Tuesday, November 7, 2017

CBSE Class 7 - Mathematics - Fractions and Decimals NCERT Exercise 2.6 (#eduvictors) (#cbsenotes) (#ncertanswers)

Mathematics - Fractions and Decimals - NCERT Exercise 2.6

CBSE Class 7 - Mathematics - Fractions and Decimals NCERT Exercise 2.6 (#eduvictors) (#cbsenotes) (#ncertanswers)

Question 1: Find:
(i) 0.2 ✕ 6 (ii) 8 ✕ 4.6 (iii) 2.71 ✕ 5
(iv) 20.1 ✕ 43 (v) 0.05 ✕ 7 (vi) 211.02 ✕ 4
(vii) 2 ✕ 0.86


Answer:
(i) 0.2 ✕ 6 = 1.2
(ii) 8 ✕ 4.6 = 36.8
(iii) 2.71 ✕ 5 = 13.55
(iv) 20.1 ✕ 4 = 80.4
(v) 0.05 ✕ 7 = 0.35
(vi) 211.02 ✕ 4 = 844.08
(vii) 2 ✕ 0.86 = 1.72




Question 2: Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Answer:
Given:
Length of rectangle = 5.7 cm and
Breadth of rectangle = 3 cm
Area of rectangle = Length ✕ Breadth
= 5.7 ✕ 3 = 17.1 cm 2

Thus, the area of rectangle is 17.1 cm 2 .


Question 3: Find
(i)   1.3 ✕ 10 (ii) 36.8 ✕ 10 (iii) 153.7 ✕ 10
(iv)  168.07 ✕ 10 (v) 31.1 ✕ 100 (vi) 156.1 ✕ 100
(vii) 3.62 ✕ 100 (viii) 43.07 ✕ 100 (ix) 0.5 ✕ 10
(x)   0.08 ✕ 10 (xi) 0.9 ✕ 100 (xii) 0.03 ✕ 1000

Answer:
(i) 1.3 ✕ 10 = 13.0
(ii) 36.8 ✕ 10 = 368.0
(iii) 153.7 ✕ 10 = 1537.0
(iv) 168.07 ✕ 10 = 1680.7
(v) 31.1 ✕ 100 = 3110.0
(vi) 156.1 ✕ 100 = 15610.0
(vii) 3.62 ✕ 100 = 362.0
(viii) 43.07 ✕ 100 = 4307.0
(ix) 0.5 ✕ 10 = 5.0
(xi) 0.9 ✕ 100 = 90.0
(x) 0.08 ✕ 10 = 0.80
(xii) 0.03 ✕ 1000 = 30.0


Question 4: A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?

Answer: ∵ In 1 litre, a two-wheeler covers a distance = 55.3 km
∴ In 10 litres, a two-wheeler covers a distance = 55.3 ✕ 10 = 553.0 km
Thus, 553 km distance will be covered by it in 10 litres of petrol.



Question 5: Find
(i) 2.5 ✕ 0.3 (ii) 0.1 ✕ 51.7 (iii) 0.2 ✕ 316.8
(iv) 1.3 ✕ 3.1 (v) 0.5 ✕ 0.05 (vi) 11.2 ✕ 0.15
(vii) 1.07 ✕ 0.02 (viii) 10.05 ✕ 1.05 (ix) 101.01 ✕ 0.01
(x) 100.01 ✕ 1.1

Answer:
(i) 2.5 ✕ 0.3 = 0.75
(ii) 0.1 ✕ 51.7 = 5.17
(iii) 0.2 ✕ 316.8 = 63.36
(iv) 1.3 ✕ 3.1 = 4.03
(v) 0.5 ✕ 0.05 = 0.025
(vi) 11.2 ✕ 0.15 = 1.680
(vii) 1.07 ✕ 0.02 = 0.0214
(viii) 10.05 ✕ 1.05 = 10.5525
(ix) 101.01 ✕ 0.01 = 1.0101
(x) 100.01 ✕ 1.1 = 110.11

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