Question

The ratio between a two - digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit place is 3 more than the digit in the tens place, what is that number

Answer 1

Solution :-

Let the ones and tens digit of a number be x and y respectively.

Case I : The ratio between a two digit number and the sum of the digits of that number is 4 : 1.

=> (10y + x)/(x + y) = 4/1

=> 10y + x = 4x + 4y

=> 10y - 4y + x - 4x = 0

=> 6y - 3x = 0 ______(i)

Case II : If the digit in the unit place is 3 more than the digit in the tens place.

=> x = y + 3 _____(ii)

Substituting the value of x in equation (i),

=> 6y - 3(y + 3) = 0

=> 6y - 3y - 9 = 0

=> 3y = 9

=> y = 3

Now, Substituting the value of y in equation (ii) we get,

=> x = 3 + 3 = 6

So, Number = 10y + x

= 10 × 3 + 6 = 36

Answer : Number = 36

Answer 2

Answer :-

Consider the ones and tens digit of a number be x and y respectively.

$\bf{\rightarrow} \: \dfrac{(10y + x)}{x + y} = \dfrac{4}{1}$

$\bf{\rightarrow} \:10y + x = 4x + 4y$

$\bf{\rightarrow} \:10y - 4y + x - 4x = 0$

$\bf{\rightarrow} \:6y - 3x = 0.........(1)$

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The digit in the unit place is 3 more than the tens place digit.

$\bf{\rightarrow} \:x = y + 3..........(2)$

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Place the value of x in the 1 Equation

$\bf{\rightarrow} \:6y - 3(y + 3) = 0$

$\bf{\rightarrow} \:6y - 3y + 9 = 0$

$\bf{\rightarrow} \:3y = 9$

$\bf{\rightarrow} \:y = 3$

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Place the value of y in (2)

$\bf{\rightarrow} \:x = 3 + 3$

$\bf{\rightarrow} \:x = 6$

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Finally, the Number is -

$\bf{\rightarrow} \:10y + x$

$\bf{\rightarrow} \:10 \times 3 + 6$

$\bf{\rightarrow} \:36$

.°. The Number is 36

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- Komolika Mukherjee ! ❤️