@scrabbler bro can you post a method for this question.. would really love to know your way of solving it
2> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?
105
102
103
104
@sachisurbhi said:The capitals of 4 partners A B C D are in the ratio of 7:8:6:5 . A and C's capitals are there in the business for the entire year.If each partner kept his money invested in the business for a period which is more than 6 months , and B and D together get 111/267 of the total profit,then for how many months is D's capital invested?a)11b)7c)9d)cant be determined
11. (edited, had answered b's time earlier)
Basically we get 8b+5d = 111 and b, d > 6 only 1 pair of integer solutions...
regards
scrabbler
Basically we get 8b+5d = 111 and b, d > 6 only 1 pair of integer solutions...
regards
scrabbler
@scrabbler said:In certain cases it can be 286, not 284...2001/7 = 285.71 so either 285 or 286 hoga...sorry for nit-picking regardsscrabbler
actually i was about to type 286 but then i thought it may be wrong.. and bro any correction made by anyone only add to the knowledge everyone gains.. so your corrections are welcomed :)
@sachisurbhi said:The capitals of 4 partners A B C D are in the ratio of 7:8:6:5 . A and C's capitals are there in the business for the entire year.If each partner kept his money invested in the business for a period which is more than 6 months , and B and D together get 111/267 of the total profit,then for how many months is D's capital invested?a)11b)7c)9d)cant be determined
Is it 11??
84x, 8px, 72x and 5qx
let the total profit be 267
=> a and c get -> 267-111 = 156
=> x(84+72) = 156
so x = 1
and 8p+5q=111
p=7 and q=11 is the only solution...
@bs0409 said:A 25 meter long wound cable is cut into 2 and 3 meter long pieces. How many different ways can this be done if also the order of pieces of different lengths is taken into account?
465?
12C1 + 11C3 + 10C5 + 9C7
OA?
regards
scrabbler
12C1 + 11C3 + 10C5 + 9C7
OA?
regards
scrabbler
@bs0409 said:A 25 meter long wound cable is cut into 2 and 3 meter long pieces. How many different ways can this be done if also the order of pieces of different lengths is taken into account?
1221?
@iLoveTorres said:@scrabbler bro can you post a method for this question.. would really love to know your way of solving it2> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
I think 103? Not sure....
Still thinking...
regards
scrabbler
Still thinking...
regards
scrabbler
@bs0409 said:A 25 meter long wound cable is cut into 2 and 3 meter long pieces. How many different ways can this be done if also the order of pieces of different lengths is taken into account?
2x+3y=25
x,y=(11,1) , (8,3) , (5,5) (2,7) ..
order of pieces bole toh?
consider the first case.. 11,1 ... he makes 1 cut of 2m then 1 of 3m then 10 more 2ms.
again 11,1 case only he cuts 2 of 2m then 1 of 3m and then 9 more of 2m??
aise karna h? ....
x,y=(11,1) , (8,3) , (5,5) (2,7) ..
order of pieces bole toh?
consider the first case.. 11,1 ... he makes 1 cut of 2m then 1 of 3m then 10 more 2ms.
again 11,1 case only he cuts 2 of 2m then 1 of 3m and then 9 more of 2m??
aise karna h? ....
@amresh_maverick said:1> last two digits of 1/5^9032> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
OA :
1> 08
2> 103
1> 08
2> 103
@ChirpiBird said:2x+3y=25x,y=(11,1) , (8,3) , (5,5) (2,7) .. order of pieces bole toh?
So if he makes 12 cuts, 11 of 2m and 1 of 3m, the order is also imp so we have to arrange the 12 cuts which we can do in 12! / (11!*1!) = 12C1 ways. (Or, choose which 1 out of the 12 is to be 3 m)
If he makes 11 cuts, 8 of 2m and 3 of 3m, we have to choose which 3 are 3m in 11C3 ways. And so on...
@joyjitpal
regards
scrabbler
If he makes 11 cuts, 8 of 2m and 3 of 3m, we have to choose which 3 are 3m in 11C3 ways. And so on...
@joyjitpal
regards
scrabbler
@scrabbler said:465?12C1 + 11C3 + 10C5 + 9C7OA?regardsscrabbler
@joyjitpal said:1221?
@ChirpiBird
OA-465......
Find the value of k such that sum of square of the roots of the quadartic equation x^2-8x+k=0 is 40
a)12
b)2
c)5
d)8
@bs0409 said:A 25 meter long wound cable is cut into 2 and 3 meter long pieces. How many different ways can this be done if also the order of pieces of different lengths is taken into account?
Is it 465??
3x+2y = 25
(x,y) = (1,11);(3,8);(5,5);(7,2)
(1,11) -> 12c1 ways = 12
(3,8) -> 11c3 ways = 165...
(5,5) -> 10c5 ways = 252
(7,2) -> 9c7 ways = 36
hence, total = 465...
@bs0409 said:@ChirpiBirdOA-465......Find the value of k such that sum of square of the roots of the quadartic equation x^2-8x+k=0 is 40a)12b)2c)5d)8
12
@bs0409 said:@ChirpiBirdOA-465......Find the value of k such that sum of square of the roots of the quadartic equation x^2-8x+k=0 is 40a)12b)2c)5d)8
12?
@bs0409 said:@ChirpiBirdOA-465......Find the value of k such that sum of square of the roots of the quadartic equation x^2-8x+k=0 is 40a)12b)2c)5d)8
12??
did it with options...
@scrabbler said:So if he makes 12 cuts, 11 of 2m and 1 of 3m, the order is also imp so we have to arrange the 12 cuts which we can do in 12! / (11!*1!) = 12C1 ways. (Or, choose which 1 out of the 12 is to be 3 m)If he makes 11 cuts, 8 of 2m and 3 of 3m, we have to choose which 3 are 3m in 11C3 ways. And so on...@joyjitpal regardsscrabbler
i hate Pnc ....(also Probability). 
thanks anyway.
thanks anyway.
@bs0409 said:@ChirpiBirdOA-465......Find the value of k such that sum of square of the roots of the quadartic equation x^2-8x+k=0 is 40a)12b)2c)5d)8
12?
@joyjitpal said:12
@iLoveTorres said:12?
@Logrhythm said:12??did it with options...
OA-12
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
A)3/20
B)29/34
C)47/100
D)13/102