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Answers
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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Hope it helps
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