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3.14
At 130 years.
find out what is the problem and taking the bad out always works.
Answer: 66.67 km approx.
Solution:
Let the first train A move at u km/h.
Let the second train B move at v km/h.
Let the distance between two trains be d km
Let the speed of bee be b km/h
Therefore, the time taken by trains to collide = d/(u+v)
Now putting all the known values into the above equation, we get,
u = 50 km/hr
v = 70 km/hr
d = 100 km
b = 80 km/hr
Therfore, the total distance travelled by bee
= b*d/(u+v)
= 80 * 100/(50+70)
= 66.67 km (approx)
Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32
after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z
After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32
Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning
(x+2)^2 -x^2 = 84
X=20
So (20,22)
Sum= 42
10, 14, 28, 32, 64, 68, 132
132
500 ways
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
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
3days