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Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6 PDF Download

Note: The questions provided in this document are similar to the questions that were asked in the actual Olympiad exam. So, we recommend you study these for your Olympiad preparation

Logical Reasoning

Q1: Select the one that does not belong with the others.
(a) 81
(b) 36
(c) 47
(d) 16 

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Ans: (c)

  • The numbers 81, 36, and 16 are all perfect squares (9x9, 6x6, and 4x4 respectively).
  • However, 47 is a prime number, meaning it cannot be divided evenly by any number other than 1 and itself.
  • This makes 47 the odd one out in this group.
  • Thus, the correct answer is (c) 47.

Q2: In a line of 60 students, Amit holds the 30th spot from the left side. What is his rank from the right side?
(a) 32nd
(b) 31st
(c) 29th
(d) 33rd

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Ans: (b)

  • To find Amit's position from the right, we can use the total number of students and his position from the left.
  • There are 60 students in total, and Amit is 30th from the left.
  • His position from the right can be calculated as: 60 - 30 + 1 = 31.
  • Thus, Amit is in the 31st position from the right end.

Q3: In a specific coding system, if GREAT is represented as JTFCW, how will SIGHT be encoded in the same system?
(a) VLJKW
(b) TJHIU
(c) VKHJW
(d) RHHGS

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Ans: (c)

  • The code shifts each letter in the word GREAT to the next letter in the alphabet, with a specific pattern.
  • For example, G becomes J, R becomes T, and so on.
  • Applying the same pattern to SIGHT, S shifts to V, I to K, G to H, H to J, and T to W.
  • Thus, SIGHT is encoded as VKHJW.

Q4: If ‘L’ represents ‘×’, ‘M’ represents ‘÷’, ‘N’ represents ‘–’ and ‘Q’ represents ‘+’, what is the result of 85 M 17 L 4 Q 18 N 5?
(a) 25
(b) 28
(c) 30
(d) 33

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Ans: (d)

  • First, replace the symbols with their respective operations: 85 ÷ 17 × 4 + 18 – 5.
  • Next, perform the operations in the correct order (BODMAS): start with division and multiplication from left to right.
  • Calculate 85 ÷ 17, which equals 5.
  • Then, multiply 5 by 4 to get 20.
  • Now, add 20 and 18 to get 38.
  • Finally, subtract 5 from 38, resulting in 33.

Q5: How many pairs of letters exist in the word CREATOR that have the same number of letters between them as they do in the English alphabet?
(a) Two
(b) Three
(c) Four
(d) None

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Ans: (a)

  • To find the pairs, we need to look at the letters in the word CREATOR and see if the distance between them matches their positions in the English alphabet.
  • For example, the letters C and E are two letters apart in the alphabet, and they are also two letters apart in the word.
  • After checking all possible pairs, we find that there are two pairs that meet this criterion.
  • Thus, the correct answer is two pairs of letters.

Q6: Pointing to a girl in the photograph, Arun said, “She is the granddaughter of the elder brother of my father.” How is that girl related to Arun?
(a) Sister
(b) Cousin
(c) Niece
(d) Mother

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Ans: (c)

  • The girl is the granddaughter of Arun's father's elder brother, which means she is the daughter of Arun's uncle.
  • This makes her Arun's cousin, but since the question states she is a granddaughter, it indicates she is the daughter of Arun's sibling.
  • Thus, the girl is Arun's niece, as she is the daughter of Arun's brother or sister.
  • Therefore, the correct relationship is that the girl is Arun's niece.

Q7: Count the number of triangles in the given figure.
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6(a) 12 
(b) 16 
(c) 22 
(d) 24

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Ans: (c)
The triangles formed areMaths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6 A, B, C, D, E, F, G, H, I, J, K, L, CD, EJ, FG, IH, GH, FI, DE, CJ, JEFI, DEFG
∴ Total number of triangles formed = 22

Q8: Count the number of cubes in the given figure.
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6

(a) 56 
(b) 55 
(c) 54 
(d) 52

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Ans: (a)
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6Number of cubes in the Layer 1 = 9
Number of cubes in the Layer 2 = 15
Number of cubes in the Layer 3 = 16
Number of cubes in the Layer 4 = 16
∴ Total number of cubes in the given figure = 9 + 15 + 16 + 16 = 56

Q9: Two positions of a dice are shown. If two dots are on the bottom, then how many dots will be on the top?
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6(a) 1 
(b) 3 
(c) 6 
(d) 5

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Ans: (b)
Number of dots on the opposite faces are (1, 4); (2, 3); (5, 6).

Q10: Group the given figures into three classes using each figure only once.
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6(a) 1,5, 6; 2, 3, 4; 7, 8, 9 
(b) 1,4, 7; 2, 5, 8; 3, 6, 9 
(c) 1, 2, 3; 4, 5, 8; 6, 7, 9 
(d) 1, 3, 5; 2, 4, 8; 6, 7, 9

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Ans: (a)

1, 5, 6 - Two similar shapes one contained in other. 
2, 3, 4 - One shape with two intersecting lines in interior. 
7, 8, 9 - Two different shapes one circum-scribing the other.

Q11: Rohit walked 25 metres towards South. Then he turned to his left and walked 20 metres. He then turned to his left and walked 25 metres. He again turned to his right and walked 15 metres. At what distance is he from the starting point and in which direction? 
(a) 35 metres East 
(b) 35 metres North 
(c) 40 metres East 
(d) 60 metres East

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Ans: (a)
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6Distance between starting point and final point =(20+15)m = 35 m 
So, Rohit is 35 metres far away from starting point and in East direction.

Q12: Which of the following elements satisfies the Venn diagram given here?
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6(a) Judge, Thief, Criminal 
(b) Father, Mother, Parents 
(c) Teacher, College, Student 
(d) Polygons, Rectangles, Squares

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Ans: (a)
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6

Q13: Pia walks a distance of 4 metres towards South. Then she turns to the left and walks 3 metres. After this she turns to the right and walks 4 metres. In which direction is she facing now? 
(a) North-East 
(b) South 
(c) North 
(d) South-West

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Ans: (b)
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6So, Pia is facing South.

Q14: In a class of 35 students, Kunal is placed seventh from the bottom whereas Sonali is placed ninth from the top. Pulkit is placed exactly in between the two. What is Kunal's position from Pulkit? 
(a) 9 
(b) 10 
(c) 11 
(d) 13

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Ans: (b)
Number of students between Kunal and Sonali = 35−(7+9)=19
Clearly, there are 9 students between Kunal and Pulkit, as well as Pulkit and Sonali. 
So, Kunal is at 10th position from Pulkit.

Q15: In a certain code, TELEPHONE is written as ENOHPELET. How is ALI GATOR written in that code? 
(a) ROTAGILA 
(b) ROTAGAIL 
(c) ROTAGILE 
(d) ROTEGILA

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Ans: (a)
The letters of the word are written in a reverse order to obtain the code. So, code for ALIGATOR is ROTAGILA.

Mathematical Reasoning

Q16: According to the 2011 census, the population of India is 1,210,193,422. Which of the following options represents this number in the International System of Numeration?
(a) One hundred twenty one crore one lakh ninety three thousand four hundred twenty two
(b) One billion two hundred ten million one hundred ninety three thousand four hundred twenty two
(c) Twelve billion ten million one hundred ninety three thousand four hundred twenty two
(d) One hundred twenty one crore nineteen lakh three thousand four hundred twenty two

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Ans: (b)

  • Understanding the number: The population figure 1,210,193,422 can be broken down into parts to understand its value in the International System of Numeration.
  • International System: In this system, the number is expressed in terms of billions and millions. Here, 1,210 million is equivalent to 1 billion and 210 million.
  • Correct representation: Therefore, the correct option that represents this number is "One billion two hundred ten million one hundred ninety three thousand four hundred twenty two."
  • Other options: The other options either use the Indian numbering system or misrepresent the value, making them incorrect.

Q17: What number must be taken away from 67.74 to arrive at 24.48?
(a) 33.46
(b) 38.46
(c) 33.26
(d) 43.26 

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Ans: (d)

  • To find the number to subtract, we can set up the equation: 67.74 - x = 24.48.
  • Rearranging gives us x = 67.74 - 24.48.
  • Calculating this, we find x = 43.26.
  • Thus, 43.26 is the value that needs to be subtracted from 67.74 to get 24.48.

Q18: What is the total of the largest and the smallest 5-digit number that can be created using the digits 0, 7, 5, 2, and 1 (each digit used only once)?
(a) 58,467
(b) 85,467
(c) 76,467
(d) 65,467

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Ans: (b)

  • To find the largest 5-digit number, arrange the digits in descending order: 7, 5, 2, 1, 0, which gives 75210.
  • For the smallest 5-digit number, arrange the digits in ascending order: 0 cannot be the first digit, so we start with 1, followed by 0, 2, 5, and 7, resulting in 10257.
  • Now, add the two numbers: 75210 + 10257 = 85467.
  • Thus, the sum of the greatest and smallest 5-digit numbers is 85,467.

Q19: Which of the following indicates the greatest increase in temperature?
(a) 0°C to 10°C
(b) – 4°C to 8°C
(c) –15°C to –8°C
(d) –7°C to 1°C 

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Ans: (b)

  • The temperature change is calculated by subtracting the initial temperature from the final temperature.
  • For option (b), the change is from –4°C to 8°C, which is a rise of 12 degrees.
  • In comparison, the other options show smaller increases: (a) 10 degrees, (c) 7 degrees, and (d) 8 degrees.
  • Thus, the option with the maximum rise in temperature is indeed (b) – 4°C to 8°C.

Q20: Which of the following pairs of numbers are co-prime?
(a) 39, 91
(b) 161, 192
(c) 385, 462
(d) 189, 243

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Ans: (b)

  • Co-prime numbers are those that have no common factors other than 1.
  • To check if 161 and 192 are co-prime, we find their greatest common divisor (GCD).
  • The GCD of 161 and 192 is 1, meaning they share no common factors.
  • Thus, 161 and 192 are co-prime, while the other pairs have common factors.

Q21: What is the minimum value that can be assigned to * so that the number 653*47 is divisible by 11?
(a) 9
(b) 6
(c) 7
(d) 1

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Ans: (d)

  • To determine the value of *, we need to check the divisibility rule for 11, which states that the difference between the sum of the digits in odd positions and the sum of the digits in even positions should be either 0 or a multiple of 11.
  • For the number 653*47, the digits in odd positions are 6, 3, 4 (1st, 3rd, and 5th), and the digits in even positions are 5, *, 7 (2nd, 4th, and 6th).
  • Calculating the sums: Odd positions = 6 + 3 + 4 = 13; Even positions = 5 + * + 7 = 12 + *.
  • The difference is |13 - (12 + *)| = |1 - *|. For this to be divisible by 11, * must be 1.

Q22: Determine the value of 2a + 5b, given that a : 3 = 72 : 27 = 40 : b.
(a) 20
(b) 75
(c) 91
(d) None of these

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Ans: (c)

  • First, we can find the value of a and b using the given ratios. From the ratio a : 3 = 72 : 27, we can set up the equation a/3 = 72/27.
  • Cross-multiplying gives us a = (3 * 72) / 27, which simplifies to a = 8.
  • Next, using the ratio 40 : b, we have 40/b = 72/27. Cross-multiplying gives us b = (40 * 27) / 72, which simplifies to b = 15.
  • Now, substituting a = 8 and b = 15 into the expression 2a + 5b gives us 2(8) + 5(15) = 16 + 75 = 91.

Q23: Which of the following represents the most reduced form of 161/299?
(a) 7/13
(b) 3/11
(c) 11/17
(d) 17/23

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Ans: (a)

  • The fraction 161/299 needs to be simplified to its lowest terms.
  • To do this, we find the greatest common divisor (GCD) of the numerator (161) and the denominator (299).
  • The GCD of 161 and 299 is 23, so we divide both by 23.
  • This gives us 7/13, which is the simplest form of the fraction.

Q24: Which of the following combinations of whole numbers will satisfy the equation?
________ × ________ + ________ = 49
(a) 7, 8, 7
(b) 5, 8, 9
(c) 7, 8, 10
(d) 9, 8, 5

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Ans: (b)

  • To solve the equation, we need to find numbers that fit into the format: first number × second number + third number = 49.
  • Testing option (b) with 5 × 8 + 9 gives us 40 + 9 = 49, which is correct.
  • Other options do not satisfy the equation when calculated.
  • Thus, the correct combination is 5, 8, 9.

Q25: Calculate the result of DXXXIX + CCCLXVII – DCLXXXI.
(a) CCCXXV
(b) DCLVI
(c) DCCXII
(d) CCXXV 

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Ans: (d)

  • Roman Numerals to Arabic Numbers: 
    DXXXIX = 539

    CCCLXVII = 367

    DCLXXXI = 681

    Perform Calculation:

    Addition: 539 + 367 = 906

    Subtraction: 906 − 681 = 225

    Convert 225 to Roman Numerals:

    200=CC, 20=XX, 5=V → CCXXV

    Correct Answer: (d) CCXXV.

Q26: If x and y are both odd numbers, which of the following results in an even number?
(a) x + y
(b) x + y + 2
(c) xy
(d) xy + 1

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Ans: (b)

  • When you add two odd numbers (x and y), the result is always even. This is because odd + odd = even.
  • However, if you add 1 to that sum (x + y + 1), it becomes odd.
  • The product of two odd numbers (xy) is also odd.
  • Adding 2 to the product (xy + 2) results in an odd number as well.

Q27: Which of the following sets of numbers are not in proportion?
(a) 28, 42, 30, 45
(b) 15, 21, 25, 35
(c) 21, 35, 32, 48
(d) 24, 40, 12, 20

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Ans: (c)

  • The concept of proportion means that the ratios of the numbers should be equal.
  • In option (c), the ratios do not maintain equality, making them not proportional.
  • For example, comparing the ratios of the first two numbers (21 and 35) with the last two (32 and 48) shows a discrepancy.
  • Thus, option (c) is the only set that fails to meet the proportionality condition.

Q28:  _______ is used to draw and measure an angle.  
(a) Protractor 
(b) Ruler  
(c) Compasses 
(d) None of these 

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Ans: (a)

Q29: Which of the following angles cannot be constructed using ruler and compasses?  
(a) 75° 
(b) 15°  
(c) 135° 
(d) 85° 

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Ans: (d)

Q30: If we bisect AB = 18 cm, then what will be the  length of each part?  
(a) 9 cm 
(b) 13.2 cm  
(c) 12 cm 
(d) None of these 

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Ans: (a) If we bisect AB, that means we divide it  into two equal parts.  
∴ Length of each part = 18/2 = cm = 9 cm

Q31: PQ of length 7.5 cm. From P, an arc of 5.9 cm is  cut off. Find the length of RQ. 
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6(a) 1.4 cm 
(b) 1.6 cm  
(c) 2.5 cm 
(d) 1.8 cm

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Ans: (b) 
RQ = PQ – PR = (7.5 – 5.9) cm = 1.6 cm

Q32: A geometry box has  
(a) ruler 
(b) compasses  
(c) divider 
(d) All of these 

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Ans: (d)

Q33: A line segment PQ is 10.2 cm long. If A is the  mid point of PQ, then the length of AP is  
(a) 4.8 cm 
(b) 8.2 cm  
(c) 5.1 cm 
(d) 5.2 cm

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Ans: (c) 
Since A is the mid point of PQ.
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6

Q34: Identify the pair that is not in proportion.  
(a) 2 : 3 & 4 : 9 
(b) 3 : 2 & 6 : 4  
(c) 5 : 7 & 10 : 14 
(d) 6 : 7 & 18 : 21

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Ans: (a) 
Product of means = 3 × 4 = 12  
Product of extremes = 2 × 9 = 18  
Since, product of means ≠ product of  extremes  
Hence, ratios 2 : 3 and 4 : 9 are not in  proportion. 

Q35: Express 500 in Roman numeral.  
(a) CX 
(b) DD  
(c) D 
(d) XXX  

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Ans: (c)

Everyday Mathematics

Q36: Ansh multiplied 160 by 89 instead of multiplying by 79. How much was his answer greater than the correct answer?
(a) 1600
(b) 16000
(c) 10060
(d) 10640

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Ans: (a)

  • First, calculate the correct answer by multiplying 160 by 79, which equals 12640.
  • Next, calculate the incorrect answer by multiplying 160 by 89, which equals 14240.
  • Now, find the difference between the incorrect answer and the correct answer: 14240 - 12640 = 1600.
  • Thus, Ansh's answer was greater by 1600 than the correct answer.

Q37: Rehan purchased a notebook for ₹ 14//1/3, a geometry box for ₹ 57//1/2, and a Mathematics book for ₹ 71//2/3. What total amount does he need to pay the shopkeeper?
(a) ₹ 123//1/2
(b) ₹ 143//1/2
(c) ₹ 123//1/6
(d) ₹ 143//1/6

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Ans: (b)

  • To find the total cost, we need to add the prices of all items: ₹ 14//1/3 + ₹ 57//1/2 + ₹ 71//2/3.
  • First, convert each mixed fraction to an improper fraction for easier addition.
  • After performing the addition, we find that the total amount Rehan should pay is ₹ 143//1/2.
  • This means he needs to give the shopkeeper ₹ 143 and a half rupee.

Q38: Arun intends to contribute a total of ₹4200 to two individuals, X and Y, in the ratio of 5:7. What will be the respective amounts received by X and Y?
(a) ₹ 2450, ₹ 1750
(b) ₹ 1750, ₹ 2450
(c) ₹ 1950, ₹ 2300
(d) ₹ 2300, ₹ 1950

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Ans: (b)

  • To find the amounts for X and Y, first, we need to determine the total parts in the ratio 5:7, which is 5 + 7 = 12 parts.
  • Next, we calculate the value of one part by dividing the total amount ₹4200 by 12, resulting in ₹350 per part.
  • Now, for X, who gets 5 parts: 5 * ₹350 = ₹1750.
  • For Y, who receives 7 parts: 7 * ₹350 = ₹2450.
  • Thus, X will receive ₹1750 and Y will receive ₹2450, confirming the ratio of 5:7.

Q39: Amisha runs 7 km to the South, then returns in the opposite direction and runs 9 km. If South is considered a negative integer and North a positive integer, what is her position relative to the starting point?
(a) 16 km
(b) –2 km
(c) –16 km
(d) 2 km

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Ans: (d)

  • Amisha starts by running 7 km South, which is represented as -7.
  • Then, she runs 9 km North, which is represented as +9.
  • To find her final position, we calculate: -7 + 9 = 2.
  • Thus, her position from the initial point is 2 km North.

Q40: A rectangular mat is placed on a square floor that measures 26 m on each side. The length of the mat is 18 m, and its breadth is two-thirds of its length. Calculate the area of the floor that is not covered by the mat.
(a) 460 m2
(b) 270 m2
(c) 410 m2
(d) None of these

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Ans: (a)

  • The area of the square floor is calculated as side × side, which is 26 m × 26 m = 676 m2.
  • The length of the mat is 18 m, and its breadth is two-thirds of the length, which is 18 m × (2/3) = 12 m.
  • The area of the rectangular mat is length × breadth, which is 18 m × 12 m = 216 m2.
  • To find the area of the floor that is not covered by the mat, subtract the area of the mat from the area of the floor: 676 m2 - 216 m2 = 460 m2.

Q41: Sushant went to bed at 10 p.m. and got up at 7 a.m. How many right angles did the hour hand of the clock form during this time?
(a) 2
(b) 3
(c) 1
(d) 4

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Ans: (b)

  • The hour hand of a clock completes a full circle (360 degrees) in 12 hours.
  • In this time frame from 10 p.m. to 7 a.m., the hour hand moves from the 10 to the 7, which is 9 hours.
  • Every hour, the hour hand makes 2 right angles (90 degrees) at the 3 and 9 positions.
  • During these 9 hours, the hour hand makes a total of 3 right angles, specifically at 12 a.m., 3 a.m., and 6 a.m.

Q42: A container X has 24.25 litres of water and another container Y has 17.75 litres of water. Both containers X and Y together are enough to fill one-fifth of a tank Z. What is the total capacity of the tank Z?
(a) 110 L
(b) 160 L
(c) 210 L
(d) 260 L 

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Ans: (c)

  • First, calculate the total amount of water in containers X and Y: 24.25 L + 17.75 L = 42 L.
  • This total (42 L) represents one-fifth of the capacity of tank Z.
  • To find the total capacity of tank Z, multiply 42 L by 5 (since 42 L is one-fifth): 42 L * 5 = 210 L.
  • Thus, the total capacity of tank Z is 210 L.

Q43: In a university, students can select from three streams: Science, Commerce, and Arts. During a specific session, a total of 1,54,676 students were enrolled across all streams. If 1,26,136 students were enrolled in both Science and Commerce, how many students were enrolled in the Arts stream during that session?
(a) 28,540
(b) 45,017
(c) 65,017
(d) 35,017

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Ans: (a)

  • To find the number of students in the Arts stream, we need to subtract the total number of students in Science and Commerce from the overall total.
  • The total number of students is 1,54,676 and those in Science and Commerce together is 1,26,136.
  • So, the calculation is: 1,54,676 - 1,26,136 = 28,540.
  • This means there are 28,540 students enrolled in the Arts stream.

Q44: Three boys start walking from the same location. Their step lengths are 63 cm, 70 cm, and 77 cm respectively. What is the least distance each boy should walk so that they all complete their steps evenly?
(a) 6930 cm
(b) 6000 cm
(c) 7000 cm
(d) 6520 cm

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (a)

  • To find the minimum distance that all boys can walk in complete steps, we need to calculate the Least Common Multiple (LCM) of their step lengths: 63, 70, and 77.
  • The LCM is the smallest number that is a multiple of all three step lengths.
  • Calculating the LCM gives us 6930 cm, which means all boys can walk this distance in whole steps.
  • Thus, the answer is 6930 cm, as it allows each boy to complete their steps without any fractions.

Q45: To create a particular dress, a fabric measuring 4 m 25 cm is required. Calculate the total fabric length needed to produce 3 of these dresses.
(a) 12.75 m
(b) 13.75 m
(c) 16.25 m
(d) 19.25 m

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (a)

  • First, convert the length of the cloth from meters and centimeters to just meters: 4 m 25 cm = 4.25 m.
  • Next, multiply this length by the number of dresses: 4.25 m × 3 = 12.75 m.
  • This gives the total length of cloth required to make 3 dresses.
  • Thus, the answer is 12.75 m.

Achievers Section

Q46: Which of the following statements is accurate?
(a) 42 and 49 are co-prime numbers.
(b) 376277 is divisible by 3.
(c) LCM of 102, 170 and 136 is 2040.
(d) None of these.

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (c)

  • The LCM (Least Common Multiple) of numbers is the smallest number that is a multiple of each of the numbers. In this case, the LCM of 102, 170, and 136 is indeed 2040.
  • To find the LCM, we can use the prime factorization method or the method of listing multiples, and in this case, it confirms that 2040 is the correct answer.
  • Option (a) is incorrect because 42 and 49 share a common factor (7), so they are not co-prime.
  • Option (b) is incorrect as 376277 is not divisible by 3 (the sum of its digits is not divisible by 3).
  • Option (d) is also incorrect since option (c) is true.

Q47: What does the following number line represent?  
Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6
(a) Addition of 5 and 6  
(b) Subtraction of 5 from 11  
(c) Multiplication of 11 and 5  
(d) All of these 
 

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (a)

Q48: 1935 × 782 × 15 =  _______.  
(a) 1935 × 43 × 252  
(b) 15 × 982 × 768  
(c) 46 × 1935 × 255  
(d) 148 × 596 × 482  

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (c) 
We have, 1935 × 782 × 15  
= 1935 × 46 × (17 × 15)  
= 1935 × 46 × 255

Q49: A number is completely divisible by 7 but when  it is divided by 2, 3, 4, 5 and 6 respectively, it  leaves a remainder of 1. The smallest such  possible number is _______.  
(a) 301 
(b) 302  
(c) 308 
(d) 315 

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (a) 
LCM of 2, 3, 4, 5 and 6 = 60  
Since we are looking for a number that is  divisible by 2, 3, 4, 5 and 6 and that number  + 1 being divisible by 7. 
On finding the  multiple of 60 such that the sum of multiple  and 1 is divisible by 7.  
We get the no. = 300 + 1 = 301

Q50: Find the product of 475 × 102?  
(a) 10900 
(b) 48450  
(c) 12500
(d) 13256  

Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6  View Answer

Ans: (b) 
475 × 102 is written as  
475 × (100 + 2) = 475 × 100 + 475 × 2  [∴ a × (b + c) = a × b + a × c]  
47500 + 950 = 48450

The document Maths Olympiad Previous Year Paper -1 | Maths Olympiad Class 6 is a part of the Class 6 Course Maths Olympiad Class 6.
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FAQs on Maths Olympiad Previous Year Paper -1 - Maths Olympiad Class 6

1. What is the format of the Maths Olympiad for Class 6 students?
Ans.The Maths Olympiad for Class 6 typically consists of multiple-choice questions that assess students' logical reasoning, mathematical reasoning, and everyday mathematics skills. The exam usually has a time limit of 60 minutes and includes a variety of topics from the Class 6 syllabus.
2. How can I prepare effectively for the Class 6 Maths Olympiad?
Ans.To prepare effectively for the Class 6 Maths Olympiad, students should practice previous year question papers, focus on understanding concepts rather than rote memorization, and solve a variety of problems from each topic. Joining a study group or enrolling in coaching classes can also be beneficial.
3. What topics are usually covered in the Class 6 Maths Olympiad?
Ans.The Class 6 Maths Olympiad covers a range of topics including arithmetic, geometry, fractions, decimals, percentages, ratios, and basic algebra. Logical reasoning and problem-solving skills are also essential components of the exam.
4. How do I register for the Class 6 Maths Olympiad?
Ans.Students can register for the Class 6 Maths Olympiad through their school, as many schools participate in the Olympiad program. Alternatively, students can visit the official website of the organizing body to find registration details and deadlines.
5. What is the scoring system for the Class 6 Maths Olympiad?
Ans.The scoring system for the Class 6 Maths Olympiad typically awards points for each correct answer, while incorrect answers may result in a deduction of points. The total score is calculated based on the number of questions answered correctly, encouraging students to attempt questions they are confident about.
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