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Shweta Basu from Amity international School, Gurgaon
Subject :IMO    Class : Class 7

Ans 1:

G. Priyanka from Chennai Public School, Chennai
Class : Class 8
15. 11+12+13+14+x/5 = 13 11+12+13+14+x = 65 50+x = 65 x= 65-50 x = 15 as simple as that

Ans 2:

Harshita Kumari from St. Joseph's Convent High School, Patna
Class : Class 7
Average=sum of the no. ÷ no. of numbers....11+12+13+14+x=13÷5,50+x÷5=13,50+x=13×5,50+x=65,x=65-50=15......answer = 15

Ans 3:

Tejas Shende from Kendriya Vidyalaya, IIT Campus, Powai, Mumbai, Mumbai
Class : Class 8
15

Ans 4:

ACHYUT SINGH from DELHI PUBLIC SCHOOL MANDLA ROAD , Jabalpur
Class : Class 8
(11+12+13+14+x)DIVIDED BY 5 = 13 (55+x)/5=13 55+x=13x5 55+x=65 x=65-55 x=15 therefore,the value of x is 15.

Ans 5:

Manan Khandelwal from Podar International School, CIE, Nerul, Navi Mumbai, Maharashtra,(400706) , Mumbai
Class : Class 7
Most probably 15

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Shweta Basu from Amity international School, Gurgaon
Subject :IMO    Class : Class 7
4:5

Ans 1:

Sanchay from BIS Dwarka, Dwarka
Class : Class 8
Let, 3rd no. = x, Let, 2nd no.= 20/100 (also N1), Let, 1st no. = 50/100 (also N2), Then, Ratio of 1st to 2nd no. = (50/100)/(20/100) =1/2(i.e. 50/100)*5(i.e. 1/{20/100}=100/20} OR Ratio of 2nd no. to 1st no. =Vice versa of (N1:N2)= 2:5

Ans 2:

Harshita Kumari from St. Joseph's Convent High School, Patna
Class : Class 7
Let the no. Be x 5(4+x)-20/8=10 20+5x-20/8=10 5x/8=10 5x=10×8 5x=80 X=80/5 X=16 According to me.

Ans 3:

Manan Khandelwal from Podar International School, CIE, Nerul, Navi Mumbai, Maharashtra,(400706) , Mumbai
Class : Class 7
Equation: ((5(4+n))- 20)/8= 10 = (5(4+n))- 20= 10x8 = 5(4+n)= (10x8)+ 20 =4+n= (10x8+20)/5 =n= ((10x8+20)/5)- 4 =n= (100/5)- 4 =n= 20- 4 n=16

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Shweta Basu from Amity international School, Gurgaon
Subject :IMO    Class : Class 7
30

Ans 1:

Manan Sharma from Naini Valley School, Kathgodam
Class : Class 7
30

Ans 2:

Sagar Shivprasad Rajoriya from St Joseph's English High School, Buldana
Class : Class 7
30 trees

Ans 3:

Harshita Kumari from St. Joseph's Convent High School, Patna
Class : Class 7
Let the required no. Be x 21x=630. ,. X=630÷21 ,X=30

Ans 4:

Manan Khandelwal from Podar International School, CIE, Nerul, Navi Mumbai, Maharashtra,(400706) , Mumbai
Class : Class 7
630/21= 30 There would be 30 trees in one row.

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MEENAKSHI SINGH from Raghav Global School, Noida, Noida
Subject :IMO    Class : Class 5

Ans 1:

Abhimanyu from MVN School Sector 17, Faridabad
Class : Class 7
100 crores

Ans 2:

Manish Narwat from Modern School, Faridabad
Class : Class 8
100 crores

Ans 3:

Krishang from Podar International School, Mumbai
Class : Class 6
100

Ans 4:

Soundharya from ISWK, Muscat
Class : Class 8
100 crores

Ans 5:

Ananya from Vibgyor High, Bengaluru
Class : Class 6
100 crores

Ans 6:

Atulya tripathy from Delhi Public School, sector 45, Gurgaon
Class : Class 10
100 CRORES

Ans 7:

Aaditya Biju from Indian Learners Own Academy, Kuwait
Class : Class 6
100 crores

Ans 8:

UVAIS ZUFAR from DON BOSCO SCHOOL OF EXCELLENCE, Chennai
Class : Class 8
100 crores

Ans 9:

praveen prabhu from private international english school, abudhabi
Class : Class 5
100 crores

Ans 10:

Ananya from Vibgyor High, Bengaluru
Class : Class 6
100 crores

Ans 11:

Krishang from Podar International School, Mumbai
Class : Class 6
100

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Shweta Basu from Amity international School, Gurgaon
Subject :IMO    Class : Class 6
4:5

Ans 1:

Aakash from Dpsrohini, New Delhi
Class : Class 6
16

Ans 2:

Aarya M Shah from Navrachana Vidyani Vidyalaya, Vadodara
Class : Class 4
4:5

Ans 3:

Aarya M Shah from Navrachana Vidyani Vidyalaya, Vadodara
Class : Class 4
Biggest : 7,430 Smallest: 3,047

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