The sum of the two digit number is7. the number formed by reversing the digit is 45 more than the original number. find the original no.
Answers
let the 1st no be x
2nd no be y
according to the question,
x+y=7 ------------ 1
on reversing the digits ,
10y+x= 10x+y+45
10y - y +x - 10x = 45
9y - 9x = 45
9 (y-x)= 45
y-x = 45/9
y-x = 5
-x + y =5 ------------2
we can find the value of any one variable using elimination method,
so now from 1 and 2 ,
x+y=7
+(-x+y=5)
----------------
2y = 12
y = 12/2
y = 6
now we can substitute the value of y in eq 1
x+y=7
x+ 6=7
x=7-6
x=1
hence , the no is 16
hope this ans helps you :)
Answer:
→ The original number is 16 .
Step-by-step explanation:
Let the unit's digit of the original number be x .
And, the ten's digit of the original number be y .
Now, A/Q,
→ Sum of the two digits number is 7 .
∵ x + y = 7 ............(1) .
Original number = 10x + y .
Number obtained on reversing the digits = 10y + x .
A/Q,
→ The number obtained on reversing the digit is 45 more than the original number .
∵ 10x + y + 45 = 10y + x .
⇒ 10x - x + y - 10y = - 45 .
⇒ 9x - 9y = - 45 .
⇒ 9( x - y ) = - 45 .
⇒ x - y = - 45/9 .
∵ x - y = -5 ...........(2) .
On substracting equation (1) and (2), we get
x + y = 7 .
x - y = -5 .
- + +
________
⇒ 2y = 12 .
⇒ y = 12/2 .
∴ y = 6.
On putting the value of 'y' in equation (1), we get
∵ x + y = 7 .
⇒ x + 6 = 7 .
⇒ x = 7 - 6 .