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.
cousin, daughter of maternal uncle
2.01
( d ) Neeraj is taller than Nilesh
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
answer : 36 –Explaination:
X= Chickens
Y = Qty per day
X*Y*9= 12 *.9 *Y*30
x=36
zero – as one letter cant be wrong – atleast two are to be wrong
(1/2)*x*y
lcm (12,18,24,32,40) – 1
1440-1 = 1439
39 hrs 15 mins..
note:
he will not be having .45 mins for his lunch on saturday..
16
Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.
Summation of 301 – summation of 99
=41550
Summation of n=((n*n+1)/2)
Ans is 10%
Printer price = x;
Computer price = 3x
Total overall price = 20x + 60 x 3x = 200x
Printer percent = (20x/200x ) x 100 = 10%
40