the digit of a two digit no. differ by 3 . if the digits are interchanged and the resulting number is added to the original nymber we get the 143. what can be the original no.
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Hey
Here is your answer,
Let the tens place digit be a and units place digit be b.
So, the actual number will be 10a + b
Digits after interchanging = 10b+a
a- b = 3 ....(i)
10a + b + 10b + a = 143
11a + 11b = 143
a+ b = 13....(ii)
Adding equation (i) and (ii)
we get :
2a =. 16
a= 8
Putting value of a in equation (i) we get:
a - b = 3
8 - b = 3
b = 8 - 3
b = 5
Putting value of a and b in 10a + b
we get :
10(8) + 5 = 85
The original number is 85.
Hope it helps you!
Here is your answer,
Let the tens place digit be a and units place digit be b.
So, the actual number will be 10a + b
Digits after interchanging = 10b+a
a- b = 3 ....(i)
10a + b + 10b + a = 143
11a + 11b = 143
a+ b = 13....(ii)
Adding equation (i) and (ii)
we get :
2a =. 16
a= 8
Putting value of a in equation (i) we get:
a - b = 3
8 - b = 3
b = 8 - 3
b = 5
Putting value of a and b in 10a + b
we get :
10(8) + 5 = 85
The original number is 85.
Hope it helps you!
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