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Subject :IMO    Class : Class 5

Ans 1:

Class : Class 8
please someone help to solve it

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Subject :IMO    Class : Class 7

Ans 1: (Master Answer)

Class : Class 1
The correct answer is C.

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 4

Ans 2:

Class : Class 4

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Subject :IMO    Class : Class 4

Ans 1:

Class : Class 4

Ans 2:

Class : Class 4
your answer is wrong the correct answer is ans3

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 5
Answer is B.Add only animals

Ans 2:

Class : Class 8
[B]

Ans 3:

Class : Class 8
[B]

Ans 4:

Class : Class 8
[B]

Ans 5:

Class : Class 5

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Subject :IMO    Class : Class 3

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Subject :IMO    Class : Class 4

Ans 1:

Class : Class 4
A

Ans 2:

Class : Class 4
A

Ans 3:

Class : Class 6
a

Ans 4:

Class : Class 4

Ans 5:

Class : Class 4
answer is A

Ans 6:

Class : Class 7
a+b=70g 10a+10b+10c=3200 10(a+b+c)=3200 a+b+c=320 c=320-70=250

Ans 7:

Class : Class 8

Ans 8:

Class : Class 5
A

Ans 9:

Class : Class 8
250 GRAMS . FIRST DIVIDE 3200 BY 10 . ANS =320 THEN SUBTRACT 70 FROM 320. ANS =250 WEIGHT OF THE CYLINDER =250

Ans 10:

Class : Class 6
If large square = A small square =B cylinder=C then, A+B=70gm 10A+10B=700gm 10C=3200gm-700gm =2500gm C=2500/10 =250

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Subject :IMO    Class : Class 3

Ans 1: (Master Answer)

Class : Class 1
The correct answer is D.

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 5
how did we arrive this answer can you help

Ans 2:

Class : Class 7

Ans 3:

Class : Class 3
58

Ans 4:

Class : Class 6
58 cm

Ans 5:

Class : Class 1
I thought it was C

Ans 6:

Class : Class 5
Answer is C that is boundary of the rectangle

Ans 7:

Class : Class 10
Longer side of the rectangle paper = 7 + 6 + 4 + 4 + 4 = 25 cm
Smaller side of the rectangle paper = 4 cm
Total length (perimeter) = 25 + 25 + 4 + 4 = 58 cm

Ans 8:

Class : Class 3

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Subject :IMO    Class : Class 8

Ans 1:

Class : Class 9
answer is told to be D not B;

Ans 2:

Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

Ans 3:

Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

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