Question

anita attempted to find the average of ten two digit numbers.one of the numbers a,was the reverse of one of the others.if these two numbers are replaced by the two digit numbers, c and the reverse of c, anita would find the average of these numbers to be 6.6 more than the average of teh original set of numbers.find the value of (sum of the digits of c) - (sum of the digits of a).
a.3 b.4 c.6 d.2

Answer 1

Answer:

Sum of digit of c - sum of digit of a = 6

Step-by-step explanation:

Let say average of ten two digit numbers = x

=> Sum = 10x

Let say a  = mn  => a reverse = nm

Value of a + a reverse = 10m + n + 10n + m  = 11(m+n)

let say c = pq then c reverse = qp

Value of c +c reverse = 11(p+q)

New sum = 10x - 11(m+n) +  11(p+q)

New Average =(10x - 11(m+n) +  11(p+q))/10

New Average = x + 6.6

(10x - 11(m+n) +  11(p+q))/10 = x + 6.6

=> 10 -  11(m+n) +  11(p+q) = 10x + 66

=> 11(p+q) -  11(m+n)  = 66

=> (p+q) -  (m+n)  = 6

=> Sum of digit of c - sum of digit of a = 6