Solution:

If you are thinking of how to cut in such a way that you are able to make seven equal pieces, you are thinking in the wrong direction. You should give more stress on how to make the transaction with the worker.

You have to break the gold bar two times so you have the following sizes of the bar:

1/7 ---- Bar A
2/7 ---- Bar B
4/7 ---- Bar C

When the day 1 ends:
Give Bar A (You remain with Bar B and Bar C; Worker possess - Bar A)

When the day 2 ends:
Give Bar B and take the Bar A back (You remain with Bar A and Bar C; Worker possess - Bar B)

When the day 3 ends:
Give Bar A (You remain with Bar C; Worker possess - Bar A and Bar B)

When the day 4 ends:
Give Bar C and take back Bar A and Bar B (You remain with Bar A and Bar B; Worker possess - Bar C)

When the day 5 ends:
Give Bar A (You remain with Bar B; Worker possess - Bar A and Bar C)

When the day 6 ends:
Give Bar B and take back Bar A (You remain with Bar A; Worker possess - Bar B and Bar C)

When the day 7 ends:
Give Bar A (You remain with no bar; Worker possess - Bar A, Bar B and Bar C)

Thus by the end of seven days, you have taken care of everything.

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:

MAIL + C = CLAIM
EXIT + S = EXIST
GOSH + T = GHOST
CITY + H = ITCHY
NEWT + I = TWINE
DENY + E = NEEDY
MINI + M = MINIM

The seven letters are CSTHIEM which can be rearranged to form CHEMIST.

Solution:

This is quite a tricky question. To solve it you must turn the grid upside down and then you will have the following numbers:

9 1 9
1 9 1
9 1 9
1 9 1

Now if you select a 9 and three 1, the sum will be 12.

It's quite easy to solve but what matters is if you can approach the possibility of it.

Solution:

Old password - Out of Date
New Password - Different

Both of them are given in the statements if you read closely.

Solution:

Though the question is very simple, most of the people answer it wrong. The real time will be 9:30. If you don't believe us, take a clock with 9:30 time in front of the mirror and look for yourself.

Solution:

Spring, summer, autumn and winter.

Solution:

According to the question
19W + 21B = 562 (where W stands for white bricks and B stands for black bricks)

The easiest way to solve this problem would be putting in the values for W, till we achieve an answer which is divisible by 21.
Why? Because it is must for the statement to be true.

If W = 1
21B = 562 - 19 = 543 i.e. not divisible

If W = 2
21B = 562 - 38 = 524 i.e. not divisible

If W = 3
21B = 562 - 57 = 505 i.e. not divisible

If W = 4
21B = 562 - 76 = 486 i.e. not divisible

If W = 5
21B = 562 - 95 = 467 i.e. not divisible

If W = 6
21B = 562 - 114 = 448 i.e. not divisible

If W = 7
21B = 562 - 133 = 429 i.e. not divisible

If W = 8
21B = 562 - 152 = 410 i.e. not divisible

If W = 9
21B = 562 - 171 = 391 i.e. not divisible

If W = 10
21B = 562 - 190 = 372 i.e. not divisible

If W = 11
21B = 562 - 209 = 353 i.e. not divisible

If W = 12
21B = 562 - 228 = 334 i.e. not divisible

If W = 13
21B = 562 - 247 = 315 i.e. divisible

Thus W must be 13 and if W = 13
B = 315/21 = 15

Thus to form that tower, 13 white bricks and 15 black bricks will be required.

Solution:

Let us move backwards.

On Thursday, I had $125.
On Wednesday, I had $125 (1 / one half) = $250
On Tuesday I had $250 (1 / two third) = $375
On Monday, I had $375 (1 / three quarters) = $500

Thus my salary is $500.

Solution:

This one is a tricky series and here we shall tell you a common way to solve the series problem when fractions are involved. You will find that so many series can be solved the way this one is solved.

Just look at the numerator and denominator separately

Numerators = 1, 3, 5, 7
Two is added to each one, thus the next number will be 9.

Denominators = 1, 2, 4, 8
Every next term is the double of previous, thus the next term will be 16.

So, the next term in the series will be 9/16.