SOLUTION: store is having a sale on jelly beans and almonds. For 7 pounds of jelly beans and 9 pounds of almonds, the total cost is $35. For 5 pounds of jelly beans and 3 pounds of almonds,
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Question 1091336: store is having a sale on jelly beans and almonds. For 7 pounds of jelly beans and 9 pounds of almonds, the total cost is $35. For 5 pounds of jelly beans and 3 pounds of almonds, the total cost is $19. Find the cost for each pound of jelly beans and each pound of almonds. Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website! Imagine one customer making the first purchase of 7 pounds of jelly beans and 9 pounds of almonds for $35; and imagine a second customer making the second purchase of 5 pounds of jelly beans and 3 pounds of almonds for $19.
Now another customer comes along and makes a purchase that is exactly 3 times what the second customer bought -- 15 pounds of jelly beans and 9 pounds of almonds, for $57. That purchase has the same number of pounds of almonds as the first customer; so the difference in what they paid must be because of the different numbers of pounds of jelly beans.
The third customer paid $22 more than the first, and the difference between their purchases was an extra 8 pounds of jelly beans. So the cost for one pound of jelly beans must be $22 divided by 8, or $2.75.
Knowing the price for each pound of jelly beans, we can use the purchase of any one of the three customers to determine the cost for each pound of almonds.
The first customer bought 7 pounds of jelly beans, at $2.75 per pound, for a total of $19.25. The cost of her 9 pounds of almonds was therefore $35 - $19.25 = $15.75. And so the cost of each pound of almonds was $15.75 divided by 9, or $1.75.
So the cost of one pound of jelly beans is $2.75, and the cost for one pound of almonds is $1.75.
The preceding discussion was just a way of explaining in real world terms an algebraic solution, which looks like the following.
Let x = cost of one pound of jelly beans
Let y = cost of one pound of almonds
Then
(1)
(2)
(3) (multiplying the second equation by 3)
(4) (subtracting equation (1) from equation (3))
(5)
(6) (substituting (5) into (1))
(7)
(8)
(9)
Those formal algebraic steps exactly followed the informal solution shown previously.