@amresh_maverick said:yaar samjha karo x=Z=w=1, y=4
amresh_maverick 😃
@iLoveTorres said:didi y=1 & y=4? yeh kya kar diya?
sry, typo tha.
amresh_maverick 😃
From 4 gentleman and 4 ladies a committee of 5 is to be formed which consist of a president,vice-president and three secretaries.What will be the number of ways of selecting the commitee with atleast 3 women such that atleast one woman hols the post of president or vice-president?
420
610
256
512
I'm having trouble with this one.
420
610
256
512
I'm having trouble with this one.
@perseverance07 said:remainder when 2^2 + 22^2 + 222^2 + 2222^2 + ....(222...49twos)^2 is divided by 9?ans options: 2,5,6,7 ?
6
@amresh_maverick said:OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
9:20:18??
@amresh_maverick said:OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
9:20:18??
@perseverance07 said:remainder when 2^2 + 22^2 + 222^2 + 2222^2 + ....(222...49twos)^2 is divided by 9?ans options: 2,5,6,7 ?
Er...6?
@amresh_maverick said:OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
10:23:15??
@amresh_maverick said:9 :20 :18OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
@perseverance07 said: remainder when 2^2 + 22^2 + 222^2 + 2222^2 + ....(222...49twos)^2 is divided by 9?ans options: 2,5,6,7 ?
OA) 6
OA) 6
@perseverance07 said:remainder when 2^2 + 22^2 + 222^2 + 2222^2 + ....(222...49twos)^2 is divided by 9?ans options: 2,5,6,7 ?
(2^2+4^2+6^2+8^2 + 5(1^2+3^2+5^2+7^2+0^2+2^2+4^2+6^2+8^2))%9 = 6
@Logrhythm said:i am getting 2... :/(4^2+6^2+8^2 + 5(1^2+3^2+5^2+7^2+0^2+2^2+4^2+6^2+8^2))%9 = 2
Add 2^2 = 4 na..
@Dexian said:ye kaise kara?
beaker A -> p+3p+2p = 6p
beaker B ->2q+q+5p = 8q
6p+8q = x [where x represents the total quantity in C]...
HCF(6,8) must perfectly divide x....which is not satisfied in that option....
@amresh_maverick said:OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
9:20:18 ?
@amresh_maverick said:OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
this is surely the answer
@Sybar said:From 4 gentleman and 4 ladies a committee of 5 is to be formed which consist of a president,vice-president and three secretaries.What will be the number of ways of selecting the commitee with atleast 3 women such that atleast one woman hols the post of president or vice-president?420610256512I'm having trouble with this one.
512??