Coin Riddles With Answers

If we are considering an ideal case and we have to calculate the maximum score, we will assume that the batsman hits six on every ball. But then, at the end of every over, the strike will change. But we have to make sure, he gets to play maximum number of balls to score maximum runs. Thus, the batsman will hit six in first five balls and will choose to take 3 runs in the last ball so that he retains the strike in the following over.

Total runs per over in that case = 6 * 5 + 3 = 33.

But in the last over (50th over), he don’t have to worry about keeping the strike and thus, he will hit six even on the last ball. Therefore, the maximum runs that a batsman can score:

33 * 49 + 6 * 6 = 1617 + 36 = 1653

If you start arranging the numbers alphabetically, the numbers 8 (eight) will come first and the number 2202 (two thousand two hundred and two) will come at last.

50%

The next toss of the coin is not dependent upon the previous ten tosses. Thus, the probability is 50 percent only.

L8

This is a bit tricky questions. To solve you must look at the series upside down. It is actually
91, 90, 89, 88, __, 86

So we need 87 in the place of blank. To achieve 87 in an upside down position, the correct answer will be L8.

Thus, the missing item in the series is L8.

The back seat of the last coach will be the best. The reason is because, the train will be accelerating after the stop and thus, it will be much faster when the back of the train enters the tunnel than when the front of the train enters. Thus you will have to spend less time in the tunnel.

The number 40 (Forty) is the only number which when written in English has all the letters in the same order as in the alphabets.

It is possible if you are using Roman Numerals.
9 = IX
10 = X
11 = XI
Now it is all possible.

146900

In such cases, we calculate from the last thing. Thus in this case, let us start counting from the 9th temple. Since, he must have offered 100 coins while climbing down stairs, it means that he must have offered Rs. 100 to god on the ninth temple and offered Rs. 100 while climbing up steps. Thus it is clear that he had Rs. 300 in his pocket before climbing the steps of the 9th temple.

Similarly, we will calculate for all the temples backwards.
Before eight temple: (300+100)*2 + 100 = 900
Before seventh temple: (900+100)*2 + 100 = 2100
Before Sixth temple: (2100+100)*2 + 100 = 4300
Before fifth temple: (4300+100)*2 + 100 = 8900
Before fourth temple: (8900+100)*2 + 100 = 18100
Before third temple: (18100+100)*2 + 100 = 36,500
Before second temple: (36500+100)*2 + 100 = 73300
Before first temple: (73300+100)*2 + 100 = 146900

Therefore he had Rs. 146900 initially.

The parking number is 87. You will know why if you invert the image and then see.

98

ab+a(a-1)
7*8+7(7-1)=56+42