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21 Aug 2024, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
68% (02:49) correct
32%
(03:08)
wrong
based on 151
sessions
History
Date
Time
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Not Attempted Yet
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
A) 6
B) 10
C) 11
D) 15
E) 23.5
21 Aug 2024, 02:01
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First k days:
\(k(5n+20)=500\)
\(5nk + 20k = 500\)
\(nk + 4k = 100\) [equation 1]
Second k days:
As one is adding 5 more workers, the initial 5n+20 daily average becomes 5(n+5)+20
\(k[5(n+5)+20]= 750\)
\(k(5n + 45)=750\)
\(5nk + 45k = 750\)
\(nk + 9k = 150\) [equation 2]
Solving for k:
Subtract [equation 1] from [equation 2]:
\(5k = 50\)
\(k = 10\)
Solving for n:
Plugging k = 10 into [equation 1]:
\(10n + 40 = 100\)
\(10n = 60\)
\(n = 6\)
Plus the 5 workers which were added in the second set of k days:\(6+5=11\)
ANSWER C
\(k(5n+20)=500\)
\(5nk + 20k = 500\)
\(nk + 4k = 100\) [equation 1]
Second k days:
As one is adding 5 more workers, the initial 5n+20 daily average becomes 5(n+5)+20
\(k[5(n+5)+20]= 750\)
\(k(5n + 45)=750\)
\(5nk + 45k = 750\)
\(nk + 9k = 150\) [equation 2]
Solving for k:
Subtract [equation 1] from [equation 2]:
\(5k = 50\)
\(k = 10\)
Solving for n:
Plugging k = 10 into [equation 1]:
\(10n + 40 = 100\)
\(10n = 60\)
\(n = 6\)
Plus the 5 workers which were added in the second set of k days:\(6+5=11\)
ANSWER C
21 Aug 2024, 06:20
Kudos
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5n + 20 is Average production / day, where n = no. of workers
In k days 500 units were produced,
500/k = 5n + 20 .................................(1)
5 workers added, in another k days the cumulative total was 1250 units, therefore 750 more in the another k days were produced
750/k = 5*(n+5) + 20 .........................(5 workers added)
750/k = 5n + 25 +20
750/k = 500/k + 25 ...........................(from 1)
k = 10
Subs. k in (1), we get n = 6
5 workers were added later,
n + 5 = 11
Answer. C
In k days 500 units were produced,
500/k = 5n + 20 .................................(1)
5 workers added, in another k days the cumulative total was 1250 units, therefore 750 more in the another k days were produced
750/k = 5*(n+5) + 20 .........................(5 workers added)
750/k = 5n + 25 +20
750/k = 500/k + 25 ...........................(from 1)
k = 10
Subs. k in (1), we get n = 6
5 workers were added later,
n + 5 = 11
Answer. C
