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Posts: 39
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Difficulty:
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Question Stats:
56% (02:35) correct
44%
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If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task? What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?
A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
alexBLR
Joined: 17 Jan 2010
Last visit: 19 Jan 2018
Posts: 105
Concentration: General Management, Strategy
Schools: Kellogg '14 (M) Ross '14 (A) Booth '14 (D)
GPA: 3.78
WE:Engineering (Manufacturing)
Schools: Kellogg '14 (M) Ross '14 (A) Booth '14 (D)
Posts: 105
answer is B:
If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task? What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?
Take the time necessary to complete the job by both Dev and Tina to be x.
From the problem statement, time necessary to complete the job by Dev alone is: x+20
From the problem statement, time necessary to complete the job by Tina alone is: x+5
Rate of work when Dev and Tina work together: 1/x
Rate of work when Dev works alone: 1/(20+x)
Rate of work when Tina works alone: 1/(5+x)
The rate of work when Tina and Dev works together is equal to sum of the rates when Tina and Dev work alone:
1/x=1/(20+x)+ 1/(5+x)
When simplified the equation becomes:
X^2=100
X can only be the positive as we talk about tome, so x=10
The necessary ratio is T(Dev)/T(Tina)=(x+20)/(x+5)=30/15=2:1
If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task? What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?
Take the time necessary to complete the job by both Dev and Tina to be x.
From the problem statement, time necessary to complete the job by Dev alone is: x+20
From the problem statement, time necessary to complete the job by Tina alone is: x+5
Rate of work when Dev and Tina work together: 1/x
Rate of work when Dev works alone: 1/(20+x)
Rate of work when Tina works alone: 1/(5+x)
The rate of work when Tina and Dev works together is equal to sum of the rates when Tina and Dev work alone:
1/x=1/(20+x)+ 1/(5+x)
When simplified the equation becomes:
X^2=100
X can only be the positive as we talk about tome, so x=10
The necessary ratio is T(Dev)/T(Tina)=(x+20)/(x+5)=30/15=2:1
jlgdr
Let's say when they work together, they take t hrs to complete the work.
Dev takes 20 hrs extra when working alone because Tina did not work for t hrs and that work Dev had to do. Ratio of time take by Dev/Tina to do the same work = 20/t
Tina takes 5 hrs extra when working alone because Dev did not work with her for t hrs and hence his part of work, Tine did in 5 hrs. Ratio of time taken by Dev/Tina to do the same work = t/5
20/t = t/5 (The ratio of time taken by Dev and Tina for the same work will always be the same irrespective of the amount of work)
t = 10
Ratio of time taken by Dev/Tina = 20:10 = 2:1
Answer (B)
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General Discussion
Posts: 1
Let time taken by Dev to complete the work alone be x days and that by Tina be y days
Work done by Dev in 1 day = 1/x
Work done by Tina in 1 day = 1/y
Work done by Dev & Tina in 1 day = 1/x + 1/y = (x+y)/xy
Thus working together they will complete the work in xy/(x+y) days
Acc. to the ques :
1) "If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina'
x - xy/(x+y) = 20
=> (x^2)/(x+y) = 20.............(i)
2) "If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev"
y - xy/(x+y) = 5
=> (y^2)/(x+y) = 5..............(ii)
Dividing (i) by (ii)
x^2/y^2 = 4
=> x/y = 2
Ans : B
Hope it helps
Work done by Dev in 1 day = 1/x
Work done by Tina in 1 day = 1/y
Work done by Dev & Tina in 1 day = 1/x + 1/y = (x+y)/xy
Thus working together they will complete the work in xy/(x+y) days
Acc. to the ques :
1) "If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina'
x - xy/(x+y) = 20
=> (x^2)/(x+y) = 20.............(i)
2) "If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev"
y - xy/(x+y) = 5
=> (y^2)/(x+y) = 5..............(ii)
Dividing (i) by (ii)
x^2/y^2 = 4
=> x/y = 2
Ans : B
Hope it helps
jusjmkol740
Solved it in a minute! Here's my solution:
Description of the formula used:
If A and B complete a task in x days, then A alone takes=x+a days and B alone take x+b days
Formula: x=\sqrt{ab}
Plugging in the numbers:
x=\sqrt{20*5}
= 10
=>A=20+10=30 and B=5+10=15
=>a/b=2/1
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,296
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
Schools: Wharton '17 (A) HBS '17 (WA$)
Posts: 1,296
jusjmkol740
Is there any way to solve this problem in a time efficient way?
Experts please advice
Cheers
J
Again the funda of Rate * Time = Work
Time (Both) = T
For Dev:
Rate (Both) =T
Rate * Time = Work
Rate * T+20 = 1
Rate (Dev) = 1/(T+20)
For Tina:
Rate * Time = Work
Rate * T+5 = 1
Rate (Tina) = 1/(T+5)
Combined Rates = 1/(T+20) + 1/(T+5) = 1/T
Solving for it we will get you will get T=10
We want to know what is Time(D)/Time(T)= (T+20)/(T+5)=2:1
Time (Both) = T
For Dev:
Rate (Both) =T
Rate * Time = Work
Rate * T+20 = 1
Rate (Dev) = 1/(T+20)
For Tina:
Rate * Time = Work
Rate * T+5 = 1
Rate (Tina) = 1/(T+5)
Combined Rates = 1/(T+20) + 1/(T+5) = 1/T
Solving for it we will get you will get T=10
We want to know what is Time(D)/Time(T)= (T+20)/(T+5)=2:1
Schools: Cambridge'16
Posts: 405
Guessing strategy says that it is B or E. Because guy works slower that girl, it must be B. Otherwise you can do algebra as written above
B
B
If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task? What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?
A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
let t=D and T's time working together
ratio of D's time working alone to T's time working alone=(t+20)/(t+5)
plugging in answer options, only 2:1 works
(t+20)/(t+5)=2/1
t=10
t+20=30
t+5=15
30:15=2:1
B
A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
let t=D and T's time working together
ratio of D's time working alone to T's time working alone=(t+20)/(t+5)
plugging in answer options, only 2:1 works
(t+20)/(t+5)=2/1
t=10
t+20=30
t+5=15
30:15=2:1
B
mvictor
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Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,116
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
time work together: x
time dev = x+20
time tina = x+5
1/x+20 + 1/x+5 = 1/x
x+20+x+5/(x+20)(x+5) = 1/x
2x+25/(x+20)(x+5) = 1/x
cross multiply
2x^2 +25x = x^2 +5x+20x + 100
rearrange:
x^2 - 100 = 0
factor:
(x+10)(x-10)=0
x=10 or x=-10, since we are talking about hours, -10 is not a solution, thus x=10.
time for dev=10+20=30
time tina = 10+5=15
time dev/time tina = 30/15 = 2/1
time dev = x+20
time tina = x+5
1/x+20 + 1/x+5 = 1/x
x+20+x+5/(x+20)(x+5) = 1/x
2x+25/(x+20)(x+5) = 1/x
cross multiply
2x^2 +25x = x^2 +5x+20x + 100
rearrange:
x^2 - 100 = 0
factor:
(x+10)(x-10)=0
x=10 or x=-10, since we are talking about hours, -10 is not a solution, thus x=10.
time for dev=10+20=30
time tina = 10+5=15
time dev/time tina = 30/15 = 2/1
mvictor
I think there is a typo. Should be 20x -100
mvictor
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,116
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
Jargalan
where exactly? I don't see it
mvictor
This:
2x^2 +25x = x^2 +5x+25x-100
shouldn't it be
2x^2 +25x = x^2 +5x +20x +100 ?
mvictor
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,116
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
Jargalan
yes, you are correct! nice catch!
2x^2 +25x = x^2 +5x+20x + 100 is the correct form
i will have it edited now!
ScottTargetTestPrep
Expert
Target Test Prep Representative
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Status:Founder & CEO
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Location: United States (CA)
Expert reply
Posts: 22,738
jusjmkol740
We can let x = the time it takes for Dev and Tina working together to complete the task. Thus, the time it takes Dev to work alone = x + 20 and the time it takes Tina to work alone = x + 5. Thus, their combined rate can be equated as follows:
1/(x + 20) + 1/(x + 5) = 1/x
Let’s first get a common denominator of (x + 20)(x + 5) for the left side of the equation:
[(x + 5) + (x + 20)]/[(x + 20)(x + 5)] = 1/x
Combining like terms in the numerator on the left side of the equation yields:
(2x + 25)/[(x + 20)(x + 5)] = 1/x
Cross-multiplying, we obtain:
x(2x + 25) = (x + 20)(x + 5)
2x^2 + 25x = x^2 + 25x + 100
x^2 = 100
x = 10
Since x = 10, the time it takes Dev to work alone = 10 + 20 = 30 and the time it takes Tina to work alone = 10 + 5 = 15. Thus, the ratio of the time taken by Dev to that taken by Tina if each worked alone is 30/15 = 2/1.
Answer: B
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VeritasPrepKarishma
Responding to a pm:
Quote:
If they both work together, say they take t hrs. So Dev is working for t hours and Tina is working for t hrs. They complete the work in this way.
What happens when Dev works alone? He works for t hrs but Tina doesn't. So he works for another 20 hrs. Why? Because Tina did not work for t hrs. So whatever work Tina did in t hrs is now done by Dev in 20 hrs.
So, time taken by Dev to do a certain work/Time taken by Tina to do the same work = 20/t
Note that this is inverse of the ratio of their rates (since work done is the same)
In exactly the same way Tina does Dev's work in 5 hrs so
Time taken by Dev to do a certain work/Time taken by Tina to do the same work = t/5
Note that this too is inverse of the ratio of their rates (since work done is the same)
Hence, 20/t = t/5
On solving this, you get answer (B)
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Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 11 Sep 2026
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Location: India
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
jusjmkol740
Given:
1, If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task.
2. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task.
Asked: What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?
Let the time taken by Dev and Tina be d & t respectively
Q: d/t = ?
Time taken by Dev and Tina to complete the task = dt/(d+t)
1, If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task.
d - dt/(d+t) = 20 (1)
2. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task.
t - dt/(d+t) = 5 (2)
d - t = 15
d = t + 15. (3)
t - (t+15)t/(2t+15) = 5
2t^2 + 15t - t^2 - 15t = 10t + 75
t^2 -10t -75 =0
t^2 - 15 t + 5t - 75 =0
(t+5)(t-15) = 0
t = 15
d = t + 15 = 30
d/t = 30/15 = 2
d:t = 2:1
IMO B
VeritasKarishma
Quote:
You can think about it in terms of ratio fo speeds.
Say Dev and Tina are working together and they take t hrs to complete the work (as per their speeds). Some part of work is done by Dev (as per his speed) and rest is done by Tina (as per her speed). That is how they both finish it in t hrs.
Now, when Dev works alone, he takes t hrs + another 20 hrs to finish the same work. Dev takes 20 hrs extra when working alone because Tina did not work for t hrs. So whatever work Tina was doing in t hrs, Dev did it in 20 hrs.
Ratio of time taken by Dev/Tina = 20/t
Ratio of speeds when work done is same is inverse of ratio of time taken.
Ratio of speeds of Dev/Tina = t/20
On the same lines, when Tina works alone, she takes 5 hrs extra.
Ratio of time taken by Dev/Tina = t/5
Ratio of speeds of Dev/Tina = 5/t
Since speeds of Dev and Tina are constant, this ratio will not change.
t/20 = 5/t
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VeritasKarishma
Hi VeritasKarishma -- is there a Veritas blog post on this type of solving strategy specifically or on this concept ?
I need to solve more problems to inculcate this on D day (under 2 minute time pressure)
Any problems you can suggest with this strategy or where i can pick up some theory on this strategy ?
Posts: 259
Probably not the best approach, but this is what i did
1 = 1/a(t+20) --->1/a = 1/(t+20)
1 = 1/b(t+5) -----> 1/b = 1/(t+5)
1 = (1/a+1/b)t
1 = [1/(t+20) + 1/(t+5)]t
1 = t/(t+20) + t/(t+5)
plugin some obvious values (i.e t=10
1 = 1/3+2/3 ---> confirmed t=10
20+10:5+10
30:15
2:1
1 = 1/a(t+20) --->1/a = 1/(t+20)
1 = 1/b(t+5) -----> 1/b = 1/(t+5)
1 = (1/a+1/b)t
1 = [1/(t+20) + 1/(t+5)]t
1 = t/(t+20) + t/(t+5)
plugin some obvious values (i.e t=10
1 = 1/3+2/3 ---> confirmed t=10
20+10:5+10
30:15
2:1
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