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P+C+M=80
P+C=70
Therefore M=10
Given , P+M=90
If M=10
P=80
Very exited
D
13 and 15
C is grand child of A
Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:
Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.
Now you have 4 litres of water in jug A.
39/50
Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s
Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs
#Train length = 125Mtrs
Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s
Sp = D/t
=139M/10s
=13.9M/s
Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
19=
13
240+x
⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.
10000
one-legged =5% of 10000=500
remaining=10000-500=9500
barefooted=9500/2=4750
remaining people= 9500-4750=4750
hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged
9500 / 2 = 4750 bare foot
minium no of shoes = 4750*2 + 500*1 = 10000
5
b=15,l=10 or b=10,l=15