each coin can have two outcomes either head or tell... so if u particular any outcomes. in the case , 3rd as head or tell. other 9 outcomes can be both head or tell. so total of 512 outcomes are there in which third outcome is head or tail

since on left side,,,you have 1/5 with power of 35.. other nos. does not mean. here both sides will be equal when both sides have 1/5 raised to power 35; so this can be done without solving as 10 = 2*5

speed for the going journey =12; return journey takes double time, mean speed becomes half, so speed here is =6; so average speed when distance for going and returning journey is same(2ab)/a+b; a= speed for going journey;( in this case = 12) b= speed for returning journey;( in this case = 6) hence a...

we are told f/g just to make the confuse n that we need to use f/g for getting solution. once you use the factor f/g, you r surely gonna waste lot of your time

3,4,5,6,7,8,9,10,11(here the average is 7); when the average becomes 9 , it means you move to the two nos. to right of 3,4......cause there 2 nos. are making ur average less... cause of eliminating these two numbers( 3,4) and putting (12,13) , we are adding total of 12+13-3-4=18; which is divided am...

number system

what is the no. of zeros at the end of the expression
(6!)^6! * (7!)^7!*(8!)^8!*(9!)^9!

functions

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

parcentage

A project scheduled to be carried out over a single fiscal year has a budget of $12,600, divided into 12 equal monthly allocations. At the end of the fourth month of that fiscal year, the total amount actually spent on the project was $4580. By how much was the project over its budget?