To achieve balance on a seesaw, the principle of moments applies, meaning load multiplied by load arm equals effort multiplied by effort arm. The adult exerts double the downward force of the child. Therefore, to balance the turning effects, the adult sits at a distance d from the central fulcrum, wRead more
To achieve balance on a seesaw, the principle of moments applies, meaning load multiplied by load arm equals effort multiplied by effort arm. The adult exerts double the downward force of the child. Therefore, to balance the turning effects, the adult sits at a distance d from the central fulcrum, while the child sits on the opposite side at a distance 2d from the fulcrum.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
Area of the rectangular region = length x breadth = 3 m x 2 m = 6 m². For the circular region, diameter = 1 m, so radius r = 1/2 m = 0.5 m. Area of the circle = π x r² = π x (1/2)² = π / 4 m² (or approximately 3.1416 / 4 = 0.7854 m²). Probability of landing inside the circle: P = Area of circle / ToRead more
Area of the rectangular region = length x breadth = 3 m x 2 m = 6 m².
For the circular region, diameter = 1 m, so radius r = 1/2 m = 0.5 m.
Area of the circle = π x r² = π x (1/2)² = π / 4 m² (or approximately 3.1416 / 4 = 0.7854 m²).
Probability of landing inside the circle:
P = Area of circle / Total area of rectangle
P = (π / 4) / 6 = π / 24 (approximately 0.131 or 13.1%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
The correct sample space is (iv) {0, 1, 2, 3}. When tossing three coins, the number of heads can be 0 (TTT), 1 (HTT, THT, TTH), 2 (HHT, HTH, THH) or 3 (HHH). Why the others fail: (i) {1, 2, 3} fails because it misses 0 heads. (ii) {0, 1, 2} fails because it misses 3 heads. (iii) {0, 1, 2, 3, 4} failRead more
The correct sample space is (iv) {0, 1, 2, 3}.
When tossing three coins, the number of heads can be 0 (TTT), 1 (HTT, THT, TTH), 2 (HHT, HTH, THH) or 3 (HHH).
Why the others fail:
(i) {1, 2, 3} fails because it misses 0 heads.
(ii) {0, 1, 2} fails because it misses 3 heads.
(iii) {0, 1, 2, 3, 4} fails because getting 4 heads from 3 coins is impossible.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
The experiment involves two simultaneous random actions: Tossing a coin gives 2 possible outcomes: Heads (H) or Tails (T). Drawing a card from numbers 1 to 6 gives 6 possible outcomes: {1, 2, 3, 4, 5, 6}. Combining each coin outcome with each card number gives a sample space with 2 x 6 = 12 elementsRead more
The experiment involves two simultaneous random actions:
Tossing a coin gives 2 possible outcomes: Heads (H) or Tails (T).
Drawing a card from numbers 1 to 6 gives 6 possible outcomes: {1, 2, 3, 4, 5, 6}. Combining each coin outcome with each card number gives a sample space with 2 x 6 = 12 elements: S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
(i) With replacement: First draw branches to 1, 2, 3, 4. Each branches to 1, 2, 3, 4. S1 = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}. (ii) Without replacement: First draw branches to 1, 2, 3, 4. Each branches to remaining 3 numbeRead more
(i) With replacement: First draw branches to 1, 2, 3, 4. Each branches to 1, 2, 3, 4.
On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
To achieve balance on a seesaw, the principle of moments applies, meaning load multiplied by load arm equals effort multiplied by effort arm. The adult exerts double the downward force of the child. Therefore, to balance the turning effects, the adult sits at a distance d from the central fulcrum, wRead more
To achieve balance on a seesaw, the principle of moments applies, meaning load multiplied by load arm equals effort multiplied by effort arm. The adult exerts double the downward force of the child. Therefore, to balance the turning effects, the adult sits at a distance d from the central fulcrum, while the child sits on the opposite side at a distance 2d from the fulcrum.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessSuppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?
Area of the rectangular region = length x breadth = 3 m x 2 m = 6 m². For the circular region, diameter = 1 m, so radius r = 1/2 m = 0.5 m. Area of the circle = π x r² = π x (1/2)² = π / 4 m² (or approximately 3.1416 / 4 = 0.7854 m²). Probability of landing inside the circle: P = Area of circle / ToRead more
Area of the rectangular region = length x breadth = 3 m x 2 m = 6 m².
For the circular region, diameter = 1 m, so radius r = 1/2 m = 0.5 m.
Area of the circle = π x r² = π x (1/2)² = π / 4 m² (or approximately 3.1416 / 4 = 0.7854 m²).
Probability of landing inside the circle:
P = Area of circle / Total area of rectangle
P = (π / 4) / 6 = π / 24 (approximately 0.131 or 13.1%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessThree coins are tossed and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space? (i) {1, 2, 3} (ii) {0, 1, 2} (iii) {0, 1, 2, 3, 4} (iv) {0, 1, 2, 3}
The correct sample space is (iv) {0, 1, 2, 3}. When tossing three coins, the number of heads can be 0 (TTT), 1 (HTT, THT, TTH), 2 (HHT, HTH, THH) or 3 (HHH). Why the others fail: (i) {1, 2, 3} fails because it misses 0 heads. (ii) {0, 1, 2} fails because it misses 3 heads. (iii) {0, 1, 2, 3, 4} failRead more
The correct sample space is (iv) {0, 1, 2, 3}.
When tossing three coins, the number of heads can be 0 (TTT), 1 (HTT, THT, TTH), 2 (HHT, HTH, THH) or 3 (HHH).
Why the others fail:
(i) {1, 2, 3} fails because it misses 0 heads.
(ii) {0, 1, 2} fails because it misses 3 heads.
(iii) {0, 1, 2, 3, 4} fails because getting 4 heads from 3 coins is impossible.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessList the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.
The experiment involves two simultaneous random actions: Tossing a coin gives 2 possible outcomes: Heads (H) or Tails (T). Drawing a card from numbers 1 to 6 gives 6 possible outcomes: {1, 2, 3, 4, 5, 6}. Combining each coin outcome with each card number gives a sample space with 2 x 6 = 12 elementsRead more
The experiment involves two simultaneous random actions:
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessA box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments: (i) A ball is drawn and the number is recorded. Then the ball is returned and a second ball is drawn and recorded. (ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball. (iii) What are the sizes of these two sample spaces?
(i) With replacement: First draw branches to 1, 2, 3, 4. Each branches to 1, 2, 3, 4. S1 = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}. (ii) Without replacement: First draw branches to 1, 2, 3, 4. Each branches to remaining 3 numbeRead more
(i) With replacement: First draw branches to 1, 2, 3, 4. Each branches to 1, 2, 3, 4.
S1 = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}.
(ii) Without replacement: First draw branches to 1, 2, 3, 4. Each branches to remaining 3 numbers.
S2 = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}.
(iii) Size of first sample space = 4 x 4 = 16. Size of second sample space = 4 x 3 = 12.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
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