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Sushanth Buttagandla from Sentia, Hyderabad
Subject :IMO    Class : Class 5

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Akshath Ramanathan S from Sri Chaitanya Techno School, Bangalore
Class : Class 5
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Amogh Uppuluri from Epistemo Global High school , Hyderabad
Class : Class 5

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Dyanesh from Lakshmi School, Madurai
Class : Class 5
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Saipriya from Sankara, Chennai
Class : Class 7
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Devashrit Panda from DAV Public School, MIDNAPORE
Subject :IMO    Class : Class 4

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Devashrit Panda from DAV Public School, MIDNAPORE
Class : Class 8
d-36 x 1unit, how 36x2 in the answer

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Devendra Pandey from Modern Public School, Jhumri Telaiya
Subject :IMO    Class : Class 9

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Aadvik saini from Mvn sector 17, Faridabad
Subject :IMO    Class : Class 5

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Rebant Jain from Jayshree Periwal High School, Jaipur
Subject :IMO    Class : Class 3

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Aditya KJ from Global Indian International School, Balewadi, Pune
Class : Class 7
A

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Dhanya Shree K Y from Vssips, Bangalore
Class : Class 3
a

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Soham from Dav, Ahmedabad
Subject :IMO    Class : Class 5

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Vivaana Ahuja from jamnabai Narsee school, mumbai
Subject :IMO    Class : Class 3

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Divyansh Giri from MPVM Ganga Gurukulam, Prayagraj
Subject :IMO    Class : Class 6

Match the following.


A(i) → (r), (ii) → (s), (iii) → (q), (iv) → (p)
B(i) → (p), (ii) → (q), (iii) → (r), (iv) → (s)
C(i) → (s), (ii) → (p), (iii) → (q), (iv) → (r)
D(i) → (r), (ii) → (p), (iii) → (q), (iv) → (s)


Solution(i) - 26 39 - 49 - 60 = - 96. Additive inverse is 96.(ii) 2000 516 - 517 - 1999 = 0. Predecessor is - 1.(iii) 89 units to right of - 7 is - 7 89 = 82.(iv) - 89 x = - 85 => x = 4 = -(-4)

Ans 1:

Divyansh Giri from MPVM Ganga Gurukulam, Prayagraj
Class : Class 9
Solution(i) - 26 39 - 49 - 60 = - 96. Additive inverse is 96.(ii) 2000 516 - 517 - 1999 = 0. Predecessor is - 1.(iii) 89 units to right of - 7 is - 7 89 = 82.(iv) - 89 x = - 85 => x = 4 = -(-4)

Ans 2:

Divyansh Giri from MPVM Ganga Gurukulam, Prayagraj
Class : Class 9

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Ishan Tripathi from Podar International School , Pune
Subject :IMO    Class : Class 3

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Dishamrutha from indian school of excellence, bangalore
Subject :IMO    Class : Class 6

Arrange the following steps of construction of an angle 120° in a correct order.
Steps of construction:
(1) Draw a ray PQ.
(2) With B as a centre, again draw an arc to intersect the first arc at C.
(3) With P as a centre and a convenient radius, draw an arc to intersect at A.
(4) Join CP to get the required angle. ∠CPQ measures 120°.
(5) With A as a centre, draw an arc with the same radius to intersect the previous arc at B.

A(1)-(4)-(2)-(3)-(5)
B(1)-(3)-(5)-(2)-(4)
C(1)-(2)-(3)-(5)-(4)
D(1)-(3)-(2)-(5)-(4)


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