User Forum

Subject :IMO    Class : Class 3

Class : Class 8
[B]

Class : Class 5

Class : Class 8
[B]

Class : Class 5

Ans 5:

Class : Class 8
[B]

Subject :IMO    Class : Class 3

Class : Class 3

Ans 2:

Class : Class 3
how easy it was

Subject :IMO    Class : Class 4

Class : Class 4

Class : Class 9
a

Class : Class 4

Ans 4:

Class : Class 7
a+b=70g 10a+10b+10c=3200 10(a+b+c)=3200 a+b+c=320 c=320-70=250

Class : Class 8

Class : Class 4
A

Ans 7:

Class : Class 6
If large square = A small square =B cylinder=C then, A+B=70gm 10A+10B=700gm 10C=3200gm-700gm =2500gm C=2500/10 =250

Ans 8:

Class : Class 8
250 GRAMS . FIRST DIVIDE 3200 BY 10 . ANS =320 THEN SUBTRACT 70 FROM 320. ANS =250 WEIGHT OF THE CYLINDER =250

Class : Class 5
A

Ans 10:

Class : Class 4
A

Subject :IMO    Class : Class 3

Class : Class 1

Subject :IMO    Class : Class 3

Ans 1:

Class : Class 5
Answer is C that is boundary of the rectangle

Class : Class 10

Ans 3:

Class : Class 5
how did we arrive this answer can you help

Class : Class 4

Class : Class 6
58 cm

Ans 6:

Class : Class 1
I thought it was C

Class : Class 3

Class : Class 3
58

Ans 9:

Class : Class 10
Longer side of the rectangle paper = 7 + 6 + 4 + 4 + 4 = 25 cm
Smaller side of the rectangle paper = 4 cm
Total length (perimeter) = 25 + 25 + 4 + 4 = 58 cm

Subject :IMO    Class : Class 8

Ans 1:

Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

Ans 2:

Class : Class 9
(a^2-b^2)/ab - (ab-b^2)/(ab- a^2) #Using special product = (a+b)(a-b)/ab - (ab-b^2)/(ab- a^2) #Removing out a and b common = (a+b)(a-b)/ab - b(a-b)/a(b-a) #LCM of both denominators i.e. ab and a(b-a) is ab(b-a) #Hence we multiply first equation by (b-a) and second equation by b = (a+b)(a-b)(b-a)/ab(b-a) - b^2(a-b)/ab(b-a) #Taking the denominator common = [(a+b)(a-b)(b-a) - b^2(a-b)]/ab(b-a) #Taking (b-a) common in the numerator =[(b-a)(a^2-b^2-b^2)]/ab(b-a) #Cancelling out (b-a) from numerator and denominator =a^2 - 2 b^2 / ab Hence B is the answer.

Ans 3:

Class : Class 8
answer is told to be D not B;

Subject :IMO    Class : Class 3

Ans 1:

Class : Class 10
Total area of uncut square is 144. 4 rectangles of (3 x 2) area 6 are cut. Which means 24 square unit area is cut. That leaves 144 - 24 = 120 square cm area. So answer is B

Ans 2:

Class : Class 5
Total area of the square=12 x 12 = 144 cm2 Area of one rectangle cut = 3 x 2 = 6 cm2 Area of 4 rectangles cut = 4 x 6 = 24 cm3 Area of the square after cutting = 144 - 24 = 120 cm2

Ans 3:

Class : Class 6
Its was a square 12X12=144sq cm. cut part area= 2x3=6 sq cm. total cut area = 6x4= 24 sq cm. Area of remaining sheet = 144-24 = 120. so, B is the answer.

Ans 4:

Class : Class 4
Answer: (B) Area of square = 12*12 = 144 sq cm Area of 1 rectangle cut = 3*2 = 6 sq cm Area of 4 rectangles cut = 4*6 = 24 sq cm Area of remaining paper = 144 - 24 = 120 sq cm

Ans 5:

Class : Class 7
the area of square paper sheet without cutting corners is Breadth x lenght = 12 x 12 = 144, now the area of the cut rectangle is breadth x length = 3 x 2 = 6, and there are four corners cutted. therefore 6 x 4 =24, now minus the cut corners area from total area. so 144-24 =120. therefore the answer is B

Ans 6:

Class : Class 10

Solution
Area of original square paper sheet = 12 x12 = 144cm2
Area of 1 rectangle cut = 3 x 2 = 6 cm2
Area of 4 such rectangles - 6 x 4 = 24 cm2
Area of paper left = 144 - 24 = 120 cm2

Ans 7:

Class : Class 10
Area of original square paper sheet = 12 x12 = 144cm2
Area of 1 rectangle cut = 3 x 2 = 6 cm2
Area of 4 such rectangles - 6 x 4 = 24 cm2
Area of paper left = 144 - 24 = 120 cm2

Class : Class 3

Ans 9:

Class : Class 4

Subject :IMO    Class : Class 3

Class : Class 5

Class : Class 6
Got it

Class : Class 5

Ans 4:

Class : Class 8
how will i solve this problem?

Ans 5:

Class : Class 6
how will we solve this?

Ans 6:

Class : Class 3
Let Janak points = J | Adil A = J + 2 | Rohit R = (9/5) * J | Adil + Janak + Rohit = 59 | A + J + R = 59 | Replace as above | (J + 2) + J + (9/5)*J = 59 | 2J + 2 + 9J/5 = 59 | To remove 5 in denominator, multiply entire equation by 5 | 10J + 10 + 9J = 295 | 19J = 295 - 10 | 19J = 285 | J = 15 | Hence Janak's points = 15 | => Adil = J + 2 = 17 points |

Ans 7:

Class : Class 3
how did you get the answer

Ans 8:

Class : Class 4
Shouldnt it be 15? The equation is (X 2) X [(9/5)X] = 59

Subject :IMO    Class : Class 3

Ans 1:

Class : Class 3

Subject :IMO    Class : Class 3

Ans 1:

Class : Class 4
The level of the question is certainly not for Class 3. It involves use of simultaneous equations. Answer is 35 soldiers.

Class : Class 6
B

Class : Class 3

Ans 4:

Class : Class 5
The way I taught my child was to try with each option, that is, start with option A, assume 40 is the number of soldiers, then calculate no of legs of soldiers, then minus that from total no. of legs, and see if the resulting number gives u the required number of horses you need in order to get a total of 45 horses and soldiers. Repeat the same for option B, C and D. Answer is B. This is how we did it: Assume 35 as no. of soldiers 35x2 = 70 legs of soldiersM Minus 70 from 110 (total legs)= 40 legs 40 horses legs = 40 divided by 4 =10 horses Therefore 35 soldiers and 10 horses = 45 soldiers and horses totally

Ans 5:

Class : Class 3
Hey you don't understand

Class : Class 3

Ans 7:

Class : Class 3
it is not a class 3 question understand that

Ans 8:

Class : Class 6
The answers in options needs to be considered while answering the question. The total number of horses and soldier in the troop is 45. Let us consider option A with 40 soldiers = 80 human legs so the balance legs are 30 which is not possible for horses as they have 4 legs. Going by option B= 35 soldiers = 70 human legs. So the balance legs remains 40, which is possible for 10 horses. Now the total number of horses and soldiers are 45 so 35 soldiers+10 horses = 45 and hence is a right answer. The balance options can be checked in similar way and will be found incorrect.