Logic Development Problems

1. Logic Development Problems

2. Puzzle #1 • After-Dinner Drink • Alok, Bharat, and Chandra often eat dinner out. • Each orders either coffee or tea after dinner. • If Alok orders coffee, then Bharat orders the drink that Chandra orders. • If Bharat orders coffee, then Alok orders the drink that Chandra does’nt order. • If Chandra orders tea, then Alok orders the drink that Bharat orders. • Who do you know always orders the same drink after dinner?

4. Puzzle #1 • A orders B orders C orders • 1 Coffee Coffee Coffee • 2 Coffee teatea • 3 tea Coffee Coffee • 4 tea tea tea • 5tea Coffee tea • 6 tea teatea

5. Puzzle #1 • From second IFcondition , 1st and 5th row are eliminated. From third IF condition, 2nd and 5th are eliminated. • So A always the same drink Tea

7. Puzzle #2 • Can you divide the number 45 into four parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied by the third part and the fourth part is divided by 2. All the four results should be the same number.

9. Puzzle #2 • Let a, b, c, d be the four parts. Then a+b+c+d=45. It is given that • a+2 = b-2 = 2c = d/2; • a=b-4; • c = (b-2)/2; • d =2(b-2); • b-4 + b + (b-2)/2 + 2(b-2) = 45;.

10. Puzzle #2 • B=12. • A= b-4 =8 • C=5 • D=20

12. Puzzle #3 • There was a man who had six sons. The man possessed a huge amount of gold, which he hid carefully in a building consisting of a number of rooms. In each room there were a number of lockers; this number of lockers was equal to the number of rooms in the building. Each locker contained a number of golden coins that equaled the number of lockers per room. When the man died, one locker was given to the old loyal servant. The remainder of the gold coins has to • Divided fairly equal among all six sons. How can you say that such an even distribution is possible in this case or will there be an uneven distribution

14. Puzzle #3 • Let the number of rooms be N. This means that per room there are N lockers with N coins each. In total there are N×N×N = N3 coins. One locker with N coins goes to the servant. For the six brothers, N3 - N coins remain.

15. Puzzle #3 • This can be written as : N(N2 - l), or N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers.

16. Puzzle #3 • One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division!

18. Puzzle #4 • A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that. After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.Find X and Y. (1 Rupee = 100 Paise).

20. Puzzle #4 • Let the original check be for A Rupees and B paise or (100A + B) paise.John actually received B Rupees and A paise or (100B + A) paiseAfter spending 20 paise, he has double the amount he wanted to withdraw.

21. Puzzle #4 • Hence, the equation is • 2 * (100X + Y) = 100Y + X - 20200X + 2Y = 100Y +X - 20199X - 98Y = -20 • 98Y-199x=20

22. Puzzle #4 • Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1

23. Puzzle #4 • Case I :Y=2XSolving two equations simultaneously • 98Y - 199X = 20Y - 2X = 0We get X = - 20/3 & Y = - 40/2

24. Puzzle #4 • Case II :Y=2X+1 • Solving two equations simultaneously98Y - 199X = 20Y - 2X = 1We get X = 26 & Y = 53 • Now, its obvious that he wanted to withdraw Rs. 26.53