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.
x/y =5/7…………………………..(1)
(x-25)/(y-25)=35/39
59x=35y+600
dividing b.t.s by y and substituting (x/y) value from equation (1)
y=84
Now, substitue y value in equation (1)
x = 60
30/70*100=42.87
saying 20%alcohol in 15 litre solution.
It means (20/100)*15=3 litres
Now 5 litre of water is added not alcohol.
So it means only the volume of solution is increases by 5 not alcohol.
It means 15+5=20
Now % of alcohol is
(3/20)*100=15%.
1 through 24, divisible by 2 in descending order.
That leaves you with:
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
8th place FROM the bottom means your answer will be 16.
500 ways
Given:
Principle Amount(P):₹5000
S.I:₹16500
Rate of Interest(R):15%
Find:T
SI=P*T*R/100
T=SI*100/P*R
T=16500*100/5000*15
T=22 YEARS
114
The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.
Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)
= 158
Number of odd prime numbers = 10
We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd
prime numbers)
Average =
= 15.8
∴ The Average of first 10 prime numbers which are odd is 15.8
6/11 or 54.54%
8men=6days
Xmen=1/2days
M1=8,D1=6
M2=X,D2=1/2
Formula:
M1*D1=M2*D2
8*6=X*(1/2)
X=8*6*2
X=96
Answer=96
If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.