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Algebra

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by sud21 » Sun Sep 20, 2015 12:59 pm
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 1 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?

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by theCEO » Sun Sep 20, 2015 2:50 pm
sud21 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 1 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?

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bt = bat and bl = ball

2bt + 4bl = 160 -equation 1
1bt + 6bl = 160 -equation 2

mutilple equation 2 time 2
2bt + 12bl = 320 -equation 3

subtract equation 1 from equation 3
2bt + 12bl = 320
-
2bt + 4bl = 160
=
8bl = 160; bl = 20

plug in this value for bl in any equation to get bt
2bt + 4bl = 160
2bt + (4*20) = 160; bt = 40

the following equation sums up what the question is asking for
(what is the greatest integer value of x that satisfies the following)
40x + 20x <= 240
60x <= 240
x <= 4

ans = c
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by GMATGuruNY » Sun Sep 20, 2015 4:19 pm
To make the solution easier to follow, let's replace the word bat with helmet.
sud21 wrote:A sporting good store sells one type of baseball helmet and one type of baseball. The cost for 2 helmets and 4 balls is $160. The cost for 1 helmet and 6 balls is $160, as well. If someone were to buy an equal number of helmets and balls, at most how many helmets can he purchase if he has a budget of $240 for the purchase?

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Let h - the cost of a helmet and b = the cost of a ball.

The cost for 2 helmets and 4 balls is $160:
2h + 4b = 160.

The cost for 1 helmet and 6 balls is $160
h + 6b = 160.

Since replacing 1 helmet with 2 balls does not change the total cost, the cost of 1 helmet must be equal to the cost of 2 balls:
h = 2b.

Substituting h = 2b into h + 6b = 160, we get:
2b + 6b= 160
8b = 160
b = 20.

Since b=20 and h=2b, h=40.
Since b=20 and h=40, we get:
b+h = 60.
4b + 4h = 240.

Thus, a budget of $240 allows for the purchase of 4 bats and 4 helmets.

The correct answer is C.
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by [email protected] » Sun Sep 20, 2015 7:32 pm
Hi sud21,

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

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by Matt@VeritasPrep » Thu Sep 24, 2015 11:08 am
One last approach:

x = cost of a bat
y = cost of a ball

2x + 4y = 160

and

x + 6y = 160

We're asked (indirectly) to find the value of (x + y).

Since 2x + 4y = x + 6y, we have x = 2y. Hence 2x + 4y is really 4x, so 4x = 160, and x = 40.

We know x = 2y, so y = 20. This gives us x + y = 60, and we can buy FOUR bats and FOUR balls for $240.
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