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For making a table it needs human labour 3 times the labour [#permalink]
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Emdadul28 wrote:
For making a table it needs human labour 3 times the labour needed to make a bench. 6 workers can make 36 benches and 5 tables in 12 days. How many days will 10 worker need to make 61 benches and 8 tables?


Work of making 1 table = 3 * Work of making 1 bench
Work of making 5 tables = Work of making 15 benches
Work of making 8 tables = Work of making 24 benches

6 workers...........36 benches+15 benches = 51 benches..............12 days
10 workers.........61 benches + 24 benches = 85 benches.............?? days

No of days = 12 * (6/10) * (85/51) = 12days

Originally posted by KarishmaB on 21 Nov 2016, 00:04.
Last edited by KarishmaB on 17 Oct 2022, 02:13, edited 1 time in total.
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Re: For making a table it needs human labour 3 times the labour [#permalink]
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Emdadul28 wrote:
For making a table it needs human labour 3 times the labour needed to make a bench. 6 workers can make 36 benches and 5 tables in 12 days. How many days will 10 worker need to make 61 benches and 8 tables?


Use the following formula:
\(\frac{(workers)(time)}{output} = \frac{(workers)(time)}{output}\)

Let each bench = 1 pound and each table = 3 pounds (implying 3 times the labor).

Output 1: 36 benches and 5 tables = 36*1 + 5*3 = 51 pounds.
Output 2: 61 benches and 8 tables = 61*1 + 6*3 = 85 pounds.

Since 6 workers take 12 days for Output 1, and we must determine the time needed by 10 workers for Output 2, we get:
\(\frac{6*12}{51} = \frac{10t}{85}\)
\(\frac{2*12}{17} = \frac{2t}{17}\)
\(12 = t\)
12 days
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Re: For making a table it needs human labour 3 times the labour [#permalink]
GMATGuruNY wrote:
Emdadul28 wrote:
For making a table it needs human labour 3 times the labour needed to make a bench. 6 workers can make 36 benches and 5 tables in 12 days. How many days will 10 worker need to make 61 benches and 8 tables?


Use the following formula:
\(\frac{(workers)(time)}{output} = \frac{(workers)(time)}{output}\)

Let each bench = 1 pound and each table = 3 pounds (implying 3 times the labor).

Output 1: 36 benches and 5 tables = 36*1 + 5*3 = 51 pounds.
Output 2: 61 benches and 8 tables = 61*1 + 6*3 = 85 pounds.

Since 6 workers take 12 days for Output 1, and we must determine the time needed by 10 workers for Output 2, we get:
\(\frac{6*12}{51} = \frac{10t}{85}\)
\(\frac{2*12}{17} = \frac{2t}{17}\)
\(12 = t\)
12 days



Hello GmatGuruNy,

Can you please explain how is that formula derived or any alternate way.

I tried maintaining bench and table work in equal units and then keeping effort and days on one side and then used the unitary approach for work and effort.
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prabsahi06 wrote:
Hello GmatGuruNy,

Can you please explain how is that formula derived


\(\frac{(workers)(time)}{output} = \frac{(workers)(time)}{output}\)

Numerator:
Workers and time are INVERSELY PROPORTIONAL.
If the number of workers doubles, the time required to produce the same output is cut in half.
Values that are inversely proportional MULTIPLY to yield a constant value:
(workers)(time) = k

Denominator:
Workers and output are DIRECTLY PROPORTIONAL.
If the number of workers doubles, the output also doubles.
Time and output are also DIRECTLY PROPORTIONAL.
If the time triples, the output also triples.
Values that are inversely proportional DIVIDE to yield a constant value:
\(\frac{workers}{output} = k\)
\(\frac{time}{output} = k\)

Putting it all together, we get:
\(\frac{(workers)(time)}{output} = k\)
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Re: For making a table it needs human labour 3 times the labour [#permalink]
GMATGuruNY wrote:
prabsahi06 wrote:
Hello GmatGuruNy,

Can you please explain how is that formula derived


\(\frac{(workers)(time)}{output} = \frac{(workers)(time)}{output}\)

Numerator:
Workers and time are INVERSELY PROPORTIONAL.
If the number of workers doubles, the time required to produce the same output is cut in half.
Values that are inversely proportional MULTIPLY to yield a constant value:
(workers)(time) = k

Denominator:
Workers and output are DIRECTLY PROPORTIONAL.
If the number of workers doubles, the output also doubles.
Time and output are also DIRECTLY PROPORTIONAL.
If the time triples, the output also triples.
Values that are inversely proportional DIVIDE to yield a constant value:
\(\frac{workers}{output} = k\)
\(\frac{time}{output} = k\)

Putting it all together, we get:
\(\frac{(workers)(time)}{output} = k\)



Thanks GMATGuru!!

It's very clear :)
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