Official Quant thread for CAT 2013

MBA CAT 2024
30861
@jaspunit said:
If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?148157
15
30862
@jaspunit said:
@catahead in case of nanogan also?
No diagonal in a nanogon pass through centre...? To not possible!

regards
scrabbler
30864
@jaspunit said:
@scrabbler in which figures no of int, pts possible then only in sq, pentagon, hexagon???
No no there will be many point of int in a nonagon as well. Just that none of them will be the CENTRE of the figure. I am not sure of the answer to your original question, would have to think about it...will get back if I find a decent solution.

regards
scrabbler
30866
@jaspunit said:
If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?148157
RK/(R+K) = 12

R*K = 12(R + K)

R*( K - 12 ) = 12K

K > 12 and (K-12) is a factor 0f 12K

So, K can be 13, 14, 15, 16, 18, 20, 21 and 24 --> 8 values
A total of 16 - 1 ( 24,24 )= 15 Values
30867
If 4th Feb 2000 is a Friday, then 12th Nov of that year will be? I am getting the answer as Monday , but the correct answer is Sunday.Can anyone explain? I am getting 31 odd days in between. So fri->mon
30868
@jaspunit said:
how to find number of points of intersection of the diagonals of a regular figure,we can find small ones easily like - square - 1hexagon - 15,any short cut?if we have to find - of regular octagon?
BTW how is hexagon 15? It is 13 I guess (excluding the vertices) as the three long diagonals all meet in a single point not 3C2 points.

regards
scrabbler
30871
@patdet said:
If 4th Feb 2000 is a Friday, then 12th Nov of that year will be? I am getting the answer as Monday , but the correct answer is Sunday.Can anyone explain? I am getting 31 odd days in between. So fri->mon
How 31? 30 days na?

Leap year so Feb gives 1 extra so
Feb +1
Mar +3
Apr +2
May +3
Jun +2
Jul +3
Aug +3
Sep +2
Oct +3
and 8 days to go from 4th to 12th
So total = 30.

regards
scrabbler
30872
@jaspunit said:
@scrabblerThe number of points of intersection of the diagonals of a regular hexagon is:a)10b)15c)18d)19this is the exact question with options
It would be 19, including the vertices of the hexagon.

regards
scrabbler
30874
@scrabbler said:
How 31? 30 days na?Leap year so Feb gives 1 extra soFeb +1Mar +3Apr +2May +3Jun +2Jul +3Aug +3Sep +2Oct +3and 8 days to go from 4th to 12thSo total = 30.regardsscrabbler
Yep, my bad, I was not doing it from 4th to 4th of each month
thanks :)
30875

plz help me with these qstns

find largest value of n such that (n^3 + 100) s divisible by (n+10)?

what is the largest value of 11! that s one bigger than 6 ?

plz tell how to approach ?

30876
@jaspunit said:
@scrabblerThe number of points of intersection of the diagonals of a regular hexagon is:a)10b)15c)18d)19this is the exact question with options
getting 19 (counted manually)


30878
@nits_rocker said:
plz help me with these qstns

find largest value of n such that (n^3 + 100) s divisible by (n+10)?


10 ??
30879
@Dexian no dude....answer s 890
30880
@nits_rocker said:
plz help me with these qstnsfind largest value of n such that (n^3 + 100) s divisible by (n+10)?what is the largest value of 11! that s one bigger than 6 ?plz tell how to approach ?
n^3 + 100 = k(n+10)

k = (n^3 + 100)/(n+10) = n^2 - 10n -100 + 900/(n+10) ---- (By Long Division Method)

From here, n should be 890..
30881
@nits_rocker said:
plz help me with these qstnsfind largest value of n such that (n^3 + 100) s divisible by (n+10)?what is the largest value of 11! that s one bigger than 6 ?plz tell how to approach ?
Could not understand
30882
@paridhi11890 said:
is thr a funda of borrowing also involved.. wat if the shopkeeper is not nice enough?
a little clarification with a new approach..hope u find it useful.

U ask for 45 toffees from d shopkeeper and u can account 4 each toffee either in cash or wid wrapper no borrowing wali prblm then . viz-

when u ask d shopkeeper to give 45 toffees together then 45 toffees k 45 wrapper honge .u give him 10rs(30 toffees) and d 45 wrappers(15 toffees).. and finally u manage to buy 45 toffees in the given amount.
30883
n^3 + 100/(n + 10) = n^2 - 10n + 100 - 900/(n+10 ) (perform simple division) now for this to be an integer n+10=900 hence n = 890
30884
@nits_rocker N^3+100= (N^2-10n+100) * (n+10) -900
If IF Thiss IS INteger then divisible
That means -900รท / 10+n
You can get values of n
Some simple values are n=2 or 5

But largest value =890 when they divide an get 1
30885
@ScareCrow28 thanks for the first one....second is
what is the largest factor of 11! that s one bigger than 6 ?
30886
@nits_rocker Answer for second
39, 916, 800 =11! is largest Factor 11!
Itself IS largest factor