MA 115 QUANTITATIVE REASONING FINAL EXAM (FALL 2017) FORM 11) A truck manufactur
ID: 3148928 • Letter: M
Question
MA 115 QUANTITATIVE REASONING FINAL EXAM (FALL 2017) FORM 11) A truck manufacturer has 93 cars of a new model to be apportioned to four dealerships. The manufacturer decides to apportion the based on the number of trucks each dealership sold in the previous year. The number of trucks sold by each dealership is shown below. Dealership A Annual Sales 5,100 22.400 400Total 14,600 3,900 3,200 a. Determine the standard divisor. (3pts) b. Determine each dealership's standard quota. (4pts) c. Determine each dealership's apportionment using Webster's method. (5pts)Explanation / Answer
(a) STANDARD DIVISOR = TOTAL POPULATION /TOTAL SEAT OVER ENTIRE POPULATION
= 14600 / 93 = 156.9892
(b)STANDARD QUOTA (DEALER A)= 5,100 / 156.9892 = 32.4863 ,(DEALER B) = 3900 / 156.9892 = 24.8425
(DEALER C ) =3200 / 156.9892 = 20.3836
(DEALER D) = 2400 / 156.9892 = 15.2877
(c) NOW EACH DEALER`S APPORTIONMENT BY WEBSTER`S METHOD IS GIVEN AS
DEALER POPULATION(CAR SOLD IN YEAR) QUOTA INITIAL FINAL
A 5,100 32.4863 32 33
B 3,900 24.8425 25 25
C 3,100 20.3836 20 20
D 2,400 15.2877 15 15 ------------------ ----------------- --------------- -------------- 14,600 92 93
TOTAL INNITIAL` 92` BUT AVAILABLE HENCE STANDARD DIVISOR MUST BE DECREASE CALLED MODIFIED DIVISOR LET IT IS 156.5
THEN QUOTA OF EACH WILL BE
DEALER` A ` = 5100/156.5 = 32.6923 = 33 DEALER` B` = 3900/156.5 = 24.9201 = 25
DEALER `C` = 3200/ 156.5 = 20.447 = 20 DEALER `C` = 2400 / 156/5 = 15.3355 = 15
HERE SUM EQUALS` 931 EXACTLY AVAILABLE, HENCE FINAL ALLOTTMENT
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