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A shop produces sarongs

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A shop produces sarongs

by vinni.k » Sun Feb 26, 2012 7:04 am
A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1250. How many workers were part of the latter production run?

A. 6
B. 10
C. 11
D. 15
E. 23.5

Answer is C

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by pemdas » Sun Feb 26, 2012 11:06 am
one-day average production, 5n+20
the first k days' production, k(5n+20)=500
the second k days' production, 2k(5n+20+25)=1250 [spoiler]>>>5 workers added means increase 5n by 5*5 or 25[/spoiler]
find the value of (n-5)? [spoiler]>>>to preserve the function of n from k we subtract 5 from n[/spoiler]

solution: 2 equations with 2 variables
{k(5n+20)=500
{2k(5n+45)=1250, divide both sides to get 2(5n+45)/(5n+20)=1250/500

2(5n+45)/(5n+20)=5/2, 4(5n+45)=5(5n+20), 20n+180=25n+100, 5n=80, n=16 and n-5=16-5=11

c
vinni.k wrote:A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1250. How many workers were part of the latter production run?

A. 6
B. 10
C. 11
D. 15
E. 23.5

Answer is C

Thanks & Regards
Vinni
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by Anurag@Gurome » Sun Feb 26, 2012 11:22 pm
vinni.k wrote:A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1250. How many workers were part of the latter production run?

A. 6
B. 10
C. 11
D. 15
E. 23.5

Answer is C

Thanks & Regards
Vinni
Daily average production = 5n + 20
In the first k days, 500 = (5n + 20)* k
500 = 5nk + 20k ...Equation 1

In the next k days, 1250 - 500 = [5(n + 5) + 20]k
750 = 5nk + 25k + 20k
750 = 5nk + 45k ...Equation 2

Subtracting equations 1 and 2, 250 = 25k or k = 10
So, 500 = 5n * 10 + 20 * 10
500 = 50n + 200
300 = 50n
n = 6

Therefore required number of workers = 6 + 5 = 11

The correct answer is C.
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by GMATGuruNY » Mon Feb 27, 2012 4:54 am
vinni.k wrote:A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1250. How many workers were part of the latter production run?

A. 6
B. 10
C. 11
D. 15
E. 23.5

Answer is C

Thanks & Regards
Vinni
In the first k days, the number of units produced = 500.
The total number of units produced over the entire run = 1250.
Thus, the number of units produced during the second run of k days = 1250-500 = 750.

We can plug in the answers, which represent the number of workers for the second run.
When the correct answer is plugged in, the number of days for each run will be the same.

Answer choice C: 11 workers.
Average daily output for 11 workers = 5(11)+20 = 75.
Time to produce 750 units at 75 units per day = 750/75 = 10 days.
Since 5 workers are added for the second run, the number of workers for the first run = 11-5 = 6.
Average daily output for 6 workers = 5(6)+20 = 50.
Time to produce 500 units at 50 units per day = 500/50 = 10 days.
Success!
The number of days for each run is the same: k=10.

The correct answer is C.
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by vinni.k » Tue Feb 28, 2012 9:46 pm
Thanks. Appreciate your replies :D

Regards
Vinni
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