Students can Download Chapter 2 Fractions and Decimals Ex 2.6, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find :
i) 0.2 × 6
Solution:
0.2 × 6 = 1.2

ii) 8 × 4.6
Solution:
8 × 4.6 = 36.8

iii) 2.71 × 5
Solution:
2.71 × 5 = 13.55

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

iv) 20.1 × 4
Solution:
20.1 × 4 = 80.4

v) 0.05 × 7
Solution:
0.05 × 7 = 0.35

vi) 211.02 × 4
Solution:
211.02 × 4 = 844.08

vii) 2 × 0.86
Solution:
2 × 0.86 = 1.72

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 2.
Find the area of a rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm
Breadth = 3 cm
∴ Area of the rectangle = l × b
= 5.7 × 3
= 17.1 sq. cms.

Question 3.
Find :
i) 1.3 × 10
Solution:
1.3 × 10 = 13.0

ii) 36.8 × 10
Solution:
36.8 × 10 = 368.0

iii) 153.7 × 10
Solution:
153.7 × 10 = 1537

iv) 168.07 × 10
Solution:
168.07 × 10 = 1680.7

v) 31.1 × 100
Solution:
31.1 × 100 = 311.00

vi) 156.1 × 100
156.1 × 100 = 15610.0

vii) 3.62 × 100
3.62 × 100 = 362.00

viii) 43.07 × 100
43.07 × 100 = 4307.00

ix) 0.5 × 10
0.5 × 10 = 5.0

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

x) 0.08 × 10
0.08 × 10 = 0.80

xi) 0.9 × 100
0.9 × 100 = 90.0

xii) 0.03 × 1000
0.03 × 1000 = 30.00

Question 4.
A wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered by a two-wheeler in one litre petrol = 55.3 kms
∴ Distance covered by a two-wheeler in 10 1 petrol = 55.3 × 10 = 553.0 kms.

Question 5.
Find :
i) 2.5 × 0.3
Solution:
2.5 × 0.3 = 0.75

ii) 0.1 × 51.7
Solution:
0.1 × 51.7 = 5.17

iii) 0.2 × 316.8
Solution:
0.2 × 316.8 = 63.36

iv) 1.3 × 3.1
Solution:
1.3 × 3.1 = 14.031

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

v) 0.5 × 0.05
0.5 × 0.05 = 0.025

vi) 11.2 × 0.15
Solution:
11.2 × 0.15 = 1 .680

vii) 1.07 × 0.02
Solution:
1.07 × 0.02 = 0.0214

viii) 10.05 × 1.05
Solution:
10.05 × 1.05 = 10.5525

ix) 101.01 × 0.01
Solution:
101.01 × 0.01 = 1.01011

x) 100.01 × 1.1
Solution:
100.01 × 1.1 = 110.011

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