Hi swerve,

While most Test Takers would probably do "System" Algebra to solve this question (which is a perfectly acceptable approach to this question), there's also a pattern-matching shortcut built into it that provides another option.

We're told that two sets of items that can be purchased for $160
2 bats + 4 balls = $160
1 bat + 6 balls = $160

Looking at this data, we can see that "trading" 1 bat will get you 2 balls at the same price. Thus, the price of 1 bat = the price of 2 balls.

We're asked to buy the SAME number of bats and balls with $240....

$240 = (1.5)($160)

Using that first equation, we can multiply all of the terms by 1.5, which gives us...

2(1.5) bats + 4(1.5) balls = $160(1.5)

3 bats + 6 balls = $240

Noting the earlier "trade" option, we can "trade" 2 balls for 1 bat....this gives us....

4 bats + 4 balls = $240

So the MOST bats that can be purchased under these conditions is 4.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

swerve wrote:A sporting good store sells one type of baseball bat and one type of baseball. The cost for 2 bats and 4 balls is $160. The cost for one bat and 6 balls is $160, as well. If someone were to buy an equal number of bats and balls, at most how many bats can he purchase if he has a budget of $240 for the purchase?

A. 2
B. 3
C. 4
D. 5
E. 6

We can let b = the cost for a ball and t = the cost for a bat, thus:

2t + 4b = 160

t + 2b = 80 (1)

and

t + 6b = 160 (2)

Subtracting equation one from equation two, we have:

4b = 80

b = 20

So t = 40

We can let n = the number of bats and balls, and create the equation:

20n + 40n = 240

60n = 240

n = 4

Answer: C