bump-edy bump

This is a tough one. My guess would be that since this an equation with three variables, you need three different equations to solve it.

x= no. of attendees
y= amount paid
z= total

the three equations are

xy=z (original)
(x-.75)(y+100) = z (Statement 1)
(x+1.50)(y-100)=z (Statement 2)

You can use original to put y in terms of x andz.
You can use Statement 1 to put z in terms of x.
Statement 2 would then be an equation featuring only x. So you could solve for number of people there.

Obviously it would be a pain in the ass to solve everything -- is there a rule that correlates number of variables to number of equations needed to solve? That would be easier.

This q was earlier posted.

The ans is C here.

We would have two variables and combining two equations from the two statements we can get the no. of people.

The equations are :

(x-.75)(y+100)=xy
(x+1.5)(y-100)=xy

assume 'x' as admission charge which is same for every one.
assume 'y' as no. of people.And 'xy' is the amount that is generated for total no. of people charged with same admission fee of 'x' for each.