A student wrote nine 3 digit numbers on board. If he had written reverse of 2 of the numbers a and b instead of actual numbers,average would have been 99 less.The sum of the average of the hundreds and unit's digit of a and that of b is 9.5. Find the sum of hundreds digits of a and
b.
Answers
Answer:
14
Step-by-step explanation:
9 3 digit numbers
Let say average of 9 number = X
Sum of 9 numbers = 9X
after reversing average would be X - 99
Sum of 9 Numbers = 9(X - 99)
Let say = a = LMN reverse of a = c = NML
b = PQR reverse of b = d = RQP
9(X - 99) = 9X - a - b + c + d
a + b = c + d + 9*99
100L + 10M + N + 100P + 10Q + R = 100N + 10M + L + 100R + 10Q + P + 99*9
99L + 99P = 99N + 99R + 99*9
L + P = N + R + 9
=> N + R = L + P - 9 - eq 1
Average of hundred & Unit digit of a = (L + N)/2
Average of hundred & Unit digit of b= (P + R)/2
Sum = (L + N + P + R)/2 = 9.5
=> L + N + P + R = 19
=> N + R = 19 - (L + P) - eq 2
Equating both
L + P - 9 = 19 - (L + P)
=> 2(L + P) = 28
=> L + P = 14
Sum of hundred's digit of a & b = 14