Digits are 0,1,2,3,4,5,6,7,8,9. So no. of digits are 10

First all possible case => 9(0 excluded) x 10 x 10 = 900

Second no repetition allowed =>9 x 9 x 8 = 648

Third all digits are same => 9 (111,222,333,444,555,666,777,888,999)

Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

Sum of 4 digit numbers = (2+4+6+8) x P33 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x P23 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x P13 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x P03 x (1) = 20 x 1 x 1 = 20 

Adding All , Sum = 147320

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C121×C320+C321×C120= 50540 ways.

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

The wages of labourers in a factory increases in the ratio 22:25 and there was a reduction in the number of labourers in the ratio 15:11. Find the original wage bill if the present bill is Rs 5000 ?
Ratio of increase of wages = 22:25
Ratio of decrease of labourers = 15:11
Compound ratio of wages of labourers = 22 x 15 : 25 x 11 = 330:275
Final bill = Rs. 5000
For 275 ratio wages = Rs. 5000
For 1 ratio wages = 5000/275
For 330 ratio wages = 5000/275 x 330 = Rs. 6000

1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320

We know that zero can't be in hundreds place. But let's assume that our number could start with zero.

The formula to find sum of all numbers in a permutation is

111 x no of ways numbers can be formed for a number at given position x sum of all given digits

No of 1 s depends on number of digits

So,the answer us

111 x 20 x (0+1+2+3+4+5) = 33300

We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula.

Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0.

This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number.

So the sum for this is 11 x 4 x (1+2+3+4+5).
=660

Hope u understood why we use 4. Each number can be formed in 4x1 ways

So, the final answer is 33300-660 = 32640

Since floor starts form the 1 therefore 2 will be the 1st floor.
Form 1 to 10 only 4,6,8,9,10 are not prime.
Form 11 to 20 only 14 ,16,18 are not prime also non of their digits are prime.
From 21 to 30 each number has a prime number so it must no be consider.
In between 40 to 49 we have to consider only 40,44,46,48,49.
50 to 59 follow either of the condition so we have too neglect it.
In between 60 to 68 60,64,66,68.
total option = 17.