Question

a two digit no. is equal to 7 times the sum of its digits the no. formed by reversing its digit is less than the original no. by 18 find the original no.

Answer 1

Let the digits in ones place be x and ten’s place be y
The original number = 10y + x
Number formed by interchanging the digits = 10x + y
Given (10y + x) = 7(x + y)
Þ 10y + x = 7x + 7y
Þ 7x + 7y – x – 10y = 0
Þ 6x – 3y = 0
Þ 2x – y = 0 → (1)
It is also given that on reversing the digits the number obtained is 18 less than the original number
Hence 10x + y = (10y + x) – 18
Þ 10x + y – 10y – x = – 18
Þ 9x – 9y = – 18
Þ x – y = – 2 → (2)
Subtract (2) from (1)
2x – y = 0
x – y = – 2
-------------
x = 2
Put x = 2 in 2x – y = 0
2(2) – y = 0
∴ y = 4
Therefore the original number is 42.

Answer 2

Answer:

42

Step-by-step explanation:

let original number be (10x+y)

given ,

(10x+y) = 7(x+y)

=> 10x +y = 7x + 7y

=> 10x-7x = 7y -y

=> 3x = 6y

=> x = 2y................1

also , (10x+y) = (10y+x) +18

=> 10x + y = 10y + x + 18

=> 10x-x = 10y-y +18

=> 9x = 9y + 18

=> x = y +2

.

putting value of x from 1

we get,

2y = y + 2

=> y = 2

x = 2y

=> x = 4

so ,the number is (10x+y) = (10×4+2)

number is 42