International Maths Olympiad Forum By SOF Olympiad Trainer

User Forum

Subject :IMO    Class : Class 7

Ans 1:

Class : Class 8
Let tom has Rs. X Let Jasmine has Rs y A.T.Q x+y = 41 x = 41-y 1/4 * x = 2 + 1/7*y 1/4*(41-y) = 2 + 1/7*y 41/4 - y/4 = 2 + y/7 41/4 - 2 = y/7 + y/4 33/4 = 11y/28 33/4*28/11 = y 21 = y x+21=41 X = 41-21 = 20

Ans 2:

Class : Class 8
Tom and jasmine have Rs 41 altogether . let, Jasmine has Rs x.. therefore tom has Rs( 41-x) {the rest of the money} according to the question, 1/4 of Tom's money is 2 more than 1/7 of jasmine's money. therefore, 1/4 x (41-X) = (1/7 x X)+2 or, ( 41-X)/4=X/7+2 or, (41-X)/4=(X+14)/7 or; (7x41)-7X=4X+(4x14) or; 287-7X=4X+56 or; 287-56=4X+7X or, 231=11X: 11X=231 X=231/11 =21 THE ANSWER IS 21 OR TOM HAS RS 21................

Post Your Answer

Subject :IMO    Class : Class 7

Ans 1:

Class : Class 7
80 Rs 5 note and 10 Rs 10 note I have solved by this equation 5x + 10(90-x) = 500

Ans 2:

Class : Class 8
@ Subarno Maji, that's the correct answer

Post Your Answer

Subject :IMO    Class : Class 8

Ans 1:

Class : Class 10
There exist two such numbers : (1-sqrt(5))/2 and (1+sqrt(5))/2 This can be taken out by making a quadratic equation and using the discriminant formula, which is: x=(b*sqrt(b^2-4ac))/2a

Post Your Answer

Subject :IMO    Class : Class 8

Ans 1:

Class : Class 8
115929

Ans 2:

Class : Class 8
21436587 is exactly divisible by

Ans 3:

Class : Class 8
99 hehe

Post Your Answer

Subject :IMO    Class : Class 2

Ans 1:

Class : Class 4
90

Ans 2:

Class : Class 3
90

Post Your Answer

Subject :IMO    Class : Class 6

Post Your Answer

Subject :IMO    Class : Class 8

Ans 1:

Class : Class 8
The point where all 3 altitudes of a triangle intersect is called a Orthocenter.

Post Your Answer

Subject :IMO    Class : Class 8

Post Your Answer

Subject :IMO    Class : Class 8

Ans 1:

Class : Class 8
Yes, because of the simple reason, that other polygons have edges! Circle has the same distance from the center at each and every point, whereas the distance from the center to the edges is generally lesser than that to the vertices in other polygons Thus, more area is covered in case of circles.

Ans 2:

Class : Class 8
yes, it was mentioned in the maths trivia a few days ago. but idk the reason.

Post Your Answer

Subject :IMO    Class : Class 4

Ans 1:

Class : Class 6
you have to practice atleast 2 to 3 hours in a day maths , follow the syllabus.

Post Your Answer