A hardware supplier manufactures three kinds of clamps, type A, B, and C. Production restrictions force it to make 20 more type C clamps than the total of the other types and twice as many as B clamps as type A clamps. The shop must produce 380 clamps per day. How many of each type are made per day? How many type A clamps are produced?
Why did the clown become a hardware supplier? Because he wanted to become a master of "clamping"! Now, let's solve this puzzle.
Let's assume that the number of type A clamps produced per day is x.
According to the given information, the number of type B clamps produced per day is 2x (twice as many as type A clamps).
And the number of type C clamps per day is (x + 2x + 20) = 3x + 20 (20 more than the combined total of type A and type B clamps).
Since the total number of clamps produced per day is 380, we can write the equation:
x + 2x + (3x + 20) = 380
Combining like terms:
6x + 20 = 380
Subtracting 20 from both sides:
6x = 360
Dividing both sides by 6:
x = 60
So, the number of type A clamps produced per day is 60.
To find the number of type B clamps:
2x = 2 * 60 = 120
And the number of type C clamps:
3x + 20 = 3 * 60 + 20 = 200
Therefore, the hardware supplier produces 60 type A clamps, 120 type B clamps, and 200 type C clamps per day.
Let's denote the number of type A clamps as A, type B clamps as B, and type C clamps as C.
According to the given information:
1. The production of type C clamps is 20 more than the total of the other types. So, we have the equation: C = (A + B) + 20.
2. The shop must produce 380 clamps per day. Therefore, we have the equation: A + B + C = 380.
3. The production of type C clamps is twice as many as B clamps as type A clamps. So, we have the equation: C = 2B.
Now, let's solve the system of equations to find the values of A, B, and C.
Substitute equation 3 into equation 1:
2B = A + B + 20
B = A + 20
Substitute equation 3 into equation 2:
A + B + 2B = 380
A + 3B = 380
Substitute B = A + 20 into A + 3B = 380:
A + 3(A + 20) = 380
A + 3A + 60 = 380
4A = 320
A = 80
Substitute A = 80 into B = A + 20:
B = 80 + 20
B = 100
Substitute A = 80 into C = 2B:
C = 2(100)
C = 200
Therefore, the number of type A clamps produced per day is 80.
To solve this problem, we need to set up a system of equations based on the given information. Let's assign variables to each type of clamp:
Let's say the number of type A clamps produced per day is 'A', type B clamps is 'B', and type C clamps is 'C'.
From the information given, we know the following:
1) The production restriction is that the shop must produce a total of 380 clamps per day:
A + B + C = 380
2) The shop needs to make 20 more type C clamps than the total of the other types (A and B combined):
C = (A + B) + 20 (Equation 1)
3) The shop is forced to manufacture twice as many type B clamps as type A clamps:
B = 2A (Equation 2)
Now we have a system of three equations:
A + B + C = 380
C = (A + B) + 20
B = 2A
To find the values of A, B, and C, we need to solve these equations simultaneously.
Let's substitute equation 2 into equations 1 and 3:
C = (A + 2A) + 20 (Substituting B = 2A into equation 1)
C = 3A + 20 (Simplifying Equation 1)
B = 2A (Equation 3)
Now we can substitute these new expressions for B and C into the first equation:
A + B + C = 380
A + (2A) + (3A + 20) = 380 (Replacing B with 2A, and C with 3A + 20)
6A + 20 = 380 (Combining like terms)
6A = 360 (Subtracting 20 from both sides)
A = 60 (Dividing both sides by 6)
So, the number of type A clamps produced per day is 60.
Now we can substitute A = 60 into equations 2 and 3 to find the values of B and C:
B = 2A
B = 2(60)
B = 120
C = 3A + 20
C = 3(60) + 20
C = 180 + 20
C = 200
Therefore, the number of type A clamps produced per day is 60, type B clamps is 120, and type C clamps is 200.
"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.