[khanna_sumit]

Originally Posted by Neo2000

Can you please explain how you got 34 places? Am getting more than 34. Am I re-counting ??

unit's place 666,676,686,696; 10 places in block of 700, 10 in 800 and 10 in 900.

[mancee]

Originally Posted by Neo2000

Can you please explain how you got 34 places? Am getting more than 34. Am I re-counting ??

May I explain?

In hundred's place, 666 to 699- 34 times.

In ten's place,
666-669- 4 times
760-769- 10 times, similarly 10+10 more for 8 and 9.
Total= 34 times.

In unit's place,
count 66 to 99. Its 34.

Hope it helps.

[sanjit saha]

i m waiting 4 that kind of input. i m not getting leave from office , so tensed but i know when amit sir is there, he will help to handle the situation.just 25 days left for the d day.
please help.
sanjit.

[MrAnderson]

<snip></snip>

[rider.royal]

Originally Posted by MrAnderson

<snip></snip>

how come???????:whatthat:

[catologic]

This a godd thread. Keep up the good work Amit

[Amit@Career_Avenues]

Originally Posted by khanna_sumit

(1).

method is same 300/2 + 300/3 + 300/5 - 300/6 - 300/10 - 300/15 + 300/30
= 220 numbers upto 300 are divisible by 2,3 or 5. means 80 are not divisible.

Similarly from 1 to 100, 26 are not divisible by 2,3 or 5. hence answer is
80-26 =54.

Be careful here. I think you mean 1 to 99 and not 1 to 100. But anyway correct answer.

(2). It depends on that how many times digit 6 occurs at unit's, ten's or hundred's place. that happend 34 times for each of the places. at unit's place this change from 6 to 9 yields an increase of 3, at ten's place of 30 and at hundred'd place of 300. so total change is 34(3+30+300).

(3) let m be the number of men and n be the number of cakes consumed in first hour.
so by given conditions
mn + (m+6)(n-5) + (m+12)(n-10) + (m+18 )(n-15) + (m+24)(n-20) = 545
solving we get
mn + 12n - 10m = 289
or (m+12)(n-10) = 169 possible only for (m+12) = 13 and (n-10) = 13.
m = 1; n=23.

so in 4th hour. (m+18 )(n-15) = 19*8 = 152.

Perfect. Hope others got this too.

Another way of doing 3 is to take hour 3 men as x and cakes per men as y. So we get (x-12)(y+10) + (x-6)(y+5) + xy + (x+6)(y-5) + (x+12)(y-10) = 545.
This method would be lot easier on calculations.

Cheers
Amit

[Amit@Career_Avenues]

Here is a question which would bring out a good concept.

7. If X, Y and Z are real numbers such that X+Y+Z = 5 and XY +YZ + XZ = 3, then what is the largest value that X can have ?
1. 5/3 2. Root 19 3. 13/3 4. None of these

[P R E M]

Great going AMIT! keep up the good work MENTOR-AH-DATOR

[khanna_sumit]

Originally Posted by Amit@Career_Avenues

Here is a question which would bring out a good concept.

7. If X, Y and Z are real numbers such that X+Y+Z = 5 and XY +YZ + XZ = 3, then what is the largest value that X can have ?
1. 5/3 2. Root 19 3. 13/3 4. None of these

y + z = 5 - x
xy + yz + zx = 3
or x(y + z) + yz = 3
or x(5 - x) + yz = 3
or yz = 3 - x(5 - x)
applying A.M. >= G.M.
y + z >= 2*sqrt(yz)
or (5 - x)^2 >= 4(3 - 5x +x^2)
or 25 + x^2 - 10x >= 12 - 20x + 4*x^2
or 3*x^2 - 10x -13 <= 0
or (3x - 13)(x + 1) <=0
0r -1 <= x <= 13/3
so maximum value of x is 13/3 and minimum is -1.