KSEEB Solutions for Class 5 Maths Chapter 2 Division

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:

KSEEB Solutions for Class 5 Maths Chapter 2 Division 1

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

KSEEB Solutions

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

KSEEB Solutions for Class 5 Maths Chapter 2 Division 2
Quotient = 8
Remainder =0

2. 36 ÷ 3

KSEEB Solutions for Class 5 Maths Chapter 2 Division 3
Dividend = (divisor × quotient) + Remainder
36 = (3 × 12) +0
36 = 36 + 0
36 = 36
Quotient = 12
Remainder =0

3. 55 ÷ 4

KSEEB Solutions for Class 5 Maths Chapter 2 Division 4
Dividend = (divisor × quotient) + Remainder
55 = (4 × 13) + 3
55 = 52 + 3
55 = 55
quotient = 13
Remainder = 3

4. 72 ÷ 7

KSEEB Solutions for Class 5 Maths Chapter 2 Division 5
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

KSEEB Solutions for Class 5 Maths Chapter 2 Division 6
Dividend = (divisor × quotient) + Remainder
232 = (4 × 58) + 0
232 = 232 + 0
232 = 232
quotient = 58
Remainder = 3

KSEEB Solutions

2. 474 ÷ 6

KSEEB Solutions for Class 5 Maths Chapter 2 Division 7
Dividend = (divisor × quotient) + Remainder
474 = (6 × 79) + 0
474 = 474 + 0
474 = 474
quotient = 79 3
Remainder =0

3. 255 ÷ 11

KSEEB Solutions for Class 5 Maths Chapter 2 Division 8
Dividend = (divisor × quotient) + Remainder
255 = (11 × 23) + 2
255 = 253 + 2
255 = 255
quotient = 23
Remainder = 2

4. 527 ÷ 12

KSEEB Solutions for Class 5 Maths Chapter 2 Division 9
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

KSEEB Solutions for Class 5 Maths Chapter 2 Division 10
Dividend = (divisor × quotient) + Remainder
1653 = (8 × 206) + 5
1653 = 1648 + 5
1653 = 1653
quotient = 206
Remainder = 5

2. 1,325 ÷ 2

KSEEB Solutions for Class 5 Maths Chapter 2 Division 11
Dividend = (divisor × quotient) + Remainder
1325 = (2 × 662) + 1
1325 = 1324 + 1
1325 = 1325
quotient = 662
Remainder = 1

3. 1,435 ÷ 15

KSEEB Solutions for Class 5 Maths Chapter 2 Division 12
Dividend = (divisor × quotient) + Remainder
1435 = (15 × 95) + 10
1435 = 1425+10
1435 = 1435
quotient = 95
Remainder =10

KSEEB Solutions

4. 2,647 ÷ 13

KSEEB Solutions for Class 5 Maths Chapter 2 Division 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

KSEEB Solutions for Class 5 Maths Chapter 2 Division 14
Dividend = (divisor × quotient) + Remainder
24658 = (2 × 12329) + 0
24658 = 24658 + 0
24658 = 24658
quotient = 12329
Remainder = 0

2. 14,005 ÷ 7

KSEEB Solutions for Class 5 Maths Chapter 2 Division 15
Dividend = (divisor × quotient) + Remainder
14005 = (7 × 2000) + 5
14005 = 14000 + 5
14005 = 14005
quotient = 2000
Remainder = 5

3. 32,745 ÷ 10

KSEEB Solutions for Class 5 Maths Chapter 2 Division 16
Dividend = (divisor × quotient) + Remainder
32745 = (10 × 3274) + 5
32745 = 32740 + 5
32745 = 32745
quotient = 3274
Remainder =5

KSEEB Solutions

4. 12,056 ÷ 12

KSEEB Solutions for Class 5 Maths Chapter 2 Division 17
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

KSEEB Solutions

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
KSEEB Solutions for Class 5 Maths Chapter 2 Division 18

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
KSEEB Solutions for Class 5 Maths Chapter 2 Division 19

KSEEB Solutions

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 Solutions for Class 5 Maths Chapter 2 Division 20

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
KSEEB Solutions for Class 5 Maths Chapter 2 Division 21

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
KSEEB Solutions for Class 5 Maths Chapter 2 Division 22

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
KSEEB Solutions for Class 5 Maths Chapter 2 Division 23

KSEEB Solutions

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