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The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.?
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The sum of 2 digit no. and no. obtained by reversing the order of digi...
Let first number be....10x+yon reversing it become... 10y+x(10a+b..is General representation for two digit number )adding both..we get 99..[(10x+y)+(10y+x)] = [11x+11y] = 99x+y = 99/11 = 9also it is given that....x-y = 3(assuming. x>y)x+y = 9....&....x-y = 3on solving these two....we will get ....x=6 & y=3hence number is 63
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The sum of 2 digit no. and no. obtained by reversing the order of digi...
Problem Statement:
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits of the number differ by 3, find the number.
Analysis:
Let the two-digit number be represented as 10a + b, where a and b are the digits.
According to the problem, the number obtained by reversing the digits is 10b + a.
Given that the digits differ by 3, we have two cases to consider:
Case 1: a > b
Case 2: b > a
Solution:
Case 1: a > b
Since the digits differ by 3, we have a = b + 3.
The sum of the two numbers is 10a + b + 10b + a = 11(a + b) = 99.
Substitute a = b + 3 into the equation: 11(b + 3 + b) = 99.
Solve for b: 22b + 33 = 99 => 22b = 66 => b = 3.
Substitute b = 3 back into a = b + 3: a = 3 + 3 = 6.
Therefore, the number is 63.
Case 2: b > a
Similarly, if b > a, we have b = a + 3.
Proceed with the same steps as in Case 1 to find the number.
Conclusion:
The two-digit number that satisfies the given conditions is 63.
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits of the number differ by 3, find the number.
Analysis:
Let the two-digit number be represented as 10a + b, where a and b are the digits.
According to the problem, the number obtained by reversing the digits is 10b + a.
Given that the digits differ by 3, we have two cases to consider:
Case 1: a > b
Case 2: b > a
Solution:
Case 1: a > b
Since the digits differ by 3, we have a = b + 3.
The sum of the two numbers is 10a + b + 10b + a = 11(a + b) = 99.
Substitute a = b + 3 into the equation: 11(b + 3 + b) = 99.
Solve for b: 22b + 33 = 99 => 22b = 66 => b = 3.
Substitute b = 3 back into a = b + 3: a = 3 + 3 = 6.
Therefore, the number is 63.
Case 2: b > a
Similarly, if b > a, we have b = a + 3.
Proceed with the same steps as in Case 1 to find the number.
Conclusion:
The two-digit number that satisfies the given conditions is 63.
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The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.?
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The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.?.
The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of 2 digit no. and no. obtained by reversing the order of digit is 99 . If the digit of no. differ by 3 ,find the no.?.
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