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ATQ,
Let, P = 10a + b, Q = 10c + d & Sum of other 3 numbers = R Given P + Q + R = 34 × 5 = 170 --- (1) P + Q – R = 10 × 5 = 50 --- (2) Solving equation 1 & 2, we have, P + Q = (170 + 50)/2 = 110, So, R = 170 – 110 = 60 & (10a + b) + (10c + d) = 110 (b + d) = 110 – 10(a + c) --- (3) Again, given that, (10b + a) + (10d + c) + R = 5 × 14.2 (10b + a) + (10d + c) + 60 = 71 (10b + a) + (10d + c) = 11 10(b + d) + (a + c) = 11 10[110 – 10(a + c)] + (a + c) = 11 1100 – 100(a + c) + (a + c) = 11 99(a + c) = 1089 (a + c) = 11 = Sum of tens digit
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