International Maths Olympiad Forum By SOF Olympiad Trainer

Anirudh Jhunjhunwala from JBCN International School, Parel, Mumbai

Subject :IMO Class : Class 7

The value of X in - 19 × (4 + (-2)) = - 19 × 4 +(- 19) × X, is

A 2
B - 19
C 4
D - 2

shunt it be 2

Ans 1:

Videep Ekbote from The Mothers International School, New Delhi

Class : Class 8

Post Your Answer

Anurag Lakhotia from ,

Subject :IMO Class : Class 7

Which one of the following is NOT true about similar figures?

A Similar figures always have the same shape.
B Similar figures always have the same size.
C Similar figures always have corresponding angles that are equal.
D Similar figures always have corresponding sides that are proportional.

how to solve it

View All Answers (1)

Ans 1: (Master Answer)

Admin from revisewise, Noida

Class : Class 1

The correct answer is B.

Submit Your Answers

Post Your Answer

T. V. Narasimha Rao from ,

Subject :IMO Class : Class 8

The sum of the interior angles of a polygon is 1440°. Find the number of sides of the polygon.

A14
B12
C10
D5

Can you please clarify the answer

View All Answers (3)

Ans 1:

Bindeshwari Prasad from ,

Class : Class 8

it would be 10. [n-2]180=1440 180n-360=1440 180n=1800 n=1800/180 = 10

Ans 2:

sanjay shete from ,

Class : Class 8

the answer should be 10 because= (n-2)*180=1440

Ans 3:

Admin from SOF, Noida

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

Submit Your Answers

Post Your Answer

Rahul Jha from Delhi Public School BN, Bengaluru

Subject :IMO Class : Class 6

What number should be subtracted from each of the numbers 54, 71, 75 and 99 so that the remainders may be proportional

(A) 1

(B) 2

(C) 3

(D) 6

Can you please explain

View All Answers (1)

Ans 1:

swagat sibsshish from davpublic school cuttack, Cuttack

Class : Class 8

Submit Your Answers

Post Your Answer

suhas from podar, kalyan, Dist thane

Subject :IMO Class : Class 4

In a zoo, there were 15 lions and peacocks altogether. The number of heads was 25 less than the number of legs. How many more peacocks were there?

(A) 20

(B) 15

(C) 10

(D) 5
how answer d came

View All Answers (2)

Ans 1:

Admin from SOF, Noida

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.

Ans 2:

Vrushti from swami vivekanand english primary school, Mumbai

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.