PM ShikSha MiTra

Question 1: Which of the following are  matter? Chair, air, love, smell, hate, almonds, thought, cold, cold drink, smell of perfume. ANSWER: Anything that occupies space and has mass is  called matter. Matter can exist in three physical states—solid, liquid, and gaseous. Chair  and almond are forms of matter in the solid state. Cold drink is a liquid state of matter. Air is a gaseous state of matter. Note : The sense of smell is not matter. However, the smell or odour of a substance is classified as matter. The smell of any substance (say, perfume) can be classified as matter. This is because, perfume is in liquid state inside the bottle. It comes out in the form of tiny droplets. These droplets of perfume are matter and spread out in the atmosphere to create a sense of smell.  This smell can be detected by our olfactory system. Question 2: Give reasons for the following observation: The smell of hot sizzling food reaches you several metres away, but to get the smell...

Chapter 2 : Linear equation in one variable         Exercise : 2.1 Question  1: Solve: x-2=7 Solution : x − 2 = 7 Transposing 2 to R.H.S, we get x = 7 + 2 = 9 Question 2: Solve: y + 3 = 10   Solution : y + 3 = 10 Shifting 3 to R.H.S, we get y = 10 – 3 = 7   Question 3: Solve: 6 = z + 2   Solution : 6 = z + 2 Shifting 2 to L.H.S, we get 6 − 2 = z z = 4   Question 4: Solve:3/7 + x =17/7   Solution : 3/7+ x =17/7 Shifting 3/7 to R.H.S, we get x = 17/7- 3/7 14/7= 2   Question 5: Solve: 6x = 12 Solution : 6x = 12 On dividing both sides by 6, we get 6x/6=12/6 x = 2   Question 6: t/5= 10 Solution : t/5= 10 t = 50   Question 7: Solve: 2x/3=18 Solution : 2x/3=18 Multiply both sides by 3/2 , we get 2x/3 × 3/2 = 18 × 3/2 x =27   Question 8: Solve: 1.6 =y/1.5   Solution : 1.6 =y/1.5 Multiply both sides by 1.5, we get 1.6×1.5 =y/1.5×1.5 2.40 = y y =2.40   Question 9: Solve: 7x – 9 = 16 Solution : 7x − 9 = 16 Shifting 9 t...

L ist of Algebraic formulas   – a 2  – b 2  = (a – b)(a + b) (a + b) 2  = a 2  + 2ab + b 2 (a – b) 2  = a 2  – 2ab + b 2 (a + b + c) 2  = a 2  + b 2  + c 2  + 2ab + 2bc + 2ca (a – b – c) 2  = a 2  + b 2  + c 2  – 2ab + 2bc – 2ca (a + b) 3  = a 3  + 3a 2 b + 3ab 2  + b 3  ;  (a + b) 3  = a 3  + b 3  + 3ab(a + b) (a – b) 3  = a 3  – 3a 2 b + 3ab 2  – b 3  = a 3  – b 3  – 3ab(a – b) a 3  – b 3  = (a – b)(a 2  + ab + b 2 ) a 3  + b 3  = (a + b)(a 2  – ab + b 2 ) (a + b) 4  = a 4  + 4a 3 b + 6a 2 b 2  + 4ab 3  + b 4 (a – b) 4  = a 4  – 4a 3 b + 6a 2 b 2  – 4ab 3  + b 4 a 4  – b 4  = (a – b)(a + b)(a 2  + b 2 ) a 5  – b 5  = (a – b)(a 4  + a 3 b + a 2 b 2  + ab 3  + b 4 ) If n is a natural number  a n  – b n  = (a – b)(a n-1 ...