@pankaj1988 said:My answer is 104. Tell your approach.
but what you have basically done is 144-120=24 so 24 boxes can have different number of oranges. therefore 128-24=104 wil have to have same number of oranges, right?
@pankaj1988 said:Raju has 128 boxes with him. He has to put atleast 120 oranges in one box and 144 at the most. Find the least number of boxes which will have the same number of oranges.
@iLoveTorres said:bhai koi approach nahi hai.. question hi nahi samajh mein aaya..
Ok..Hahaha
Solution:
Put 120 oranges in first 104 boxes.
In 105th box Put 121 oranges
In 106th box Put 122 oranges
On similar lines,
Put 144 oranges in 128th box
So there will be least 104 boxes with same number of oranges.
@Dexian said:B1 120 B2 121 B3 122 ................ B24 144B25 120.......................................B48 144B49 120.......................................B72 144B73 120.......................................B96 144B97 120.......................................B120 144B121 120 ........... B128 127ANS 5
aapka sig. bhi 5 hi hai :P
@pankaj1988 said:Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260 kms?
120
x/60 = 180 +x/90 = 1080-x/v
solve for v u will get 120
@Dexian said:arey sahi hai ya nahi ye to batao........
I don't have OA. I thought it would be 104.
But your approach seems more convincing.
What say @iLoveTorres
@pankaj1988 said:I don't have OA. I thought it would be 104.But your approach seems more convincing.What say @iLoveTorres
what he did is [128/24] where [] is the greatest integer function. this is also right in my opinion :)
@abhishek.2011 said:120x/60 = 180 +x/90 = 1080-x/vsolve for v u will get 120
shorter approach
60t = 90(t - 3) => t = 9
540/90 = (1260 - 540)/ (speed 0f C) :)
@ayushbhalotia said:Q) there are 10 stations between station A and Station B. 4 new stations have been added now. what are the number of ways in which new tickets can be issued considering only 1 way of the journey ?
@scrabbler said:91 - 55 na?n(n-1)/2 not n(n+1)/2...Edit: Sorry, if you include A and B phir tera waala makes sense.regardsscrabbler
explain plz, how n(n-1)/2 ?
Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
Triplets x,y,z are chosen from the set{1,2,3,4......24,25} such that x
Triplets x,y,z are chosen from the set{1,2,3,4....24,25} such that x
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
@amresh_maverick said:Sumit committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.
a=105
B=2
C=2
TOTAL=109 ??
B=2
C=2
TOTAL=109 ??
@techgeek2050 said:420 + 280 + 3 = 703 ?
if these are A B C ............ then the LCM will still be 840 na(IMO)
LCM is less than 840
LCM is less than 840
@Dexian said:if these are A B C ............ then the LCM will still be 840 na(IMO)LCM is less than 840
ya ryt.. sry... it should be 420 , 210 and 1