squieeez part 2 solutions

Ans: 1
Let 5 digits are X1, X2, X3, X4 and X5.
Since we are getting 10 different values, therefore we have all digits different. ( = 10)
So we Assume X1 < X2 < X3 < X4 < X5 (All different)
By given values of summation
X1 + X2 + X3 = 0 …………………………………….. (1)
X3 + X4 + X5 = 19……………………………………. (2)
X2 + X4 + X5 = 14 …………………………………… (3)
X1 + X2 + X4 = 3 …………………………………….. (4)
6(X1 + X2 + X3 + X4 + X5) = 90
Or X1 + X2 + X3 + X4 + X5 = 15……………………. (5)
=> X4 + X5 = 15
=> X3 = 4
=> X2 = -1
=> X1 = -3
=> X4 = 7
=> X5 = 8

Ans: 2
Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. Since X and Y are integers, therefore Y can only take one of the values in the set {0, 12, 24, 36, 48}, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12.
Similarly, the second occasion's equation is Y = 5X + Y/12 + 7. The only legal values here are X = 3 and Y = 24. So the time is 3:24.
Between 1:12 and 3:24, two hours and twelve minutes have elapsed.