Question

in 2 digit number the digit at unit place sq. of the at 10 place in if 18 is added to the number the gate in the change find the no.​

Answer 1

Correct Question:

In a two digit number the digit at units place is equal to the square of the digit at tens place if 18 is added to the number the digit get interchanged. find the number?

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☯ Let the digit at ten's place be x.

Therefore, The digit at ones place will be x².

And,

• The number become = 10x + x²

Now,

★ According to the Question:

• If 18 is added to the number the digit get interchanged.

➯ 10x + x² + 18 = 10x² + x

➯ 10x - x + 18 = 10x² - x²

➯ 9x + 18 = 9x²

➯ 9x² - 9x - 18 = 0

➯ 9(x² - x - 2) = 0

➯ x² - x - 2 = 0

➯ x² - 2x + x - 2 = 0

➯ x(x - 2) + 1(x - 2) = 0

➯ (x - 2)(x + 1) = 0

Either (x - 2) = 0 or (x + 1) = 0

➯ x = 2 or - 1

Since, Value of x shouldn't be negative.

➯ x = 2

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Therefore, The required number is,

➯ 10x + x²

➯ 10 × 2 + (2)²

➯ 20 + 4

➯ 24

∴ Hence, The required two digit number is 24.

Answer 2

${\bold{\sf{\underline{Correct \; question}}}}$

In a 2 digit number the digit at unit place is equal to the square of the at tens place, if 18 is added to that number the digits get interchanged. Find the number ?

${\bold{\sf{\underline{Understanding \; the \; question}}}}$

➨ This question says that in a two digit number the digit at unit place is equal to the square of the tens place , if 18 is add to that number the digits get interchanged. We have to find the original number.

${\bold{\sf{\underline{Given \; that}}}}$

➨ Two digit number the digit at unit place is equal to the square of the tens place , if 18 is add to that number the digits get interchanged

${\bold{\sf{\underline{To \; find}}}}$

➨ Original number

${\bold{\sf{\underline{Solution}}}}$

➨ Original number = 24

${\bold{\sf{\underline{Assumptions}}}}$

➨ Let tens place be a and the unit place be a²

${\bold{\sf{\underline{Full \: solution}}}}$

~ Seeming the question

➨ Tens place is a

➨ The unit place is a²

➨ Hence, the number is 10a + a²

~ Interchanging the values and add 18

➨ 10a + a² + 18 = 10a² + a

➨ 0 = 10a² + a - 10a - a² - 18

➨ 0 = 9a² - 9x - 18

~ Let's divide

➨ 0 = a² - a - 2

➨ a² - a - 2 = 0

➨ a² - 2a + a - 2 = 0

➨ a(a-2) + 1(a-2) = 0

➨ (a-2) (a-1) = 0

~ Therefore,

➨ (a-2) or (a-1) = 0

➨ a = 2 or a = -1

~ Remember : a is never negative !

~ Therefore,

➨ a = 2

• Henceforth, value of a is 2

~ Now putting the values to find original number

➨ 10a + a²

➨ 10(2) + 2²

➨ 20 + 4

➨ 24

• Henceforth, original number is 24