In the next month, Technology Firm X will hire exactly 24 new enginee

Question

In the next month, Technology Firm X will hire exactly 24 new engineers to at least partially fulfill the hiring needs of the firm's 3 engineering departments: the mechanical design department, which needs 15 new engineers; the electrical design department, which needs 12 new engineers; and the software design department, which needs 8 new engineers. Each of the engineering departments will receive at least 1 of the 24 new engineers, while none of the departments will receive more than it needs. Each of the new engineers will be employed in exactly one of the three departments.

Assume that in every case in which a department D, needs more engineers than a department Dz, D, will receive more engineers than Dz. Select the minimum number of engineers that the mechanical design department could receive, given this assumption and the information provided. And select the minimum number of engineers that the electrical design department could receive, given this assumption and the information provided. Make only two selections, one in each column.

chetan2u · Accepted Answer

1. Any department having a greater requirement will get more engineers. 
2. No one will get more than the requirement.

Now, as per the above restrictions, M>E>S
 We have to make M the minimum, so maximise E and S.  
the max possible values for E and S are: E=M-1 and S=M-2
Thus, M+M-1+M-2=24 or 3M=27 or  M=9

Next we have to find minimum value of E. 
So maximise M and S 
M can be filled up with the entire requirement, so M=15
The max value of S can be E-1.
Thus, 15+E+E-1=24 or 2E=10 or  E=5 ..

jack5397 · Answer

Hi  chetan2u ,

Thank you for the solution. I am unable to understand the highlighted part can you please elaborate a bit on this.

Thanks !

zfu4820 · Answer

Total hiring: 24 
Premisis: Mechanical > Eletical > Software 
Starting try the minimum number for Mechanical from 9
M 9+ E 8+ S 7 = 24 (M can hire 9 by following the rule M>E>S)
Then try 7 for mechanical, in order to find the minimum 
M 7+ E 6+ S 5 = 18 < 24 (  not work , because M hire 7, the largest number for E is 6 and for S is 5) 
then 9 is the minimum for Mechanical
Starting try the minimum number for Electrical from 9, using the formular of 24-E-S = M 
24- E9-S8= M7 not work, becase M must > E
then try 7
24- E7-S6 = M11 Work, but need to contiue to find the minimum 
then try 5
24- E5-S4 = M15 Work, since M=15, the maximum number for M, 
therefore this is what we want, The minimum number for Electrical is 5.

emont · Answer

if M gets 9 and E gets 5, then doesn't that mean software will get the rest (10) to make 24? and then in that case the software dept is getting more than others which is not alllowed since it only needs 8?

mustafa87 · Answer

Hitesh0701 · Answer

Here are my 2 cents,

We know that M>E>S
 Answer to part 1: Minimize the M, hence i will maximize the other two.

M+E+S=24
Requirement: 15,12 and 8
We will first allocate equal to all departments:
8+8+8=24
We know that M should receive the highest 
Hence, Add 1 to M. Now we want to reduce 1 from the other two. Can we reduce it from E?? No, E should be second highest, hence i will remove this 1 from S department.
Making solution as : 9+8+7=24
Hence, minimum to be allocated is 9 to M.

Answer to part 2: Minimize the E, hence i will maximize the other two.

Again allocate equally to all three
M+E+S
8+8+8=24
I will first maximize M because i can allocate maximum engineers over there
M's requirement is 15, hence allocate 15 to M. Remaining=9
I want to allocate 9 between E and S
E+S=9
Allocate equally, not possible
Allocate 5 to E and 4 to S. (Why E>S)
Hence my new allocation will be
15+5+4=24

KarishmaB : Please let me know if we can rightly use this approach.

demawtf · Answer

In the next month, Technology Firm X will hire exactly 24 new engineers to at least partially fulfill the hiring needs of the firm's 3 engineering departments: the mechanical design department, which needs 15 new engineers; the electrical design department, which needs 12 new engineers; and the software design department, which needs 8 new engineers. Each of the engineering departments will receive at least 1 of the 24 new engineers, while none of the departments will receive more than it needs. Each of the new engineers will be employed in exactly one of the three departments.

Assume that in every case in which a department D, needs more engineers than a department Dz, D, will receive more engineers than Dz. Select the minimum number of engineers that the mechanical design department could receive, given this assumption and the information provided. And select the minimum number of engineers that the electrical design department could receive, given this assumption and the information provided. Make only two selections, one in each column.

Minimum of M --> max of E and S, respecting M > E > S
E=M-1  (since E must be less than M) 
S=M-2 ( since S must be less than E) 
M+M-1+M-2=24
3M=27
M=9

Minimum of E --> max of M and S respecting M > E > S
M can be 15 
S=E-1
E+E-1+15=24
2E=10
E=5

hirdesh · Answer

tricky question

arunbhati · Answer

Dear Sir,

How we know that their is requirement of M>E>S

24 can possible like this also na 11 from E+ 8 from software + 5 from M

Bunuel · Answer

If I understand your doubt correctly, you are asking why M must be greater than E and E must be greater than S. It is because the problem directly states that any department needing more engineers will receive more engineers, so your example split is not allowed.

Assume that in every case in which a department D, needs more engineers than a department Dz, D, will receive more engineers than Dz.

In the next month, Technology Firm X will hire exactly 24 new enginee

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GMAT Two-part Analysis (TPA) Questions

In the next month, Technology Firm X will hire exactly 24 new enginee

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards