Aptitude - Problems on Numbers - Discussion
Discussion Forum : Problems on Numbers - General Questions (Q.No. 8)
8.
A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:
Answer: Option
Explanation:
Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
Video Explanation: https://youtu.be/lytJE8GqRvM
Discussion:
20 comments Page 1 of 2.
Pattan Bajibabu said:
4 years ago
This very easy;
First, we take two-digit number I am take two digit number is 10,
Interchange positions 01,
Add the values 10+01=11,
So 11 is divisible by 11,
So, the answer is 11.
First, we take two-digit number I am take two digit number is 10,
Interchange positions 01,
Add the values 10+01=11,
So 11 is divisible by 11,
So, the answer is 11.
(13)
Anitha Karrre said:
5 years ago
@Ramya Sana.
Of course, your answer is right but while we dividing a number whether it is the small or big number we will choose the nearest divisor it makes easily.
Suppose, 99/3=33 & again we are dividing 33 by 3 so its very complex right. So that's why we are going take the nearest number which is divisible by that number.
Of course, your answer is right but while we dividing a number whether it is the small or big number we will choose the nearest divisor it makes easily.
Suppose, 99/3=33 & again we are dividing 33 by 3 so its very complex right. So that's why we are going take the nearest number which is divisible by that number.
(1)
Ramya sana said:
7 years ago
For example if we take 27 and when we reverse it, we will get 72 and addition of these two numbers is 99 which will also divisible by 3 and 9 and 11. Then what is the answer. Anybody can help me out?
(2)
Annah said:
8 years ago
(10x+y) + (10y + x)=
For; (10x+y)
Let x be 2and y be 1, substitute in 10x +y,
(10*2) + 1=21.
For; (10y+ x) interchange.
Let x be 2 and y be 1, substitute in (10y+x),
(10*1)+ 2=12.
Then, 12+ 21=33 which is divisible by 11.
For; (10x+y)
Let x be 2and y be 1, substitute in 10x +y,
(10*2) + 1=21.
For; (10y+ x) interchange.
Let x be 2 and y be 1, substitute in (10y+x),
(10*1)+ 2=12.
Then, 12+ 21=33 which is divisible by 11.
Pooja P Kumar said:
8 years ago
@Yogita.
Whenever the product of two numbers is asked.
For example: The product of two number is such that the result is 5.. so we will take same variables i.e x because x*5/x is 5. And mostly in all other cases, we will take two variables x and y.
Whenever the product of two numbers is asked.
For example: The product of two number is such that the result is 5.. so we will take same variables i.e x because x*5/x is 5. And mostly in all other cases, we will take two variables x and y.
Yogita said:
9 years ago
@Santhosh.
I'm pretty confused that when we have to take x, y, and when it is only x for the formation of two digit number?
I'm pretty confused that when we have to take x, y, and when it is only x for the formation of two digit number?
Santosh h said:
9 years ago
11 = 10x+y = 10(1)+1.
Another ex: 25=10x+y = 10(2)+5.
Also: 255=100x+10y+z = 100(2)+10(5)+5.
Another ex: 25=10x+y = 10(2)+5.
Also: 255=100x+10y+z = 100(2)+10(5)+5.
Raja said:
10 years ago
Simple answer.
10 = x.
y = "1" for example.
10x+y = 11.
11 is a answer. So option is D.
10 = x.
y = "1" for example.
10x+y = 11.
11 is a answer. So option is D.
Kunal said:
10 years ago
If we take the no as 99. After reversing the no stays as 99.
Then new no added to original no, we get 99+99 = 198. 198 is divisible by 9 as well as 11.
So can anyone help out about this doubt.
Then new no added to original no, we get 99+99 = 198. 198 is divisible by 9 as well as 11.
So can anyone help out about this doubt.
Vinay said:
10 years ago
When Should we take a 2 digit as 10x+y. And when should we take as x+y? Anybody help?
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