While using the pattern used to square numbers with unit digit 5 for 105 - we get- - - 11 hundreds -25- Fill in the blank with the correct value-

For a number with unit digit 5, i.e. a5 (a5)^2=(10a+5)^2 ⇒10a(10a+5)+5(10a+5) ⇒100a^2+50a+50a+25 ⇒100a(a+1)+25 ⇒a(a+1)hundred+25 According to the formula, here ...

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Standard VIII

Mathematics

Finding Square of a Number with Unit's Place 5

While using t...

Question

While using the pattern used to square numbers with unit digit 5 for 105, we get: (___ x 11)hundreds + 25. Fill in the blank with the correct value.

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Solution

For a number with unit digit 5, i.e. $a 5$
$(a 5)^{2} = (10 a + 5)^{2}$
$\Rightarrow$ $10 a (10 a + 5) + 5 (10 a + 5)$
$\Rightarrow$ $100 a^{2} + 50 a + 50 a + 25$
$\Rightarrow$ $100 a (a + 1) + 25$
$\Rightarrow$ $a (a + 1) h u n d r e d + 25$
According to the formula, here $a = 10$ .

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While using the pattern used to square numbers with unit digit 5 for 105, we get: (___ x 11)hundreds + 25. Fill in the blank with the correct value.

For a number with unit digit 5, i.e. a5 (a5)2=(10a+5)2 ⇒10a(10a+5)+5(10a+5) ⇒100a2+50a+50a+25 ⇒100a(a+1)+25 ⇒a(a+1)hundred+25 According to the formula, here a=10.

For a number with unit digit 5, i.e. $a 5$
$(a 5)^{2} = (10 a + 5)^{2}$
$\Rightarrow$ $10 a (10 a + 5) + 5 (10 a + 5)$
$\Rightarrow$ $100 a^{2} + 50 a + 50 a + 25$
$\Rightarrow$ $100 a (a + 1) + 25$
$\Rightarrow$ $a (a + 1) h u n d r e d + 25$
According to the formula, here $a = 10$ .