twobagels
A charity sells 140 benefit tickets for a total of $2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?
A. $782
B. $986
C. $1158
D. $1219
E. $1449
Chinese GMAT practice book
Let \(x=\) no. of tickets sold at full price.
Let \(y=\) no. of ticksts sold at half price.
Let \(f\) be the full price dollar value.
Let \(\frac{f}{2}\) be the half price dollar value.
Hence we need the value of \(x(f)\)
we know :
\(x+y=140\) .....(I)
\(x(f) + y(\frac{f}{2})= 2001\) ....(II)
\(x( \frac{f}{2})+y (\frac{f}{2})=140 (\frac{f}{2})\).... Mutiplying (I) with \(\frac{f}{2}\) ....(III)
\(x(f) + y(\frac{f}{2})= 2001\) ....(IV)
IV -III
\(x(f)-x( \frac{f}{2})= 2001 - 140 (\frac{f}{2})\)
\(x( \frac{f}{2}) = 2001 - 140 (\frac{f}{2})\)
\(x(f) = 4002 - 140(f)\)
Now \(140(f)\) is always going to end with a zero.
Hence \(4002 -140(f) \)will always end with a \(2\)
There is only one option that ends with a \(2\)
Ans A
Hope it's clear.