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sai ramakrishna GUDIMETLA  from AMM MATRICULATION HIGHER SECONDARY SCHOOL, Chennai
Subject :IMO    Class : Class 4

Ans 1:

RAJESH DEWANGAN from ,
Class : Class 4
c

Ans 2:

Shivam Dave from Jamnabai Narsee School, Gandhinagar
Class : Class 7
The correct answer is 'B'. As you have been asked - 84 thousands 60 hundreds 50 tens = 90,150 16 thousands 82 hundreds 28 tens 19 ones = 24,499 Now over here we have not been given an equation symbol. So, we assume that 90,150 - 24,999 = 65,651 Therefore the answer is 'B: 65,651'

Ans 3:

Shivam Dave from Jamnabai Narsee School, Gandhinagar
Class : Class 7
Check, has been corrected. The equation symbol has been placed -

Ans 4:

Admin from SOF, Noida
Class : Class 10
There was a minus sign in between which was missing. It has been corrected.
First Number = 84000+6000+150 = 90150
Second Number = 16000+ 8200 + 280 + 19 = 24499
Difference = 90150 - 24499 = 65651

Ans 5:

AARAV GULATI from Vivek Summit, New Delhi
Class : Class 4
b

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arnav makani from The Gurukul sector-20, Panchkula
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
The correct answer is B.

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arnav makani from The Gurukul sector-20, Panchkula
Subject :IMO    Class : Class 6

Ans 1:

Sanket Kulkarni from Euroschool, Pune
Class : Class 8

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arnav makani from The Gurukul sector-20, Panchkula
Subject :IMO    Class : Class 6

Ans 1:

Agney Gurukul from ,
Class : Class 6
A

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arnav makani from The Gurukul sector-20, Panchkula
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
The correct answer is A.

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Talaha Modak from Vishwajyot High School, Panvel
Subject :IMO    Class : Class 7

Ans 1:

Tamajit Das from Bhavan's Tripura Vidya Mandir, Agartala
Class : Class 7
C . Can also be the answer.It is also not the net of cone

Ans 2:

kunal from Av vidya mandir school , cheeka
Class : Class 7
D is answer because it is not a net of the given figure

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Prabhav Agarwal from Lawrence High School, Bengaluru
Subject :IMO    Class : Class 6

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sanjay shete from ,
Subject :IMO    Class : Class 8

Ans 1:

tanishka Bondhare from Saraswati international school , Hingoli
Class : Class 8
Thanks

Ans 2:

Samedh from Tree House High School, Pune
Class : Class 9
Quantity of mixture = 400 ml. Quantity of pure alcohol in it = 15 % of 400 = 15/100 * 400 = 60 ml Let's consider 'x ml' of pure alcohol was added to the mixture. After adding, Quantity of mixture = (400 + x) ml Quantity of pure alcohol in it = (60 + x) ml Strength = (60 + x)/(400 + x) = 32/100 => 100(60 + x) = 32(400 + x) => 6000 + 100x = 12800 + 32x => 100x - 32x = 12800 - 6000 => 68x = 6800 => x = 6800/68 => x = 100 Hence answer is 100 ml

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Prabhav Agarwal from Lawrence High School, Bengaluru
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
The correct answer is B.

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Aditya Muthukumar from Alpha Cambridge International School, Trichirappalli
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1
The correct answer is D.

Ans 2:

Kashish Kathuria from Delhi Public School, Bangalore
Class : Class 9
THE CORRECT ANSWER IS D

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