@bodhi_vriksha said:It is 9/20 of 250 = 112.5Team BV - Kamal Lohia
sir, plz elaborate , how in a single line
Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
@albiesriram ,@nramachandran ,@abhishek.2011 
my approach is little smaller hope it helps
consider sales 100 , then increased by 50% = 150 ->then sales decreased by 17.5%= 82.5
therefore total decrease 67.5-> 67.5/150*100 =45 %
so 250 *.45 =112.5
@amresh_maverick said:sir, plz elaborate , how in a single line
see it is 45% of 250 which is 9/20 of 250
@Joey_Sharma said:Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
is it 286?not sure
@Joey_Sharma said:Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
EDIT:-
.?@raopradeep said:@albiesriram ,@nramachandran ,@abhishek.2011my approach is little smaller hope it helps consider sales 100 , then increased by 50% = 150 ->then sales decreased by 17.5%= 82.5therefore total decrease 67.5-> 67.5/150*100 =45 %so 250 *.45 =112.5
sales increases by 50% and COLLECTIONS FELL BY 17.5%. but u have taken again then sales decreased by 17.5% , could not get this ?
@raopradeep said:is it 286?not sure
i dont have the answer. i m getting a=47 and b=4 for its the min.. i.e. a+b=51.
@amresh_maverick said:sales increases by 50% and COLLECTIONS FELL BY 17.5%. but u have taken again then sales decreased by 17.5% , cold not get this ?
sales is equal to revenue that generated after all this ups n downs hope i cleared ur doubt if it still there let me tell
@Joey_Sharma said:i dont have the answer. i m getting a=47 and b=4 for its the min.. i.e. a+b=51.
yes you are right it satisfying the conditions
One from my side: Nimai and Nitai, two extremely intelligent brothers, are playing a special game, on a 4 by 4 square grid (containing 16 identical squares) which is further divided into four 2 by 2 sub grids, by moving alternately.
In every column and row of the grid each of the letter A, B, C, D is to be written exactly once. Also in each of the 2 by 2 sub grid each of the letter A, B, C, D is to be written exactly once.
Nitai, being elder, starts the by putting one letter in a grid and then Nimai does the same and so on. The player who puts the last letter in the grid will win. Who wins the game?
Be careful that suppose a situation arises as First column(from left), first row(top) contains A and in Second column's second, third and fourth row contains B, C, D in any order, then Second Column's first row will remain empty. :)
Team BV - Kamal Lohia
@Joey_Sharma said:Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
11k = a + 13b -------- (1)
13n = a + 11b ---------(2)
11k - 13n = 2b [when (1) - (2)] --------------------------(3)
169n - 121k = 2a [when , (2)*13 - (1)*11] -----------------------(4)
156n - 110k = 2*(b + a) [when , (3) + (4) ]
put n = k = 1
b + a = 23
@raopradeep said:yes you are right it satisfying the conditions
@albiesriram
do u have any approach for this question? i generated an equation from the given data then i had to use h&t; to get a and b.
do u have any approach for this question? i generated an equation from the given data then i had to use h&t; to get a and b.
Find the sum of all possible values of A + C + F + J if each letter denotes a single distinct digit and A + BC + DEF = GHIJ.
via - kamal/vineet sir
@jain4444 said:11k = a + 13b -------- (1) 13n = a + 11b ---------(2) 11k - 13n = 2b [when (1) - (2)] --------------------------(3)169n - 121k = 2a [when , (2)*13 - (1)*11] -----------------------(4) 156n - 110k = 2*(b + a) [when , (3) + (4) ]put n = k = 1 b + a = 23
put n=k=1 in eq 1 & 2 and u will get b=-1. so not possible as a,b are +ve integers.
@jain4444 said:11k = a + 13b -------- (1) 13n = a + 11b ---------(2) 11k - 13n = 2b [when (1) - (2)] --------------------------(3)169n - 121k = 2a [when , (2)*13 - (1)*11] -----------------------(4) 156n - 110k = 2*(b + a) [when , (3) + (4) ]put n = k = 1 b + a = 23
bad approach B comes out to be -ve
@bodhi_vriksha sir any approach other than hit and trial ?
@jain4444 @nole
http://www.pagalguy.com/forums/quantitative-ability-and-di/xtreme-maths-t-19967/p-635681/r-639600
This is the solution posted by Vineet more than 6 years back. Check it.
Team BV - Kamal Lohia
http://www.pagalguy.com/forums/quantitative-ability-and-di/xtreme-maths-t-19967/p-635681/r-639600
This is the solution posted by Vineet more than 6 years back. Check it.
Team BV - Kamal Lohia
@Joey_Sharma said:Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.
Mine version of solution -
We need to find smallest sum a + b such that
a + 11b = 0 mod 13, and a + 11b = 0 mod 11
Now analyse them individually,
a + 11b ≡ a - 2b ≡ 0 mod 13
Also 6(a - 2b) = 6a - 12b ≡ 0 mod 13
i.e. 6a + b ≡ 0 mod 13.
Similarly a + 13b ≡ a + 2b ≡ 0 mod 11
Also 6(a + 2b) = 6a + 12b ≡ 0 mod 11
i.e. 6a + b ≡ 0 mod 11.
Combining the two we get that
6a + b ≡ 0 mod 143
So minimum value of 6a + b should be 143 as both a and b are positive integers.
Clearly, for least value of a + b, a should be maximum possible.
So checking the solution for 6a + b = 143, we get that b = 5 will satisfy. And thus a = 23.
Finally min(a + b) = 23 + 5 = 28 :)
Team BV - Kamal Lohia
We need to find smallest sum a + b such that
a + 11b = 0 mod 13, and a + 11b = 0 mod 11
Now analyse them individually,
a + 11b ≡ a - 2b ≡ 0 mod 13
Also 6(a - 2b) = 6a - 12b ≡ 0 mod 13
i.e. 6a + b ≡ 0 mod 13.
Similarly a + 13b ≡ a + 2b ≡ 0 mod 11
Also 6(a + 2b) = 6a + 12b ≡ 0 mod 11
i.e. 6a + b ≡ 0 mod 11.
Combining the two we get that
6a + b ≡ 0 mod 143
So minimum value of 6a + b should be 143 as both a and b are positive integers.
Clearly, for least value of a + b, a should be maximum possible.
So checking the solution for 6a + b = 143, we get that b = 5 will satisfy. And thus a = 23.
Finally min(a + b) = 23 + 5 = 28 :)
Team BV - Kamal Lohia