Time speed, work, Alligations and mixtures problems

[eagles_dare13]

brilliant approach..
but can ne 1 pls answer hitman's question?

Originally Posted by catdog

Hi buddy there s standard method for solving this problem .....i will try to explain it

assume 1/(x^2-6x+2)=k

now => 1=kx^2-6kx+2k
=>kx^2-6kx+2k-1=0

Now for this equation to have real roots b^2-4*a*c>=0.....

=>4k(7k+1)>=0

=> k>=0 or k<=-1/7
now u can easily infer that max value of k can b infinity ...and min value is -infinity ............-1/7 that u got ur ans is max value of f(x) when f(x) is -ve


u can solve this all in max 2 mins if u r in practise

[anishag]

Here is a vedic funda to solve work and time questions. I am giving you an example but you can use this approach to solve any question which again depends upon your mind.

AS I HAVE TRIED TO EXPLAIN THIS IN VERY SIMPLE WAYS IT SEEMS THAN THERE ARE TOO MANY STEPS BUT ONCE YOU MASTER THIS YOU WILL BE ABLE TO SOLVE THESE TYPE OF QUESTION WITHIN SECONDS.

suppose there are 40 men, 25 women, 10 boys working 8 hrs a day complete a job in 10 days so how many days will be taken by 20 men, 30 women, 15 boys working 6 hrs a day.

to solve this type of question follow following simple steps :

1) take the value to be found i.e days. and write the previous value i.e. 10 days.

2) now we need to multilpy this value with fractions of other values which we have.

3) To create fraction just remember a simple thing if the REQUIRED VALUE is less than what we have than it should be multiplied with less than 1 fraction and if the REQUIRED VALUE is more than the present value than it should be multiplied with more than 1 fraciton.

4) Now we had 40 men and now we have 20 men so the days taken will obviuosly be more so the more than 1 fraction would be 40/20. same ways for women less than 1 fraction 25/30 and for boys again less than 1 fraction 10/15 and for days more than 1 fraction 10/6.

5) just multiply all this fraction with previous days and cross-cut all the values and you will get the answer.

10 X 40/20 X 25/30 10/15 X 10/6

SOLVE THIS TO GET YOUR ANSWER.

THE RULE GIVEN IN POINT # 4 IS A GOLDEN RULE WHICH CAN BE USED ANYWHERE.

ANISH

[eagles_dare13]

hi catdog,
please explain why hte differentiation method of finding maxima minima is not working in this case, as pointed out by hitman.

[roarkhoward]

folks do visit www.ptindia.com and find numerous brain teasers....
cheers,
roark

[catdog]

Originally Posted by eagles_dare13

hi catdog,
please explain why hte differentiation method of finding maxima minima is not working in this case, as pointed out by hitman.

Hi eagle


i think differentiation method will also work fine but u shud differentiate 1/(x^2-6x+2) to get the correct answer .........if u approachis that if g(x)=1/f(x)...then at min(f(x) we shud get max(g(x))....is not correct since this is applicable only if f(x) and g(x) both assume +ve values only ..............hope that helps!!!!!!!!

[ankursanghi]

Originally Posted by anishag

Here is a vedic funda to solve work and time questions. I am giving you an example but you can use this approach to solve any question which again depends upon your mind.

AS I HAVE TRIED TO EXPLAIN THIS IN VERY SIMPLE WAYS IT SEEMS THAN THERE ARE TOO MANY STEPS BUT ONCE YOU MASTER THIS YOU WILL BE ABLE TO SOLVE THESE TYPE OF QUESTION WITHIN SECONDS.

suppose there are 40 men, 25 women, 10 boys working 8 hrs a day complete a job in 10 days so how many days will be taken by 20 men, 30 women, 15 boys working 6 hrs a day.

to solve this type of question follow following simple steps :

1) take the value to be found i.e days. and write the previous value i.e. 10 days.

2) now we need to multilpy this value with fractions of other values which we have.

3) To create fraction just remember a simple thing if the REQUIRED VALUE is less than what we have than it should be multiplied with less than 1 fraction and if the REQUIRED VALUE is more than the present value than it should be multiplied with more than 1 fraciton.

4) Now we had 40 men and now we have 20 men so the days taken will obviuosly be more so the more than 1 fraction would be 40/20. same ways for women less than 1 fraction 25/30 and for boys again less than 1 fraction 10/15 and for days more than 1 fraction 10/6.

5) just multiply all this fraction with previous days and cross-cut all the values and you will get the answer.

10 X 40/20 X 25/30 10/15 X 10/6

SOLVE THIS TO GET YOUR ANSWER.

THE RULE GIVEN IN POINT # 4 IS A GOLDEN RULE WHICH CAN BE USED ANYWHERE.

ANISH

Hi Anish,

I have a little doubt with this approach.

For e.g. a man works M units/hour, woman W units/hour, boy B units/hour.

Then the first eqn would be
(40M + 25W + 10B) * 8 * 10.

The second eqn would be
(20M + 30W + 15B) * 6 *x

Since the above two have to be equal to find x (the number of days required)

we get (40M + 25W + 10B) * 8 * 10 = (20M + 30W + 15B) * 6 *x

Your way of calculating gives 10 X 40/20 X 25/30 10/15 X 10/6

Now, from the above, your calculation doesn't seem to follow. Basically, you have assumed that each man, woman and boy work equally.
Please correct me if I am wrong. If you think I am right, please make it a point to verify your fundae before posting on a public forum as it has a wide audience.

Disclaimer: This post doesn't mean to hurt feelings. Members are welcome to contribute.

[Naresh Reddy]

Originally Posted by ankursanghi

Hi Anish,

I have a little doubt with this approach.

For e.g. a man works M units/hour, woman W units/hour, boy B units/hour.

Then the first eqn would be
(40M + 25W + 10B) * 8 * 10.

The second eqn would be
(20M + 30W + 15B) * 6 *x

Since the above two have to be equal to find x (the number of days required)

we get (40M + 25W + 10B) * 8 * 10 = (20M + 30W + 15B) * 6 *x

Your way of calculating gives 10 X 40/20 X 25/30 10/15 X 10/6

Now, from the above, your calculation doesn't seem to follow. Basically, you have assumed that each man, woman and boy work equally.
Please correct me if I am wrong. If you think I am right, please make it a point to verify your fundae before posting on a public forum as it has a wide audience.

Disclaimer: This post doesn't mean to hurt feelings. Members are welcome to contribute.

hi ankur..
you are right one needs to be careful before posting some funda..But you were too harsh in your tone.We learn though mistakes and forum gives us all teh same opportunity.

about the current prob .i think the method mentioned by anish was not right. you cant multiply the no.of men ,women& boys. you can only add them ..because theyare all only one part of the problem of mandays..i.e worker..you can multiply man(all workers) fraction,hour fraction ,days fraction etc..but not among components of same part...

the problem whether effieciency of men,women& boys will effect the problem..but the initial method itself is wrong..
SOLn..
for any efficiency
x =[(40M + 25W + 10B) /(20M + 30W + 15B)]*8 /6*10
its same as mentioned by ankur
now for same efficiency
x=[(40+ 25 + 10) /(20 + 30 + 15)]*8 /6*10 =75/65*8/6*10 ..

[hitman]

Originally Posted by catdog

Hi buddy there s standard method for solving this problem .....i will try to explain it

assume 1/(x^2-6x+2)=k

now => 1=kx^2-6kx+2k
=>kx^2-6kx+2k-1=0

Now for this equation to have real roots b^2-4*a*c>=0.....

=>4k(7k+1)>=0

=> k>=0 or k<=-1/7
now u can easily infer that max value of k can b infinity ...and min value is -infinity ............-1/7 that u got ur ans is max value of f(x) when f(x) is -ve


u can solve this all in max 2 mins if u r in practise

thanx mate.
its a gud approach.
but regardin this qn i still have a doubt .
actually i checked by puttin few values for x.
x=3 y=-1/7 ( as rightly pointed by you , its the max value of f(x) when f(x) is -ve )
x=6 y=1/2
x=7 y=1/9
similarly for higher values of x, y keeps on decreasing. ie y< 1/2 for x>6
so can we say the max value is 1/2.
plz help me out . i'm really confused.

[minali]

sorry guys .......
i didnt calculate da roots i just concentrated on the aproach .u r right

Originally Posted by vin

i dont Think the ans is correct ..yep u will get three roots but only one wud be real i.e x=0 and remaining wud be imaginary

[rama_1978]

Hi Friends,

This is a problem I came across and was wondering if my approach is correct. Sorry I dont know the correct answer to this one.

Problem:
For every 9 leaps of a rabbit, a hound takes 4 leaps. In each leap the rabbit covers 1 meter while the hound covers 2+1/3 meters. In how many leaps would the hound catch the rabbit if the rabbit has a start of 16 meters.

Options:
1. 172
2. 152
3. 192
4. None of these.

My reasoning:
In 9 leaps the rabbit covers 9 meters. In 4 leaps the hound covers 9+1/3 meters. Thus the difference is 1/3 meters.
Therefore, number of leaps = 16x4/(1/3)= 16x3x4 = 192 leaps.

Could anyone please tell me if my logic and answer is correct?

Thanks,
-Rama.