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Answer
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1.
in which at the center there common part of two triangle is hexagon so from he center of the circle draw 6 triangle in hexagon.
so total there is equilateral 12 triangle
after that their find altitude of one of the main triangle that is
(sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude
so altitude is=6sqrt(3)
now the small triangle below that altitude having the altitude is
(sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude
sp altitude=2sqrt(3)
so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
so radius is = 4sqrt(3)
remaining area (ans) = pi * r * r – 12* (sqrt(3)/4) * a * a where a=4
so ans is=67.58
2. m+d+I = 74
M=46+i+d
I=0.4*D
46+d+d0.4d+d+0.4d=74
D=10
6. APPLY PERIMETER FORMULA 2(Length+breadth)
8. A, B are common to all Three. And another is common
1 2 3
A B A B A B
Now one is common in 1 and 2
One is common in 2 and 3
One is common in 3 and 1
So total no of common is 4(a+b+1+1)
9.
Calculate after 5 sec the fly will come to no 1.
So as 5 is a factor of 60 so fly will be in 1 after 60 min.
Directions for Q. 1 to Q. 5: Refer the data:
L=12,N=11,J=27,M=14,K=36
Given,
K+M+J+L+N=100 ———(i)
K,M,J,L,N>10 ———(ii)
K?M?N?J?L ———(iii)
K+M=J+L+N ———(iv)
L=3 + X^3 , where X is an integer ———(v)
J= Y^3, where Y is an integer (Modify the question here it should be like that) ———(vi)
K=Z^2 or Z^3, where Z is an integer ———(vii)
Now from the ques value of J can be 27 or 64
but if we take 64 then eqn iv would not satisfy so J=27
Again considering eqn ii, iv & ii we can say that L+N = 23 & K+M=50
Now from eqn v the value of L must be 11, thus L=11
That conclude N = 12
Now from the question part 3,
The difference of numbers collected by L & M was:
(1) 3 (2) 2 (3) 5 (4) 9
Considering the fact M-L=any of those values
Now M=L+any of those values
M=11+any of those values
Now considering each values, M= 14, value =3; K=50-14 =36 which is a square of integer, thus satisfies
again M=13, value = 2; K=37 which is neither square or cube of any integer
M=16, value = 5; K=34 which is neither square or cube of any integer
M=20, value = 9; K=30 which is neither square or cube of any integer
Thus M = 14 & K = 36.
Answer
As you can see this question is not completed. We didn’t get the full question, so we tried to solve as much as we can. You are requested to submit your solution also as a comment here or in our Facebook page. www.fb.com/TechCSE
1.
in which at the center there common part of two triangle is hexagon so from he center of the circle draw 6 triangle in hexagon.
so total there is equilateral 12 triangle
after that their find altitude of one of the main triangle that is
(sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude
so altitude is=6sqrt(3)
now the small triangle below that altitude having the altitude is
(sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude
sp altitude=2sqrt(3)
so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
so radius is = 4sqrt(3)
remaining area (ans) = pi * r * r – 12* (sqrt(3)/4) * a * a where a=4
so ans is=67.58
2. m+d+I = 74
M=46+i+d
I=0.4*D
46+d+d0.4d+d+0.4d=74
D=10
6. APPLY PERIMETER FORMULA 2(Length+breadth)
8. A, B are common to all Three. And another is common
1 2 3
A B A B A B
Now one is common in 1 and 2
One is common in 2 and 3
One is common in 3 and 1
So total no of common is 4(a+b+1+1)
9.
Calculate after 5 sec the fly will come to no 1.
So as 5 is a factor of 60 so fly will be in 1 after 60 min.
Directions for Q. 1 to Q. 5: Refer the data:
L=12,N=11,J=27,M=14,K=36
Given,
K+M+J+L+N=100 ———(i)
K,M,J,L,N>10 ———(ii)
K?M?N?J?L ———(iii)
K+M=J+L+N ———(iv)
L=3 + X^3 , where X is an integer ———(v)
J= Y^3, where Y is an integer (Modify the question here it should be like that) ———(vi)
K=Z^2 or Z^3, where Z is an integer ———(vii)
Now from the ques value of J can be 27 or 64
but if we take 64 then eqn iv would not satisfy so J=27
Again considering eqn ii, iv & ii we can say that L+N = 23 & K+M=50
Now from eqn v the value of L must be 11, thus L=11
That conclude N = 12
Now from the question part 3,
The difference of numbers collected by L & M was:
(1) 3 (2) 2 (3) 5 (4) 9
Considering the fact M-L=any of those values
Now M=L+any of those values
M=11+any of those values
Now considering each values, M= 14, value =3; K=50-14 =36 which is a square of integer, thus satisfies
again M=13, value = 2; K=37 which is neither square or cube of any integer
M=16, value = 5; K=34 which is neither square or cube of any integer
M=20, value = 9; K=30 which is neither square or cube of any integer
Thus M = 14 & K = 36.
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