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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 6
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 4

Ans 3:

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

Ans 4:

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

Ans 5:

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 6:

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 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 8:

Class : Class 3
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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

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 6
Got it

Ans 2:

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

Ans 3:

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 4:

Class : Class 6
how will we solve this?

Ans 5:

Class : Class 5
how will i solve this problem?

Ans 6:

Class : Class 3
how did you get the answer

Ans 7:

Class : Class 5

Ans 8:

Class : Class 5

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Subject :IMO    Class : Class 3

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 3
Explain the answer properly

Ans 2:

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 3:

Class : Class 3

Ans 4:

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.

Ans 5:

Class : Class 3
Hey you don't understand

Ans 6:

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

Ans 7:

Class : Class 6
B

Ans 8:

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

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Subject :IMO    Class : Class 3

Ans 1:

Class : Class 9
B312

Ans 2:

Class : Class 3
Answer is D as per question

Ans 3:

Class : Class 4
correct answer is B

Ans 4:

Class : Class 2
solution is (24*5) (48*4)=312hence answer is B

Ans 5:

Class : Class 3
D

Ans 6:

Class : Class 1
in the question each is missing.. the balls are for each box

Ans 7:

Class : Class 7
D

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Subject :IMO    Class : Class 5

Ans 1: (Master Answer)

Class : Class 1
The correct answer is D.

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Subject :IMO    Class : Class 5

Ans 1: (Master Answer)

Class : Class 1
The correct answer is B.

Ans 2:

Class : Class 5
the correct answer is b

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Subject :IMO    Class : Class 5

Ans 1:

Class : Class 9
option D 31 since it is a prime number

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Subject :IMO    Class : Class 4

Ans 1:

Class : Class 7
C

Ans 2:

Class : Class 4
A- EMI 1970 per month 3 years = 36 months Total Amount Paid in three years 36 x 1970 = 70920 B - Rajat Paid Advance - 35000 Total Amount Paid A +B 70920+35000= 105920 Answer is C

Ans 3:

Class : Class 4

Ans 4:

Class : Class 4

Ans 5:

Class : Class 8
3 years = 36 months, and EMI paid each month= 1970/-, so the total amount paid in EMI is 36 x 1970, which is 70920. Therefore, total value of the car is 35000 + 70920 = 105920/-.

Ans 6:

Class : Class 4
3years=36 monthsEMIper month=1970EMI in three year=72920Rajat paid in advance=35000Total amount paid=70920 35000=105920......[C] is the answer............................

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Subject :IMO    Class : Class 6

Ans 1:

Class : Class 8
I THINK C. IS NOT THE NET OF THE SOLID

Ans 2:

Class : Class 7
C is the answer of this question

Ans 3:

Class : Class 7
good question

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