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Friday 5 May 2017
CBSE Class 6 Maths - Playing With Numbers - NCERT Exercise 3.7 (#cbsenotes)
Playing With Numbers
EXERCISE 3.7
CBSE Class 6 Maths
Q1: Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
Answer:
The maximum weight = HCF of 75 and 69.
Factors of 75 = 3 ✕ 5 ✕ 5
Factors of 69 = 3 ✕ 69
HCF = 3
∴ the required weight is 3 kg.
Q2: Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Thursday 4 May 2017
CBSE Class 7 Maths - Integers - NCERT Exercise 1.2 (#cbsenotes)
Integers
EXERCISE 1.2
CBSE Class 7 Maths
Q1: 1. Write down a pair of integers whose:
(a) sum is –7
(b) difference is –10
(c) sum is 0
Answer:
(a) One such pair whose sum is -7 = -4 + (-3) = -7
(b) One such pair whose difference is -10 = -3 - 7 = 10
(c) One such pair whose sum is 0 = -5 + 5 = 0
Wednesday 3 May 2017
Tuesday 2 May 2017
CBSE Class 6 Maths - Playing With Numbers - NCERT Exercise 3.6 (#cbsenotes)
Playing With Numbers
EXERCISE 3.6
CBSE Class 6 Maths
Q1: Find the HCF of the following numbers:
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Answer:
(a) Factors of 18 = 2 × 3 × 3
Factors of 48 = 2 × 2 × 2 × 2 × 3
HCF (18, 48) = 2 × 3 = 6
(b) Factors of 30 = 2 × 3 × 5
Factors of 42 = 2 × 3 × 7
HCF (30, 42) = 2 × 3 = 6
(c) Factors of 18 = 2 × 3 × 3
Factors of 60 = 2 × 2 × 3 × 5
HCF (18, 60) = 2 × 3 = 6
Sunday 30 April 2017
CBSE Class 10 - Physics - Electricity - Important Points To Remember (#cbsenotes)
Electricity - Important Points To Remember
CBSE Class 10 - Physics
① Electric current is defined as the amount of charge flowing through a particular cross section area in a unit time.
② The SI unit of electric current is ampere.
③ A stream of electrons moving through a conductor constitutes an electric current. In a conventional current, the direction of current is taken opposite to the direction of flow of electrons i.e. flow of positive charges from +ve terminal to -ve terminal.
④ 1 A is the flow of 6.25 × 10¹⁸ electrons per second, or 1 coulomb per second.
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