Subject :IMO Class : Class 3

Subject :IMO Class : Class 3

## Post Your Answer

Subject :IMO Class : Class 4

## Ans 4:

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 6

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

## Post Your Answer

Subject :IMO Class : Class 3

## Post Your Answer

Subject :IMO Class : Class 8

## 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 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.

## Post Your Answer

Subject :IMO Class : Class 3

## Ans 1:

Class : Class 3

aaaaaaabbbbbbbcccccccdddddddeeeeeeffffgggghhhhhiiiiiijjjjjjkkkkkkkllllllmmmmmnnnnnnoooooopppppqqqqrrrrssssttttuuuuvvvvwwwwwxxxxxxxxxyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

## 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 5

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 4:

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

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 5:

Class : Class 10

Answer: B

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

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 6:

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 7:

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 9:

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

## Post Your Answer

Subject :IMO Class : Class 3

## 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 |

## Post Your Answer

Subject :IMO Class : Class 3

## Post Your Answer

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.

## Ans 6:

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 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.