Given statement
The sum of any two successive odd numbers is always divisible by 4.
Example:
Let any two successive odd numbers are
(i) 13, 15 (ii) 17, 19
(i) Sum = 13 + 15 = 28 is divisible by 4.
(ii) Sum = 17 + 19 = 36 is divisible by 4.
Therefore given statment is true.

Given
Wilson, Mathan and Guna can complete 1 round of a circular track in 10,15 and 20 minutes respectively.
Starting point at 7 a.m.

LCM = \(5 \times 2 \times 3 \times 2\) = 60 min = 1 hour = 1a.m.
⇒ 7a.m. + 1a.m. = 8a.m.
Therefore after 60 minutes at 8 a.m. will they meet together again at the starting point.

Given
The product of 2 two digit numbers = 300, HCF = 5
300 = \(15 \times 20=60 \times 5=100 \times 3=2 \times 150=75 \times 4\)
But, given that, two digit numbers.
Therefore 60 \(\times\) 5, 100 \(\times\) 3, 2 \(\times\) 150, 75 \(\times\) 4 is not possible.
300 = 15 \(\times\) 20

HCF = 5
Therefore numbers are 15, 20.

This is a HCF related problem. So, we need to find the HCF of 175, 245 and 385.

175 = 5 \(\times\) 5 \(\times\) 7; 245 = 5 \(\times\) 7 \(\times\) 7; 385 = 5 \(\times\) 7 \(\times\) 11
The HCF of 175, 245 and 385 is the product of the common factors 5 and 7 i.e, 5 \(\times\) 7=35
Since each box contains equal number of books, the greatest possible number of boxes = 35
The number of English books in each box = 175 \(\div \) 35 = 5
The number of Science books in each box = 245 \(\div \) 35 = 7
The number of Maths books in each box = 385 \(\div \) 35 = 11
Hence, the total number of books in each box is 5+7+11 = 23.

                   
The product of common factors of 40 and 56
= 2 \(\times\) 2 \(\times\) 2 = 8 and so, HCF (40, 56) = 8
Dividing by the common factor 2, (in 3 steps)
HCF = Product of common factors
 = 2 \(\times\) 2 \(\times\) 2 = 8