CBSE Class 10 Mathematics Puzzles Set 02

Question 1. Catching Fish: If Five fishermen catch 5 fishes in 5 minutes, how long will it take fifty fishermen to catch fifty fish?
Answer: It will take 5 minutes.
If 5 fishermen catch 5 fish in 5 minutes, it means each fisherman takes 5 minutes to catch 1 fish. Therefore, 50 fishermen working simultaneously can catch 50 fish in the exact same duration of 5 minutes.
In simple words: The time taken remains the same because all fishermen work at the same time.

Exam Tip: Do not fall into the trap of direct proportion where time increases with quantity; analyze the rate of work per individual.

Question 2. Look at the Division: One day professor Agarwal went to the blackboard and demonstrated to his astonished class that one half of eight was equal to three! What did the professor do?
Answer: The professor wrote the number 8 in Roman numerals or used a horizontal line to cut the number 8 in half horizontally.
If you write the number 8 as the Roman numeral VIII and draw a horizontal line through the middle, the top half leaves the Roman numeral III, which represents the number 3.
In simple words: Cutting the Roman numeral VIII horizontally in half leaves you with III (three).

Exam Tip: Look for alternative number systems like Roman numerals when standard arithmetic does not seem to fit the puzzle.

Question 3. How Big: Can you guess how big the number: ninth power nine?
Answer: The ninth power of nine is written mathematically as \(9^9\).
Calculating the value: \(9^9 = 387,420,489\).
This is approximately 387 million.
In simple words: It is 9 multiplied by itself 9 times, which equals 387,420,489.

Exam Tip: Remember that exponential growth increases extremely fast as the base and exponent grow larger.

Question 4. Counting Street Lights: On two sides of a street, there are 35 street lights, each one is at a distance of 30 metres from the other. The street lights on one side are arranged so that each lamp fills a gap between the two other street lights on the opposite. How long is the street?
Answer: Total number of street lights = 35
Since there are lights on both sides of the street, the number of lights on each side = \(\frac{35}{2}\) is not possible directly, meaning one side has 18 lights and the other side has 17 lights.
Number of intervals on the side with 18 lights = \(18 - 1 = 17\)
Since each interval is 30 metres, the length of the street = \(17 \times 30 = 510\) metres.
In simple words: Divide the total lights equally across the two sides, find the number of spaces between lamps on one side, and multiply by the distance per space.

Exam Tip: Always remember that for \(n\) objects in a line, there are \(n - 1\) intervals between them.

Question 5. Who covered more distance: Two friends Vijay and Ajay walk with constant speed of 100m/min. Vijay takes rest for 1 min after walking 100metres while Ajay takes rest for 3 min after walking 300 metres on a square path of side 400m. Both of them start from the same corner in opposite direction. Who covered more distance and when they meet?
Answer: Perimeter of the square path = \(4 \times 400 = 1600\) metres.
Vijay's walking time per 100m = \(\frac{100}{100} = 1\) min, plus 1 min rest = 2 min for every 100m.
Ajay's walking time per 300m = \(\frac{300}{100} = 3\) min, plus 3 min rest = 6 min for every 300m (which is 2 min per 100m on average).
Both friends move at the same average effective pace and cover the entire perimeter of 1600 metres, meeting back at the starting point after completing the full loop.
In simple words: Both cover the entire 1600-meter loop because their average walking rates and rest ratios balance out identically.

Answer Tip: Break down complex motion problems into distance, speed, and time ratios for each individual separately.

Question 6. The missing Six: Place the six numbers below into empty circles, so that both the equation are true. Use each number once and only once. (Numbers given: 1, 2, 3, 4, 5, 7)
Answer: Using the numbers 1, 2, 3, 4, 5, 7:
Equation 1: \(7 + 2 = 9\) or placing numbers such that: \(7 + 1 = 8\) (using available numbers correctly: \(7 + 1 = 8\) or similar combinations). Let us place: Row 1 (\(7, 2, 9\)), Row 2 (\(5, 3, 2\)) based on standard layouts.
Valid configuration satisfying operations:
\(7 - 4 = 3\)
\(5 + 2 = 7\) (using numbers 1, 2, 3, 4, 5, 7 to fit the grid slots).
In simple words: Arrange the given numbers into the grid slots so that the addition and subtraction equations hold true.

Exam Tip: Test small numbers first for subtraction and larger combinations for addition.

Question 7. Magic Triangle: Place the numbers 4 through 9 in the circles in such a way that every side of the triangle add up to 21.
Answer: Numbers to use: 4, 5, 6, 7, 8, 9.
Sum of all numbers = \(4 + 5 + 6 + 7 + 8 + 9 = 39\).
Since there are 3 sides and the sum of each side is 21, the total sum including corners counted twice equals \(3 \times 21 = 63\).
Difference = \(63 - 39 = 24\), which represents the sum of the three corner numbers.
Placing the correct numbers at the corners (e.g., 9, 8, 7) and filling the edges such that each side sums to 21.
In simple words: Place numbers 4 to 9 in the triangle nodes so that each of the three sides totals 21.

Exam Tip: In magic polygons, corner elements are counted in two intersecting lines, which is key to solving the totals.

Question 8. Add up: Here is an equilateral triangle. Add another equilateral triangle to it in such a way that you get five equilateral triangles.
Answer: Place a smaller equilateral triangle inside the given large equilateral triangle such that its vertices touch the midpoints of the sides of the original triangle.
This divides the central region into 1 inverted triangle and 3 corner triangles, plus the large outer triangle itself, making a total of 5 equilateral triangles.
In simple words: Draw a smaller equilateral triangle inside by connecting the midpoints of the outer triangle's sides.

Exam Tip: Midpoint division of geometric shapes is a classic method to generate self-similar sub-figures.

Question 9. Magic Sticks: Just by moving one stick, make another equation. (Original equation shown as: \(7 + 5 = 12\))
Answer: Take one matchstick from the plus sign (+) to turn it into a minus sign (-), and place that stick vertically onto the number 5 to turn it into a 6, or move a stick from the 12 to make it 11.
Correct modified equation: \(7 - 5 = 2\) (by moving the vertical stick of the plus sign to form a minus sign, and moving a stick from 12 to make it 2). Alternatively: \(7 + 6 = 13\) depending on stick configuration.
In simple words: Shift one matchstick to transform operators or digits into a new valid mathematical statement.

Exam Tip: Count the total matchsticks before and after the move to ensure no sticks are created or destroyed.

Question 10. Identical Four: Divide the adjoining figure into four identical pieces.
Answer: The trapezoid-like polygon can be divided into four congruent, scaled-down trapezoidal or triangular identical pieces by connecting the midpoints of opposite sides and drawing parallel internal lines.
In simple words: Slice the geometric shape symmetrically into four equal and matching parts.

Exam Tip: Look for lines of symmetry to divide complex polygons into identical sub-components.