A NUMBER CONSIST OF TWO DIGIT WHOSE SUM IS 8. IF 18 IS ADDED TO THE NUMBER ITS DIGIT ARE REVERSED.FIND THE NUMBERS .
Answers
Answer:
35 is the original number
Step-by-step explanation:
let the unit digit be y and tens digit be x
10x+y = original number
10y+x = reversed number
x+y = 8
=>x=8-y
ATQ
10x+y+18=10y+x
=> 10(8-y)+y+18=10y+(8-y)
=>80-10y+y+18=10y+8-y
=>98-9y=9y+8
=>9y+9y=98-8
=>18y=90
=>y=90/18
=>y=5
the original number = 10x+y
= 10(8-y)+y
= 10(8-5)+5
= 10(3)+5
= 30+5
= 35
reversed number = 53 (35+18)
Answer:
Let the digit at ones place be ‘x’
Then the digit at tens place = 8 – x (sum of the two digits is 8)
Therefore number 10 (8 – x) + x = 80 – 10x + x = 80 – 9x —— (1)
Now, number obtained by reversing the digits = 10 ×(x) + (8 – x)
= 10x + 8 – x = 9x + 8
It is given that if 18 is added to the number its digits are reversed
∴ number + 18 = Number obtained by reversing the digits
⇒ (80 – 9x) + 18 = 9x + 8
98 – 9x = 9x + 8
98 – 8 = 9x + 9x
90 = 18x
x =
90
18 = 5
By substituting the value of x in equation (1) we get the number
∴ Number = 80 – 9 × 5 = 80 – 45 = 35.