Question

a number consist of two digit the digits at tens place is two times the digit at the unit place the number formed by reversing the digit is 27 less than the original number find the original number

Answer 1

Let the unit digit be x tense be y
no=10y+x
y=2x
10x+y=10y+x-27
9x-9y+27=0
x-y+3=0
x-2x+3=0
-x+3=0
x=3
y=6

Answer 2

Answer:

The number = 63

Step-by-step explanation:

Given,

In a two-digit number, the digit at tens place is two times the digit at the unit place

The number formed by reversing the digits is 27 less than the original number

Solution:

Let x be the digit in the tens place and y be the digit in the unit place

Then the number is 10x+y

The obtained by reversing the digits = 10y+x

By the given conditions we have,

x = 2y -------------(1)

10y+x = 10x+y -27

10y +x - 10x -y +27 = 0

9y-9x+27 = 0

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

Substituting the value of x from equation(1) in equation(2)

y-2y+3 = 0

-y+3 = 0

y = 3

From equation(1), x = 2y = 2×3 = 6

The tens digit = 6 and unit digit = 3

∴ The required number is 63

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