Question

a two digit number is equal to 7 times the sum of its digit the number formed by reversing its digit is less than the original number is 18 find the original number

Answer 1

Let the digit in ones place be x and ten's place be y
Original number= 10y+x
According to the question
10y+x =7 (x+y)
10y+x = 7x+7y
7x+7y-10y-x=0
6x-3y=0
2x-y=0 .........(1)
According to the question
10x+y=10y+x-18
10x+y-10y-x = -18
9x-9y= -18
x-y= -2 .........(2)
Subtract (2) from (1), we get
x=2
Substitute the value of x in (2), we get
2-y= -2
-y=-2-2
-y = -4
y =4
Therefore the original number is 42

Answer 2

$\rule{200}2$

$\huge\tt{GIVEN:}$

A two digit number is 7 times the sum of its digits.

The number formed by reversing the digits is 18 less than the original number.

$\rule{200}2$

$\huge\tt{TO~FIND:}$

The number

$\rule{200}2$

$\huge\tt{SOLUTION:}$

Let us assume that the two digits of that number are x and y respectively.

so, The Number formed by them would be, 10x+y

It is also stated that, the number is 7 times the sumSo, we can make an equation by these points as -

$\rule{200}1$

↪(10x + y) ____(EQ.1)

↪10x + y = 7(x+y)

↪10x + y = 7x + 7y

↪10x - 7x = 7y - y

↪3x = 6y

↪x = 6/3 y

↪x = 2y ____(EQ.2)

$\rule{200}1$

The number formed by reversing the digits is, 18 less than the original number

↪10x + y - (10y + x) = 18

↪10x + y - 10y - x = 18

↪9x - 9y = 18

↪9(x-y) = 18

↪(x-y) = 18/9

↪(x-y) = 2 ____(EQ.3)

$\rule{200}1$

Now, putting the value of (EQ.1) to (EQ.2),

↪x - y = 2

↪2y - y = 2

↪y = 2

↪x = 2×2 = 4

$\rule{200}1$

Now, as we got our values,

Putting the values of (EQ.2,3) in (EQ.1)

↪THE NUMBER = (10x + y)

↪THE NUMBER = (10 × 4 + 2)

↪THE NUMBER = (40 + 2)

↪THE NUMBER = 42

$\rule{200}2$