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HOTs for Maths Olympiad - 2 - Class 5 MCQ


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10 Questions MCQ Test Math Olympiad for Class 5 - HOTs for Maths Olympiad - 2

HOTs for Maths Olympiad - 2 for Class 5 2025 is part of Math Olympiad for Class 5 preparation. The HOTs for Maths Olympiad - 2 questions and answers have been prepared according to the Class 5 exam syllabus.The HOTs for Maths Olympiad - 2 MCQs are made for Class 5 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for HOTs for Maths Olympiad - 2 below.
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HOTs for Maths Olympiad - 2 - Question 1

Fill in the blanks and select the correct option.
(i) The sum of 372 and 648 is P.
(ii) The difference between 987 and 543 is Q.
(iii) The product of 25 and 4 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 1

(i) The sum of 372 and 648 is P:
We calculate:

372 + 648 = 1020
So, P = 1020.

(ii) The difference between 987 and 543 is Q:
We calculate:
987 - 543 = 444
So, Q = 444.

(iii) The product of 25 and 4 is R:
We calculate:
25 × 4 = 100
So, R = 100.

Thus, the correct answer is A: (P)-1020, (Q)-444, (R)-100.

HOTs for Maths Olympiad - 2 - Question 2

Fill in the blanks and select the correct option.
(i) The greatest common divisor of 48 and 64 is P.
(ii) The next prime number after 17 is Q.
(iii) The square of 12 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 2

(i) The greatest common divisor of 48 and 64 is P:

The divisors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

The divisors of 64 are: 1, 2, 4, 8, 16, 32, 64.

The common divisors of 48 and 64 are: 1, 2, 4, 8, and 16.

The greatest of these common divisors is 16.

So, P = 16.

(ii) The next prime number after 17 is Q:

The prime numbers after 17 are 19, 23, etc.

The next prime number after 17 is 19.

So, Q = 19.

(iii) The square of 12 is R:

The square of 12 is 12 × 12 = 144.

So, R = 144.

Thus, the correct answer is C: (P)-16, (Q)-19, (R)-144.

HOTs for Maths Olympiad - 2 - Question 3

Fill in the blanks and select the correct option.
(i) The area of a rectangle with length 10 cm and width 5 cm is P cm².
(ii) The perimeter of a square with side length 7 cm is Q cm.
(iii) The volume of a cube with side length 3 cm is R cm³.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 3

(i) The area of a rectangle with length 10 cm and width 5 cm is P cm²:

The area of a rectangle is calculated as:
Area = Length × Width
Area = 10 cm × 5 cm = 50 cm²
So, P = 50.

(ii) The perimeter of a square with side length 7 cm is Q cm:

The perimeter of a square is calculated as:
Perimeter = 4 × Side length
Perimeter = 4 × 7 cm = 28 cm
So, Q = 28.

(iii) The volume of a cube with side length 3 cm is R cm³:

The volume of a cube is calculated as:
Volume = Side³
Volume = 3 cm × 3 cm × 3 cm = 27 cm³
So, R = 27.

Thus, the correct answer is A: (P)-50, (Q)-28, (R)-27.

HOTs for Maths Olympiad - 2 - Question 4

Fill in the blanks and select the correct option.
(i) The smallest 3-digit number divisible by 5 is P.
(ii) The number of days in a leap year is Q.
(iii) The sum of the angles in a triangle is R degrees.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 4

(i) The smallest 3-digit number divisible by 5 is P:

The smallest 3-digit number is 100.
Since 100 is divisible by 5 (100 ÷ 5 = 20), the smallest 3-digit number divisible by 5 is 100.
So, P = 100.

(ii) The number of days in a leap year is Q:

A leap year has 366 days (an extra day in February).
So, Q = 366.

(iii) The sum of the angles in a triangle is R degrees:

The sum of the angles in any triangle is always 180 degrees.
So, R = 180.

Thus, the correct answer is B: (P)-100, (Q)-366, (R)-180.

HOTs for Maths Olympiad - 2 - Question 5

Fill in the blanks and select the correct option.
(i) The number of edges in a cube is P.
(ii) The number of vertices in a cuboid is Q.
(iii) The number of faces in a pyramid with a square base is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 5

(i) The number of edges in a cube is P:

A cube has 12 edges.
So, P = 12.

(ii) The number of vertices in a cuboid is Q:

A cuboid has 8 vertices.
So, Q = 8.

(iii) The number of faces in a pyramid with a square base is R:

A pyramid with a square base has 1 square face (the base) and 4 triangular faces.
So, the total number of faces is 5.
So, R = 5.

Thus, the correct answer is A: (P)-12, (Q)-8, (R)-5.

HOTs for Maths Olympiad - 2 - Question 6

Fill in the blanks and select the correct option.
(i) The next even number after 14 is P.
(ii) The sum of 9 and 6 is Q.
(iii) The difference between 20 and 13 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 6

(i) The next even number after 14 is P:

The next even number after 14 is 16.
So, P = 16.

(ii) The sum of 9 and 6 is Q:

The sum of 9 and 6 is:
9 + 6 = 15.
So, Q = 15.

(iii) The difference between 20 and 13 is R:

The difference between 20 and 13 is:
20 - 13 = 7.
So, R = 7.

Thus, the correct answer is B: (P)-16, (Q)-15, (R)-7.

HOTs for Maths Olympiad - 2 - Question 7

Fill in the blanks and select the correct option.
(i) The result of 81 ÷ 9 is P.
(ii) The result of 49 ÷ 7 is Q.
(iii) The result of 64 ÷ 8 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 7

(i) The result of 81 ÷ 9 is P:

81 ÷ 9 = 9
So, P = 9.

(ii) The result of 49 ÷ 7 is Q:

49 ÷ 7 = 7
So, Q = 7.

(iii) The result of 64 ÷ 8 is R:

64 ÷ 8 = 8
So, R = 8.

Thus, the correct answer is A: (P)-9, (Q)-7, (R)-8.

HOTs for Maths Olympiad - 2 - Question 8

Fill in the blanks and select the correct option.
(i) The average of 8, 12, and 20 is P.
(ii) The median of 4, 15, and 9 is Q.
(iii) The mode of 2, 3, 3, 4, 4, 4, 5 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 8

(i) The average of 8, 12, and 20 is P:

The average is calculated as:
Average = (Sum of all numbers) / (Number of numbers)
Average = (8 + 12 + 20) / 3 = 40 / 3 = 13
So, P = 13.

(ii) The median of 4, 15, and 9 is Q:

To find the median, we first arrange the numbers in ascending order:
4, 9, 15
The middle number is 9, so Q = 9.

(iii) The mode of 2, 3, 3, 4, 4, 4, 5 is R:

The mode is the number that appears most frequently. In this case, 4 appears three times, more than any other number.
So, R = 4.

Thus, the correct answer is D: (P)-13, (Q)-9, (R)-4.

HOTs for Maths Olympiad - 2 - Question 9

Fill in the blanks and select the correct option.
(i) The next number in the sequence 3, 6, 9, 12, ... is P.
(ii) The total number of minutes in 3 hours is Q.
(iii) The decimal form of 3/4 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 9

(i) The next number in the sequence 3, 6, 9, 12, ... is P:

The sequence increases by 3 each time.
3 + 3 = 6, 6 + 3 = 9, 9 + 3 = 12, so the next number is 12 + 3 = 15.
So, P = 15.

(ii) The total number of minutes in 3 hours is Q:

1 hour = 60 minutes, so 3 hours = 3 × 60 = 180 minutes.
So, Q = 180.

(iii) The decimal form of 3/4 is R:

To convert 3/4 to a decimal, divide 3 by 4:
3 ÷ 4 = 0.75.
So, R = 0.75.

Thus, the correct answer is B: (P)-15, (Q)-180, (R)-0.75.

HOTs for Maths Olympiad - 2 - Question 10

Fill in the blanks and select the correct option.
(i) The number of days in February in a non-leap year is P.
(ii) The total number of sides in a pentagon is Q.
(iii) The product of 7 and 9 is R.

Detailed Solution for HOTs for Maths Olympiad - 2 - Question 10

(i) The number of days in February in a non-leap year is P:

In a non-leap year, February has 28 days.
So, P = 28.

(ii) The total number of sides in a pentagon is Q:

A pentagon has 5 sides.
So, Q = 5.

(iii) The product of 7 and 9 is R:

The product of 7 and 9 is:
7 × 9 = 63.
So, R = 63.

Thus, the correct answer is D: (P)-28, (Q)-5, (R)-63.

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