Hey I need some help with this problem. Any ideas on how to solve it? Thanks lots. : A hardware supplier manufactures three kinds of clamps, types A, B and C. Production restrictions require them to make 10 units more type C clamps than the total of the other types and twice as many type B clamps as type A. The shop must produce a total of 490 units of clamps per day. How many units of each type can be made per day?
To solve this problem, we can define the variables:
Let A = number of type A clamps
Let B = number of type B clamps
Let C = number of type C clamps
Based on the given information, we can create the following equations:
Equation 1: C = A + B + 10 (Production restrictions require 10 more type C clamps than the total of the other types)
Equation 2: B = 2A (Twice as many type B clamps as type A)
Equation 3: A + B + C = 490 (Total number of clamps to be produced per day)
Now, we have three equations with three variables. We can substitute the values from equation 2 and equation 1 into equation 3 to solve for the values of A, B, and C.
Substituting equation 1 and equation 2 into equation 3:
A + B + (A + B + 10) = 490
2A + 2B + 10 = 490
2A + 2(2A) + 10 = 490 (substituting B = 2A)
2A + 4A + 10 = 490
6A + 10 = 490
Now, we can solve for A:
6A = 490 - 10
6A = 480
A = 480 / 6
A = 80
Substituting A = 80 into B = 2A:
B = 2(80)
B = 160
Substituting A = 80 into C = A + B + 10:
C = 80 + 160 + 10
C = 250
Therefore, the number of clamps that can be made per day is:
Type A: 80 units
Type B: 160 units
Type C: 250 units
Let's set up the equations:
C=A+B+10 ....(1)
B=2A .....(2)
A+B+C=490 ....(3)
Substitute (2) in (1) to get:
C=A+2A+10=3A+10 .....(1A)
Substitute (1A) & (2) in (3) to get
A+2A+3A+10 = 490 ......(3A)
Solve for A and back-substitute in (2) and (1) to get B and C.