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x-28=(1/3)x
x= 42
50 %of 42 = 21
X = 4
The queen gave two rings to two daughter’s and she kept the
third ring in her jewel box and she presented the ring to
her third daughter along with that jewel box.
B
The answer probably lies in finding the in-centre of the
traingle. Bisect all the angles of the triangles and the
point where these angle bisectors meet gives u the point P
length is 30 breadth is 12 and height is 12
——————————
1
.——————————
1 .
1 .
1 10 .
1
1 .
1
1
1 30 .
——————————-.
1
consider right angled triangle
so the distance= square root of (30*30+10*10)
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
Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
A