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
let present age of ANAND AND BALA BE( A AND B ) RESPECTIVELY
A.T.Q
(A-10) = 1/3[B-10]. ……… – {1}
given
B = A+ 12 PUTTING IN 1
THEREFORE A-10 = 1/3 (A+2) = A+2 = 3A – 30
2A =32
A=16
Monday
7days for a week so 62÷7 remainder is 6 so
After Tuesday + 6days = monday
2
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.
Suppose the lengthier arm of weighing pan is of x cm and other arm is y cm .Also let weight if each melon be m kg.
so applying equilibrium of torque principles ,we get
case 1:-
1x-8my=0
case 2:-
2mx-1y=0
using case 1 equation , we substitute value of x into case 2 equation;
16mmy-1y=0
(16mm-1)y=0
y is length of weighing arm and cannot be 0,
therefore ,
16mm-1 =0
16mm=1
mm=1/16
m= square root (1/16)
m=+-1/4
m is weight of melon and cant be negative.
Hence m, weight of one melon is 1/4 kg
33
x,y sides of rectangle
area=xy
increased by 100% =double length
hence
new area=4xy
increased area=4xy-xy=3xy
percentage of area will be increased
by 300%
119