Question

3. The sum of a two-digit number and the
number obtained by interchanging its digits
is 110. If the original number exceeds six
times the sum of its digits by 4, find the
original number. Solve the equations by the
cross-multiplication method.
​

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.

HOPE IT HELPS YOU............................