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December 20, 2014

December 20, 2014

Posted by **ROSE** on Saturday, March 24, 2012 at 10:36pm.

- ALGEBRA -
**MathMate**, Sunday, March 25, 2012 at 8:56am"three kinds of clamps, type A, B, and C"

Let A,B,C represent the number of each.

"20 more type C clamps than the total of the other types"

C=A+B+20

"twice as many as B clamps as type A clamps."

A=2B

"produce 380 clamps per day"

A+B+C=380

So in summary, the equations are:

A+B+C=380 ...(1)

C=A+B+20 ...(2)

A=2B ...(3)

This can be solved by successive substitution as follows:

A=2B => C=(2B)+B+20=3B+20

Substitute A and C in the first equation

(2B) + B + (3B+20) = 380

Solve for B:

6B=360

B=60

therefore

A=2B = 120

C=3B+20=200

Check: A+B+C=120+60+200=380 OK.

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