Question

8. The digit at the tens place of a two-digit number is 3 times the digit at the units place. If the digits
are reversed, the new number will be 36 less than the original number. Find the number​

Answer 1

EXPLANATION.

Let the digit at ten's place be = x

Let the digit at unit place be = y

original number = 10x + y

reversing number = 10y + x

To find the number.

According to the question,

Case = 1.

The digit at the ten's place of a two digit

number is 3 times the digit at the unit place.

=> x = 3y ........(1)

Case = 2.

If the digit are reversed the new number will

be 36 less then the original number.

=> ( 10x + y) - 36 = ( 10y + x)

=> ( 10x + y) - ( 10y + x) = 36

=> 10x + y - 10y - x = 36

=> 9x - 9y = 36

=> x - y = 4 .......(2)

From equation (1) and (2) we get,

put the value of equation (1) in equation (2)

we get,

=> 3y - y = 4

=> 2y = 4

=> y = 2

put the value of y = 2 in equation (1)

we get,

=> x = 3 X 2 = 6

Therefore,

original number = 10x + y

=> 10(6) + 2 = 62.

Answer 2

Answer:

62 is the number required

Step-by-step explanation:

let the digit in tens place be = x

let the digit in unit place be = y

the digit in tens place is 3 times the digit at units place

ie : x = 3y

the required number is in the form of 10x + y

if the digits are reversed it becomes = 10y + x

which is 36 less than the original number

ie : 10x + y = 10y + x + 36

10x + y - 10y - x = 36

9x - 9y = 36

9(x - y) = 36

x - y = 36/9

x - y = 4 equ (1)

3y - y = 4

2y = 4

y = 4/2

y = 2

x = 3y

x = 3 × 2

x = 6

the required number is 10x + y

= 10(6) + 2

= 60 + 2

= 62

verification

62 - 36 = 26

Hope you get your answer