Ratio/Proportion: 3 variables

[kanha81]

I do know that this problem has been discussed on the forum before and I sincerely apologize for reposting this. I am in really some time crunch, so thanks for being patient.

I am looking for a quicker way to understand and solve this problem.

If 6 workers can build 4 cars in 2 days, then how many days would it take 8 workers to build 6 cars?
(A) 5/3
(B) 9/4
(C) 8/3
(D) 11/4
(E) 10/3

This is how I solved the problem. It's just that at times it gets confusing.
w:workers
c:cars
d:days

6w ---> 4c ---> 2d
1w ---> 4c ---> 6*2 = 12d ....(i)

Also, 6w ---> 2c ---> 1d ....(ii)

Taking (i), 1w ---> 4c ---> 12d
8w ---> 4c ---> 12/8 = 3/2 days
8w ---> 6c ---> (6*3) / (2*4) = 9/4 days

Hence, [spoiler][/spoiler]

[mals24]

There are various ways to solve this question.

Ill give you 2.

You can either change number of cars first or the number of workers.

1. change the number of cars first.

6w-->4c-->2d
6w-->6c-->x

x = 2*6/4 = 3 days

6w-->6c-->3d
8w-->6c-->y

y = 3*6/8 = 9/4 days.


2. change the number of workers first.

6w-->4c-->2days
8w-->4c--> 6*2/3 = 3/2 days

8w-->4c-->3/2 days
8w-->6c-->(6*3)/(4*2) = 9/4 days

Hope that helps

[gmat740]

Questions like this have a common concept

Number of Man*days = constant for same number of units produced.

man means workers

so,
for 4 cars, 6*2 Man-days

2 cars = (6*2)/2 = 6 man-days

so,

6 cars = 6*3 man-days = 18

But man-day is constant

(8 workers)*x days = 18

x = 9/4 days

Hope this Helps

Karan

[kanha81]

Questions like this have a common concept

Number of Man*days = constant for same number of units produced.

man means workers

Thanks Karan. This formula will be quite useful.

You can either change number of cars first or the number of workers.

1. change the number of cars first.

6w-->4c-->2d
6w-->6c-->x

x = 2*6/4 = 3 days

6w-->6c-->3d
8w-->6c-->y

y = 3*6/8 = 9/4 days.

Thanks mals. I like this way faster. Thank you both for your inputs.

[xunil56]

Questions like this have a common concept

Number of Man*days = constant for same number of units produced.

man means workers

so,
for 4 cars, 6*2 Man-days

2 cars = (6*2)/2 = 6 man-days

so,

6 cars = 6*3 man-days = 18

But man-day is constant

(8 workers)*x days = 18

x = 9/4 days

Hope this Helps

Karan

This is a great way of looking at it. Thanks.

[kanishkporwal]

It looks like you have some trouble in knowing how to solve proportions.
If they are still not cleared the just visit the link given

https://math.tutorvista.com/number-syste ... rtion.html

[GMATGuruNY]

If 6 workers can build 4 cars in 2 days, then how many days would it take 8 workers to build 6 cars?
(A) 5/3
(B) 9/4
(C) 8/3
(D) 11/4
(E) 10/3

Let rate for each worker = 1 unit per day.
Rate for 6 workers = 6 units per day.
In 2 days, the number of units produced = r*t = 6*2 = 12 units.
Since 4 cars are produced, each car = 12/4 = 3 units.
6 cars = 6*3 = 18 units.
Rate for 8 workers = 8 units per day.
Time needed = w/r = 18/8 = 9/4 days.

The correct answer is D.

[VB15]

I would like to add one more method to solve this question:

we have,
6w 4c 2d --> w is workers ; c is cars ; d is days
then,
8w 6c x

multiplying factor of first term is 8/6 or 4/3. (To take out multiplying factor: New Value/ Old Value)
multiplying factor of second term is 6/4 or 3/2.
multiplying factor of third term is x/2. (we have to find X)

by observing, we can find out the proportionality (α) between w,c,d.
More workers --> lesser days. So, (w α 1/d) {worker is inversely proportional to days}
More workers --> more cars. So, (w α c) (worker is directly proportional to cars}

we can combine above two cases,

w α c/d
taking d other side,

d α c/w {Since, we need to find days}

multiplying factor of c = 3/2, w = 4/3

d α (3/2)c / (4/3)w

d α (9/8) c/w

Since, d is directly proportional to c/w. So factor of 9/8 should come on LHS also.

(9/8)d α (9/8) c/w

Multiplying factor of d is 9/8.
Earlier we found MF of d as x/2.
so, x/2 = 9/8
x = 9/4. Ans.

PS: This might seem lengthy because of explantion, but its not. As most of the calculations are to be done in head.

[Matt@VeritasPrep]

PS: This might seem lengthy because of explantion, but its not. As most of the calculations are to be done in head.

It might be a little elaborate.

I'd say that

6 workers do 2 cars per day, so 1 worker builds (1/3) of a car per day.

We've got 8 workers, so they can jointly build (8/3) cars per day.

W = RT

6 = (8/3) * T

9/4 = T

[wgrady]

I would like to propose another solution using a direct/inverse proportion approach.
The number of days taken to build the cars is directly proportional to the number of cars and inversely proportional to the number of workers.

D = days, C = cars, W = workers

so D is proportional to C/W

and introducing the constant of proportionality K we get

D = kC/W
rearranging to make K the subject, K = DW/C = 2*6/4 = 3 i.e K = 3 man-days per car

So the number of days is given by D = 3*C/W

Thus the time for 8 worker to build 6 cars is
D = 3 * 6/8 = 18/8 = 9/4

Time required is 9/4 days