Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.

Given teacher drinks 9 ltr
Let number of boys be 'A'.
Let number of girls be 'B'.
Boy drinks 7 ltr and girl drinks 4 ltr
After class total water consumed = 42 ltr
Then,
9 + 7A + 4B = 42
=> 7A + 4B = 33
By trial and error method,
The only integers which satisfy the equation is A = 3 and B = 3
Therefore, number of girls in the class = 3.

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.

1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14

Let first no. be "k" and multiple be "l"...start value of y with 3
then,
[(k x y) + {l x (l-1)}]
Thus, we get

15 x 3 + 3 x 2 = 51
51 x 4 + 4 x 3 = 216
216 x 5 + 5 x 4 = 1100
1100 x 6 + 6 x 5 = 6630
6630 x 7 + 7 x 6 = 46452

Therefore, the missing number is 6630.

Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.

The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.