CAT 2008: Quantitative Questions a Day 101-133 - The discussions

[slam]

Originally Posted by Aarav

Why do you need to make it sooooooo COMPLEX? Simple average nahin soch sakte kya?

I was actually laughing at myself after I read your post!
Anyway, haan, soch sakte hain average se. But I was trying to avoid using the options to calculate. We can start off with 2 days and move on, to see that the avg first increases and decreases, right?

Take (1-0.9^n)/(n+1)
n = 2 => 0.19 / 3 = 0.0633
n = 3 => 0.271/4 = 0.06775
n = 4 => 0.344/5 = 0.0688
n = 5 => 0.409/6 = 0.0681 => decreasing from here.

PS: I did not use a calculator for this!

[Aarav]

Originally Posted by implex

Yet another mistake
Process correct, but thinking too much and ending up choosing the wrong one huh!!
Anyways my last qqad for 10 days !

Take complete break from Quant Implex, this will help things. Come back fresh and charged up

[Aarav]

Originally Posted by slam

I was actually laughing at myself after I read your post!
Anyway, haan, soch sakte hain average se. But I was trying to avoid using the options to calculate. We can start off with 2 days and move on, to see that the avg first increases and decreases, right?

Take (1-0.9^n)/(n+1)
n = 2 => 0.19 / 3 = 0.0633
n = 3 => 0.271/4 = 0.06775
n = 4 => 0.344/5 = 0.0688
n = 5 => 0.409/6 = 0.0681 => decreasing from here.

PS: I did not use a calculator for this!

That's right Slam, we didn't need to use calculator here. You must start answering using options also as that saves time in CAT. As long as the answer is logically derived, it is very much acceptable here.

[shashank3012]

Let total work done be x.
So,
Day---Work-------Efficiency
1)....x.....0.5x
2)....0.9x...0.63x
3)....0.81x...0.6775x
4)....0.729x...0.6878x
5)....0.6561x...0.6825x
Here,efficiency means that the work done over a period of n+1 days.here,the +1 component is taken as the off-day.
Now,efficiency starts decreasing after day 4.So,maximum number of days on the trott are 4 days.
So,option(2).

[shivam_01]

Page crashed

[shivam_01]

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Quantitative Question # 109
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Priyanka does daily work for a fixed time, and her efficiency would reduce by 10% on each subsequent day. However, when she took a break for a day her efficiency would come to normal the next day. Assume Priyanka worked for a long time, then after how many days of consecutive work does she needs to take a break to have the maximum output?

(1) 3 (2) 4 (3) 5 (4) 7 (5) none of these

need to retype poora page refresh ho gaya aur saari mehnat paani mein

My Take:

Let the efficiency of Priyanka be 1000 units

switch(No of continous days of working)
{

Case 3:
In case she works for continously 3 days
Total job done 1000 + 900 + 810 = 2710 units
Average efficiency achieved per day = 2710/4
break;

Case 4:
In case she works for continously 4 days
Total job done 1000 + 900 + 810 + 729 = 3439 units
Average efficiency achieved per day = 3439/5
break;

Case 5:
In case she works for continously 5 days
Total work done 1000 + 900 + 810 + 729 + 656.1 = 4095.1 units
Average efficiency achieved per day = 4095.1/6
break;

Case 6:
In case she works for continously 6 days
Total work done 1000 + 900 + 810 + 729 + 656.1 + 590.5 = 4685.6 units
Average efficiency per day = 4685.6/7
break;

}

Maximum Efficiency for Priyanka in case she works continuously for 4 days
Option 2.....

May IT field persons sees this and makes 4 working days for us ...

[vishrock]

let efficiency be = x%
next day = .9x%
3rd day = .81x%
4th day = .729x%
5th day = .6561 x%
6th day = .58...x%
Aftr 3 days = so output = till 3rd + 0/4 = 2.71/4 = .6775
After 4 days = till 4th day + 0/5 = 3.439/5 = .687...
After 5 days = till 5th day + 0/6 = 4.0951/6 = .682...
After 6 days = 4.67../7 = .66....
So max is Aftr 4 days.
option b

[simplifiedlogic]

Priyanka when works for 7 consecutive days generates a cumulative efficiency of approx 569.7 on the 8th day
If she takes a break on the 7th day, the cum. efficiency becomes approx 568.6 on the 8th day.
So she should take a break on the 8th day..

[tyro.novice]

Originally Posted by tyro.novice

We will consider this over cycles of work which include the days when efficiency decreases by 10% on each subsequent day and the day of rest when no work is done.
If rest taken after 3 days we consider 4 day cycles, rest taken after 4 days -- 5 day cycle and further.
The average efficiency over the n+1 day cycles continuously decreases from the case considering break after 3 days onwards.
hence option 1. (3)

my mistake
the method to solve was correct but rather than taking it 100 90 81 ..... for subsequent days made it
100 90 80



Now the correct solution
Considering the cycles over a period of n+1 days
After 3 days
Average efficiency per day= (100 + 90 + 81 + 0)/4
= 271/4=67.75
Average efficiency per day after 4 days =
(100+ 90+ 81 +72.9 + 0 ) / 5 = 343.9 / 5= 68.78


Average efficiency per day after 5 days= 409.51/6
=65.61

The efficiency then continuously decreases

Hence option 2 . After 4 days.

[tyro.novice]

Originally Posted by slam

How'd you figure this out?

By now you have figured that out for yourself. Didn't see your post -- was offline.