Ans 3:
Class : Class 10
Answer: C
Solution
If a polygon has 'n' sides, then it is divided into (n – 2) triangles. The sum of the angles of a triangle = 180°. Therefore, the sum of interior angles of a polygon having n sides is (n – 2) x 180.
(n - 2) x180 = 1440
n - 2 = 1440/180 = 8
n = 8 + 2 = 10
Solution
If a polygon has 'n' sides, then it is divided into (n – 2) triangles. The sum of the angles of a triangle = 180°. Therefore, the sum of interior angles of a polygon having n sides is (n – 2) x 180.
(n - 2) x180 = 1440
n - 2 = 1440/180 = 8
n = 8 + 2 = 10
Post Your Answer
Ans 1:
Class : Class 10
Answer : D
Solution
Let number of peacocks be p
Let number of lions be l
Number of legs = 15
Number of heads = 15 + 25 = 40
p + l = 15
2p + 4l = 40
Solving the above equations
p = 10 , l = 5
So there are 5 more peacocks than lions.
Solution
Let number of peacocks be p
Let number of lions be l
Number of legs = 15
Number of heads = 15 + 25 = 40
p + l = 15
2p + 4l = 40
Solving the above equations
p = 10 , l = 5
So there are 5 more peacocks than lions.
Ans 2:
Class : Class 4
Heads ( Lions + Peacocks) = 15, No. of Legs = 25 +15 = 40. Since total legs were 40 no. of lions has to be less than 10 (because 10 x 4 legs = 40 which means zero peacock, which is not possible...) . so if we take 5 lions and 10 peacocks, total heads = 15 & total legs = 5 x 4 + 10 x 2 = 40 legs. Thus the answer is 5 peacocks were more than lions.
Post Your Answer
Post Your Answer
Post Your Answer
Post Your Answer
Post Your Answer
Post Your Answer