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Sub 505 (Easy)|   Algebra|   Word Problems|                     
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carcass

As shown in the diagram above, a lever resting on a fulcrum has weights of w1 pounds and w2 pounds, located d1 feet and d2 feet from the fulcrum. The lever is balanced and \(w_1d_1=w_2d_2\). Suppose w1 is 50 pounds and w2 is 30 pounds. If d1 is 4 feet less than d2, what is d2, in feet?


A. 1.5

B. 2.5

C. 6

D. 10

E. 20


Attachment:
diagram.jpg

We are given that w(1) = 50, w(2) = 30, and d(1) = d(2) - 4. If we let d(2) = d, we have:

50 x (d - 4) = 30 x d

50d - 200 = 30d

20d = 200

d = 10

Answer: D
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carcass

As shown in the diagram above, a lever resting on a fulcrum has weights of w1 pounds and w2 pounds, located d1 feet and d2 feet from the fulcrum. The lever is balanced and \(w_1d_1=w_2d_2\). Suppose w1 is 50 pounds and w2 is 30 pounds. If d1 is 4 feet less than d2, what is d2, in feet?


A. 1.5

B. 2.5

C. 6

D. 10

E. 20


Attachment:
diagram.jpg

The algebraic way is to write it as 50 * (d2 - 4) = 30d2

Or d1 = d2 - 4

We can pick from the answer choices a number that makes them equal.

If we pick d2 = 10

then 30*10 = 300
50 * 6 = 300

Answer choice D
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­Can someone help me understand why we don't have to take into account the difference of units in this problem?

Thanks.
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