Subject :IMO Class : Class 4

Subject :IMO Class : Class 3

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Subject :IMO Class : Class 3

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Subject :IMO Class : Class 4

## Ans 7:

Class : Class 6

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

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Subject :IMO Class : Class 3

## Ans 2:

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

Smaller side of the rectangle paper = 4 cm

Total length (perimeter) = 25 + 25 + 4 + 4 = 58 cm

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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.

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Subject :IMO Class : Class 3

## Ans 1:

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 2:

Class : Class 3

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## Ans 3:

Class : Class 5

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 7:

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 8:

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

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Subject :IMO Class : Class 3

## Ans 8:

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 |

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Subject :IMO Class : Class 3