Exercise 12.1 : Solutions of Questions on Page Number : 197
Q1 : Evaluate
(i) 3 – 2 (ii) ( – 4) – 2 (iii)
Answer :
(i)
(ii)
(iii)
Q2 :Simplify and express the result in power notation with positive exponent.
(i)
(ii)
(iii)
(iv)
(v)
Answer :
(i) ( – 4)5 ÷( – 4)8 = ( – 4)5 – 8 (am ÷ an = am – n)
= ( – 4) – 3
(ii)
(iii)
(iv) (3 – 7 ÷ 3 – 10) × 3 – 5 = (3 – 7 – ( – 10)) × 3 – 5 (am ÷ an = am – n)
= 33 × 3 – 5
= 33 + ( – 5) (am × an = am + n)
= 3 – 2
(v) 2 – 3 × ( – 7) – 3 =
Q3 : Find the value of.
(i) (30 + 4 – 1) × 22
(ii) (2 – 1 × 4 – 1) ÷2 – 2
(iii)
(iv) (3 – 1 + 4 – 1 + 5 – 1)0
(v)
Answer :
(i)
(ii) (2 – 1 × 4 – 1) ÷ 2 – 2 = [2 – 1 × {(2)2} – 1] ÷ 2 – 2
= (2 – 1 × 2 – 2) ÷ 2 – 2
= 2 – 1+ ( – 2) ÷ 2 – 2 (am × an = am+ n)
= 2 – 3 ÷ 2 – 2
= 2 – 3 – ( – 2) (am ÷ an = am – n)
= 2 – 3 + 2 = 2 – 1
(iii)
(iv) (3 – 1 + 4 – 1 + 5 – 1)0
= 1 (a0 = 1)
(v)
Q4 : Evaluate (i)
(ii)
Answer :
(i)
(ii)
Q5 : Find the value of m for which 5m ÷5-3 = 55.
Answer :
5m ÷ 5-3 = 55
5m – (- 3) = 55 (am ÷ an = am– n)
5m+ 3 = 55
Since the powers have same bases on both sides, their respective exponents must be equal.
m + 3 = 5
m = 5 – 3
m = 2
Q6 : Evaluate (i)
(ii)
Answer :
(i)
(ii)
Q7 : Simplify. (i)
(ii)
Answer :
(i)
(ii)
Exercise 12.2 : Solutions of Questions on Page Number : 200
Q1 :Express the following numbers in standard form.
(i) 0.0000000000085 (ii) 0.00000000000942
(iii) 6020000000000000 (iv) 0.00000000837
(v) 31860000000
Answer :
(i) 0.0000000000085 = 8.5 x 10-12
(ii) 0.00000000000942 = 9.42 x 10-12
(iii) 6020000000000000 = 6.02 x 1015
(iv) 0.00000000837 = 8.37 x 10-9
(v) 31860000000 = 3.186 x 1010
Q2 :Express the following numbers in usual form.
(i) 3.02 x 10-6 (ii) 4.5 x 104
(iii) 3 x 10-8 (iv) 1.0001 x 109
(v) 5.8 x 1012 (vi) 3.61492 x 106
Answer :
(i) 3.02 x 10-6 = 0.00000302
(ii) 4.5 x 104 = 45000
(iii) 3 x 10-8 = 0.00000003
(iv) 1.0001 x 109 = 1000100000
(v) 5.8 x 1012 = 5800000000000
(vi) 3.61492 x 106 = 3614920
Q3 :Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 
m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
Answer :
(i) = 1 × 10 – 6
(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10 – 19
(iii) 0.0000005 = 5 × 10 – 7
(iv) 0.00001275 = 1.275 × 10 – 5
(v) 0.07 = 7 × 10 – 2
Q4 :In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Answer :
Thickness of each book = 20 mm
Hence, thickness of 5 books = (5 x 20) mm = 100 mm
Thickness of each paper sheet = 0.016 mm
Hence, thickness of 5 paper sheets = (5 x 0.016) mm = 0.080 mm
Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets
= (100 + 0.080) mm
= 100.08 mm
= 1.0008 x 102 mm