Page 112 - Maths Skills - 7

110 Maths

4. Simplify.

(i) (2 × 7) (ii) (3 × 4) (iii) (– 6 × – 2) 3
3
4
 −1  2  − 1 3
(iv)  ×3  (v) (a × b) x (vi)  8×  2 


 7



5. Evaluate.
   1   2

3

(i) (6 ) (ii) {(–3) } (iii)    (iv) (10 )
2 2
2 3
3 4

   5  
4

   −   2
1
(v)     (vi) (x ) (vii) (b ) (viii) {(–2) }
x 3
m n
2 a
   3  

6. Simplify.
7 
1
1
2
4
6
2
3
3
(i) ( ) ×   2 ×− ( ) ×   8 (ii) ( ) × 2 ×  − 1 10
−3

 
 
3
4
     2 
EXPRESSING LARGE NUMBERS IN THE STANDARD FORM
Here in this section, we explain the method to write a given number in scientific notation or standard form.
Standard form: Any number can be expressed as a decimal number between 1.0 and 10.0 including 1.0 multiplied
by a power of 10. Such a form is called its standard form or scientific notation. It may also be written as k × 10
n
where k is a terminating decimal and 1 ≤ k < 10 and n in an integer.
Let’s Attempt
Example 1: Write the following in scientific notation.
(i) 23 lakh (ii) 60,000,000 (iii) 0.038
Solution: (i) 23 lakh = 23,00,000 = 2.3 × 10 6 (ii) 60,000,000 = 6.0 × 10 (iii) 0.038 = 3.8 × 10 –2
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Example 2: Express the following in the standard form.
(i) 30564235 (ii) 1090856007
Solution: (i) 30564235 = 3.0564235 × 10 (ii) 1090856007 = 1.090856007 × 10 9
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Example 3: Fill in the blanks.
(i) 12 ton = _______ × 10 milligram (ii) 21 terabyte = ________ × 10 byte
(iii) 9 nanometre = _________ × 10 metre (iv) 1 micrometre = _______ × 10 metre
(v) 1 millilitre = ________ × 10 kilolitre

Solution: (i) 12 ton = 12000000000 mg = 1.2 × 10 mg [Q 1 ton = 1000000000 mg]
10
(ii) 21 terabyte = 21000000000000 bytes = 2.1 × 10 bytes [Q 1 terabyte = 1000000000000]
13
(iii) 9 nanometre = 0.0000000090 metre = 9 × 10 m [Q 1 metre = 1000000000 nanometres]
–9
(iv) 1 micrometre = 0.000001 metre = 1 × 10 metre [Q 1 metre = 1000000 micrometres]
–6
(v) 1 millilitre = 0.000001 kilolitre = 1 × 10 kilolitre [Q 1 kilo litre = 1000000 millilitres]
–6