Latest IBM Aptitude Question SOLUTION: A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the student interchanges the digits of one number say ba for ab. Due to this, the

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A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the student interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What is the difference between the two digits a and b?

Asked In IBM (12 years ago)

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Peter  (12 years ago)

Say, sum of the 9 correct numbers is S

So, (S + (10a + b))/10 - (S + (10b + a))/10 = 1.8
=> 9a - 9b = 18
=> a - b = 2.

Is this solution Helpfull? Yes (23) | No (2)  

gyayak  (11 years ago)

Let the original number be ab i.e., (10a + b).
After interchanging the digits, the new number becomes ba i.e., (10b + a).

The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be 10*1.8 more than the sum of the 10 numbers with the digits interchanged.

i.e., 10a + b = 10b + a + 18, 9a - 9b = 18, a - b = 2.

Is this solution Helpfull? Yes (8) | No (1)  

manisha  (6 years ago)

(10a+b)/10-(10b+a)/10=1.8
(9a-9b)/10=1.8
9(a-b)=18
a-b=2

Is this solution Helpfull? Yes (1) | No   

SAMAR BANERJEE  (7 years ago)

say, the original number is 10x+y aqnd after interchanging the number is 10y+x.
as per question,10(10x+y)-{9(10x+y)+(10y+x)}=1.8*2
so,9x-9y=18
x-y=2

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