Students can Download Maths Chapter 2 Division Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 5 Maths Chapter 2 Division

### KSEEB Class 5 Maths Division Revision Exercise

I. Encircle the objects in as shown in the example and write answer in given box:

II. Divide the following using repeated substruction method:

a. 12 ÷ 4

12 – 4 = 8

8 – 4 = 4

4 – 4 = 0

Subtract is done 3 times

\(\frac{12}{4}\) = 3

b. 25 ÷ 5

25 – 5 = 20

20 – 5 = 15

15 – 5 = 10

10 – 5 = 5

5 – 5 = 0

Subtract is done 5 times

25 ÷ = 5

c. 42 ÷ 7

42 – 7 = 35

35 – 7 =28

28 – 7 = 21

21 – 7 = 14

14 – 7 = 7

7 – 7 = 0

Subtract is done 6 times

d. 30 ÷ 10

30 – 10 = 20

20 – 10 = 10

10 – 10 = 0

Subtract is done 3 times

30 ÷ 10 = 3

e. 75 ÷ 15

75 – 10 = 60

60 – 15 = 45

45 – 15 = 30

30 – 15 = 15

15 – 15 = 0

Subtract is done 5 times

5 ÷ 15 = 5

### KSEEB Class 5 Maths Division Ex 2.1

I. Find the quotient and the remainder:

1. 48 ÷ 6

Quotient = 8

Remainder =0

2. 36 ÷ 3

Dividend = (divisor × quotient) + Remainder

36 = (3 × 12) +0

36 = 36 + 0

36 = 36

Quotient = 12

Remainder =0

3. 55 ÷ 4

Dividend = (divisor × quotient) + Remainder

55 = (4 × 13) + 3

55 = 52 + 3

55 = 55

quotient = 13

Remainder = 3

4. 72 ÷ 7

Dividend = (divisor × quotient) + Remainder

72 = (7 × 10) + 2

72 = 70 + 2

72 = 72

quotient = 10

Remainder = 02

II. Find the quotient and the remainder:

1. 232 ÷ 4

Dividend = (divisor × quotient) + Remainder

232 = (4 × 58) + 0

232 = 232 + 0

232 = 232

quotient = 58

Remainder = 3

2. 474 ÷ 6

Dividend = (divisor × quotient) + Remainder

474 = (6 × 79) + 0

474 = 474 + 0

474 = 474

quotient = 79 3

Remainder =0

3. 255 ÷ 11

Dividend = (divisor × quotient) + Remainder

255 = (11 × 23) + 2

255 = 253 + 2

255 = 255

quotient = 23

Remainder = 2

4. 527 ÷ 12

Dividend = (divisor × quotient) + Remainder

527 = (12 × 43) + 11

527 = 516 + 11

527 = 527

quotient = 43

Remainder = 11

III. Find the quotient and the remainder:

1. 1,653 ÷ 8

Dividend = (divisor × quotient) + Remainder

1653 = (8 × 206) + 5

1653 = 1648 + 5

1653 = 1653

quotient = 206

Remainder = 5

2. 1,325 ÷ 2

Dividend = (divisor × quotient) + Remainder

1325 = (2 × 662) + 1

1325 = 1324 + 1

1325 = 1325

quotient = 662

Remainder = 1

3. 1,435 ÷ 15

Dividend = (divisor × quotient) + Remainder

1435 = (15 × 95) + 10

1435 = 1425+10

1435 = 1435

quotient = 95

Remainder =10

4. 2,647 ÷ 13

Dividend = (divisor × quotient) + Remainder

2647 = (13 × 203) + 8

2647 = 2639 + 8

2647 = 2647

quotient = 203

Remainder = 8

IV. Find the quotient and the remainder:

1. 24,658 ÷ 2

Dividend = (divisor × quotient) + Remainder

24658 = (2 × 12329) + 0

24658 = 24658 + 0

24658 = 24658

quotient = 12329

Remainder = 0

2. 14,005 ÷ 7

Dividend = (divisor × quotient) + Remainder

14005 = (7 × 2000) + 5

14005 = 14000 + 5

14005 = 14005

quotient = 2000

Remainder = 5

3. 32,745 ÷ 10

Dividend = (divisor × quotient) + Remainder

32745 = (10 × 3274) + 5

32745 = 32740 + 5

32745 = 32745

quotient = 3274

Remainder =5

4. 12,056 ÷ 12

Dividend = (divisor × quotient) + Remainder

12056 = (12 × 1004) +8

12056 = 12048 + 8

12056 = 12056

quotient = 1004

Remainder = 8

V. Solve the following problems:

Question 1.

A tailor has 18 metres of cloth. He stitch 9 shirts from this cloth. Find the length of cloth required to stitch one shirt.

Answer:

A tailor has 18 meter of cloth

Tailor has stitch 9 shirts

The length of cloth required to stitch one shirt = \(\frac{18}{9}\) = 2 meter

Question 2.

A bike can cover 240 km with 5 litres of petrol. Find how many kilometres does it cover with 1 litre of petrol.

Answer:

A bike can cover 240 Km with 5 litres of petrol

In one litre of petrol, bike covers the distance = \(\frac{240}{5}\) = 48 km

Question 3.

25,004 bags of cement are loaded equally into 14 railway wagons. Find the number of bags in each wagon.

Answer:

25,004 bags of cement are loaded equally into 14 railway wagons

The number of bags

= \(\frac{25004}{14}\) = 1,786

Question 4.

Mr. Sridhar purchased 11 toys of same price for Rs. 946 from a toy shop. Find the cost of each toy.

Answer:

The cost of a toy =?

Mr. Sridhar purchase 11 toys of same price = Rs 946

The cost of each toy

= \(\frac{946}{11}\) = 86 Rs.

Question 5.

The annual income of a person is Rs.90,912. Find his monthly income.

Answer:

The Annual income of a peusen Rs 90,912

Monthly income = \(\frac{90912}{12}\) = 7,576 Rs

Question 6.

A car manufacturing company manufactures 14,820 cars in 13 months. Find the number of cars manufactured in one month.

Answer:

A car manufacturing company manufactures 14820 cars in 13 months. For one month, the number of cars = \(\frac{14820}{13}\) = 1,140

### KSEEB Class 5 Maths Division Ex 2.2

I. Solve

Question 1.

An orange garden has 82 orange plants. A farmer plucks 60 oranges from each plant and packs 12 oranges in each box. Calculate the number of boxes required to pack all oranges.

Answer:

An orange garden has 82 oranges . A farmer plunks 60 oranges from each plant

The number of boxes = ?

The number of oranges in each box = 12

The number of oranges in an orange garden = 410

Question 2.

15 school children hire a cab for a day’s excursion at Rs. 9 per km. If they travel a distance of 325 km, find the amount to be shared by each one of them.

Answer:

A cab of charges Rs 9 Km

The childrens travel a distance 325Km

= 325 × 9

= 2925 Km

The amount shared by each student

= 2925 ÷ 15 = Rs 195

Question 3.

Anita has a cow which yields 8 litres of milk in a day. The selling price of one litre of milk is Rs. 18. She wants to divide the amount earned in one month (30days) among her 4 sons. What amount does each one get?

Answer:

One litre of milk = Rs 18

A cow which yields 8 litres of milk in a day for 30 days, the number of litres of milk

18 × 8 = Rs. 144

The amount of milk

144 × 30 = Rs. 4320

The amount of each son 1,080

4320 ÷ 4