Question

the sum of the digits of a two digit number and the number obtained by reversing the digits is110.if the digits of the number differ by 10, find the number,how many such number are........ ✨ spammers will be reported immediately ​

Answer 1

Answer:

let the unit place digit be x and tens place digit be y

then the two-digit number will be 10y + x

and the number formed by interchanging the unit place and tens place digits will be 10x + y

according to the first condition given in the qs i.e, the sum of two numbers is 110 that is

10y + x + 10x + y = 110

=> 11x + 11y = 110

divide the above equation by 11 we get

x + y = 10

x = 10 - y ....(i)

now according to the second equation,

if 10 is subtracted from the first number i.e, the new number is 10y + x - 10

given that the new number is 4 more than 5 time the sum of its digits in the first number i.e

the sum of its digits in the first number is x + y, now 5 times of its, 5(x + y), and now 4 more that is, 4 + 5(x + y)

therefore new number = 4 + 5(x + y)

10y + x - 10 = 4 +5(x + y)

10y - 5y + x = 4 +10 +5x

5y = 14 + 4x.....(ii)

substitute the value of x from eq(i) to eq (ii)

we get , 5y = 14 + 4(10 - y)

5y = 14 + 40 - 4y

y = 6

and from eq(i)

x = 4

then the first number 10y + x = 10x6 + 4 = 64

first number is 64.

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