@krum said:triplets as in 111,222... or like the last question?
Like the Last question
I have secured 78.56 percentile in CAT 2012,96.46 percentile in Dec 12 MAT and my academics rcrd are as follows-Cls 10-84.16%,Cls 12-73.8% and B.E. in Electrical Engineering-63%.Cld u pls suggest me a list of colleges where I can apply???
@Amrofa said:Like the Last question
@Amrofa said:Like the Last question
123,135,234
2,6,10
1,7,10
3,5,10
3,8,7
5,6,7
total 8?
@shubho50 post ur query here http://pagalguy.com/forums/cat-and-related-bschools/2012-which-b-schools-should-you-apply-t-83927/p-3594168?page=393
2,6,10
1,7,10
3,5,10
3,8,7
5,6,7
total 8?
@shubho50 post ur query here http://pagalguy.com/forums/cat-and-related-bschools/2012-which-b-schools-should-you-apply-t-83927/p-3594168?page=393
@Amrofa said:CorrectAdditional QuestionHow many of the triplets formed are such that the sum of number is divisible by 9 and they do not have a 9 in them ?????
is the answer 9?
Find sum of all integers from 1 to 10^6?
@krum said:123,135,2342,6,101,7,103,5,103,8,75,6,7total 8?@shubho50 post ur query here pagalguy.com/forums/cat-and-re...
you missed 864..
and i am assuming that 123 you mentioned is 126 :)
@pirateiim478 said:Find sum of all integers from 1 to 10^6?
wouldnt this be just n(n+1)/2 = 5000005* 10^5
@pirateiim478 said:Find sum of all integers from 1 to 10^6?
As per the formula,
n/2 * (n+1),
=10^6/2 * (10^6+1),
Solving the above equation, we get
5*10^6
@cadmium said:As per the formula,n/2 * (n+1),=10^6/2 * (10^6+1),Solving the above equation, we get5*10^6
It is 5 * 10^17
@pirateiim478 said:Find number of solutions for a,b such that a+b+LCM(a,b)=89?
say H is HCF of a,b
=> it also must be a factor of LCM
=> a,b and LCM(a,b) all must be divisible by H
now 89 is a prime number i.e 89*1 hence there is no common factor other than 1 for a,b, LCM(a,b) => HCF = 1
also we know LCM*HCF = a*b
=>LCM = a*b
hence we seek a solution for a + b + ab = 89
=> we seek a solution for (1+a)(1+b) = 90, hence possible (a,b)
(1,44)
(2,29)
(4,17)
(5,14)
(8,9)
(9,8)
(14,5)
(17,4)
(29,2)
(44,1)
ATDH.
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
without constraints the number of solutions = (36+3-1)C(3-1) = C(38,2)
We consider the cases when x=y (say k)
=>2k+z = 36
so for all possible even values of z we will get a solution. There are 19 such values of z
hence number of solutions where x and y are different are C(38,2) - 19
=703 - 19 = 684
Now in half the cases x will be greater than y and remaining half it would be less than y
hence 684/2 = 342 cases
ATDH.
In which base system will be 1/5=0.333333.....
@pirateiim478 said:In which base system will be 1/5=0.333333.....
.20=(0.33333)b
.2=3/b+3/b^2 + ...
.2=3(1/b)/(1-1/b)
.2-.2/b=3/b
.2=1/b(3.2)
1/b=1/16
b=16
@Estallar12 - I agree with your equation:-
(3x/4)*100*(1/10) +(x/4)*200*1/12 = 600 , I felt there are two possible equations:-
1.(3x/4)*100*(1/10) +(x/4)*200*1/12 = 600) and
2.(x/4)*200*(1/12) +(3x/4)*100*1/10 = 600) , however the second one gives x=40 Kg , so the first one is correct because we can reach to 51 KG of weight , Is that how you thought of this equation ? , correct me If I am wrong in my thinking .
Q.
21 mango trees , 42 apple trees and 56 orange trees are to be planted in rows such that each row contains the same number of trees of one variety only. Minimum number of rows in which above trees can be planted is ?
@cadmium said:Q.21 mango trees , 42 apple trees and 56 orange trees are to be planted in rows such that each row contains the same number of trees of one variety only. Minimum number of rows in which above trees can be planted is ?
17 rows are required
HCF of 21,42,56 = 7
So 21/7+42/7+56/7 = 17 rows
Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)
(log 2 = 0.3010)
(log 2 = 0.3010)
approach please?
@negiSannu said:Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)(log 2 = 0.3010)approach please?
x=20^-37
log x = -37 log20
log x= -37(log 10+ log2)=-37(1+.3010)= -48.137=-49+0.863
so x= 10^(-49+0.863)
48 zeroes
log x = -37 log20
log x= -37(log 10+ log2)=-37(1+.3010)= -48.137=-49+0.863
so x= 10^(-49+0.863)
48 zeroes
@negiSannu said:Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)(log 2 = 0.3010)approach please?
-37log 20
-37(log2 + log 10) = -37(0.3010 +1)
=-37*1.3010 = -48.137
48 zeroes?