The sum of a 2 digit …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Test
Related Questions
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 1 answers
Posted by Hari Anand 6 months, 2 weeks ago
- 0 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Kanika . 1 month, 1 week ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 6 months ago
Let the ten's and unit's digits of the required number be x and y respectively.
Then, the number = (10x + y).
The number obtained on reversing the digits = (10y + x).
As per given condition
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99.
{tex}\therefore{/tex} (10y + x) + (10x + y) = 99
{tex}\Rightarrow{/tex} 11(x + y) = 99
{tex}\Rightarrow{/tex} x + y = 9.
The digits differ by 3
So, (x - y) = ±3.
Thus, we have
x + y = 9 ........ (i)
x - y = 3 .......... (ii)
or x + y = 9 ......... (iii)
x - y = -3 ............ (iv)
From (i) and (ii), we get x = 6, y = 3.
From (iii) and (iv), we get x = 3, y = 6.
Hence, the required number is 63 or 36.
0Thank You