@Torque024 said:Q) In a triangle ABC with angle ACB = 30 degree, median from A intersects BC at D such that Angle ADB = 45 degree. Find Angle ABC.
105 i think
@Torque024 said:@techsurge @surajmenonv
Mai aise kr rha
t*c = 77
(t,c)=(11,7),(7,11),(77,1),(1,77)
alb=3n,3n+1,3n+2
berk=4m,4m+1,4m+2,4m+3
for 11,7
alb= 33,34,35
berk= 28,29,30,31
35-28=7 will satisfy the condition
So alb=35, berk=28
Or koi combination se diff 7 nai aa rha tha
So 1, sirf 2 se possible nai
did the same thing
what i meant when u take the case 77,1 or 1,77 u get 2 cases
77*3,77*3+1,77*3+2 => 232
1*3,1*3+1,1*3+2 => 4
2 cases 4 and 232 hence 2nd should not be there
did i miss something :P
@krum said:let total quantity of sea water in the flask be Tamount of salt in it beS
after 10% is transferred, amount of water and salt left in flask =>
90/100(T-S) and 90/100(S)
amount of water and salt in beaker =>
10/100(T-S) and 10/100(S)
lets say z amount if water is now evaporated =>
S/T*M=S/10/(T/10-z)
=>S/T*M = S/(T-10z)
=>(T-10z)/T=1/M
=>1-10z/T=1/M
=>z=(1-1/M)T/10
so amount of water left in beaker = 10/100(T-S) - (1-1/M)T/10
now this is transferred into the flask
amount of water in flask = T-S - (1-1/M)T/10
amount of salt = S
given
(100+x)/100*S/T=S/[T-S - (1-1/M)T/10 + S]
=>(100+x)/100*1/T=1/[T- (1-1/M)T/10]
=>(1+x/100)*1/T=1/T*10M/(9M+1)
=>x/100=(10M-9M-1)/(9M+1)
=>x/100=(M-1)/(9M+1)
sox=100*(M-1)/(9M+1)
now what
@kinkong what is the OA anyway?
couldnt we just have put vlues of m and get x, and then check with options
@Torque024 said:Q) In a triangle ABC with angle ACB = 30 degree, median from A intersects BC at D such that Angle ADB = 45 degree. Find Angle ABC.
In triangle ADC,
sin(15)/DC = sin(30)/AD
In triangle ADB,
sin(a)/BD = sin(135-a)/AD
=> sin(135-a)/sin(a) = sin(15)/sin(30)
a = 105 degrees
I think there is a better way, but I can't think of it right now
Mukesh, Suresh and Dinesh, Travel from delhi to Mathura to attend Janmashtami. They have a bike which can carry only two persons at a time.Only Mukesh can drive the bike. Mathura is 300 Km from delhi.All of them can walk at 15 Km / Hr. All of them start the journey from delhi at same time. They are required to reach mathura at same time. If speed of bike is 60 Km/Hr Then WHAt is the shortest possible time in which call can reach at same time. ?
PS: Don't have the OA.
@techsurge said:couldnt we just have put vlues of m and get x, and then check with options
guy didn't gave any options and we have to find total amount here which gets eliminated, dunno about the rest, its from arun sharma so anything is possible
@rachit_28 said:Mukesh, Suresh and Dinesh, Travel from delhi to Mathura to attend Janmashtami. They have a bike which can carry only two persons at a time.Only Mukesh can drive the bike. Mathura is 300 Km from delhi.All of them can walk at 15 Km / Hr. All of them start the journey from delhi at same time. They are required to reach mathura at same time. If speed of bike is 60 Km/Hr Then WHAt is the shortest possible time in which call can reach at same time. ?PS: Don't have the OA.
11 hi hai. 5+3+3.
mukesh and suresh will leave delhi on byk, will travel for 180 km. then suresh will start walking alone. for 120 km, he will need 8 hrs so 8+3 on byk: 11 hrs
mukesh will return to delhi, will return to mathura with dinesh. so
3 hr+ 3 hr+ 5 hr= 11 hrs
@rachit_28 said:Mukesh, Suresh and Dinesh, Travel from delhi to Mathura to attend Janmashtami. They have a bike which can carry only two persons at a time.Only Mukesh can drive the bike. Mathura is 300 Km from delhi.All of them can walk at 15 Km / Hr. All of them start the journey from delhi at same time. They are required to reach mathura at same time. If speed of bike is 60 Km/Hr Then WHAt is the shortest possible time in which call can reach at same time. ?PS: Don't have the OA.
I don't know if it's right, but I did this.
To minimize time, I try to minimize return trips and hence only one return trip.
M,S first move in bike for time (t) hours => distance covered = 60t
By that time, D covers = 15t
Now, M leaves S and goes to pick up D.
Distance of destination for S = 300-60t
Time to reach D = 45t/75
By this time, D moves = 15*(45t/75) = 9t, Also S moves = 9t
So, distance to destination for D = 300-24t
Distance to destination for S = 300-69t
Now, they all reach together =>
(300-24t)/60 = (300-69t)/15
=> t = 25/7
Total time taken = t + (45t/75) + (300-24t)/60 = 25/7 + 15/7 + 5 - 10/7 = 30/7 + 5 = 65/7 hours ?
PS: Please check calculations, and also confirm the logic from somewhere.
@soham2208 said:I don't know if it's right, but I did this.To minimize time, I try to minimize return trips and hence only one return trip.M,S first move in bike for time (t) hours => distance covered = 60tBy that time, D covers = 15tNow, M leaves S and goes to pick up D.Distance of destination for S = 300-60tTime to reach D = 45t/75By this time, D moves = 15*(45t/75) = 9t, Also S moves = 9tSo, distance to destination for D = 300-24tDistance to destination for S = 300-69tNow, they all reach together =>(300-24t)/60 = (300-69t)/15=> t = 25/7Total time taken = t + (45t/75) + (300-24t)/60 = 25/7 + 15/7 + 5 - 10/7 = 30/7 + 5 = 65/7 hours ? PS: Please check calculations, and also confirm the logic from somewhere.
mukesh+suresh to destination 300/60 = 5 hrs
dinesh will walk for these 5hrs
distance covered by dinesh = 15*5=45
distance between mukesh and suresh 300-225
they will move towards each other and will meet at
15t+60t=255
t=3 hrs
in this 3 hrs dinesh will travel 45
remaining distance = 225-45=180
both will use bike so time taken = 180/60 = 3hrs
total 5+3+3=11
dinesh will walk for these 5hrs
distance covered by dinesh = 15*5=45
distance between mukesh and suresh 300-225
they will move towards each other and will meet at
15t+60t=255
t=3 hrs
in this 3 hrs dinesh will travel 45
remaining distance = 225-45=180
both will use bike so time taken = 180/60 = 3hrs
total 5+3+3=11
@Torque024 But they want to reach together , right ? In your case, Mukesh and Suresh reached the destination but Dinesh is not there yet..
@soham2208 said:@Torque024 But they want to reach together , right ? In your case, Mukesh and Suresh reached the destination but Dinesh is not there yet..
han bhai question interpret nai kia.
Q: 2222^3333^4444^....... div 9999 R=?
@viewpt said:Q: 2222^3333^4444^....... div 9999 R=?
9999 = 9*11*101
N = 2222^3333^4444^....
N mod 9 = (-1)^odd = 8
N mod 11 = 0
N mod 101 = 0
N = 9k + 8 = 11m = 101a
N = 1111p = 9k + 8
=> 4*(p mod 9) = 8
=> p = 9r + 2
=> N = 1111*(9r + 2) = 9999r + 2222
N mod 9999 = 2222 ?
@soham2208 said:9999 = 9*11*101N = 2222^3333^4444^....N mod 9 = (-1)^odd = 8N mod 11 = 0N mod 101 = 0N = 9k + 8 = 11m = 101aN = 1111p = 9k + 8=> 4*(p mod 9) = 8=> p = 9r + 2=> N = 1111*(9r + 2) = 9999r + 2222N mod 9999 = 2222 ?
jab se cat gaya hai, aisa lagta hai sab kuch bhool gaya hoon :D
why did i forget CRT :banghead: