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Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.2 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Std 8 Maths Practice Set 10.2 Chapter 10 Solutions Answers

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y² + 10y + 24) ÷ (y + 4)
ii. (p² + 7p – 5) ÷ (p + 3)
iii. (3x + 2x² + 4x³) ÷ (x – 4)
iv. (2m³ + m² + m + 9) ÷ (2m – 1)
v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)
vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Solution:
i. (y² + 10y + 24) ÷ (y + 4)

∴ Quotient = y + 6
Remainder = 0

ii. (p² + 7p – 5) ÷ (p + 3)

∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x² + 4x³) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x

∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m³ + m² + m + 9) ÷ (2m – 1)

∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
Write the dividend in descending order of their indices.
(x⁴ + x³ – 3x² + 3x – 12) ÷ (x² + 2)

∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)

∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x⁴ – 5x³ – 7x + 1) = (4x⁴ – 5x³ + 0x² – 7x + 1)

∴ Quotient = \(x^{3}-x^{2}-\frac{x}{4}-\frac{29}{16}\)
Remainder = \(\frac { -13 }{ 16 }\)

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions

Statistics Class 8 Maths Chapter 11 Practice Set 11.3 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.3 8th Std Maths Answers Solutions Chapter 11 Statistics.

Std 8 Maths Practice Set 11.3 Chapter 11 Solutions Answers

Exercise 11.3 Class 8 Question 1.
Show the following information by a percentage bar graph.

Solution:

Statistics for Class 8 Question 2.
Observe the following graph and answer the questions.

1. State the type of the bar graph.
2. How much percent is the Tur production to total production in Ajita’s farm?
3. Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
4. Whose percentage production of Tur is the least?
5. State production percentages of Tur and Gram in Sudha’s farm.

Solution:

1. The given graph is a percentage bar graph.
2. Percent of tur production to the total production in Ajita’s farm is 60%.
3. Production of Gram in the farm of Yash = 50%
   Production of Gram in the farm of Ravi = 30%
   ∴ Difference in the production = 50% – 30% =20%
   ∴ Yash’s production of Gram is more and by 20%.
4. Sudha’s percentage production of Tur is the least.
5. Production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.

8th Standard Statistics Question 3.
The following data is collected in a survey of some students of 10th standard from some schools. Draw the percentage bar graph of the data.

Solution:

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.3 Intext Questions and Activities

Statistics 8th Class Question 1.
Compare and discuss a percentage bar diagram and a subdivided bar diagram. Use it to learn the graphs in the subjects like Science, Geography. (Textbook pg, no. 74)
Solution:
[Students should attempt the above activity on their own.]

Maharashtra Board Class 8 Maths Solutions

• Discount and Commission Practice Set 9.1 Class 8 Maths Solutions
• Discount and Commission Practice Set 9.2 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.1 Class 8 Maths Solutions
• Division of Polynomials Practice Set 10.2 Class 8 Maths Solutions
• Statistics Practice Set 11.1 Class 8 Maths Solutions
• Statistics Practice Set 11.2 Class 8 Maths Solutions
• Statistics Practice Set 11.3 Class 8 Maths Solutions

Problems on Measurement Class 5 Problem Set 46 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 11 Problems on Measurement

Question 1.
Add :

(1) ₹ 9, 50 paise + ₹ 14, 60 paise
Solution:

50 paise + 60 paise
= 110 paise
= 1 ₹ 10 paise
∴ ₹ 24, 10 paise

(2) 6 cm 5 mm + 7 cm 9 mm
Solution:

5 mm + 9 mm
= 14 mm 14 mm
= 1 cm 4 mm
∴ 14 cm 4 mm

(3) 22 m 50 cm + 25 m 75 cm
Solution:

50 cm + 75 cm
= 125 cm
= 1 m 25 cm
∴ 48 m 25 cm

(4) 15 km 740 m + 13 km 950 m
Solution:

740 m + 950 m
= 1690 m 1690 m
= 1km 690 m
∴ 29 km 690 m

(5) 25 kg 650 g + 29 kg 770 g
Solution:

650 gm + 770 gm
= 1420 gm
= 1 kg 420 gm
∴ 55 kg 420 gm

(6) 19l 840ml + 25l 250ml
Solution:

840 ml + 250 ml
= 1090 ml
= 11 + 90 ml
∴ 45 l 90 ml

Question 2.
Subtract :

(1) ₹ 19, 50 paise – ₹ 12, 60 paise
Solution:

We cannot subtract 60 paise from 50 paise. So convert 1 ₹ into 100 paise.
₹ 6, 90 paise

∴ ₹ 6, 90 paise

(2) 24 cm 2 mm – 3 cm 8 mm
Solution:

We cannot subtract 8 mm from 2 mm. So, convert 1 cm = 10 mm

∴ 20 cm 4 mm

(3) 20 m 30 cm – 17 m 60 cm
Solution:

We cannot subtract 60 cm from 30 cm. So, convert 1 m = 100 cm

∴ 2 m 70 cm

(4) 40 km 255 m – 17 km 960 m
Solution:

We cannot subtract 960 m from 225 m. So, convert 1 km = 1000 m

∴ 22 km 265 m

(5) 35 kg 150 g – 26 kg 470 g
Solution:

We cannot subtract 470 gm from 150 gm. So, convert I kg= 1000gm

∴ 8 kg 680 gm

(6) 46 l 200 ml – 38 l 750 ml
Solution:

We cannot subtract 750 ml from 200 ml. So, convert 1 l = 1000 ml

∴ 7 l 450 ml

Word problems

Study the following examples.

Example (1) If a shopkeeper has 150 kg 500 g of rice and sells 75 kg 750 g, how much rice will be left?

74 kg 750 g of rice is left.

Example (2) A can of milk has 20 l 450 ml of milk. Another can has 18 l 800 ml. How much milk is there in the two cans altogether?

The total quantity of milk is 39l 250ml.

Example (3) At a speed of 90 km per hour, what distance will a train cover in two and a half hours?

The speed of the train is 90 kmph. That is, it travels 90 km in one hour. It travels 90 more km in the second hour.
In the next half an hour, 90 ÷ 2 = 45 km
The total distance travelled is 90 + 90 + 45 = 225 km.

Example (4) If one dress requires 3 m 25 cm of cloth, how much do 4 dresses need?

Manju’s method :
3 m 25 cm for the 1st dress
+ 3 m 25 cm for the 2nd dress
+ 3 m 25 cm for the 3rd dress
3 m 25 cm for the 4th dress
_________
12 m 100 cm
1 m is 100 cm, therefore 12 + 1 = 13 m

Example (5)
If a wire that is 9 m 50 cm long is cut into pieces of 5 cm each, how many pieces will be made?
9 m 50 cm = (900 + 50) cm
To find out how many pieces of 5 cm can be made from a wire 950 cm long, let us use division.
190 pieces will be made.

Example (6) A play started at 30 minutes past 6 in the evening and finished two and three quarter hours later. What time did the play get over?

The play got over at 15 minutes past 9 at night.

Note : The units for length, mass and capacity are written in decimal form. This makes it easy to carry out addition and subtraction of length, mass and capacity.

Units of measuring time are not in decimal form. It is a little more difficult to carry out additions and subtractions of those quantities.

Problems on Measurement Problem Set 46 Additional Important Questions and Answers

Add the following:

(1) 12 km 880 m + 7 km 620 m
Solution:

880m + 620 m = 1500 m
= 1km 500 m
∴ 20 km 500 m

(2) ₹ 62, 45 paise + ₹ 37, 55 paise
Solution:

45 paise + 55 paise
100 paise = 1 ₹
∴ 100 rupees

Subtract the following:

(1) 15 m 15 cm – 4 m 65 cm
Solution:

We cannot subtract 65 cm from 15 cm. So, convert l m = 100 cm
∴ 10 m 50 cm

(2) 29 kg 880 gm – 8 kg 900 gm
Solution:

We cannot subtract 900 gin from 880 gm. So, convert 1 kg = 1000 gm
∴ 20 kg 980 gm

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

Multiplication and Division Class 5 Problem Set 15 Question Answer Maharashtra Board

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

Std 5 Maths Chapter 4 Multiplication and Division

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

Maharashtra Board Class 5 Maths Solutions

• Multiplication and Division Problem Set 14 Class 5 Maths Solutions
• Multiplication and Division Problem Set 15 Class 5 Maths Solutions
• Multiplication and Division Problem Set 16 Class 5 Maths Solutions

Number Work Class 5 Problem Set 6 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 6 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Write the proper symbol, ‘<’ or ‘>’ in the box.
(1) 5,705 [ < ] 15,705
(2) 22,74,705 [  ] 12,74,705
(3) 35,33,302 [  ] 35,32,302
(4) 99,999 [  ] 9,99,999
(5) 4,80,009 [  ] 4,90,008
(6) 35,80,177 [  ] 35,88,172
Answer:
(1) <
(2) >
(3) >
(4) <
(5) <
(6) <

Question 2.
Solve the problems given below.

(1) The Swayamsiddha Savings Group made 3,45,000 papads while the Swabhimani Group made 2,95,000. Which group made more papads?
Answer:
Here, 3,45,000 > 2,95,000
Hence, the Swayamsiddha saving group made more papads.

(2) Children of the Primary School in Ahmadnagar District collected 2,00,000 seeds while those in Pune District collected 3,25,000. Which children collected more seeds?
Answer:
Here, 3,25,000 > 2,00,000
Hence, Pune District children collected more seeds.

(3) The number of people who took part in the Republic Day flag hoisting ceremony was 2,01,306 in Pandharpur taluka and 1,97,208 in Malshiras taluka. In which taluka did a larger number of people participate?
Answer:
Here, 2,01,306 > 1,97,208
Hence, people of Pandharpur taluka participated in larger number

(4) At an exhibition, the Annapoorna Savings Group sold goods worth 5,12,345. The Nirman Group sold goods worth 4,12,900. This figure was 4,33,000 for the Srujan Group and 5,11,937 for the Savitribai Phule group.

Which group had the largest sales?

Which group had the smallest?

Write the sales figures in ascending order.
Answer:
Among the numbers 5,12,345; 4,12,900; 4,33,000; 5,11,937

5,12,345 is largest and 4,12,900 is smallest. Hence, Annapoorna group had the largest sale and Nirman Group had the smallest sales.

Sales in ascending order

4,12,900 < 4,33,000 < 5,11,937 < 5,12,345

Introducing crores

99,99,999 is the biggest seven-digit number. On adding the number 1 to it, we get the smallest eight-digit number, 1,00,00,000. We read this number as ‘one crore’. The new place created to write this number is called the ‘crores’ place.

From the following examples, you can learn to read eight-digit numbers.

Number – Reading

8,45,12,706 – Eight crore forty-five lakh twelve thousand seven hundred and six
5,61,63,589 – Five crore sixty-one lakh sixty-three thousand five hundred and eighty-nine
6,09,04,034 – Six crore nine lakh four thousand and thirty-four

Something more

On the left of the crores place are the places for ten crores, abja, and ten abja in that order. The place value of each of these is ten times the value of the one on its right. According to the Census of the year 2011, the population of our country is 1,21,01,93,422. We read this as ‘one Abuja twenty-one crore one lakh ninety-three thousand four hundred and twenty-two.

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Write the proper symbol, ‘<‘ or ‘>’ in the box.
(1) 68,34,170 [     ] 8,43,170
(2) 5,04,132 [     ] 5,04,123
(3) 1,01,001 [     ] 1,00,101
(4) 14,55,432 [     ] 4,54,532
Answer:
(1) >
(2) >
(3) >
(4) >

Question 2.
Write the numbers in words.

(1) 15,97,21,409
Answer:
Fifteen crore, ninety-seven lakh, twenty-one thousand, four hundred and nine

(2) 99,99,99,999
Answer:
Ninety-nine crore, ninety-nine lakh, ninety- nine thousand, nine hundred and ninety nine.

(3) 7,54,21,607
Answer:
Seven crore, fifty-four lakh, twenty-one thousand, six hundred and seven.

(4) 5,16,36,854
Answer:
Five crore, sixteen lakh, thirty-six thousand, eight hundred and fifty four.

Question 3.
Write in figures.

(1) One crore, fifteen lakh, fifty-nine thousand, seven hundred and four
Answer:
1,15,59,704

(2) Sixty-five crore, seventy lakh, fifty thousand and thirty nine
Answer:
65,70,50,039

(3) Four crore, fifty-nine lakh, fourty-three thousand, five hundred and thirty four
Answer:
4,59,43,534

(4) Eighteen crore, seventy-six lakh, fifty-four thousand and one
Answer:
18,76,54,001

Question 4.
Fill in the blanks in the table below:

Answer:

Question 5.
Write the following numbers in words.
(1) 17,301
(2) 45,019
(3) 40,018
(4) 28,740
Answer:
(1) Seventeen thousand, three hundred and one.
(2) Forty-five thousand and nineteen.
(3) Forty thousand and eighteen
(4) Twenty-eight thousand seven hundred and forty

Question 6.
How many rupees do they make?
(1) 8 notes of rupees 2,000, 3 notes of rupees 100,11 notes of rupees 10.
Answer:
16,410

(2) 9 notes of rupees 2,000, 18 notes of rupees 100,18 notes of rupees 50,18 notes of rupees 10.
Answer:
20,880

(3) Write the smallest and the biggest five-digit numbers that can be made using the digits only once.
(a) 6, 8, 0,1, 9
(b) 3, 5,1,2, 8
Answer:
Smallest number : (i) 10,689 (ii) 12358
Biggest number : (i) 98,610 (ii) 85321

(4) Write the smallest and the biggest number from the following numbers.
(a) 35,798
(b) 39,785
(c) 39,587
(d) 35,789
Answer:
Smallest number : 35,789
Biggest number : 39,785

(5) Write the number from the given number which is neither biggest nor smallest.
(a) 45, 798
(b) 45, 789
(c) 45, 897
Answer:
45,798.

(6) Write the biggest and the smallest three-digit numbers that can be made using the digits 0,1, 2, 3, 4, 5, 6, 7, 8, 9 only once.
Answer:
Biggest three-digit number : 987
Smallest three-digit number : 102

Question 7.
Read the numbers and write them in words.
(1) 2,65,048
(2) 1,80,794
(3) 1,06,709
(4) 8,80,006
Answer:
(1) Two lakh sixty-five thousand and forty- eight,
(2) One lakh eighty thousand seven hundred and ninety-four.
(3) One lakh six thousand seven hundred and nine.
(4) Eight lakh eighty thousand and six.

Question 8.
Read the numbers and write them in figures.
(1) Two lakh five thousand three hundred and six.
(2) Six lakh and six
(3) Nine lakh forty thousand and thirty seven.
(4) Five lakh ninety-nine thousand and fifteen.
Answer:
(1) 2,05,306
(2) 6,00,006
(3) 9,40,037
(4) 5,99,015

Question 9.
Write six, six-digit numbers using the digits 0,.3,5,7,9,1 only once with 9 lakh fifty-seven thousand in all numbers.
Answer:
(1) 9,57,301
(2) 9,57,310
(3) 9,57,103
(4) 9,57,130
(5) 9,57,013
(6) 9,57,031

Question 9.
(A) Match the columns:

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

(B) Match the columns:

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

Question 10.
Read the numbers and write them in words.
(1) 34,87,569
(2) 70,85,039
(3) 48,07,102
(4) 67,40,960
(5) 88,00,080
(6) 40,40,004
Answer:
(1) Thirty-four lakh, eighty-seven thousand, five hundred and sixty-nine.
(2) Seventy lakh, eight-five thousand and thirty-nine.
(3) Forty-eight lakh, seven thousand, one hundred and two.
(4) Sixty-seven lakh, forty thousand, nine hundred and sixty.
(5) Eighty-eight lakh and eighty.
(6) Forty lakh, forty thousand and four.

Question 11.
Read the numbers and write them in figures.
(1) Fifty-nine lakh, seven thousand, seventeen.
(2) Twenty-two lakh, ten thousand, five hundred.
(3) Fifty-two lakh, twenty-five thousand, four hundred and fifteen.
(4) Thirty lakh, thirty thousand and thirty.
Answer:
(1) 59,07,017
(2) 22,10,500
(3) 52,25,415
(4) 30,30,030

Question 12.
Write the place value of the underlined digit.
(1) 68,03,512
(2) 3,42,157
(3) 84,52,170
(4) 79,345
(5) 38,14,093
(6) 8,10,618
(7) 35,10,387
Answer:
(1) 8,00,000
(2) 40,000
(3) 2,000
(4) 5
(5) 90
(6) 600
(7) 30,00,000

Question 13.
Write the numbers in their expanded form.
(1) 78,15,692
(2)50,95,182
(3)6,40,078
(4) 9,58,802
Answer:
(1) 70,00,000 + 8,00,000 + 10,000 + 5,000 + 600 + 90 + 2
(2) 50,00,000 + 90,000 + 5,000 + 100 + 80 + 2
(3) 6,00,000 + 40,000 + 70 + 8
(4) 9,00,000 + 50,000 + 8,000 + 800 + 2

Question 14.
Write the place name and place value of each digit in the following numbers.
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555
Answer:
(1) 27,306
(2) 1,70,425
(3) 75,68,041
(4) 55,555

Question 15.
The expanded form of the number is given. Write the number.
(1) 70,000 + 6,000 + 500 + 40 + 8
(2) 8,00,000 +-30,000 + 5,000 + 400 + 3
(3) 60,00,000 + 2,00,000 + 70 + 4
(4) 20,00,000 + 5,00,000 + 900 + 5
Answer:
(1) 76,548
(2) 8,35,403
(3) 62,00,074
(4) 25,00,905

Question 16.
Considering the number 50,43,176.
Fill in the blanks.
(1) The digit in the ten thousand place is ……………………………………….. .
(2) Place value of 1 is ……………………………………….. .
(3) The digit in the lakhs place is ……………………………………….. .
(4) Place value of 5 is ……………………………………….. .
(5) The digit 7 is in ……………………………………….. place.
Answer:
(1) 4
(2) 100
(3) 0
(4) 50,00,000
(5) tens

Question 17.
Write the proper symbols ‘<‘ or ‘>’ in the box.
(1) 12,625 [     ] 21,526
(2) 23,564 [     ] 23,546
(3) 36,60,660 [     ] 36,60,606
(4) 89,14,507 [     ] 89,15,407
Answer:
(1) <
(2) >
(3) >
(d) <

Question 18.
Solve the problems given below.
(1) Population of city A is 8,57,238 and that of city B is 8,75,461. Population of which city is more?
Answer:
city B

(2) Yearly income of Rajnikant is? 3,48,600 and that of Shashikant is? 3,46,500. Whose income is less?
Answer:
Shashikant

Question 19.
Profit of the four companies A, B, C, D is as follows.

Now, answer the following questions.
(1) Which company made maximum profit?
(2) Which company made minimum profit?
(3) Write the profit of the companies in the descending order.
Answer:
(1) B
(2) C
(3) profit of company B > D > A > C

Question 20.
In a certain election, candidates : Tavade, Patel, Chauhan, and Shinde got the votes as follows.

Now, answer the following questions.
(1) Who got the highest number of votes?
(2) Who got the least number of votes?
(3) Write the number of votes obtained in the ascending order.
Answer:
(1) Patel
(2) Shinde
(3) 34,67,008 < 37,51,386 < 43,51,239 < 48,00,173

Question 21.
Compare the following using >, < or = signs.
(1) 3,97,48,632 [     ] 3,97,58,632
(2) 1,50,15,178 [     ] 1,50,15,780
(3) 3,74,98,561 [     ] 96,42,748
(4) 30,49,75,831 [     ] 30,49,00,831
Answer:
(1) <
(2) <
(3) >
(4) >

Question 22.
Circle the correct answer:
(1) Mark periods 617231801 according to the Indian Number system.
(a) 61,72,31,801
(b) 16,172,31
(c) 617,231,801
Answer:
(a) 61,72,31,801

(2) Mark periods 90289164 according to the international Number system.
(a) 9,0289,164
(b) 902891,64
(c) 90,289,164
Answer:
(c) 90,289,164

(3) 1,00,00,000 is read as ……………………………….. .
(a) ten crore
(b) one crore
(c) hundred thousand
Answer:
(b) one crore

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

Fractions Class 5 Problem Set 23 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 23 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

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:

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 ₹

Maharashtra Board Class 5 Maths Solutions

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

Decimal Fractions Class 5 Problem Set 36 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 36 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Write the following mixed fractions in decimal form and read them aloud.

\(\text { (1) } 3 \frac{9}{10}\)
Answer:
3.9, Three-point nine.

\(\text { (2) } 1 \frac{4}{10}\)
Answer:
1.4, One point four.

\(\text { (3) } 5 \frac{3}{10}\)
Answer:
5.3, Five-point three.

\(\text { (4) } \frac{8}{10}\)
Answer:
0.8, Zero points eight.

\(\text { (5) } \frac{7}{10}\)
Answer:
0.5, Zero points five.

Hundredths

If \(\frac{1}{10}\) is divided into 10 equal parts, each part becomes \(\frac{1}{100}\) or one hundredth. Therefore, note that 1 tenth =10 hundredths, or 0.1=0.10. By multiplying \(\frac{1}{100}\) by 10 we get \(\frac{10}{100}\) = \(\frac{1}{10}\). Therefore, it is possible to create a hundredths place next to the tenths place. After creating a hundredths place we can write \(\frac{14}{100}\) as 0.14.

\(\frac{14}{100}=\frac{10+4}{100}=\frac{10}{100}+\frac{4}{100}=\frac{1}{10}+\frac{4}{100}\) meaning that when writing \(\frac{14}{100}\) in decimal form, 1 is written in the tenths place and 4 is written in the hundredths place. This fraction is written as 0.14 and is read as ‘zero point one four’. Similarly, 6 \(\frac{57}{100}\) is written as 6.57 and 50 \(\frac{71}{100}\) is written as 50.71.

While writing \(\frac{3}{100}\) in decimal form, we must remember that there is no number in the tenths place and so, we put 0 in that place, which means that \(\frac{3}{100}\) is written as 0.03.

Study how the decimal fractions in the table below are written and read.

Decimal Fractions Problem Set 36 Additional Important Questions and Answers

\(\text { (1) } 4 \frac{6}{10}\)
Answer:
4.6, Four point six. 7

\(\text { (2) } 4 \frac{6}{10}\)
Answer:
2.7, Two point seven.

\(\text { (3) } 4 \frac{6}{10}\)
Answer:
6.2, Six points two.

\(\text { (4) } 4 \frac{6}{10}\)
Answer:
21.1, Twenty-one point one.

\(\text { (5) } 4 \frac{6}{10}\)
Answer:
17.5, Seventeen points five.

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

Multiplication and Division Class 5 Problem Set 14 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 14 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 4 Multiplication and Division

Question 1.
Multiply the following:

(1) 327 × 92
Solution:
3 2 7
x
9 2
_____
6 5 4
+
2 9 4 3 0
Answer:
3 0 0 8 4

(2) 807 × 126
Solution:
8 0 7
x
1 2 6
______
4 8 4 2
+
1 6 1 4 0
+
8 0 7 0 0
Answer:
1 0 1 6 8 2

(3) 567 × 890
Solution:
5 6 7
x
8 9 0
______
0 0 0
+
5 1 0 3 0
+
4 5 3 6 0 0
Answer:
5 0 4 6 3 0

(4) 4317 × 824
Solution:
4 3 1 7
8 2 4
1 7 2 6 8
+ 8 6 3 4 0
3 4 5 3 6 0 0
Answer:
3 5 5 7 2 0 8

(5) 6092 × 203
Solution:
6 0 9 2
x
2 0 3
______
1 8 2 7 6
+
0 0 0 0 0
+
1 2 1 8 4 0 0
Answer:
1 2 3 6 6 7 6

(6) 1177 × 99
Solution:
1 1 7 7
x
9 9
1 0 5 9 3
+
1 0 5 9 3 0
Answer:
1 1 6 5 2 3

(7) 456 × 187
Solution:
4 5 6
x
1 8 7
3 1 9 2
+
3 6 4 8 0
+
4 5 6 0 0
Answer:
8 5 2 7 2

(8) 6543 × 79
Solution:
6 5 4 3
x
7 9
5 8 8 8 7
+
4 5 8 0 1 0
Answer:
5 1 6 8 9 7

(9) 2306 × 832
Solution:
2 3 0 6
x
8 3 2
______
4 6 1 2
+
6 9 1 8 0
+
1 8 4 4 8 0 0
______
Answer:
1 9 1 8 5 9 2

(10) 6429 × 509
Solution:
6 4 2 9
x
5 0 9
______
5 7 8 6 1
+
0 0 0 0 0
+
3 2 1 4 5 0 0
Answer:
3 2 7 2 3 6 1

(11) 4,321 × 678
Solution:
4 3 2 1
x
6 7 8
_____
3 4 5 6 8
3 0 2 4 7 0
2 5 9 2 6 0 0
_______
Answer:
2 9 2 9 6 3 8

(12) 20,304 × 87
Solution:
2 0 3 0 4
x
8 7
1 4 2 1 2 8
+
1 6 2 4 3 2 0
_________
Answer:
1 7 6 6 4 4 8

Question 2.
As part of the ‘Avoid Plastic’ campaign, each of 745 students made 25 paper bags. What was the total number of paper bags made ?
Solution:
7 4 5 Number of students
x
2 5 bags made by each
___________
3 7 2 5
+
1 4 9 0 0
__________
1 8 6 2 5
__________

Answer:
18,625 bags made

Question 3.
In a plantation, saplings of 215 medicinal trees have been planted in each of the 132 rows of trees. How many saplings are there in the plantation altogether ?
Solution:
2 1 5 Saplings in each now
x
1 3 2 Number of rows
__________
4 3 0
+
6 4 5 0
+
2 1 5 0 0
__________
2 8 3 8 0
__________

Answer:
Altogether there is 28,380 saplings.

Question 4.
One computer costs 27,540 rupees. How much will 18 such computers cost ?

Solution:
2 7 5 4 0 Cost of 1 computer
x
1 8 No. of computers
__________
2 2 0 3 2 0
+
2 7 5 4 0 0
___________
4 9 5 7 2 0
__________

Answer:
4,95,720 rupees cost of 18 computers.

Question 5.
Under the ‘Inspire Awards’ scheme, 5000 rupees per student were granted for the purchase of science project materials. If 154 students in a certain taluka were covered under the scheme, find the total amount granted to that taluka.

Solution:
₹ 5 0 0 0 Granted per student
x 1 5 4 Number of students
__________
2 0 0 0 0
+
2 5 0 0 0 0
+
5 0 0 0 0 0
______________
₹ 7 7 0 0 0 0
______________

Answer:
7,70,0000 granted totally

Question 6.
If a certain two-wheeler costs 53,670 rupees, how much will 35 such two-wheelers cost ?
Solution:
5 3 7 6 0 Cost of 1 two-wheeler
x 3 5 No. of two-wheelers
___________
2 6 8 8 0 0
+
1 6 1 2 8 0 0
_____________
1 8 8 1 6 0 0
______________

Answer:
18,81,600 is the total cost of 35- two-wheelers

Question 7.
One hour has 3,600 seconds. How many seconds do 365 hours have ?
Solution:
3 6 0 0 Seconds of 1 hour
x
3 6 5 No. of hours
_________
1 8 0 0 0
+
2 1 6 0 0 0
+
1 0 8 0 0 0 0
______________
1 3 1 4 0 0 0
______________

Answer:
13,14,000 seconds for 365 hours.

Question 8.
Frame a multiplication word problem with the numbers 5473 and 627 and solve it.

Solution:
Cost of one mobile is 5,473. What is the cost of such 627 mobiles?
5 4 7 3 Cost of 1 mobile
x 6 2 7 Number of mobiles
__________
3 8 3 1 1
+
1 0 9 4 6 0
+
3 2 8 3 8 0 0
_____________
3 4 3 1 5 7 1
_____________

Answer:
34,31,571 cost for 627 mobiles.

Question 9.
Find the product of the biggest three-digit number and the biggest four-digit number.

Solution:
9 9 9 9 Biggest four digit no.
x 9 9 9 Biggest three-digit no.
_________
8 9 9 9 1
+
8 9 9 9 1 0
+
8 9 9 9 1 0 0
_____________
9 9 8 9 0 0 1
_____________

Answer:
99,89,001

Question 10.
One traveller incurs a cost of 7,650 rupees for a certain journey. What will be the cost for 26 such travellers?
Solution:
7 6 5 0 Cost of one traveller
x 2 6 No. of travellers
______
4 5 9 0 0
+
1 5 3 0 0 0
_____________
1 9 8 9 0 0
_____________

Answer:
1,98,900 cost of 26 travellers.

Pairing off objects from two groups in different ways

(1) Ajay wants to travel light. So he took with him three shirts – one red, one green and one blue and two pairs of trousers – one white and one black. How many different ways does he have of pairing off a shirt with trousers?

Writing ‘S’ for shirt and ‘T’ for trousers, the possible different pairs are :

(2) Suresh has three balls of different colours marked A, B and C and three bats marked P, Q and R. He wishes to take only one bat and one ball to the playground. In how many ways can he pair off a ball and a bat to take with him?

How many different pairs have been shown here?

(3) The three friends, Sanju, John and Ali went to the fair. A shop there, had five different types of hats. Each of the boys had photos taken of himself, wearing every type of hat, in turn. Find how many photographs were taken in all.

How many different pairs were formed ? That is, how many photos were taken ?

Take two collections, each containing the given number of objects. Make as many different pairs as possible, taking one object from each collection every time. Thus, complete the table below.

What does this table tell us ?
The number of different pairs formed by pairing off objects from two groups is equal to the product of the number of objects in the two groups.

Division
Teacher : You have learnt some things about division. For example, we know that division means making equal parts of a given number, or, subtracting a number repeatedly from a given number. What else do you know ?
Shubha : We know that we get two divisions from one multiplication. From 9 × 4 = 36, we get the divisions 36 ÷ 4 = 9 and 36 ÷ 9 = 4.
Teacher : Very good! Right now, there’s nothing new to learn about division. Only the number of digits in the dividend and the divisor will grow. Tell me what is 354 ÷ 6 ?
Sarang : 354 = 300 + 54. 300 divided by six is 50. And 54 ÷ 6 = 9. Hence the quotient is 50 + 9 = 59.
Teacher : Right! Now let’s learn, step by step, how to divide a four-digit number by a one-digit number. So now, divide 4925 by 7 and tell me the quotient and the remainder.
Shubha : We cannot divide 4 thousands by 7 into whole thousands. Now, 4 Th = 40 H. So let us instead take the 40 hundreds together with 9 hundreds and divide 49 hundreds. 49 ÷ 7 = 7. So, everyone gets 7 hundreds. Now, we cannot divide 2T equally among 7 people. So we must write 0 in the tens place in the quotient. Then on dividing 25 by seven, we get quotient 3 and the remainder is 4. Thus, the answer is quotient 703, remainder 4.

Teacher : Very good ! Now divide 7439 by 9.
Sarang : It’s difficult to do this mentally. I’ll write it down on paper. The quotient is 826 and the remainder, 5.
Teacher : We use the same method to divide a four-digit number by a two-digit number. If necessary, we can prepare the table of the divisor before we start.

Study the solved examples shown below.
Example (1)

Quotient 170, Remainder 4

Example (2)

Quotient 305, Remainder 23

Example (3) Divide. 9842 ÷ 45

We can prepare the 45 times table to do this division.
But when the divisor is a big number, we can solve the example by first guessing what the quotient will be. Let us see how to do that.
We have 0 in the thousands place in the quotient.
Now, to guess the quotient when dividing 98 by 45, look at the first digits in both – the dividend and the divisor. These are 9 and 4, respectively.
Dividing 9 by 4, we will get 2 in the quotient. Let us see if 2 times 45 can be subtracted from 98. 45 × 2 = 90. 90 < 98. So, we write 2 in the hundreds place in the quotient.
Next, dividing 84 by 45 we can easily see that as 90 > 84, we have to write 1 in the tens place in the quotient.
Now, we have to divide 392 by 45. As 3 < 4, let us look at 39, the number formed by the first 2 digits, to guess the next digit in the quotient.
4 × 9 = 36 and 36 < 39. Let us check if the next digit can be 9. 45 × 9 = 405 and 405 > 392. Therefore, 9 cannot be the next digit in the quotient.
Let us check for 8. 45 × 8 = 360. 360 < 392. So, we write 8 in the units place of the quotient.
We subtract 8 × 45 from 392 and complete the division.
The quotient is 218 and the remainder, 32.

Example (4)
If 35 kilograms of wheat cost 910 rupees, what is the rate of wheat per kg?

Weight of wheat in kg × rate of wheat per kg = cost of wheat Hence, 35 × rate per kg = 910
Therefore, when we divide 910 by 35, we will get the per kg rate of wheat.
The rate per kilogram of wheat is 26 rupees.

Multiplication and Division Problem Set 14 Additional Important Questions and Answers

Multiply the following:

(1) 2132 x 231
Solution:
2 1 3 2
x
2 3 1
2 1 3 2
+
6 3 9 6 0
+
4 2 6 4 0 0
____________
Answer:
4 9 2 4 9 2

(2) 1863 x 432
Solution:
1 8 6 3
x
4 3 2
3 7 2 6
+
5 5 8 9 0
+
7 4 5 2 0 0
___________
Answer:
8 0 4 8 1 6

Solve the following word problems:

(1) A factory manufactures 34,796 pairs of socks in one hour. How many pairs will the factory manufacture in one day?
Solution:
3 4 7 9 6 Pairs of socks
x 2 4 No. of hours
______
1 3 9 1 8 4
+
6 9 5 9. 2 0
_____________
8 3 5 1 0 4
_____________

Answer:
8,35,104 pairs of socks manufactured in one day

(2) There are 375 toffees in a box. How many toffees will be there in 632 such boxes?
Solution:
3 7 5 No. of toffees
x 6 3 2 No. of boxes
_______
7 5 0
+
1 1 2 5 0
+
2 2 5 0 0 0
___________
2 3 7 0 0 0
_____________

Answer:
There will be 2,37,000 toffees.

(3) There are 144 articles in a gross. How many articles are there in 2174 gross?
Solution:
2 1 7 4 No. of gross
x 1 4 4 Articles in I gross
______
8 6 9 6
+
8 6 9 6 0
+
2 1 7 4 0 0
_____________
3 1 3 0 5 6
_____________

Answer:
There are 3,13,056 articles in 2174 gross.

Maharashtra Board Class 5 Maths Solutions

• Multiplication and Division Problem Set 14 Class 5 Maths Solutions
• Multiplication and Division Problem Set 15 Class 5 Maths Solutions
• Multiplication and Division Problem Set 16 Class 5 Maths Solutions

Number Work Class 5 Problem Set 4 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 2 Number Work Problem Set 4 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 2 Number Work

Question 1.
Read the numbers and write them in words.
(1) 25,79,899
(2) 30,70,506
(3) 45,71,504
(4) 21,09,900
(5) 43,07,854
(6) 50,00,000
(7) 60,00,010
(8) 70,00,100
(9) 80,01,000
(10) 90,10,000
(11) 91,00,000
(12) 99,99,999
Answer:
(1) Twenty-five lakh, seventy-nine thousand, eight hundred and ninety-nine.
(2) Thirty lakh, seventy thousand, five hundred and six.
(3) Forty-five lakh, seventy-one thousand, five hundred and four.
(4) Twenty-one lakh, nine thousand, nine hundred.
(5) Forty-three lakh, seven thousand, eight hundred and fifty-four.
(6) Fifty lakh.
(7) Sixty lakh and ten.
(8) Seventy lakh and one hundred.
(9) Eighty lakh and one thousand
(10) Ninety lakh and ten thousand
(11) Ninety-one lakh
(12) Ninety-nine lakh, ninety-nine thousand, nine hundred and ninety-nine.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.
Pune : ₹ 94,29,408
Nashik : ₹ 61,07,187
Nagpur : ₹ 46,53,570
Ahmadnagar : ₹ 45,43,159
Aurangabad : ₹ 37,01,282
Yavatmal : ₹ 27,72,348
Sindhudurg : ₹ 58,49,651
Answer:
Rupees ninety-four lakh, twenty-nine thousand, four hundred and eight.
Rupees sixty-one lakh, seven thousand, one hundred and eighty-seven
Rupees forty-six lakh, fifty-three thousand, five hundred and seventy.
Rupees forty-five lakh, forty-three thousand one hundred and fifty-nine.
Rupees thirty-seven lakh, one thousand, two hundred and eighty-two.
Rupees twenty-seven lakh, seventy two thousand, three hundred and forty-eight.
Rupees fifty-eight lakh, forty-nine thousand, six hundred and fifty-one.

The expanded form of a number and the place value of digits

Teacher : Look at the place value of each of the digits in the number 27,65, 043.

Hamid : When we write the place values of the digits as an addition, we get the expanded form of the number. So, the expanded form of the number 27,65,043 is 20,00,000 + 7,00,000 + 60,000 + 5,000 + 0 + 40 + 3.

Teacher : Now tell me the expanded form of 95,04,506.

Soni : 90,00,000 + 5,00,000 + 0 + 4,000 + 500 + 0 + 6.

Teacher : Good! It can also be written as 90,00,000 + 5,00,000 + 4,000 + 500 + 6. Now write the number from the expanded form that I give you. 4,00,000 + 90,000 + 200

Asha : Here, we have 4 in the lakhs place, 9 in the ten thousands place and 2 in the hundreds place. There are no digits in the ten thousands place and in the tens and units places. Hence, we write 0 in those places. Therefore, the number is 4,90,200.

Teacher : Tell me the place value of the underlined digit in the number 59,30,478.
Soni : The underlined digit is 5. The digit is in the ten lakhs place. Hence, its place value is 50,00,000 or fifty lakhs.

Roman Numerals Problem Set 4 Additional Important Questions and Answers

Question 1.
Read the numbers and write them in words:

(1) 80,91,001
Answer:
Eighty lakh, ninety-one thousand and one.

(2) 50,50,505
Answer:
Fifty lakh, fifty thousand, five hundred and five.

(3) 68,06,086
Answer:
Sixty-eight lakh, six thousand and eighty- six.

Question 2.
Given below are the deposits made in the Women’s Co-operative Credit Societies of some districts. Read those figures.

(1) Thane : 75,14,365
Answer:
Rupees seventy-five lakh, fourteen thousand, three hundred and sixty-five.

(2) Jalgaon : 39,42,180
Answer:
Rupees thirty-nine lakh, forty-two thousand, one hundred and eighty.

(3) Kalyan : 37,40,509
Answer:
Rupees thirty-seven lakh, forty thousand, five hundred and nine.

(4) Kolhapur: 16,05,430
Answer:
Rupees sixteen lakh, five thousand, four hundred and thirty.

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

Addition and Subtraction Class 5 Problem Set 11 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 11 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Subtract the following:

(1) 8,57,513 – 4,82,256
Solution:

Answer:
3,75,257

(2) 13,17,519 – 10,07,423
Solution:

Answer:
3,10,096

(3) 68,34,501 – 23,57,823
Solution:

Answer:
44,76,678

(4) 45,43,827 – 12,05,938
Solution:

Answer:
33,37,889

(5) 70,12,345 – 28,64,547
Solution:

Answer:
41,47,798

(6) 38,01,213 – 37,54,648
Solution:

Answer:
46,565

Study the following word problem.

In 2001, the population of a city was 21,43,567. In 2011, it was 28,09,878. By how much did the population grow?

The population grew by 6,66,311.

Addition and Subtraction Problem Set 11 Additional Important Questions and Answers

Subtract the following:

(1) 53,14,018 – 43,14,019
Solution:

Answer:
9,99,999

(2) 67,05,136 – 34,56,789
Solution:

Answer:
32,48,347

Maharashtra Board Class 5 Maths Solutions

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions 
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions