no of units producedno of units of component A requiredprice per.pdf

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no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16.

Education

2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june

no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16800
june
1250
14
17500
total direct labor cost in process department -April to june
48300
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2

1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED

2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500

12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16800
june
1250
14
17500
total direct labor cost in process department -April to june
48300
Solution
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8

2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April
1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-

total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000
May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April

1000
14
14000
May
1200
14
16800
june
1250
14
17500
total direct labor cost in process department -April to june
48300
no of units produced
no of units of component A required
price per unit
total= no of units*no of unit of component*price per unit
7-
cost of component A
1200
2
1.25
3000
cost of component B
1200
3
0.8
2880
total budgeted cost for may 2018
5880
no of units produced
cost per unit of A = 2*1.25
cost per unit of b = 3*.8
total cost =(no of units*cost of A)+(no of units*cost of B)
8-
April

1000
2.5
2.4
4900
May
1200
2.5
2.4
5880
June
1250
2.5
2.4
6125
total
16905
9-
INVENTORY OF COMPONENT A IN APRIL
UNITS PRODUCED
1000
PER UNIT, UNITS OF COMPONENT a REQUIRED
2
UNITS OF COMPONENT A REQUIRED
2000
VALUE OF INVENTORY =2000*1.25
2500
10-
total direct labor cost in process department -April to june
Month
units produced
labor cost per unit in process department = labor hour per unit* labor cost per hour
total labor cost in process department
April
1000
8
8000

May
1200
8
9600
june
1250
8
10000
total direct labor cost in process department -April to june
27600
11-
total direct labor cost in june
no of units
labor cost per unit in process department = labor hour per unit* labor cost per hour
labor cost per unit in assembly department = labor hour per unit* labor cost per hour
total = no of units produced*labor cost per unit in process department +no of units*labor cost in
assembly department
1250
8
6
17500
12-
Month
units produced
direct labor cost per unit = labor cost in process +labor cost in assembly
total
April
1000
14
14000
May
1200
14
16800
june
1250

14
17500
total direct labor cost in process department -April to june
48300

More from karan8801

omparative historical research is a method of social science that examines historical events in order to create explanations that are valid beyond a particular time and place, either by direct comparison to other historical events, theory building, or reference to the present day.[1]Generally, it involves comparisons of social processes across times and places. It overlaps with historical sociology. While the disciplines ofhistory and sociology have always been connected, they have connected in different ways at different times (see \'Major researchers\' below). This form of research may use any of several theoretical orientations. It is distinguished by the types of questions it asks, not the theoretical framework it employs (see \'Illustrations\' below) Some commentators have identified three waves of historical comparative research.[2] The first wave of historical comparative research concerned how societies came to be modern, i.e. based on individual and rational action, with exact definitions varying widely. Some of the major researchers in this mode were Alexis de Tocqueville,[3]Karl Marx,[4]Emile Durkheim,[5]Max Weber,[6] and W.E.B. Du Bois.[7] The second wave reacted to a perceived ahistorical body of theory and sought to show how social systems were not static, but developed over time.[8] Notable authors of this wave include Barrington Moore, Jr.,[9]Theda Skocpol,[10]Charles Tilly,[11]Michael Mann,[12] and Mark Gould.[13] Some have placed the Annales school and Pierre Bourdieu in this general group, despite their stylistic differences.[14] The current wave of historical comparative research sociology is often but not exclusively post-structural in its theoretical orientation. Influential current authors include Julia Adams,[15] Anne Laura Stoler,[16]Philip Gorski,[17] and James Mahoney.[18] There are four major methods that researchers use to collect historical data. These are archival data, secondary sources, running records, and recollections. The archival data, or primary sources, are typically the resources that researchers rely most heavily on. Archival data includes official documents and other items that would be found in archives, museums, etc. Secondary sources are the works of other historians who have written history. Running records are ongoing series of statistical or other sorts of data, such as census data, ship\'s registries, property deeds, etc. Finally recollections include sources such as autobiographies, memoirs or diaries.[19] There are four stages, as discussed by Schutt, to systematic qualitative comparative historical studies: (1) develop the premise of the investigation, identifying events, concepts, etc., that may explain the phenomena; (2) choose the case(s) (location- nation, region) to examine; (3) use what Theda Skocpol has termed as \"interpretive historical sociology\" and examine the similarities and the differences; and (4) based on the information gathered, propose a causal explanation for the pheno.

omparative historical research is a method of social science that e.pdf

karan8801

1 .D 2.BSolution 1 .D 2.B.pdf

karan8801

where is it .pdf

karan8801

The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated. Solution The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated..

The name of the reaction is aldol condesation (Cl.pdf

karan8801

Step1 3 moles of H2O react with Cr metal = 2moles.pdf

karan8801

sp2 - sp2 .pdf

karan8801

Part A neutralizes acids - base Part B produces.pdf

karan8801

Lewis Acid Base Reaction .pdf

karan8801

KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf

karan8801

It is a reaction for secondary amines .Option 2 i.pdf

karan8801

fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly. Solution fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly..

fluorene is mobile phase and fluorenone is statio.pdf

karan8801

equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ) Solution equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ).

equivalent mass is 27.9 . so Molecular massmole.pdf

karan8801

There is no solutionSolutionThere is no solution.pdf

karan8801

This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed. Solution This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed..

This is a case of allopatric speciation. In this case lice from gori.pdf

karan8801

d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total. Solution d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total..

d. 8 The bonding electrons are the electrons taki.pdf

karan8801

The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans are considered just as one population observed at different times. The multi regional models predict that at least some level of genetic continuity can be observed from archaic Neanderthal forms to the contemporary Hominins forms and up to today’s Europeans. The results have not matched with the view that the Neanderthal were genetically related with the anatomically modern ancestors of current Europeans or contributed to the present day human gene pool. Solution The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans .

The genetical changes of single locus are responsible for divergent .pdf

karan8801

The corrects answers areThr and Asn are polar amino acids.Isole.pdf

karan8801

Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic. Solution Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic..

Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf

karan8801

selected palnts xerophyte verus mesophyte ========================================================= A xerophyte is a types of plant that has adjusted to get by in a domain with minimal fluid water, for example, a forsake or an ice-or snow-canvassed locale in the Alps or the Arctic. The morphology and physiology of xerophytes are differently adjusted to moderate water, and usually likewise to store substantial amounts of water, amid dry periods. Different species might be adjusted to survive long stretches of parching of their tissues, amid which their metabolic action may viably close down. Plants with such morphological and physiological adjustments are xeromorphic. Xerophytic plants may have comparable shapes, structures, and structures and look fundamentally the same as, regardless of the possibility that the plants are not firmly related, through a procedure called concurrent development. For instance, a few types of desert flora (individuals from the family Cactaceae), which advanced just in the Americas, may seem like Euphorbias, which are dispersed around the world. A random types of caudiciforms, plants with swollen bases that are utilized to store water, may likewise show such likenesses. Xerophytic plants can have less general surface territory than different plants, so diminishing the range that is presented to the air and lessening water misfortune by vanishing. Xerophytes can have littler leaves or less branches than different plants. A case of leaf surface decrease are the spines of a desert flora. A case of compaction and diminishment of spreading are the barrel desert flora. Different xerophytes may have their leaves compacted at the base, as in a basal rosette, which might be littler than the plant\'s blossom. This adjustment is displayed by some Agave and Eriogonum species, which can be discovered developing close Death Valley. A few xerophytes have minor hairs on their surface to give a wind break and decrease wind current, along these lines diminishing the rate of dissipation. At the point when a plant surface is secured with minor hairs, it is called tomentose. In a still domain, the regions under the leaves/spines where transpiration is occurring structure a little limited environment that is more soaked than typical with water vapor. In the event that this is not overwhelmed by wind, the water vapor potential angle is diminished as is transpiration. Subsequently, in a windier circumstance, this confinement is not held thus the angle stays high, which helps the loss of water vapor. Spines trap a layer of dampness furthermore moderate air development over tissues. =================== Mesophytes are earthbound plants which are adjusted to neither an especially dry nor especially wet environment. A case of a mesophytic living space would be a country calm glade, which may contain goldenrod, clover, oxeye daisy, and Rosa multiflora. Mesophytic plants have unbending, tough, openly expanded stems and stringy, all around create.

selected palntsxerophyte verus mesophyte========================.pdf

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no of units producedno of units of component A requiredprice per.pdf

1. no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5

2. 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june

3. no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2

4. 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED

5. 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500

6. 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 Solution no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8

7. 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10-

8. total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April

9. 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April

10. 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000

11. May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250

12. 14 17500 total direct labor cost in process department -April to june 48300

no of units producedno of units of component A requiredprice per.pdf

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