Question

6. The digit at the tens place of a two-digit number is four times that is the units place. If the digits
reversed, the new number will be 54 less than the original number. Find the number​

Answer 1

Let the unit digit of original no. is y and ten's digit is x

hence the two digit no.= 10x+y

now according to the question

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

if the digits are reversed the new no. become 10y+x

Hence

10y+x=10x+y-54.............................(2)

Put the value of x from 1 in 2 we get

10y+4y=40y+y-54

27y=54

y=2

Therefore x=8

Answer 2

Given:-

• Sum of digits of the two digits number is four times that in the unit's place.
• If the digits are reversed the new number will be 54 less that the original number.

To find:-

• Find the original number..?

Solutions:-

• Let the digits at unit place be y.
• Let the digits at ten's place be x.

Number = 10x + y

Sum of digits of the two digits number is four times that in the unit's place.

=> x + y = 4y

=> x = 4y - y

=> x = 3y ......(i).

If the digits are reversed the new number will be 54 less that the original number.

Number obtained by reversing the digits = 10y + x

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

=> 10y + x = 10x + y - 54

=> 54 = 10x + y - 10y - x

=> 54 = 9x - 9y

=> 54 = 9(x - y)

=> 54/9 = x - y

=> 6 = x - y .........(ii).

Putting the value of x from Eq (i). in Eq (ii).

=> 6 = x - y

=> 6 = 3y - y

=> 6 = 2y

=> y = 6/2

=> y = 3

Putting the value of y in Eq (ii).

=> 6 = x - y

=> 6 = x - 3

=> -x = -6 - 3

=> -x = -9

=> x = 9

So, Number = 10x + y

=> 10(9) + 3

=> 90 + 3

=> 93

Hence, the original number is 93.