Question

Sum of digits of the two digit number is four times that in the units place if the digit are reversed the new number will be 54 less than the original number find the number

Answer 1

GIVEN:

• Sum of digits of the two digit number is four times that in the units place.
• If the digit are reversed the new number will be 54 less than the original number.

TO FIND:

• What is the original number ?

SOLUTION:

Let the digit at unit's place be 'y' and the digit at ten's place be 'x'

◘ NUMBER = 10x + y

CASE:- 1

◆ Sum of digits of the two digit number is four times that in the units place.

$\rm{\hookrightarrow x + y = 4y }$

$\rm{\hookrightarrow x = 4y - y }$

$\bf{\hookrightarrow x = 3y....1)}$

CASE:- 2

◆ If the digit are reversed the new number will be 54 less than the original number.

▣ Number obtained by reversing the digits = 10y + x

▣ Number obtained by reversing the digits = 10x + y - 54

According to question:-

$\rm{\hookrightarrow 10y + x = 10x + y - 54}$

$\rm{\hookrightarrow 54 = 10x +y- 10y - x }$

$\rm{\hookrightarrow 54 = 9x - 9y }$

$\bf{\hookrightarrow 6 = x - y ....2)}$

Put the value of 'x' from equation 1 in equation 2

$\rm{\hookrightarrow 6 = 3y - y }$

$\rm{\hookrightarrow 6 = 2y}$

$\rm{\hookrightarrow \cancel\dfrac{6}{2} = y }$

$\bf{\hookrightarrow 3 = y }$

Put the value of 'y' in equation 2

$\rm{\hookrightarrow 6 = x - 3}$

$\rm{\hookrightarrow 6 + 3 = x }$

$\bf{\hookrightarrow 9 = x }$

• Number = 10x + y
• Number = 10(9)+3
• Number = 90 + 3
• Number = 93

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