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.
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.
40
Ans 635.5
4000
1 year 5 % 200 rs int
Total 4200
2 year 5 % 210 ra int
Total 4410
4410-2210 = 2200
1 year 5% 110
Total 2310
2 year 5 % 115.5 rs int
Total 2425.5
Total interest paid
200 +210+ 110+115.5. = 635.5
madras
Mohammad Gauri
Take 5 pills from jar 1, 4 pills from jar2,…. and 1 pill
from jar 5 and put altogether in the scale. the ideal
weight should be (1+2+3+4+5)x10= 150gms. for eg if the jar2
is contaminated then the weight will be 146 gms. so
depending on the amount of weight loss we can identify the
contaminated jar.
Chi-ca-go
4155
Answer:270
Explanation: Formula is SI= P*R*T/100
Where P is principal amount
R is Rate
T is Time
So, SI= 500*6*9/100
i.e SI= 27000/100
SI= 270
0
12.5