Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

98

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

Solution:

146900

Explanation:
Whenever you face such type of questions, it is wise to begin from the last thing. Here in this question the last thing will be the 9th temple. He climbed down 100 steps and thus you know, he had Rs. 100 before beginning climbing down. Thus, he must have offered Rs. 100 to the god in that temple too (he offered half of the total amount). Also, he must have dropped Rs. 100 while climbing the steps of the ninth temple. This means that he had Rs. 300 before he begand climbing the steps of the ninth temple.

Now, we will calculate in the similar manner for each of the temples backwards.
Before the devotee climbed the eight temple: (300+100)*2 + 100 = 900
Before the devotee climbed the seventh temple: (900+100)*2 + 100 = 2100
Before the devotee climbed the Sixth temple: (2100+100)*2 + 100 = 4300
Before the devotee climbed the fifth temple: (4300+100)*2 + 100 = 8900
Before the devotee climbed the fourth temple: (8900+100)*2 + 100 = 18100
Before the devotee climbed the third temple: (18100+100)*2 + 100 = 36,500
Before the devotee climbed the second temple: (36500+100)*2 + 100 = 73300
Before the devotee climbed the first temple: (73300+100)*2 + 100 = 146900

Therefore, the devotee had Rs. 146900 with him initially.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

40

Explanation
The number 40 (Forty) is the only number in one million numbers whose letters are arranged in the order they are arranged in English alphabets as well.

Solution:

L8

Explanation:
To know why, just read the number series upside down and you will find that the numbers are 91, 90, 89, 88, __, 86
So L8 when read upside down reads as 87.

Solution:

If you analyze the numbers and arrange them alphabetically, you will find that the number 8 (eight) will be the first one to occur and number 2202 (two thousand two hundred and two) will be the last one.

Solution:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653

Solution:

101/125

Explanation:
This is an extremely tricky question. You have to find the probability where all of them have same number in mind. But if you try to do that, you will have to do a lot of calculations and utilize your brain plenty much. However, the trick to do this question is something opposite. We are not implying that you won’t get answer by directly calculating the probabilities, but it will take much time.

However, if you just go on calculating the negative probabilities, then you can simply subtract it from 1 to find the positive probability. Let us begin the operations and not confuse you further more.

Let us take two friends in the beginning. The probability that Adam and Bella don’t have the same number in their respective mind will be 4/5.

Now let us include Christopher in the scene. The probability that Christopher don’t have a digit same as either Adam or Bella will be 3/5.

Let us now include Drew as well. The probability that she does not have any number in mind which is same as either Adam or Bella or Christopher will be 2/5.

Now the combined probability that none of them has the same number in mind will be:
4/5 * 3/5 * 2/5 = 24/125

Now the probability that they have same number in mind will be:
1 – 24/125 = 101/125