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

Ans 1:

Kaustubh M Sawale from Campion school, Bhopal
Class : Class 3
A is the right answer

Ans 2:

SAMEER ANAND from ST MICHAEL'S HIGH SCHOOL, PATNA
Class : Class 4
A is the right answer

Ans 3:

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

Ans 4:

Aaryaman Pradhan from De paul School , Berhampur
Class : Class 4
a is the correct answer

Ans 5:

Kaustubh M Sawale from Campion school, Bhopal
Class : Class 3
B is the right answer as the date he will join is 14th Oct

Ans 6:

Anirudh Bhattacharjee from THE LEXICON INTERNATIONAL SCHOOL KALYANI NAGAR PUNR, Pune
Class : Class 4
ans will be b because holidays till 13th but he will join on 14th

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

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

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periwinklemohite 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

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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:

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 Geethanjali Vidyalaya, BANGALORE
Subject :IMO    Class : Class 9

Ans 1:

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

Ans 2:

Arpit Gupta from Pkr Jain , Ambala
Class : Class 9
C option is correct......as the first is false and rest are true.

Ans 3:

Tejas from Geethanjali Vidyalaya, 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 4:

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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Elena from NPS, Bnagalore
Subject :IMO    Class : Class 5

Ans 1:

Pranav M S from JSS Public school, Mysore
Class : Class 5
It is too easy 80-60+15-2=33

Ans 2:

Shruti Rajshree from NPS Indiranagar, Bengaluru
Class : Class 5
Yeah A is the answer

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Kaustubh Jha from Mount litera zee school, Patna
Subject :IMO    Class : Class 5

Ans 1:

Pranav M S from JSS Public school, Mysore
Class : Class 5
1836

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Aryaveer Agrawal from DELHI PUBLIC SCHOOL, Ahmedabad
Subject :IMO    Class : Class 8

Ans 1:

Abhigyan from ST MICHAEL SCHOOL, Durgapur
Class : Class 8
B. Is the correct answer

Ans 2:

Khyathi Komalan from Vaels international school, Chennai
Class : Class 9
The answer is (B) Using Unitary Method- Angle for 15,000 rupees= 360 degrees Angle for x rupees=60 degrees If money spent increases, degree of angle used increases, Therefore it is direct variation- x/1500=60/360 x=1/6 x 15,000 x= 2500 rupees

Ans 3:

sagar kumar from D A V public school , NOIDA
Class : Class 8
B

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