Question

Using $a^2-b^2=(a+b)(a-b)$, find
i) ${101}^2-{99}^2$
ii) $(10.3{)}^2-(9.7{)}^2$

• A. (i) $300$ (ii) $12$
• B. (i) $400$ (ii) $12$
• C. (i) $400$ (ii) $20$
• D. None

Answer 1

The correct option is B (i) $400$ (ii) $12$

$a^2-b^2=(a+b)(a-b)$

(1)

If $a=101$ and $b=99$

${101}^2-{99}^2=(101+99)(101-99)$

$=200\times 2=400$

(2)

If $a=10.3$ and $b=9.7$

$(10.3{)}^2-(9.7{)}^2=(10.3+9.7)(10.3-9.7)$

$=20\times 0.6=12$