SOLUTION: A hardware supplier manufactures three kinds of​ clamps, types​ A, B, and C. Production restrictions force it to make 10 more type C clamps than the total of the other

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Question 1124862: A hardware supplier manufactures three kinds of​ clamps, types​ A, B, and C. Production restrictions force it to make 10 more type C clamps than the total of the other types and twice as many type B clamps as type A clamps. The shop must produce 310 clamps per day. How many of each type are made per​ day?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.
Take off for a minute 10 clamps of type C.


Then the rest amount of clamps C will be exactly equal to sum of types A and B,

and altogether (C-10), A and B will comprise 300 clamps.


Hence,  C = 150 + 10, while A + B = 150.


We also are given that the number of clamps of the type B is twice as many as that of type A.


Hence, we just get the


Answer.  A = 50,  B = 100  and  C = 160.

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The problem is SO SIMPLE, that equations are not needed.


Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


(1) An informal solution , using logical reasoning and a bit of mental arithmetic....

Make those 10 "extra" type C clamps first.
That leaves 300 more clamps to make; and of those the number of type C clamps is the same as the total of the types A and B clamps.
That means 150 more type C clamps (for a total of 160), and a total of 150 clamps of types A and B.
Then twice as many type B as type A means 100 type B and 50 type A.

Answer: 50 type A, 100 type B, 160 type C.

Using formal algebra....

(1) A+B+C = 310
(2) C = A+B+10
(3) B = 2A

There are many algebraic paths from these three equations to a solution. Here is one -- which, it turns out, doesn't look much at all like the informal solution above.

Using (3), substitute 2A for B in (1) and (2):
(4) 3A+C = 310
(5) C = 3A+10

Using (4), substitute 3A+10 for C in (5):
(6) 3A+3A+10 = 310

Solve for A:
6A+10 = 310
6A = 300
A = 50

Substitute A=50 in (5):
150+C = 310
C = 160

Substitute A=50 and C=160 in (1):
50+B+160 = 310
B = 100

You can see that, in this problem, the informal solution is much easier than the one using formal algebra....
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