( e ) None of these
5 litre jug contains 4 litres of salt water solution
with 15% of salt, that means it has 4000*.15 = 600 ml salt
in it.
If 1.5 litres solution is spills out, remaining solution
is 2.5 litres, then the salf content is 2500*.15 = 375ml.
Now, the jug is filled to full capacity with water i.e.,
now the jug has 5 litre solution in it.
Now, the salt content is 375/5000 = 7.5%
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
Placing three trees in triangle and placing the fourth tree in center
1972
17:3
45
let the amount of money be x
cloths 1/3 X x=rs.x/3
balance = x- x/3 = 2x/3
food = 1/5 X 2x/3 = 2x/15
balance = 2x/3 – 2x/15= 8x/15
travel = 1/4 X 8x/15 = 2x/15
now he has 100 rupees
2x/5 = 100
2x= 500
x = 500/2
x = 250
373
Ans =18
Explanation:
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X – 4)
bullets each.
But it is given that,after they shot 4 bullets each, total
number of bullets remaining is equal to the bullets each
had after division i.e. X
Therefore, the equation is
3 * (X – 4) = X
3 * X – 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127
B. 32%