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coders to system admins

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by uptowngirl92 » Fri Jul 17, 2009 5:23 pm
A certain IT dept. of fewer than 15 people hires coders and system administrators.Coders are paid $55000 per yr. on average,while systems administrators are paid an average yearly slary od $45000.What is the ratio of coders to system admins?

1.If two of the coders were made systems admins instead,the yrly payroll for the IT dept. would be $535000.

2.If syestems admins salaries were reduced by one third,and coders salary were increased to $58000 the dept. would save $5700 in yrly payroll.

[spoiler]Ans"D[/spoiler]

Guys messed up this one.Detailed expl. please!
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Source: — Data Sufficiency |
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by gmat740 » Fri Jul 17, 2009 5:37 pm
Lets kick of all the Zero's because all the figures are multiple of 1000(just to deal with numbers easily)

Coder Admin
55/ yr 45/yr
x 15-x

x = number of coders

We need to find coders/Admin = x/15-x

St(I)=>
coder Admin
x-2 (15-x) +2

55*(x-2) + 45*(17-x) = 535
so, we can have the value of x and hence the ratio coders/Admin can be calculated

St(II) =>

(55+3)*x + (45/3)*(15-x) +57 = 55*x +45(15-x)

58x + 15*(15-x) + 57 = 55*x +45(15-x)
Hence we can find x from here as well.

It looks lengthy,just for the sake of explanation but actually it is a two step question.

Hope this Helps
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by adamsmith2009 » Sat Jul 18, 2009 10:00 pm
gmat 740,

The question states there are less than 15 coders+admin but you're taking 15 as the total. Is this the correct approach or am I misunderstanding the question?
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by tohellandback » Sun Jul 19, 2009 1:51 am
uptowngirl,
are you sure it is 5700 dollars and not 57000 dollars in the second statement.?
plz check and then I will post the explanation
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by gmat740 » Sun Jul 19, 2009 2:05 am
adamsmith2009 wrote:gmat 740,

The question states there are less than 15 coders+admin but you're taking 15 as the total. Is this the correct approach or am I misunderstanding the question?
Well While solving this one, the same question came to my mind. But if we do that way, we will bring in inequalities and not the equalities.

I am from engineering background so I solve questions from my instincts. But I would advice it would be better at times, not to over analyse the words of a question.

By the way, what is the source of the question??
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by PussInBoots » Sun Jul 19, 2009 11:14 pm
gmat740 you cannot just assume that htere are 15 total. This is a HUGE and wrong assumption

a = admins
c = coders

(1) => (a+2) * 45 + (c - 2) * 55 = 535
(a+2) * 9 + (c-2) * 11 = 107
only when (c-2) = 4 we have valid solution c = 6 and a = 5

(2) 3,000*c - 15,000*a = -5,700
3*c -15*a = -5.7 IMPOSSIBLE

In case it's 57000:

3,000 * c - 15,000 * a = 57 000
3c - 15a = 57
c - 5a = 19
c = 19 + 5a > 15 IMPOSSIBLE

IMO A
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by tohellandback » Sun Jul 19, 2009 11:38 pm
PussInBoots wrote:gmat740 you cannot just assume that htere are 15 total. This is a HUGE and wrong assumption

a = admins
c = coders

(1) => (a+2) * 45 + (c - 2) * 55 = 535
(a+2) * 9 + (c-2) * 11 = 107
only when (c-2) = 4 we have valid solution c = 6 and a = 5

(2) 3,000*c - 15,000*a = -5,700
3*c -15*a = -5.7 IMPOSSIBLE

In case it's 57000:

3,000 * c - 15,000 * a = 57 000
3c - 15a = 57
c - 5a = 19
c = 19 + 5a > 15 IMPOSSIBLE

IMO A
IMO A too.
if in the second statement it is 57000.
pussinboots, you have got the second equation wrong


let x- no of coders
y- sys admins
1) (x-2)55000+(y+2)45000=535000
55000x+45000y=555000
55x+45y=555
11x+9Y=111
only possibility x=6,y=5 SUFFICIENT

2) 30000y+58000x=55000x+45000y-57000
15000y-3000x=57000
15y-3x=57
5y-x=19
possibilities, y=4,x=1 and y=5,x=2
INSUFF
The powers of two are bloody impolite!!
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by PussInBoots » Mon Jul 20, 2009 12:10 am
Forgot the (-1), thanks for correction.
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by Domnu » Mon Jul 20, 2009 5:23 pm
I agree with D. Here's why:

Let A be the number of system administrators and C be the number of coders. For both parts, A + C = 15.

1) We know that 55*C + 45*A = 535. Solve for A, C and subtract out two system administrators and add two coders, and find the ratio (remember... don't bother actually solving this out). We know that this is solvable since both "lines" don't have the same slope.

2) We know that (55*C + 45*A) - (58*C + 30*A) = 5.7. Same story as earlier. So we're done.

Remember that you don't actually have to solve out the equations.
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by tohellandback » Mon Jul 20, 2009 5:56 pm
Domnu wrote:I agree with D. Here's why:

Let A be the number of system administrators and C be the number of coders. For both parts, A + C = 15.

1) We know that 55*C + 45*A = 535. Solve for A, C and subtract out two system administrators and add two coders, and find the ratio (remember... don't bother actually solving this out). We know that this is solvable since both "lines" don't have the same slope.

2) We know that (55*C + 45*A) - (58*C + 30*A) = 5.7. Same story as earlier. So we're done.

Remember that you don't actually have to solve out the equations.
I don't think so.
1) You can't suppose A+C=15. its clearly mentioned that total is fewer than 15
2) your equation 2 will not give the value of A and C even if you suppose A+C=15
3) A is sufficient not because you can solve two equations in two variables but because the solution is limited by <15 and there is only one solution that satisfies.
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by Domnu » Mon Jul 20, 2009 7:00 pm
Ah, I see. I didn't read the question properly :mrgreen:

I guess that it's the answer since there is only one integral case of A, C?
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by uptowngirl92 » Mon Sep 07, 2009 1:55 pm
Yes sorry there's a typo..it's indeed 57,000.
the OA: is D..wats the consensus guys??????
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