In an auto parts manufacturing company, 2 types of carburetors (Type A

This question fails to provide details on the points system including
> point for Type A carburetor
> point for Type B carburetor
> point (must be negative) for spoilage
> point (must be negative) for major defects
> point (must be negative) for minor defects

Without the implementation rules, this question is currently unsolvable.

A = No of Type A
B = No of Type B
S = Spoilage
MAD = Major Defect
MID = Minor Defect

The generic equation would be
((A*3)+(B*2)) - ((S*\(\frac{3}{4}\))+(MAD*\(\frac{1}{2}\))+(MID*\(\frac{1}{4}\)))

1.
((53*3)+(42*2)) - ((6*\(\frac{3}{4}\))+(3*\(\frac{1}{2}\))+(2*\(\frac{1}{4}\))) = 236.5 > 200. Incentivized

2.
((62*3)+(16*2)) - ((16*\(\frac{3}{4}\))+(9*\(\frac{1}{2}\))+(12*\(\frac{1}{4}\))) = 198.5 < 200. Neither Incentivized nor Penalized

3.
((28*3)+(48*2)) - ((27*\(\frac{3}{4}\))+(19*\(\frac{1}{2}\))+(7*\(\frac{1}{4}\))) = 148.5 < 150. Penalized

4.
((29*3)+(49*2)) - ((26*\(\frac{3}{4}\))+(18*\(\frac{1}{2}\))+(7*\(\frac{1}{4}\))) = 154.75 > 150, but less than 200. Neither Incentivized nor Penalized

5.
((40*3)+(41*2)) - ((6*\(\frac{3}{4}\))+(3*\(\frac{1}{2}\))+(2*\(\frac{1}{4}\))) = 195.5 150, but less than 200. Neither Incentivized nor Penalized
_________________

Competition mode is off now.

OA is:

Incentivized: Steve assembled 53- Type A & 42 - Type B No-defect carburetors. There were 6 spoilage, 3 major defects, & 2 minor defects carburetors.

Penalized: Hector assembled 28- Type A & 48 - Type B No-defect carburetors. There were 27 spoilage, 19 major defects, & 7 minor defects carburetors.

Hi Sr1994

You might need to revise the calculation of the above highlighted part.

Thank you!

\sqrt{}


Updated! Thank you sir!
_________________