Math, asked by tulsikasera3978, 1 year ago

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

Answered by amitnrw
1

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

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