Question

2. A number consists of two digits. The digit in the ten's place
is Greater than unit place by 4 sum of square of digits
of the number is 15 less than the number. Find the number​

Answer 1

$\bf{\huge{\underline{\boxed{\underline{\sf{\red{Answer:}}}}}}}$

$\bf{\underline{\underline{\sf{\purple{Given:}}}}}$

• A number consists of two digits.
• The digit in the ten's place
• is greater then the digit in unit place by 4
• Sum of square of digits of the number is 15 less than the number.

$\bf{\underline{\underline{\sf{\purple{To\:find:}}}}}$

• The number.

$\bf{\underline{\underline{\sf{\purple{Solution:}}}}}$

Let the digit in the tens place be x

Let the digit in the units place be y.

Original number = 10x + y

$\bf{\underline{\underline{\sf{\green{As\:per\:first\:condition:}}}}}$

• The digit in the ten's place
• is greater then the digit in unit place by 4

Representing the condition mathematically.

=> Tens place = units place + 4

=> x = y + 4 ----> 1

$\bf{\underline{\underline{\sf{\green{As\:per\:second\:condition:}}}}}$

• Sum of square of digits of the number is 15 less than the number.

Representing the second condition mathematically.

=> x² + y² = 10x + y - 15

Substitute value of x from equation 1,

=> (y + 4)² + y² = 10 (y + 4) + y - 15

Simplify the bracket on LHS using the identity : (a + b)²

=> y² + 2 (y) (4) + 4² + y² = 10y + 40 + y - 15

=> y² + 2y (4) + 16 + y² = 10y + y + 40 - 15

=> y² + 8y + 16 + y² = 11y + 25

=> y² + y² + 8y - 11y = 25 - 16

=> 2y² - 3y = 9

=> 2y² - 3y - 9 = 0

Solving the equation further using factorization method.

=> 2y² - 6y + 3y - 9 = 0

=> 2y ( y - 3) + 3 ( y - 3) = 0

=> ( y - 3) ( 2y + 3) = 0

=> y - 3 = 0 OR 2y + 3 = 0

=> y = 3 OR 2y = - 3

=> y = 3 OR y = $\large\frac{-3}{2}$

y = $\large\frac{-3}{2}$ is not acceptable.

•°• y = 3

$\bf{\large{\underline{\boxed{\sf{\blue{Digit\:in\:the\:tens\:place\:=\:x\:=\:y\:+\:4=\:3+\:4\:=\:7}}}}}}$

$\bf{\large{\underline{\boxed{\sf{\blue{Digit\:in\:the\:units\:place\:=\:y\:=\:3}}}}}}$

$\bf{\large{\underline{\boxed{\sf{\blue{Original\:Number\:=\:10x\:+\:y\:=\:10\times\:7\:+3\:=73}}}}}}$

$\bf{\huge{\underline{\boxed{\underline{\sf{\red{Verification:}}}}}}}$

For first case :-

• The digit in the ten's place
• is greater then the digit in unit place by 4

Tens place = x = 7

Units place = y = 3

=> x = y + 4

=> 7 = 3 + 4

=> 7 = 7

LHS = RHS.

For second case :-

• Sum of square of digits of the number is 15 less than the number.

Digit in tens place = x = 7

Digit in units place = y = 3

Number = 73 = 10x + y

=> x² + y² = 10x + y - 15

=> 7² + 3² = 73 - 15

=> 49 + 9 = 73 - 15

=> 58 = 58

LHS = RHS

Hence verified.

Answer 2

$\huge{\boxed{\underline{\blue{Answer :-}}}}$

$\sf{\large{\underline{\boxed{\red{Digit\:in\:the\:tens\:place\:=\:x\:=\:3+\:4\:=\:7}}}}}$

$\sf{\large{\underline{\boxed{\red{Digit\:in\:the\:units\:place\:=\:y\:=\:3}}}}}$

$\sf{\large{\underline{\boxed{\red{Original\:Number\:=\:10x\:+\:y\:=\:10\times\:7\:+3\:=73}}}}}$

Assuming :-

Let the digit in the tens place be x

Let the digit in the units place be u

Original number = 10x + y

$\huge{\bf{\underline{\underline{\blue{Case \: 1:-}}}}}$

The digit in the ten's place

is greater then the digit in unit place by 4

A.T.Q

Tens place = units place + 4

x = y + 4 .......(1)

$\huge{\bf{\underline{\underline{\blue{Case \:2:-}}}}}$

Sum of square of digits of the number is 15 less than the number.

A.T.Q

x² + y² = 10x + y - 15

Substitute value of x from equation 1,

______________[Put Values]

(y + 4)² + y² = 10 (y + 4) + y - 15

Using Identity

$\huge{\boxed{\underline {\red{(a + b)^{2} = a^{2} + b^{2} + 2(a)(b)}}}}$

_________________[Put Values]

y² + 2 (y) (4) + 4² + y² = 10y + 40 + y - 15

y² + 8 (y) + 16 + y² = 10y + y + 40 - 15

y² + 8y + 16 + y² = 11y + 25

y² + y² + 8y - 11y = 25 - 16

2y² - 3y = 9

2y² - 3y - 9 = 0

By Factorisation

2y² - 6y + 3y - 9 = 0

2y(y - 3) + 3(y - 3) = 0

(2y + 3) (y - 3) = 0

So, y can be

y = -3/2 or y = -3