Questions Related to NCERT

(i) Hydrogen + Chlorine → Hydrogen chloride
Answer:
H₂ + Cl₂ → 2HCl

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
Answer:
3BaCl₂ + Al₂(SO₄)₃ → 3BaSO₄ + 2AlCl₃

(iii) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
2Na + 2H₂O → 2NaOH + H₂↑

When magnesium is kept in the open air for some time, it reacts slowly with the oxygen in the air to form a thin, white layer of magnesium oxide (MgO) on its surface. This layer sticks tightly to the metal and prevents the fresh magnesium underneath from coming in direct contact with oxygen.

Before burning magnesium in an experiment, this oxide layer must be removed by rubbing the ribbon with sandpaper or a file. This exposes the shiny surface of pure magnesium, which reacts easily and quickly with oxygen when heated.

When clean magnesium is burned, it reacts vigorously with oxygen to form magnesium oxide, producing a bright white flame and intense heat.

Reasons for cleaning the ribbon:

• To remove the oxide layer (MgO).
• To ensure smooth and complete combustion.
• To allow accurate observation of the reaction.

(i) Not correct.
Reason:
When two coins are tossed, the actual outcomes are: HH, HT, TH, TT — a total of 4 outcomes.
Two heads = HH → 1 outcome
Two tails = TT → 1 outcome
One of each = HT or TH → 2 outcomes
So the probability of:
Two heads = 1/4
Two tails = 1/4
One of each = 2/4 = 1/2
Since these grouped outcomes are not equally likely, assigning a probability of 1/3 to each is incorrect.

(ii) Correct.
Reason:
When a die is thrown, possible outcomes are: 1, 2, 3, 4, 5, 6
Odd numbers = 1, 3, 5 → 3 outcomes
Even numbers = 2, 4, 6 → 3 outcomes
So, the probability of getting an odd number = 3/6 = 1/2.
This reasoning is correct because the outcomes are equally likely.

Total possible outcomes when a die is thrown twice:
= 6 × 6 = 36

i) 5 will not come up either time

Let’s count how many outcomes include 5:
5 appears in the first throw: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes
5 appears in the second throw: (1,5), (2,5), (3,5), (4,5), (6,5) → 5 outcomes
(We already counted (5,5) once, so don’t count it again.)

So, total outcomes where 5 appears at least once = 6 + 5 = 11
Therefore, outcomes where 5 does not appear at all = 36 − 11 = 25
Answer (i): Probability = 25 / 36

(ii) 5 will come up at least once

From above, favorable outcomes = 11
(Where 5 appears in at least one position)
Answer (ii): Probability = 11 / 36

Total number of possible outcomes when a coin is tossed 3 times:
Each toss has 2 outcomes: Head (H) or Tail (T)
The outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
So, total outcomes are 8

Favorable outcomes (Hanif wins):
Hanif wins if the result is either HHH or TTT
Number of winning outcomes = 2

Unfavorable outcomes (Hanif loses) = Total outcomes − Winning outcomes
= 8 − 2 = 6

Probability of an event = (Number of favorable outcomes) ÷ (Total number of outcomes)

So, The probability that Hanif loses = 6 ÷ 8 = 3/4

Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 − 20 = 124

(i) Probability that Nuri will buy the pen (i.e., it is good):

Probability =  Number of good pens / Total number of pens
=>  124 / 144
=> 31 / 36

(ii) Probability that Nuri will not buy the pen (i.e., it is defective):

Probability = Number of defective pens / Total number of pens
=>  20 / 144
=>  5 / 36

A box contains 90 discs numbered from 1 to 90. One disc is drawn at random.

Total number of outcomes = 90

(i) Probability of getting a two-digit number:

Two-digit numbers are from 10 to 90
Total two-digit numbers = 81
Probability = 81 / 90 = 9 / 10

(ii) Probability of getting a perfect square number:

Perfect square numbers between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81
Total perfect squares = 9
Probability = 9 / 90 = 1 / 10

(iii) Probability of getting a number divisible by 5:

Numbers divisible by 5 from 1 to 90 are: 5, 10, 15, ..., 90
Total such numbers = 18
Probability = 18 / 90 = 1 / 5

(i) A lot of 20 bulbs contains 4 defective bulbs.

So, total bulbs = 20
Number of defective bulbs = 4

Probability that the bulb drawn is defective = 4 / 20 = 1 / 5

(ii) The first bulb drawn is not defective and is not replaced.

That means:
One good bulb is removed.
Remaining bulbs = 19
Remaining good bulbs = 16 − 1 = 15 (since total good bulbs were 16 = 20 − 4)

Probability that the second bulb drawn is not defective = 15 / 19

Total number of pens = 12 defective + 132 good = 144

Number of good pens = 132

Probability of taking out a good pen = Number of good pens / Total number of pens
= 132 / 144
=>  11 / 12 

Hence answer is : 11/12

(i) What is the probability that the card is the queen?

There is only 1 queen in the 5 cards.
So,
Probability = 1/5

(ii) If the queen is drawn and put aside,

The remaining cards are: 10, J, K, A   i.e 4 cards remaining
Let’s now find:

(a) Probability that the second card picked is an ace
Only 1 ace left in 4 cards.
So,
Probability = 1/4

(b) Probability that the second card picked is a queen
Since the queen has already been drawn and put aside, there is no queen left.

So,
Probability = 0

(i) Probability of getting a king of red colour:

There are 2 red kings — King of hearts and King of diamonds.
Favorable outcomes = 2
So,
Probability = 2/52 = 1/26

(ii) Probability of getting a face card:

Face cards = Jacks, Queens, Kings
Each suit has 3 face cards 
Total face cards = 4 suits × 3 = 12
So,
Probability = 12/52 = 3/13

(iii) Probability of getting a red face card:

Red suits = Hearts  and Diamonds
Each red suit has 3 face cards 
Total red face cards = 2 × 3 = 6
So,
Probability = 6/52 = 3/26

(iv) Probability of getting the jack of hearts:

Favorable outcomes = 1
So,
Probability = 1/52

(v) Probability of getting a spade:

Spade suit has 13 cards
So,
Probability = 13/52 = 1/4

(vi) Probability of getting the queen of diamonds:

Only one card  Queen of diamonds
Favorable outcomes = 1
So,
Probability = 1/52

A standard die has 6 faces numbered: 1, 2, 3, 4, 5, 6
Total number of outcomes = 6

(i) Probability of getting a prime number:
Prime numbers between 1 and 6 are: 2, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

(ii) Probability of getting a number lying between 2 and 6:
"Between 2 and 6" means: 3, 4, 5 (not including 2 and 6)
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

(iii) Probability of getting an odd number:
Odd numbers between 1 and 6 are: 1, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1 to 8. All outcomes are equally likely. So, the total number of outcomes = 8.

(i) Probability that it will point at 8:
        Favorable outcome = 1 (only the number 8)
        Probability = 1/8

(ii) Probability that it points at an odd number:
         Odd numbers = 1, 3, 5, 7
         So favorable outcomes is 4
         Probability = 4/8 = 1/2

(iii) Probability that it points at a number greater than 2:
          Numbers greater than 2 = 3, 4, 5, 6, 7, 8 
          So favorable outcomes is 6
          Probability = 6/8 = 3/4

(iv) Probability that it points at a number less than 9:
           All numbers (1 to 8) are less than 9 
          So favorable outcomes is 8
          Probability = 8/8 = 1 

Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13

Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)
Probability that the fish is male = 5 / 13

Total number of coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180 coins

Formula:
Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)

(i) Probability that the coin will be a 50p coin:
             Number of 50p coins = 100
             P(50p coin) = 100 / 180 = 5/9

(ii) Probability that the coin will not be a ₹5 coin:
             Number of coins that are not ₹5 = 100 + 50 + 20 = 170
             P(not ₹5 coin) = 170 / 180 = 17/18

The box contains: 
Red marbles = 5
White marbles = 8
Green marbles = 4
So total marbles = 5 + 8 + 4 = 17

Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes

(i) Probability that the marble is red:
Number of red marbles = 5
P(red) = 5 / 17

(ii) Probability that the marble is white:
Number of white marbles = 8
P(white) = 8 / 17

(iii) Probability that the marble is not green:
Number of marbles that are not green = 5 (red) + 8 (white) = 13
P(not green) = 13 / 17

A bag contains 3 red balls and 5 black balls.
Total number of balls = 3 + 5 = 8
Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes

(i) Probability that the ball drawn is red:
Number of red balls = 3
P(red) = 3 / 8

(ii) Probability that the ball drawn is not red:
Number of balls that are not red (i.e., black balls) = 5
P(not red) = 5 / 8

Probability of an event happening = 1 – Probability of the event not happening

So,
Probability that 2 students have the same birthday = 1 – Probability that they do not have the same birthday
=>  1 – 0.992
=>  0.008

(i) Probability of taking out an orange flavoured candy:
Since there are no orange flavoured candies in the bag,
So Probability of orange flavoured candy is = 0

(ii) Probability of taking out a lemon flavoured candy:
All candies are lemon flavoured, so the probability is
P(lemon flavoured candy) = 1

If P(E) = 0.05, then the probability of ‘not E’ is:

P(not E) = 1 – P(E)
= 1 – 0.05
= 0.95

Hence  ‘not E’ is: 0.95