5.
He makes 4 cigarettes with the 16 butts. Then he smokes them and has 4 more butts to make another cigarette.

5.
A = 50, B = 10.

C- (35, 45)

30 + 40 + 50 + x + y = 200 =>
x + y = 80 => x = 35 and y = 45

34

11

Average = Total Sum/Total numbers in case of 3 numbers: 29 = sum/3 , so sum = 87 in case of two numbers 38 = sum/2 so sum = 76, so the 3rd number= 87-76 = 11

3/4

Let's take a look at the explanation of the riddle.

As per the total laps in the race are 50 and the driver has completed 12 1/2 laps. This means, we have to subtract 12 1/2 from the total 50 laps. This equal to 37 1/2 or 37.5

50 - 12 1/2 = 37 1/2 or 37.5

Now, we need to calculate the fractional part of the race remains. For this, we need to divide the remaining laps by total laps, that is, 37 1/2 divide by 50 or 37.5 divided by 50 which will be equal to 0.74 or 3/4.

37 1/2 / 50 = 0.74 or 3/4

4600

12 = 552/x * 100 x= 552/12 * 100 = 4600

The facts in this riddle are clear: There is an initial $30 charge. It should have been $25, so $5 must be returned and accounted for. $3 is given to the 3 friends, $2 is kept by the bellhop - there you have the $5. The trick to this riddle is that the addition and subtraction are done at the wrong times to misdirect your thinking - and quite successfully for most. Each of the 3 friends did indeed pay $9, not $10, and as far as the friends are concerned, they paid $27 for the night. But we know that the clerk will tell us that they were charged only $25 and when you add the $3 returned with the $2 kept by the bellhop, you come up with $30.

8 feet.

You just need to divide your dollar by the price of the wire/ foot
1/0,125 = 8

36.

First of all we know that the boys are going to be in the 1st,3rd, and 5th seats. So the girls must be in the 2nd, 4th, and 6th seats.
So the first question we should ask ourselves is how many ways can 3 boys be arranged in 3 seats.
Lets say our boys are named A, B, and C inspiring names I know...
We can have ABC, ACB, BAC, BCA, CAB, or CBA 6 ways to arrange the boys.
Now for each way there is to arrange a the boys, there are six ways to arrange the girls (same arguement for finding the number of ways to arrange the boys).

9642 and 87531.
The digits of the requested two numbers obviously form descending sequences. Furthermore, if you have two pairs of numbers with equal sums, the pair of which the numbers have the smallest absolute difference, is the one of which the numbers have the largest product. Using this knowledge, the two numbers can easily be constructed by placing the digits one by one, starting with 9 and ending with 1:

9

8 -> 96

87 -> 964

875 -> 9642

8753 -> 9642

87531
Conclusion: the requested two numbers are 9642 and 87531 (and the product of these two numbers is 843973902).

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