cm

Question 4.
Write how many metres in decimal form.

(1) 110 cm =
Answer:
1.1m

(2) 60 cm =
Answer:
0.6 m

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

(1) 0.1 cm =
Answer:
1 mm

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

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

(2) 100 mm =
Answer:
10.0 cm

Class 5 Maths Solution Maharashtra Board

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

Number Work Class 5 Problem Set 2 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 2 Number Work

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

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

Class 5 Maths Solution Maharashtra Board

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

Circles Class 5 Problem Set 31 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 7 Circles

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

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

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

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

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

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

Chapter 7 Circles Problem Set 31 Additional Important Questions and Answers

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

(2) 2.5 cm
Answer:

(3) 3.3 cm
Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 5 Maths Solution Maharashtra Board

• 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

Fractions Class 5 Problem Set 22 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 5 Fractions

Question 1.
Add the following:

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

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

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

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

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

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

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

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

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

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

Question 2.
Subtract the following:

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


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

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

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

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

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

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

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

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

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

A fraction of a collection and a multiple of a fraction

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

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

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

Thrice 5 – \(\frac{1}{3}\) times 15

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

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

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

Addition and Subtraction Problem Set 13 Additional Important Questions and Answers

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

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

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

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

Subtract the following:

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

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

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

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

Class 5 Maths Solution Maharashtra Board

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

Patterns Class 5 Problem Set 53 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 15 Patterns

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

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

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

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

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

Think about it.

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

Activity

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

Patterns in floor tiles

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

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

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

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

Note the pattern and complete the design.

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

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Solve the following :

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

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

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

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

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

Question 6.
Match the columns

A	B
(1) Third square number	(a) 15
(2) Fourth triangular number	(b) 36
(3) Number neither square nor triangular	(c) 16
(4) Number is both square as well as triangular number	(d) 35

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

Class 5 Maths Solution Maharashtra Board

• 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

Number Work Class 5 Problem Set 5 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 2 Number Work

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

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

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

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

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

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

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

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

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

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

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

An interesting dice game

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

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

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

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

Bigger and smaller numbers

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

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

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

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

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

Roman Numerals Problem Set 5 Additional Important Questions and Answers

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

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

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

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

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

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

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

Question 4.
The expanded form write the number.

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

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

Class 5 Maths Solution Maharashtra Board

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

Measuring Time Class 5 Problem Set 44 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 10 Measuring Time

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

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

Examples of time measurement

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

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

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

Hr	Min
15 – 11	30 00
4	30

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

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

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

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

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

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

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

Measuring Time Problem Set 44 Additional Important Questions and Answers

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

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

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

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

Class 5 Maths Solution Maharashtra Board

• 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

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

Class 5 Maths Solution Maharashtra Board

• 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:

Class 5 Maths Solution Maharashtra Board

• 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