Problem Solving for 780+ Aspirants.

sureshbala wrote:Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...

Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?

A. 48 B. 60 C. 56 D. 40 E. None of these

I am getting the answer for this as 70 days. So E. What is the OA?

This question can be answered in the limited time only if working step by step according to the basic ratio equation: Rate = Work/Time.

If the entire group were working together, we could have used the equation with the following parameters:
Time = T hours, Work = 1 project, thus Rate = 1/T.

However, the entire group worked at this rate only for 14 hours.
Time = 14, Rate = 1/T (as before), and thus the part of work managed in this time is Work1 = 14/T project.

When the size of the group is reduced by 0.25, the rate also reduced the same:
Time = T hours, Rate = 3/(4T), so the portion of the work is now: Work2 = 3/4 project.

We know that once the second phase is done, the entire project is completed so we can use the next equation:

Work1 + Work2 = 14/T + 3/4 = 1 project.

Solving this for T we come up with: T = 56.

But we need T + 14 = 70.
Therefore, E.
comments?

sureshbala wrote:Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...

Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?

A. 48 B. 60 C. 56 D. 40 E. None of these

My initial thought was, hmm, I guess sureshbala hasn't read this:
https://en.wikipedia.org/wiki/The_Mythical_Man-Month

but that aside:
initial project duration estimate = t
initial staffing level = s

remaining work after 14 days = (0.75s * t)
work accomplished after 14 days = (s * 14)
total work to be done (from original estimate) = (s * t)

so:

(0.75s * t) + (s * 14) = (s * t)
=> 0.75t + 14 = t
=> t = 56

That was the original duration estimate, so 14 days have to be added to get total duration, 70 days. That's E.

sureshbala wrote:Folks, finally here is the solution....Thought of being more clear on this problem, so uploaded the image....Have a look at this....

AFAIK, Trigonometry is not tested on the GMAT, correct? I just want to make sure that I'm studying the right stuff.

No, trigonometry is not tested on the GMAT

awesomeusername wrote:

sureshbala wrote:Folks, finally here is the solution....Thought of being more clear on this problem, so uploaded the image....Have a look at this....

AFAIK, Trigonometry is not tested on the GMAT, correct? I just want to make sure that I'm studying the right stuff.

You are correct ...Trigonometry is not tested. but it would help if you remember the basic trigonometric values of Sin and Cos. i used it to make the calculation fast...in fact this can be answered from the rule that sides are in the ratio of the Sin of the corresponding angles.

cjb wrote:

sureshbala wrote:Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...

Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?

A. 48 B. 60 C. 56 D. 40 E. None of these

My initial thought was, hmm, I guess sureshbala hasn't read this:
https://en.wikipedia.org/wiki/The_Mythical_Man-Month

but that aside:
initial project duration estimate = t
initial staffing level = s

remaining work after 14 days = (0.75s * t)
work accomplished after 14 days = (s * 14)
total work to be done (from original estimate) = (s * t)

so:

(0.75s * t) + (s * 14) = (s * t)
=> 0.75t + 14 = t
=> t = 56

That was the original duration estimate, so 14 days have to be added to get total duration, 70 days. That's E.

Hey, I didn't know about this...I am sure you will agree with me once you go through my solution...

Let us say x men can finish the project in y days

Given that all the x men worked initially for 14 days. After this (3/4)x men worked for y days, which means in this case 3/4 of the work is done. So initially in the first 14 days all x men working together have done 1/4 of the work. Hence all the x men can do the work in 14x4 = 56 days.

Coming to this case the work is done in 14+56 = 70.

Here is the next one....

How many integers x satisfy the equation (x^2-x-1)^(x+2) = 1 ?

A. 2 B. 4 C. 5 D. 3 E. None of these


OA:B

1 = a^b has only three options:
a. a can be anything but b has to be 0. This means that (x^2-x-1)^(x+2) = 1 gives us x+2 = 0, with x = -2.
b. a can be 1 and b can be anything. This means that x^2 - x - 1 = 1, with x^2 - x - 2 = 0. This has two solutions x = 2 and x = -1.
c. a can be -1 and b has to be even. This means x^2 - x - 1 = -1, with x^2 - x = 0. Solutions to this one will be 0 and 2. In both cases, x + 2 is even (corresponding to 2 and 4), which fits the initial restriction of b being even.

Now let's count: -2, 2, -1, 0. This means that there are 4 solutions, with my guess B. I'd really like to see an OA in your questions, it is one of the RULES of posting: provide an OA using the spoiler function.

DanaJ wrote:1 = a^b has only three options:
a. a can be anything but b has to be 0. This means that (x^2-x-1)^(x+2) = 1 gives us x+2 = 0, with x = -2.
b. a can be 1 and b can be anything. This means that x^2 - x - 1 = 1, with x^2 - x - 2 = 0. This has two solutions x = 2 and x = -1.
c. a can be -1 and b has to be even. This means x^2 - x - 1 = -1, with x^2 - x = 0. Solutions to this one will be 0 and 2. In both cases, x + 2 is even (corresponding to 2 and 4), which fits the initial restriction of b being even.

Now let's count: -2, 2, -1, 0. This means that there are 4 solutions, with my guess B. I'd really like to see an OA in your questions, it is one of the RULES of posting: provide an OA using the spoiler function.

Dear DanaJ, I will post the OA from now. In fact I have edited my previous post also.

You have provided the correct answer. The only mistake (which didn't effect the answer) you made is in the bold format above.

Solutions to this one will be 0 and 1. But if x=1, x+2 will not be even and hence this is eliminated.

The above explanation given by you is absolutely perfect and I think we can move ahead to the next query.

Yeah, I guess it's too early in the morning....

sureshbala wrote: Hey, I didn't know about this...I am sure you will agree with me once you go through my solution...

I do! 70 days is the answer I gave above.

My point, which wasn't really serious, was just that person-days of effort in software projects are not readily interchangeable. That's the main message of the book I referred to. Obviously I knew it was a rate problem anyway, so it didn't stop me from solving it.

Actually, as a word problem it was one of the toughest ones I've seen - the equation to be solved didn't leap out at me. My first attempt was wrong, but fortunately led me to an answer that was nonsense, rather than plausible but wrong!

Got any more problems, sureshbala? Looking forward to the next one!

Here is the next question...

On Sunday John left the place A for B at 3:30 am and on the same day Carsten started from B towards A at 8:30 am. They met each other at 12:30 pm. After meeting each other they took equal amount of time to reach their respective destinations. At what time did they reach their respective destinations?

A. 7:00 pm
B. 7:30 pm
C. 6:00 pm
D. 8:00 pm
E. None of these

OA:E

sureshbala wrote:Here is the next question...

On Sunday John left the place A for B at 3:30 am and on the same day Carsten started from B towards A at 8:30 am. They met each other at 12:30 pm. After meeting each other they took equal amount of time to reach their respective destinations. At what time did they reach their respective destinations?

A. 7:00 pm
B. 7:30 pm
C. 6:00 pm
D. 8:00 pm
E. None of these

OA:E

IMO 6:30 PM

Let speed of John : x km/hr
Let Speed of Carsten : y km/hr

Distance covered by John in 9 hrs = 9x km
Distance covered by Carsten in 4 hrs = 4y Km

Also given , that after meeting travel time is same

therefore :

t1 = t2

Time taken by John to cover 4y distance : 4y/x
Time taken by Carsten to cover 9x distance : 9x/y

9x/y = 4y/x

x/y = 2/3
(ratio of speeds of John : Carsten : 2:3)
x = 2/3y

Total Distance = 9x + 4y
= 9(2/3y) + 4y
= 10y

Distance covered by John in 9hrs intially (In terms of y) = 9 (2/3y) = 6y

6y distance covered in 9 hrs

hence additional 4y will be covered in 9/6y * 4y = 6 hrs