A two digit number has sum of it's digits equal to 6.Now digits are reversed and the difference between two numbers is 18.Find the number.
Answers
Answer:
Let the 2 digit number as(10a+b)
Number obtain by reversing the digit=(10b+a)
Given,
( a+b=11)~equation-1
also,
=(10a+b)-(10b+a)=9
=10a+b-10b-a=9
=9a-9b=9
=9(a-b)=9
=(a-b)=9–9
=a-b=1
=a+b+a-b=11+1
=2a=12
=a=6
Putting the value of a in equation 1, we get
=6+b=11
=b=11–6
=b=5
Original number=10×6+5=60+5=65
So the require number is 65
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
Let the no. at ones place be x.
No. at tens place = 6-x
No. = 10(6-x) + x = 60 - 10x + x
= 60-9x
When digits are reversed.
No. at ones place = 6-x.
No. at tens place = 10x
No. = 6-x + 10x = 6 - 9x
Now,
60 - 9x - 6 - 9x = 18
54 - 18x = 18
- 18x = 18 - 54
-18x = -36 (Cancelling Negative sign from both side)
18x = 36
x = 36 / 18 = 2
One's Place = x = 2
Ten's Place = 6-x = 6 - 2 = 4
No. = 24 Ans.
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