Solution:

You may solve this question with different methods but we are only giving away one.

What you have to do in that method is look for a number x to which all the numbers from 2 to 10 divide evenly. Multiplying the numbers together (2*3*4*5*6*7*8*9*10) will give you a big number but we have to find a smaller possibility. What we can do is find the prime factors, a subset of which we can use to form a number from 2 to 10.

2*2*2*3*3*5*7 will certainly give us what we need. The product is 2520 and it is the lowest possible number which can be evenly divided by 2-10.

Now if we add 1 to the number, the result will be 2521 which will satisfy all the conditions from 2 to 10. But will not hold true in case of 11.

If you calculate, you will see that you can multiply 2520 by any integer and add 1 to it to satisfy all the conditions except the divisibility by 11. What we need is an integer Y that can be multiplied by 2520 and if we add 1 to the product, it becomes divisible by 11.

2520/11 leaves a remainder 1. Now let us keep increasing the number of 2520. Suppose if we divide two 2520 by 11, the remainder then will be 1 + 1 = 2. If we keep going on like that. Ten 2520 when divided by 11 will give 10 as remainder and if we add 1 to them, it will be completely divisible. Also it will satisfy all the conditions from 2-10.

Thus our answer is 25201.

Solution:

Fifteen Minutes.

After frying for five minutes, you can take out one steak and put the third one inside while turning the other one inside. After five more minutes, one steak is fried from both sides and you can take it out. Now put in the steak that we removed and turn the other one inside. After five more minutes, they both will also be fried from both sides.

Solution:

Put one red marble in one box and all other marbles in the 2nd box.

Probability :
50 + 24/49

Solution:

It can be down as shown below.

Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

Solution:

Just One!

example
If u get Blue rock with label as "blue+red" => this is red as all labels are wrong.
In this way u can logically solve this puzzle.

Solution:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

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:

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.