Question #231385

Saeed’s Bakery has a fire and Saeed loses some of his cost data. The bits of paper that he recovers after the fire provide the information in the following table (all the cost numbers are Rupees).

Q

AFC

AVC

ATC

 MC

25

300

250

550

 

 

 

 

 

200

50

A

B

375

 

 

 

 

 

225

75

100

225

325

 

 

 

 

 

325

100

75

C

D

 

 

 

 

 

E

125

60

270

330

 

Saeed asks you to come to his rescue and provide the missing data in the five spaces identified as ABCD, and E.


1
Expert's answer
2021-08-31T08:49:16-0400



B

ATC=TCQATC=\frac{TC}{Q}


MC=ΔTCΔQ\\MC=\frac{\Delta TC}{\Delta Q}


375=(200×B)(25×550)20025\\375=\frac{(200\times B)-(25\times550)}{200-25}


375=200B13750175\\375=\frac{200B-13750}{175}


65,625=200B13750\\65,625=200B-13750


65,625+13750=200B\\65,625+13750=200B




A

AVC=TVCQTVC=TCTFCTFC=AFC×QTC=ATC×QTC=396.875×200=79,375AVC=\frac{TVC}{Q}\\TVC=TC-TFC\\TFC=AFC\times Q\\TC=ATC\times Q\\TC=396.875\times200=79,375


TVC=79,375(50×200)=69,375\\TVC=79,375-(50\times200)=69,375


AVC=69,375200=346.875\\AVC=\frac{69,375}{200}=346.875



C

ATC=TCQTC=TFC+TVCTC=(325×100)+(325×75)=56,875ATC=56,875325=175ATC=\frac{TC}{Q}\\ TC=TFC+TVC\\TC=(325\times 100)+(325\times75)=56,875\\ATC=\frac{56,875}{325}=175


D

MC=ΔTCΔQ\\MC=\frac{\Delta TC}{\Delta Q }


=(175×325)(225×225)325225=\frac{(175\times325)-(225\times225)}{325-225}


56,87550,625100\\\frac{56,875-50,625}{100}


=6250100=62.5\\=\frac{6250}{100}=62.5



E

MC=ΔTCΔ\\MC=\frac{\Delta TC}{\Delta }


330=(270×E)(175×325)E325330=270E56875E325330E107,250=270E5687560=50375E=839.5833333330=\frac{(270\times E)-(175\times325)}{E-325}\\330=\frac{270E-56875}{E-325}\\330E-107,250=270E-56875\\60=50375\\E=839.5833333





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