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Naisha Bhatia from Cambridge court high school, Jaipur
Subject :IMO    Class : Class 4

Ans 1:

dveep jain from mindful gurukul, Mumbai
Class : Class 4
Answer is D). How? So let’s hear the question and answers fill in the blanks. • There were 40 children at a party. 14 of them were boys. __P__ fraction of children were girls. • Cost of a teddy bear is of the cost of a doll. If cost of the teddy bear is ₹ 114, then cost of the doll is ₹ __Q__ . • Difference between is __R__. • If then x = __S__. P Q R S A 7/20 76 5/8 1/4 B 13/20 76 7/8 1/4 C 7/20 171 5/8 3/4 D 13/20. 171 7/8 3/4 So the first question is. • There were 40 children at a party. 14 of them were boys. __P__ fraction of children were girls. So the answer is 26/40 or 13/20 because if there are 14 boys out of 40 children then 26 girls out of 40 children answer for the first question is 26/40 or 13/20. So option A) and C) cannot be the answer because they are not equivalent to 26/40 or 13/20. Now let’s hear the second question. • Cost of a teddy bear is of the cost of a doll. If cost of the teddy bear is ₹ 114, then cost of the doll is ₹ __Q__ . So the cost of the doll is ₹171 because cost of a teddy bear is 2/3 cost of a doll And the cost of the teddy bear is also equal to ₹114 so how get answer as ₹171 you have to divide ₹114 by 2 to get 1/3 of cost of the doll and then we ₹57, and then we multiply ₹57 by 3 to get the cost of the doll answer is ₹57 + ₹57 + ₹57 = ₹114 + ₹57 or ₹171. Now the option B) cannot be the answer because it is not equivalent to ₹171. Now you’ll are thinking that why I didn’t show the explanation of questions 3 and 4. I did not show you questions 3 and 4 because there are only 4 options and in answers none of the options none of these as their answers. Hence we have an answer which is D).

Ans 2:

Nideesha Sandeep Morajkar from Swami Vivekanand English primary School, Chembur
Class : Class 6

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Naisha Bhatia from Cambridge court high school, Jaipur
Subject :IMO    Class : Class 4

Ans 1:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
Hope you answer soon

Ans 2:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
According to BODMAS or DMAS, first / (divide),then * (multiply), then (add) and then - (subtract). So answer should be: S * R - P Q= 5 x 7 - 2 8= 35 - 2 8= 35 - 10= 25

Ans 3:

Ruthwin kumar from Yuvabharathi International School, Singapore
Class : Class 3

Ans 4:

Advika Bhatta from The Gaudium School, Hyderabad
Class : Class 4
According to BODMAS or DMAS, we first divide, then multiply, next add, and finally subtract. So following the rule we should get 25 because:P = 2Q = 8R = 7S = 55 ( S ) x 7 ( R ) - 2 ( P ) 8 ( Q ) 35 ( 5 x 7 ) - 10 ( 2 8 ) = 25Hence answer should be 25.

Ans 5:

Advika Narkar from The Shishukunj International School, Jhalaria Campus, Indore
Class : Class 8
Is there any rule for DMAS solving? Hope you answer both my questions. 1. According to......... and 2. Is there any ......................

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Anirudh from 989820Delhi Pubic school Rohini , Delhi
Subject :IMO    Class : Class 8

Ans 1:

Anshuman D from MVM, Hyderabad
Class : Class 9
Let the sum be x. Compound interest: P(1+r/100)^n-P =x(11/10)^2-x Taking x common, x[(121/100)-1] =21x/100 Simple interest:p*r*t/100 =x*10*2/100 =x/5 Compound interest - simple interest =500 21x/100-x/5=500 x/100=500 x=50000

Ans 2:

ARAVINDAN MANIKANDAN from NPS INternational, Singapore
Class : Class 8

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Trijal Kaushik from DPS Numaligarh, Golaghat
Subject :IMO    Class : Class 6

Ans 1:

Viggya Sutradhar from New look Senior Secondary school, Banswara
Class : Class 7
414720 /144 = 2880 correct answer is D

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Dheer Pandey from Sentia The Global School, Hyderabad
Subject :IMO    Class : Class 6

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Theertha sudeep from National academy for learning, Bengaluru
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(A): Let the  other number be y.
HCF × LCM = Product of numbers
11 × 7700 = 275 × y
y = 308


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Lakshya Singh from Dps kalyanpur , Kanpur
Subject :IMO    Class : Class 7

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Rishik De Sarkar from Delhi Public School Gurgaon, Gurgaon
Subject :IMO    Class : Class 3

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Riya V from NCFE, Bangalore
Subject :IMO    Class : Class 6

Ans 1: (Master Answer)

Admin from revisewise, Noida
Class : Class 1

(C): Two plots have same perimeter:
So, 60 × 3 = 2 × ( length + 30)
90 = length + 30
Length = 60 m 
Hence, area = 60 × 30 = 1800 sq. m 


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divish from Delhi Public School, gurgaon
Subject :IMO    Class : Class 5

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