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A Ramakrishna Rao from D A V Public School. Unit 8. Bhubaneswar , Bhubaneswar
Subject :IMO    Class : Class 6

Ans 1:

DHEER PANDEY from Sentia School, Hyderabad
Class : Class 8
its d 1\15

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Manvi Mundepi from Montfort School Roorkee, Roorkee
Subject :IMO    Class : Class 6

Ans 1:

Nivedhini from Vivekanada Vidyalaya Chitlapakkam, Chennai
Class : Class 7
INDIA invented zero and its inventor is Bramagupta or Aryabhatta

Ans 2:

Gauri Singh from St Johns School Marhauli, Varanasi
Class : Class 8
India invented 0 and its inventor is Aryabhatta.

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Manvi Mundepi from Montfort School Roorkee, Roorkee
Subject :IMO    Class : Class 6

Ans 1:

Aryan Mohapatra from DAV PUBLIC SCHOOL,UNIT-8, Bhubaneswar
Class : Class 6

Ans 2:

Aryan Mohapatra from DAV PUBLIC SCHOOL,UNIT-8, Bhubaneswar
Class : Class 6
The answer is A

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Rs from DPS, Delhi
Subject :IMO    Class : Class 4

Ans 1:

Vania Aravindan from Unicent School , Miyapur Hyderabad
Class : Class 4
A is the right one

Ans 2:

Vedang Bhardwaj from Lotus Valley, Gurgaon
Class : Class 4

Ans 3:

Vedang Bhardwaj from Lotus Valley, Gurgaon
Class : Class 4
Class:

Ans 4:

Roshni from Mount litera Zee School Goa, Goa
Class : Class 4
b is the correct answer

Ans 5:

Vedang Bhardwaj from Lotus Valley, Gurgaon
Class : Class 4
I did A but then I understood that holidays end at 13 but he will join on 14

Ans 6:

Vania Aravindan from Unicent School , Miyapur Hyderabad
Class : Class 4

Ans 7:

Kaustubh M Sawale from Campion school, Bhopal
Class : Class 6
A is the right answer

Ans 8:

Kaustubh M Sawale from Campion school, Bhopal
Class : Class 6
B is the right answer as the date he will join is 14th Oct

Ans 9:

Vedang Bhardwaj from Lotus Valley, Gurgaon
Class : Class 4
no b

Ans 10:

DAIWIK SAHA from DUDES N DOLLS, FARIDABAD
Class : Class 4
a is the correct answer

Ans 11:

Sameer Anand from ST MICHAEL'S HIGH SCHOOL, PATNA
Class : Class 7
A is the right answer

Ans 12:

Saipriya from Sankara, Chennai
Class : Class 5
ans is be b ma because holidays till 13th but he will join on 14th only

Ans 13:

Anirudh Bhattacharjee from THE LEXICON INTERNATIONAL SCHOOL KALYANI NAGAR PUNR, Pune
Class : Class 5
ans will be b because holidays till 13th but he will join on 14th

Ans 14:

Ryan Majumdar from Epistemo , Hyderabad
Class : Class 4
A is the correct answer.

Ans 15:

Aaryaman Pradhan from De paul School , Berhampur
Class : Class 5
a is the correct answer

Ans 16:

Aviya Daniel from HIM ACADEMY PUBLIC SCHOOL, Hamirpur
Class : Class 4

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Mehrin from DPS, Riyadh
Subject :IMO    Class : Class 2

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Champ from Reliance Foundation School, Mumbai
Subject :IMO    Class : Class 7

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Champ from Reliance Foundation School, Mumbai
Subject :IMO    Class : Class 7

Ans 1:

Varnit khurana from Saint kabir school, Hissar
Class : Class 7
As the length of side of square is 8 cm ,so it's area will be 8×8cm which is equal to 64 sq. cm.Now ,to find the area of triangle,i.e. the unshaded area ,we have to apply the formula of (1\2×base×height).Now the base is equal to the length of the side of square I.e. 8cm and the height is 3cm.Now the area of triangle is (1\2×8×3)cm =12cm.Now we have to subtract the unshaded area from the area of whole square I.e. 64 sq.cm-12 sq.cm =52sq.cm.So the answer is option B.

Ans 2:

Varnit khurana from Saint kabir school, Hissar
Class : Class 7
As the length of side of square is 8 cm ,so it's area will be 8×8cm which is equal to 64 sq. cm.Now ,to find the area of triangle,i.e. the unshaded area ,we have to apply the formula of (1\2×base×height).Now the base is equal to the length of the side of square I.e. 8cm and the height is 3cm.Now the area of triangle is (1\2×8×3)cm =12cm.Now we have to subtract the unshaded area from the area of whole square I.e. 64 sq.cm-12 sq.cm =52sq.cm.So the answer is option B.

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adnan from Birla public school , Doha
Subject :IMO    Class : Class 3

Ans 1:

Vivaan Pathak from Prudence School , Ashok Vihar, Delhi
Class : Class 4
B Question

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tanmay srivastava from DPS Noida, Noida
Subject :IMO    Class : Class 4

Ans 1:

AREEN from BOLTON SCHOOL, Secunderabad
Class : Class 5
Answer is C because: LIV=54 XCIX=99 54<99 XLIV=44 XLVI=46 44<46 CCV=205 XCV=95 205>95

Ans 2:

Krishnam Singhal from GEMS MODERN ACADEMY, DUBAI
Class : Class 3
bad question

Ans 3:

Advika Narkar from Vibgyor High Airoli, Navi Mumbai
Class : Class 4
C C C THE ANSWER IS C

Ans 4:

AREEN from BOLTON SCHOOL, Secunderabad
Class : Class 5
Ans is C because LIV=54 & XCIX=99 XLIV=44 & XLVI=46 CCV=205 & XCV=95 Therefore, LIVXCV

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Tejas from GV, BANGALORE
Subject :IMO    Class : Class 9

Ans 1:

Nipun Abhilash from National Public School, Banashankhari, Bangalore
Class : Class 9
The answer is C . B is wrong.Ans:FTF

Ans 2:

deepa shet from National public school , Shimoga
Class : Class 9
Yes it is correct

Ans 3:

Mihir Rupani from Allen Career Institute, Rajkot
Class : Class 9
Yes the correct answer is option C because first is false and rest are.

Ans 4:

Arpit Gupta from Pkr Jain , Ambala
Class : Class 10
C option is correct......as the first is false and rest are true.

Ans 5:

Tejas from GV, BANGALORE
Class : Class 10
The answer is (C) --- F,T,T. Here's why:The first one is false anyway, as the '12' side is equal to (not less than, as must be) the sum of the other two '4' and '8' sides.The second one is TRUE (not false!) as I can easily construct a 90^ angle and 45^ angle with ruler and compass; and simply bisect that in-between angle to get a 67.5 angle.Even the third one is TRUE (not false) as construction of a triangle allows using a protractor for angles (that's how R.D.Sharma and my teacher do it) and ANY such triangle can be constructed. In short, my answer is correct. I marked the same, but the olympiad trainer says that it is (B) --- all are false.

Ans 6:

from d.g.v. sr. sec. school, rohtak
Class : Class 9
answer should be FTF

Ans 7:

Sairam from Swami Vivekanand High School, Mumbai
Class : Class 10
Actually, the answer should be FTF which is not given in the options. Considering statement 1, the sides of the triangle are 4cm, 8cm, and 12cm. But according to inequalities of triangles, sum of any to sides of a triangle is greater than the third side. But 4+8=12; i.e. sum of the two sides is equal to the third side. So statement 1 is FALSE. Considering statement 2, it is possible to construct an angle of 67.5 only using a ruler and compass. Taking statement 3, without a base, we can't construct a triangle.

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