Total letters in the word PEACE = 5, so total possible outcomes = 5. (i) Favourable letters for 'a P, E or C' are P, E, C, E (1 P, 2 Es and 1 C). Number of favourable outcomes = 1 + 2 + 1 = 4. Probability(P, E or C) = 4 / 5 = 0.8 (or 80%). (ii) Letters that are not E are P, A and C. Number of favourRead more
Total letters in the word PEACE = 5, so total possible outcomes = 5.
(i) Favourable letters for ‘a P, E or C’ are P, E, C, E (1 P, 2 Es and 1 C).
Number of favourable outcomes = 1 + 2 + 1 = 4.
Probability(P, E or C) = 4 / 5 = 0.8 (or 80%).
(ii) Letters that are not E are P, A and C.
Number of favourable outcomes = 3.
Probability(not an E) = 3 / 5 = 0.6 (or 60%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
Total number of cases = 1000. (i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%). (ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%). (iii) ForRead more
Total number of cases = 1000.
(i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%).
(ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%).
(iii) For tyre lasting more than 14000 km, frequency = 445. Probability = 445 / 1000 = 0.445 (or 44.5%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
Element A has one valence electron, so it loses one electron to achieve an octet, forming a positive cation A+. Element B has six valence electrons and requires two electrons to complete its outer octet, forming a negative anion B2-. The transfer of electrons creates an ionic bond between them. BalaRead more
Element A has one valence electron, so it loses one electron to achieve an octet, forming a positive cation A+. Element B has six valence electrons and requires two electrons to complete its outer octet, forming a negative anion B2-. The transfer of electrons creates an ionic bond between them. Balancing their positive and negative charges gives the chemical formula A2B.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
Element X contains six valence electrons, meaning it requires two additional electrons to attain a stable octet configuration. Two X atoms achieve stability by sharing two pairs of electrons, resulting in a double covalent bond. When X reacts with element Y having two valence electrons, they share oRead more
Element X contains six valence electrons, meaning it requires two additional electrons to attain a stable octet configuration. Two X atoms achieve stability by sharing two pairs of electrons, resulting in a double covalent bond. When X reacts with element Y having two valence electrons, they share or transfer two electrons, yielding the stable compound Y double bond X or YX.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
Option (iii) is the correct answer. In this combination, two ferric ions Fe3+ contribute a positive charge calculated as two multiplied by plus three, totaling plus six. Three oxide ions O2- contribute a negative charge calculated as three multiplied by minus two, totaling minus six. This perfectlyRead more
Option (iii) is the correct answer. In this combination, two ferric ions Fe3+ contribute a positive charge calculated as two multiplied by plus three, totaling plus six. Three oxide ions O2- contribute a negative charge calculated as three multiplied by minus two, totaling minus six. This perfectly satisfies the requirement of having a total 6+ positive charge and 6- negative charge.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking. (i) What is the probability that it is a P, E or C? (ii) What is the probability that it is not an E?
Total letters in the word PEACE = 5, so total possible outcomes = 5. (i) Favourable letters for 'a P, E or C' are P, E, C, E (1 P, 2 Es and 1 C). Number of favourable outcomes = 1 + 2 + 1 = 4. Probability(P, E or C) = 4 / 5 = 0.8 (or 80%). (ii) Letters that are not E are P, A and C. Number of favourRead more
Total letters in the word PEACE = 5, so total possible outcomes = 5.
(i) Favourable letters for ‘a P, E or C’ are P, E, C, E (1 P, 2 Es and 1 C).
Number of favourable outcomes = 1 + 2 + 1 = 4.
Probability(P, E or C) = 4 / 5 = 0.8 (or 80%).
(ii) Letters that are not E are P, A and C.
Number of favourable outcomes = 3.
Probability(not an E) = 3 / 5 = 0.6 (or 60%).
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 tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts: (i) Less than 4000 km, (ii) Between 4000 and 14000 km, (iii) More than 14000 km.
Total number of cases = 1000. (i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%). (ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%). (iii) ForRead more
Total number of cases = 1000.
(i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%).
(ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%).
(iii) For tyre lasting more than 14000 km, frequency = 445. Probability = 445 / 1000 = 0.445 (or 44.5%).
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 particular element A has one electron in its third shell. There is another element B with six electrons in its second shell. (i) How many electrons does A tend to give or take to become stable? (ii) What kind of ion would it form? (iii) How many electrons does B tend to give or take to become stable? (iv) What kind of ion would it form? (v) If A and B were to combine, what kind of bond would be formed? (vi) What would be the formula for the compound thus formed?
Element A has one valence electron, so it loses one electron to achieve an octet, forming a positive cation A+. Element B has six valence electrons and requires two electrons to complete its outer octet, forming a negative anion B2-. The transfer of electrons creates an ionic bond between them. BalaRead more
Element A has one valence electron, so it loses one electron to achieve an octet, forming a positive cation A+. Element B has six valence electrons and requires two electrons to complete its outer octet, forming a negative anion B2-. The transfer of electrons creates an ionic bond between them. Balancing their positive and negative charges gives the chemical formula A2B.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-9/
See lessAn element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure of the molecule it would form. (iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Element X contains six valence electrons, meaning it requires two additional electrons to attain a stable octet configuration. Two X atoms achieve stability by sharing two pairs of electrons, resulting in a double covalent bond. When X reacts with element Y having two valence electrons, they share oRead more
Element X contains six valence electrons, meaning it requires two additional electrons to attain a stable octet configuration. Two X atoms achieve stability by sharing two pairs of electrons, resulting in a double covalent bond. When X reacts with element Y having two valence electrons, they share or transfer two electrons, yielding the stable compound Y double bond X or YX.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-9/
See lessYou want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6-. Which of the following combinations gives the correct number of ions? (i) 2 Al3+ and 3 Cl- (ii) 3 Mg2+ and 1 PO4 3- (iii) 2 Fe3+ and 3 O2- (iv) 3 Ca2+ and 2 SO4 2-
Option (iii) is the correct answer. In this combination, two ferric ions Fe3+ contribute a positive charge calculated as two multiplied by plus three, totaling plus six. Three oxide ions O2- contribute a negative charge calculated as three multiplied by minus two, totaling minus six. This perfectlyRead more
Option (iii) is the correct answer. In this combination, two ferric ions Fe3+ contribute a positive charge calculated as two multiplied by plus three, totaling plus six. Three oxide ions O2- contribute a negative charge calculated as three multiplied by minus two, totaling minus six. This perfectly satisfies the requirement of having a total 6+ positive charge and 6- negative charge.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-9/
See less