Question

. Solve by Expanding the brackets:-
(i) 12 x 106 (ii) 105 x 101 (iii) 3 x 102

Answer 1

Answer:

(12×106)=1272. (102×3)=306 (105×101)=10605

Answer 2

$\underline { \pink { Distributive \: Property: }}$

$\boxed { \blue { a(b+c) = a \times b + a\times c }}$

$i ) 12 \times 106 \\= 12 (100 + 6 ) \\= 12 \times 100 + 12 \times 6 \\= 1200 + 72 \\= 1272$

$\therefore \red{ 12 \times 106} \green {= 1272}$

$\underline { \pink { By \: Algebraic\: Identity }}$

$\boxed { \blue {(x+a)(x+b) = x^{2} + (a +b)x + ab }}$

$\red{ii) Value \:of \: 105 \times 101 } \\= ( 100 + 5 )( 100 + 1 ) \\= 100^{2} + ( 5 + 1 )\times 100 + 5 \times 1 \\= 10000 + 600 + 5 \\= 10605$

$\therefore \red{ 105 \times 101} \green {= 10605}$

$iii ) 3 \times 102 \\= 3 (100 + 2 ) \\= 3 \times 100 + 3 \times 2 \\= 300 + 6 \\= 306$

$\therefore \red{ 3 \times 102} \green {= 306}$

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