Tag: Class 10th Maths NCERT Solutions

Exercise-14.1, Class 10th, Maths, Chapter 14, NCERT

1. Complete the following statements:

(i) Probability of an event $E$ + Probability of the event ‘not $E$ = $\underline{1}$ .

(ii) The probability of an event that cannot happen is $\underline{0}$ . Such an event is called $\underline{an impossible event}$ .

(iii) The probability of an event that is certain to happen is $\underline{1}$ . Such an event is called $\underline{a sure (certain) event}$ .

(iv) The sum of the probabilities of all the elementary events of an experiment is $\underline{1}$ .

(v) The probability of an event is greater than or equal to $\underline{0}$ and less than or equal to $\underline{1}$ .
(That is, $0 \leq P (E) \leq 1$

2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Answer: Not necessarily equally likely. Starting depends on many factors (battery, fuel, etc.). We cannot assume equal chance.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer: Not necessarily equally likely. The probabilities depend on the player’s skill, distance, defence, etc.

(iii) A trial is made to answer a true–false question. The answer is right or wrong.

Answer: If the answer is chosen at random (a pure guess), the two outcomes (right/wrong) are equally likely (each probability $1 / 2$ ). If the student knows the answer (or mostly knows), they are not equally likely. So: equally likely only when guessing at random.

(iv) A baby is born. It is a boy or a girl.

Answer: For practical school-problems we assume the two outcomes are approximately equally likely (each about $1 / 2$ ). In reality there is a slight natural bias, but the experiment is normally treated as equally likely.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution: A (fair) coin is symmetric and has two equally likely outcomes — Head or Tail. Thus each team has equal probability $1 / 2$ of being chosen; this makes the choice unbiased and fair.

4. Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$ (B) $- 1.5$ (C) $15 %$ (D) $0.7$

Solution: A $\frac{2}{3}$ is valid ( $\approx 0.666$ ), C $15 % = 0.15$ valid, D $0.7$ valid. (B) $- 1.5$ is impossible (probabilities cannot be negative).
Answer: (B) $- 1.5$ cannot be a probability.

5. If $P (E) = 0.05$ , what is the probability of ‘not $E$ ?

Solution: $P (not E) = 1 - P (E) = 1 - 0.05 = 0.95$

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Solution: (i) Orange flavoured: impossible ⇒ probability $0$ .
(ii) Lemon flavoured: certain ⇒ probability $1$ .

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is $0.992$ . What is the probability that the 2 students have the same birthday?

Solution: Probability same birthday $= 1 - 0.992 = 0.008.$

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Solution: Total $= 3 + 5 = 8$
(i) $P (red) = \frac{3}{8}$
(ii) $P (not red) = \frac{5}{8}$

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

Solution: Total $= 5 + 8 + 4 = 17$
(i) $P (red) = \frac{5}{17}$
(ii) $P (white) = \frac{8}{17}$
(iii) $P (not green) = 1 - P (green) = 1 - \frac{4}{17} = \frac{13}{17}$

10. A piggy bank contains hundred 50p coins, fifty $₹ 1$ coins, twenty $₹ 2$ coins and ten $₹ 5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a $₹ 5$ coin?

Solution: Total coins $= 100 + 50 + 20 + 10 = 180$
(i) $P (50p) = \frac{100}{180} = \frac{10}{18} = \frac{5}{9}$
(ii) $P (not ₹5) = 1 - P (₹5) = 1 - \frac{10}{180} = 1 - \frac{1}{18} = \frac{17}{18}$

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution: Total $= 5 + 8 = 13$ . So $P (male) = \frac{5}{13}$

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, \dots, 8$ (all equally likely). What is the probability that it will point at

(i) $8$ ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution: Total outcomes $= 8$
(i) $P (8) = \frac{1}{8} .$
(ii) Odd numbers $= 1, 3, 5, 7$ ⇒ $4$ outcomes ⇒ $P = \frac{4}{8} = \frac{1}{2}$
(iii) Numbers $> 2$ are $3, 4, 5, 6, 7, 8$ ⇒ $6$ outcomes ⇒ $P = \frac{6}{8} = \frac{3}{4}$
(iv) Numbers $< 9$ are $1$ to $8$ ⇒ all outcomes ⇒ $P = 1$

13. A die is thrown once. Find the probability of getting

(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Solution: Faces $= 1, 2, 3, 4, 5, 6$ Total $= 6$
(i) Primes: $2, 3, 5$ ⇒ $3$ outcomes ⇒ $P = \frac{3}{6} = \frac{1}{2} .$
(ii) “Lying between 2 and 6” (interpreted as strictly between): $3, 4, 5$ ⇒ $3$ outcomes ⇒ $P = \frac{3}{6} = \frac{1}{2}$

(iii) Odd numbers: $1, 3, 5$ ⇒ $3$ outcomes ⇒ $P = \frac{3}{6} = \frac{1}{2}$

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds

Solution: Deck has 52 cards; each equally likely. Face cards = J,Q,K (3 per suit), total 12.
(i) Red kings: two red suits (hearts, diamonds) ⇒ 2 kings ⇒ $P = \frac{2}{52} = \frac{1}{26}$
(ii) Face card: $12$ cards ⇒ $P = \frac{12}{52} = \frac{3}{13}$
(iii) Red face card: red suits have $3$ face cards each ⇒ $6$ cards ⇒ $P = \frac{6}{52} = \frac{3}{26}$
(iv) Jack of hearts: $1$ card ⇒ $P = \frac{1}{52}$
(v) A spade: $13$ spades ⇒ $P = \frac{13}{52} = \frac{1}{4}$
(vi) Queen of diamonds: $1$ card ⇒ $P = \frac{1}{52}$

15. Five cards—the ten, jack, queen, king and ace of diamonds—are well-shuffled face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution: Initially 5 cards, all equally likely.
(i) $P (queen) = \frac{1}{5}$
(ii) After queen removed, remaining cards $= 4$ (ten, jack, king, ace).
(a) $P (ace) = \frac{1}{4}$
(b) $P (queen) = 0$ (queen already removed).

16. 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random. Determine the probability that the pen taken out is a good one.

Solution: Total $= 12 + 132 = 144.Good pens$ $= 132$
$P (good) = \frac{132}{144} = \frac{11}{12} \approx 0.9167.$

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution: (i) $P (defective) = \frac{4}{20} = \frac{1}{5}$
(ii) Initially good bulbs $= 20 - 4 = 16$ . If the first drawn is not defective (i.e. a good one) and not replaced, remaining bulbs $= 19$ and remaining good $= 15$ . So $P (not defective second) = \frac{15}{19}$

18. A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5

Solution: Total outcomes $= 90$

(i) Two-digit numbers from 10 to 90 inclusive ⇒ count = $90 - 10 + 1 = 81$
$P = \frac{81}{90} = \frac{9}{10}$

(ii) Perfect squares $\leq 90$ : $1^{2}, 2^{2}, \dots, 9^{2} = 1, 4, 9, 16, 25, 36, 49, 64, 81$ ⇒ 9 numbers.
$P = \frac{9}{90} = \frac{1}{10}$

(iii) Numbers divisible by 5 up to 90: $5, 10, \dots, 90$ . Count $= 90 / 5 = 18$
$P = \frac{18}{90} = \frac{1}{5} .$

19. A child has a die whose six faces show the letters as given below:

$A B C D E A$

The die is thrown once. What is the probability of getting (i) $A$ ? (ii) $D$ ?

Solution: There are 6 faces; $A$ appears twice.
(i) $P (A) = \frac{2}{6} = \frac{1}{3}$
(ii) $D$ appears once ⇒ $P (D) = \frac{1}{6}$

20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter $1$ m?* (Area method)

Solution: Rectangle area = $3 m \times 2 m = 6 m^{2}$ (from figure labels). Circle radius $r = \frac{1}{2}$ , area $= π r^{2} = π {(\frac{1}{2})}^{2} = \frac{π}{4}$
Probability $= \frac{area of circle}{area of rectangle} = \frac{π / 4}{6} = \frac{π}{24}$ (Approx.) $\frac{π}{24} \approx 0.1309$

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it ?
(ii) She will not buy it ?

Solution: Good $= 144 - 20 = 124$ Total $= 144.$
(i) $P (buy) = \frac{124}{144} = \frac{31}{36}$
(ii) $P (not buy) = \frac{20}{144} = \frac{5}{36}$

22. Refer to Example 13 (two dice).

(i) Complete the following table:

Event ‘Sum on 2 dice’: $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$
Probability: $\frac{1}{36}, \frac{2}{36}, \frac{3}{36}, \frac{4}{36}, \frac{5}{36}, \frac{6}{36}, \frac{5}{36}, \frac{4}{36}, \frac{3}{36}, \frac{2}{36}, \frac{1}{36}$

(ii) A student argues that ‘there are 11 possible outcomes (2 through 12). Therefore, each has probability $1 / 11$ . Do you agree?

Solution: (i) Completed as above — probabilities come from counting ordered pairs $(i, j)$ with $i, j \in {1, \dots, 6}$ (ii) No — incorrect. The sums are not equally likely: e.g., sum 7 can occur in 6 ways $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ while sum 2 occurs only as $(1, 1)$ . There are 36 equally likely ordered outcomes, and each sum’s probability equals (number of ordered pairs giving that sum)/36, not $1 / 11$ .

23. A game consists of tossing a one-rupee coin 3 times. Hanif wins if all the tosses give the same result (three heads or three tails), and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution: Total outcomes $= 2^{3} = 8$ . Winning outcomes = {HHH, TTT} ⇒ 2 outcomes ⇒ $P (win) = \frac{2}{8} = \frac{1}{4}$ . Thus $P (lose) = 1 - \frac{1}{4} = \frac{3}{4}$

24. A die is thrown twice. What is the probability that

(i) $5$ will not come up either time?
(ii) $5$ will come up at least once?

Solution: Treat two throws as independent. Probability a single throw is not 5 is $\frac{5}{6}$ .

(i) Neither time ⇒ ${(\frac{5}{6})}^{2} = \frac{25}{36}$

(ii) At least once $= 1 -$ probability of none $= 1 - \frac{25}{36} = \frac{11}{36}$

25. Which of the following arguments are correct and which are not? Give reasons.

(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $1 / 3$ .

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $1 / 2$ .

Solution: (i) Incorrect. Two tossed coins produce 4 equally likely ordered outcomes: HH, HT, TH, TT. Grouping them gives three events {HH}, {HT or TH}, {TT}, but these grouped events are not equally likely: middle event (“one of each”) has probability $2 / 4 = 1 / 2$ , while HH and TT each have probability $1 / 4$ . So the claim $1 / 3$ is wrong.

(ii) Correct. For a fair die the three odd faces ${1, 3, 5}$ and three even faces ${2, 4, 6}$ are each 3 outcomes out of 6. So $P (odd) = \frac{3}{6} = \frac{1}{2}$

October 12, 2025
Exercise-13.3, Class 10th, Maths, Chapter 13, NCERT
Q1

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units) — Number of consumers
65–85 : 4
85–105 : 5
105–125 : 13
125–145 : 20
145–165 : 14
165–185 : 8
185–205 : 4

Solution
1. Compute class-marks $x$ and $f$ :
Class f class-mark x f·x

65–85 4 75 300

85–105 5 95 475

105–125 13 115 1495

125–145 20 135 2700

145–165 14 155 2170

165–185 8 175 1400

185–205 4 195 780 --------------------------------- Total 68 Σf x = 9320
1. Mean
$\overset{ˉ}{x} = \frac{Σ f x}{Σ f} = \frac{9320}{68} = 137.0588 \approx 137.06 units$
1. Median
$N = 68$ . $N / 2 = 34.Cumulative frequencies:$

65–85: 4

85–105: 9

105–125: 22

125–145: 42

← median class (since 34 lies in this class)

Median class: $125 - 145$ .
$l = 125, h = 20, f = 20, c.f. before = 22$

Median formula:

$Median = l + \frac{\frac{N}{2} - c.f. before}{f} \times h = 125 + \frac{34 - 22}{20} \times 20$

$= 125 + 12 = 137.0 units$
1. Mode
Modal class = class with largest frequency = $125 - 145$ with $f_{1} = 20$
$f_{0} = 13$ (preceding), $f_{2} = 14$ (succeeding), $l = 125, h = 20$

$Mode = l + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h = 125 + \frac{20 - 13}{40 - 13 - 14} \times 20$

$= 125 + \frac{7}{13} \times 20 = 125 + 10.7692 \approx 135.77 units$

Comparison / Interpretation:
Mean $\approx 137.06$ , Median $= 137.0$ , Mode $\approx 135.77$ — all three measures are very close, indicating a fairly symmetric distribution around about 136–137 units.

Q2

If the median of the distribution given below is $28.5$ , find the values of $x$ and $y$ .

Class interval — Frequency
0–10 : 5
10–20 : $x$
20–30 : 20
30–40 : 15
40–50 : $y$
50–60 : 5
Total = 60

Solution

Median is $28.5$ → it lies in class $20 - 30$ . Use median formula:

Here $N = 60$ , so $N / 2 = 30$ . Median class: $20 - 30$ with $l = 20, h = 10, f = 20$ . c.f. before = frequency up to previous class = $5 + x$ .

Median:

$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10 = 20 + \frac{25 - x}{20} \times 10 = 20 + \frac{25 - x}{2}$

So

$28.5 = 20 + \frac{25 - x}{2} \Rightarrow \frac{25 - x}{2} = 8.5 \Rightarrow 25 - x = 17 \Rightarrow x = 8.$

Now total frequencies:

$5 + x + 20 + 15 + y + 5 = 60$

Substitute $x = 8$ :

$5 + 8 + 20 + 15 + y + 5 = 60 \Rightarrow 53 + y = 60 \Rightarrow y = 7.$

Answer: $x = 8, y = 7$

Q3

A life insurance agent found the following data for distribution of ages of 100 policy holders (cumulative ‘below’ form). Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years) — Number of policy holders (Below …)
Below 20 : 2
Below 25 : 6
Below 30 : 24
Below 35 : 45
Below 40 : 78
Below 45 : 89
Below 50 : 92
Below 55 : 98
Below 60 :100

Solution

First convert to class frequencies (by differences):

20–25: 6 − 2 = 4

25–30: 24 − 6 = 18

30–35: 45 − 24 = 21

35–40: 78 − 45 = 33

40–45: 89 − 78 = 11

45–50: 92 − 89 = 3

50–55: 98 − 92 = 6

55–60:100 − 98 = 2 (And below 20 group: 2)

Total $N = 100$ . Median position $= N / 2 = 50$ . Cumulative frequencies:

Below20: 2

20–25: 2+4 = 6

25–30: 6 +18 = 24

30–35: 24 +21 = 45

35–40: 45 +33 = 78

← median class (since 50th lies here)

Median class = $35 - 40$ . Use: $l = 35, h = 5, f = 33, c.f. before = 45$

$Median = 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{5}{33} \times 5$

$= 35 + \frac{25}{33} \approx 35 + 0.7576 = 35.76 years (approx)$

Q4

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre; the data:

Length (mm) — Number of leaves
118–126 : 3
127–135 : 5
136–144 : 9
145–153 : 12
154–162 : 5
163–171 : 4
172–180 : 2

Find the median length of the leaves.
(Hint: convert to continuous classes: 117.5–126.5, 126.5–135.5, …, 171.5–180.5.)

Solution

Use continuous class limits (class-width $h = 9$ ):

117.5–126.5 : 3 (cf = 3)

126.5–135.5 : 5 (cf = 8)

135.5–144.5 : 9 (cf = 17)

144.5–153.5 :12 (cf = 29)

← median class (since N/2 = 20)

153.5–162.5 : 5 (cf = 34)

162.5–171.5 : 4 (cf = 38)

171.5–180.5 : 2 (cf = 40)

$N = 40, N / 2 = 20.$ Median class: $144.5 - 153.5$
$l$ c.f. before $= 17$

$Median = l + \frac{\frac{N}{2} - c.f. before}{f} \times h = 144.5 + \frac{20 - 17}{12} \times 9$

$= 144.5 + \frac{3}{12} \times 9 = 144.5 + 2.25 = 146.75 mm$

Q5

The following table gives the distribution of the life time of 400 neon lamps. Find the median life time.

Life time (hours) — Number of lamps
1500–2000 : 14
2000–2500 : 56
2500–3000 : 60
3000–3500 : 86
3500–4000 : 74
4000–4500 : 62
4500–5000 : 48

Solution

Total $N = 400, N / 2 = 200$ . Cumulative frequencies:

1500–2000 : 14

2000–2500 : 14+56 = 70

2500–3000 : 70+60 = 130

3000–3500 : 130+86 = 216

← median class (since 200th lies here)

Median class = $3000 - 3500$ $l = 3000, h = 500, f = 86, c.f. before = 130$

$Median = 3000 + \frac{200 - 130}{86} \times 500 = 3000 + \frac{70}{86} \times 500$

$= 3000 + 0.8139535 \times 500$

$= 3000 + 406.9767 \approx 3406.98 hours$

Answer can be rounded: $3407 hours (approx)$

Q6

100 surnames were picked; frequency distribution of number of letters:

Number of letters — Number of surnames
1–4 : 6
4–7 : 30
7–10 : 40
10–13 : 16
13–16 : 4
16–19 : 4

Determine the median number of letters and find the mean number of letters.

Solution
1. Median
$N = 100, N / 2 = 50.$ Cumulative frequencies:

1–4 : 6

4–7 : 6+30 = 36

7–10: 36+40 = 76

← median class (50th lies here)

Use continuous limits with class-width $h = 3$ (e.g. 0.5–4.5, 4.5–7.5, 7.5–10.5 …). Median class interval in continuous form = $7.5 - 10.5$ So $l = 7.5, h = 3, f = 40,$ c.f. before = 36.

$Median = 7.5 + \frac{50 - 36}{40} \times 3 = 7.5 + \frac{14}{40} \times 3 = 7.5 + 1.05 = 8.55 letters (approx)$
1. Mean
Class-marks $x$ : 2.5, 5.5, 8.5, 11.5, 14.5, 17.5
Frequencies $f$ : 6, 30, 40, 16, 4, 4

Compute $f \cdot x$ :

2.5·6 = 15

5.5·30 = 165 (total 180)

8.5·40 = 340 (total 520)

11.5·16 = 184 (total 704)

14.5·4 = 58 (total 762)

17.5·4 = 70 (total 832)

Σf x = 832

$\overset{ˉ}{x} = \frac{Σ f x}{Σ f} = \frac{832}{100} = 8.32 letters$
October 12, 2025
Exercise-13.2, Class 10th, Maths, Chapter 13, NCERT

Question 1

The following table shows the ages of the patients admitted in a hospital during a year:

Age (years) 5–15 15–25 25–35 35–45 45–55 55–65

No. of patients 6 11 21 23 14 5

Find the mean and mode of the data. Compare the two measures.

Solution:

Total frequency (Σf) = 6 + 11 + 21 + 23 + 14 + 5 = 80

Class marks (x): 10, 20, 30, 40, 50, 60

f × x = 60 + 220 + 630 + 920 + 700 + 300 = 2830

Mean = Σ(fx) / Σf = 2830 / 80 = 35.38 years

Modal class = 35–45
l = 35, h = 10, f₁ = 23, f₀ = 21, f₂ = 14

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 35 + [(23 − 21) / (46 − 21 − 14)] × 10
= 35 + (2 / 11) × 10 = 36.82 years

Interpretation:
Mean = 35.38 years, Mode = 36.82 years.
Both are close, showing most patients are around 35–37 years old.

Question 2

Lifetimes (in hours) of 225 electrical components:

Lifetime (hours) 0–20 20–40 40–60 60–80 80–100 100–120

Frequency 10 35 52 61 38 29

Find the modal lifetime.

Solution:

Modal class = 60–80
l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 60 + [(61 − 52) / (122 − 52 − 38)] × 20
= 60 + (9 / 32) × 20
= 60 + 5.625 = 65.63 hours

Question 3

Monthly household expenditure of 200 families:

Expenditure (₹) 1000–1500 1500–2000 2000–2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000

No. of families 24 40 33 28 30 22 16 7

Find the mode and mean monthly expenditure.

Solution:

Modal class = 1500–2000
l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 1500 + [(40 − 24) / (80 − 24 − 33)] × 500
= 1500 + (16 / 23) × 500 = 1500 + 347.83 = ₹1847.83

Class marks (x): 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750

Σ(fx) = 532500, Σf = 200

Mean = Σ(fx)/Σf = 532500 / 200 = ₹2662.50

Interpretation:
Mean (₹2662.50) > Mode (₹1847.83), showing the data is right-skewed (some families spend much more).

Question 4

Teacher–student ratio (students per teacher):

Students per teacher 15–20 20–25 25–30 30–35 35–40 40–45 45–50 50–55

No. of states/U.T. 3 8 9 10 3 0 0 2

Find the mode and mean of this data.

Solution:

Modal class = 30–35
l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3

Mode = 30 + [(10 − 9) / (20 − 9 − 3)] × 5
= 30 + (1 / 8) × 5 = 30.63

Class marks (x): 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5

Σ(fx) = 1022.5, Σf = 35

Mean = 1022.5 / 35 = 29.21

Interpretation:
Mean = 29.21, Mode = 30.63 → Both indicate roughly 30 students per teacher.

Question 5

Runs scored by top batsmen in one-day internationals:

Runs scored 3000–4000 4000–5000 5000–6000 6000–7000 7000–8000 8000–9000 9000–10000 10000–11000

No. of batsmen 4 18 9 7 6 3 1 1

Find the mode of the data.

Solution:

Modal class = 4000–5000
l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9

Mode = 4000 + [(18 − 4) / (36 − 4 − 9)] × 1000
= 4000 + (14 / 23) × 1000
= 4000 + 608.7 = 4608.7 runs

Question 6

Number of cars passing a spot during 100 three-minute intervals:

Number of cars 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80

Frequency 7 14 13 12 20 11 15 8

Find the mode.

Solution:

Modal class = 40–50
l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11

Mode = 40 + [(20 − 12) / (40 − 12 − 11)] × 10
= 40 + (8 / 17) × 10
= 40 + 4.71 = 44.71 cars

Final Answers Summary

Question Mean Mode

1 35.38 36.82

2 — 65.63

3 2662.50 1847.83

4 29.21 30.63

5 — 4608.7

6 — 44.71

October 12, 2025
Exercise-13.1, Class 10th, Maths, Chapter 13, NCERT

Q1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants: 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14
Number of houses: 1, 2, 1, 5, 6, 2, 3

Solution (Direct method — numbers are small so direct is simplest).
Class marks $x_{i}$ :

$1, 3, 5, 7, 9, 11, 13$

Frequencies $f_{i}$ :

$1, 2, 1, 5, 6, 2, 3$

Compute $f_{i} x_{i}$ :

$1 \cdot 1 = 1, 2 \cdot 3 = 6, 1 \cdot 5 = 5, 5 \cdot 7 = 35, 6 \cdot 9 = 54, 2 \cdot 11 = 22, 3 \cdot 13 = 39$

$Σ f_{i} x_{i} = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162$
$Σ f_{i} = 20.$

Mean $\overset{ˉ}{x} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{162}{20} = 8.1$

Answer: $8.1 plants per house$

Q2.
Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (₹): 500–520, 520–540, 540–560, 560–580, 580–600
Number of workers: 12, 14, 8, 6, 10

Find the mean daily wages.

Solution (Direct / class-mark method).
Class marks $x_{i}$ : $510, 530, 550, 570, 590$ .
Frequencies $f_{i}$ : $12, 14, 8, 6, 10$ .

Compute $f_{i} x_{i}$ :

$12 \cdot 510 = 6120, 14 \cdot 530 = 7420, 8 \cdot 550 = 4400, 6 \cdot 570 = 3420, 10 \cdot 590 = 5900$

$Σ f_{i} x_{i} = 6120 + 7420 + 4400 + 3420 + 5900 = 27260$
$Σ f_{i} = 50.$

Mean $\overset{ˉ}{x} = \frac{27260}{50} = 545.2$

Answer: $₹ 545.20 per day (mean)$

Q3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency $f$ .

Daily pocket (₹): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Number of children: 7, 6, 9, 13, $f$ , 5, 4

Solution.
Class marks $x_{i}$ : $12, 14, 16, 18, 20, 22, 24$ . Frequencies as above.

Compute known part of $Σ f_{i} x_{i}$ :

$7 \cdot 12 = 84, 6 \cdot 14 = 84, 9 \cdot 16 = 144, 13 \cdot 18 = 234, 5 \cdot 22 = 110, 4 \cdot 24 = 96$

Sum of known products $= 84 + 84 + 144 + 234 + 110 + 96 = 752$

Term with unknown frequency: $f \cdot 20 = 20 f$

Total frequency $= 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$

Given mean $= 18$ , so

$\frac{752 + 20 f}{44 + f} = 18.$

Solve:

$752 + 20 f = 18 (44 + f) = 792 + 18 f$ $20 f - 18 f = 792 - 752 \Rightarrow 2 f = 40 \Rightarrow f = 20.$

Answer: $f = 20$

Q4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women.

Number of heartbeats per minute: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86
Number of women: 2, 4, 3, 8, 7, 4, 2

Solution (Direct method with class marks).
Class marks $x_{i}$ : $66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5$
Frequencies $f_{i}$ : $2, 4, 3, 8, 7, 4, 2$

Compute $f_{i} x_{i}$ :

$2 \cdot 66.5 = 133.0 4 \cdot 69.5 = 278.0 3 \cdot 72.5 = 217.5 8 \cdot 75.5$

$= 604.0 7 \cdot 78.5 = 549.5 4 \cdot 81.5 = 326.0 2 \cdot 84.5 = 169.0$

$Σ f_{i} x_{i} = 133 + 278 + 217.5 + 604 + 549.5 + 326 + 169 = 2277.0$
$Σ f_{i} = 30.$

Mean $\overset{ˉ}{x} = \frac{2277.0}{30} = 75.9$

Answer: $75.9 heartbeats per minute$

Q5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes: 50–52, 53–55, 56–58, 59–61, 62–64
Number of boxes: 15, 110, 135, 115, 25

Find the mean number of mangoes kept in a packing box.

Solution (Direct method).
Class marks $x_{i}$ : $51, 54, 57, 60, 63$
Frequencies $f_{i}$ : $15, 110, 135, 115, 25$

Compute $f_{i} x_{i}$ :

$15 \cdot 51 = 765 110 \cdot 54 = 5940 135 \cdot 57 = 7695 115 \cdot 60 = 6900 25 \cdot 63 = 1575$

$Σ f_{i} x_{i} = 765 + 5940 + 7695 + 6900 + 1575 = 22875$
$Σ f_{i} = 15 + 110 + 135 + 115 + 25 = 400$

Mean $\overset{ˉ}{x} = \frac{22875}{400} = 57.1875$

Answer: $57.1875 \approx 57.19 mangoes per box.$

(Method used: direct class-mark method — frequencies are large but class marks are simple.)

Q6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (₹): 100–150, 150–200, 200–250, 250–300, 300–350
Number of households: 4, 5, 12, 2, 2

Find the mean daily expenditure on food.

Solution (Direct method).
Class marks $x_{i}$ : $125, 175, 225, 275, 325$ .
Frequencies $f_{i}$ : $4, 5, 12, 2, 2$

Compute $f_{i} x_{i}$ :

$4 \cdot 125 = 500, 5 \cdot 175 = 875, 12 \cdot 225 = 2700, 2 \cdot 275 = 550, 2 \cdot 325 = 650$

$Σ f_{i} x_{i} = 500 + 875 + 2700 + 550 + 650 = 5275$
$Σ f_{i} = 25$

Mean $\overset{ˉ}{x} = \frac{5275}{25} = 211.0$

Answer: $₹ 211.00 per day (mean)$

Q7.
To find out the concentration of SO $_{2}$ in the air (in ppm), the data was collected for 30 localities:

Concentration (ppm): 0.00–0.04, 0.04–0.08, 0.08–0.12, 0.12–0.16, 0.16–0.20, 0.20–0.24
Frequencies: 4, 9, 9, 2, 4, 2

Find the mean concentration of SO $_{2}$ .

Solution (Direct method with small decimals).
Class marks $x_{i}$ : $0.02, 0.06, 0.10, 0.14, 0.18, 0.22$
Frequencies $f_{i}$ : $4, 9, 9, 2, 4, 2$

Compute $f_{i} x_{i}$ :

$4 \cdot 0.02 = 0.08 9 \cdot 0.06 = 0.54 9 \cdot 0.10 = 0.90 2 \cdot 0.14 = 0.28 4 \cdot 0.18 = 0.72 2 \cdot 0.22 = 0.44$

$Σ f_{i} x_{i} = 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96$
$Σ f_{i} = 30$

Mean $\overset{ˉ}{x} = \frac{2.96}{30} = 0.098666 \dots$

Answer: $0.09867 ppm (approx)$

Q8.
A class teacher has the following absentee record of 40 students for the whole term. Find the mean number of days a student was absent.

Number of days: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40
Number of students: 11, 10, 7, 4, 4, 3, 1

Solution (Direct method).
Class marks $x_{i}$ : $3, 8, 12, 17, 24, 33, 39$
Frequencies $f_{i}$ : $11, 10, 7, 4, 4, 3, 1$

Compute $f_{i} x_{i}$ :

$11 \cdot 3 = 33, 10 \cdot 8 = 80, 7 \cdot 12 = 84, 4 \cdot 17 = 68, 4 \cdot 24 = 96, 3 \cdot 33 = 99, 1 \cdot 39 = 39$

$Σ f_{i} x_{i} = 33 + 80 + 84 + 68 + 96 + 99 + 39 = 499$
$Σ f_{i} = 40$

Mean $\overset{ˉ}{x} = \frac{499}{40} = 12.475$

Answer: $12.475 days (mean) \approx 12.48 days$

Q9.
The following table gives the literacy rate (in %) of 35 cities. Find the mean literacy rate.

Literacy rate (%): 45–55, 55–65, 65–75, 75–85, 85–95
Number of cities: 3, 10, 11, 8, 3

Solution (Direct method).
Class marks $x_{i}$ : $50, 60, 70, 80, 90$
Frequencies $f_{i}$ : $3, 10, 11, 8, 3$

Compute $f_{i} x_{i}$ :

$3 \cdot 50 = 150, 10 \cdot 60 = 600, 11 \cdot 70 = 770, 8 \cdot 80 = 640, 3 \cdot 90 = 270$

$Σ f_{i} x_{i} = 150 + 600 + 770 + 640 + 270 = 2430$
$Σ f_{i} = 35.$

Mean $\overset{ˉ}{x} = \frac{2430}{35} = 69.428571 \dots$

Answer: $69.42857 % \approx 69.43 %$

October 12, 2025
Exercise-12.1, Class 10th, Maths, Chapter 12, NCERT

Q1.

Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:
Volume of each cube = 64 cm³
⇒ side = ∛64 = 4 cm

Dimensions of cuboid = 8 cm × 4 cm × 4 cm

Surface area = 2(lb + bh + hl)
= 2(8×4 + 4×4 + 8×4)
= 2(32 + 16 + 32)
= 2×80 = 160 cm²

Q2.

A vessel is a hollow hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:
Radius (r) = 7 cm
Height of cylinder (h) = 13 − 7 = 6 cm

Inner surface area = 2πrh + 2πr² + πr²
= 2πrh + 3πr²
= 2×(22/7)×7×6 + 3×(22/7)×7×7
= 264 + 462 = 726 cm²

Q3.

A toy is in the form of a cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:
r = 3.5 cm, h = 15.5 − 3.5 = 12 cm
Slant height (l) = √(h² + r²) = √(12² + 3.5²) = 12.5 cm

Total surface area = πrl + 2πr²
= (22/7)×3.5×12.5 + 2×(22/7)×3.5×3.5
= 137.5 + 77 = 214.5 cm²

Q4.

A cubical block of side 7 cm is surmounted by a hemisphere. Find the greatest diameter of the hemisphere and the surface area of the solid.

Solution:
Greatest diameter = side of cube = 7 cm
Radius r = 3.5 cm

Surface area = 6a² − a² + 2πr²
= 5a² + 2πr²
= 5×7² + 2×(22/7)×3.5²
= 245 + 77 = 322 cm²

Q5.

A hemispherical depression is cut out from one face of a cubical block of edge l, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:
Edge = a = l
Radius = a/2

Surface area = 6a² − πr² + 2πr²
= 6a² + πr²
= 6a² + π(a²/4)
= a²(6 + π/4)

∴ Surface area = 6a² + (πa² / 4)

Q6.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter is 5 mm. Find its surface area.

Solution:
r = 2.5 mm, h = 14 − 2r = 9 mm

Surface area = 2πrh + 4πr²
= 2×(22/7)×2.5×9 + 4×(22/7)×2.5²
= (990/7) + (550/7)
= (1540/7) = 220 mm²

Q7.

A tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of the cylindrical part are 2.1 m and 4 m respectively. The slant height of the top is 2.8 m. Find the area of the canvas required to make the tent, excluding the base. Also, find the cost of the canvas at ₹500 per m².

Solution:
r = 2 m, h = 2.1 m, l = 2.8 m

Area = 2πrh + πrl
= 2π×2×2.1 + π×2×2.8
= 8.4π + 5.6π = 14π
= 14×3.14 = 43.96 ≈ 44 m²

Cost = 44×500 = ₹22,000

Q8.

From a solid cylinder (height 2.4 cm, diameter 1.4 cm), a conical cavity of the same height and diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution:
r = 0.7 cm, h = 2.4 cm
l = √(r² + h²) = √(0.7² + 2.4²) = 2.5 cm

Total surface area = 2πrh + πrl + πr²
= 2×(22/7)×0.7×2.4 + (22/7)×0.7×2.5 + (22/7)×0.7²
= 3.36π + 1.75π + 0.49π = 5.6π
= 5.6×3.14 = 17.6 ≈ 18 cm²

Q9.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder whose height is 10 cm and base radius 3.5 cm. Find the total surface area of the article.

Solution:
r = 3.5 cm, h = 10 cm

Total surface area = 2πrh + 4πr²
= 2×(22/7)×3.5×10 + 4×(22/7)×3.5²
= 70π + 49π = 119π
= 119×(22/7) = 374 cm²

October 12, 2025
Exercise-9.1, Class 10th, Maths, Chapter 9, NCERT

Q 1.

Question:
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the rope makes an angle 30° with the ground.

Solution:
Let the height of the pole = AB m, and rope = AC = 20 m.
∠ACB = 30°, right Δ ABC.

$\sin 30 ° = \frac{A B}{A C} \Rightarrow \frac{1}{2} = \frac{A B}{20} \Rightarrow A B = 10 m .$

Height of the pole = 10 m.

Q 2.

Question:
A tree breaks due to storm and the broken part bends so that the top touches the ground making 30° with it. The distance between the foot of the tree and the point where top touches ground = 8 m. Find the height of the tree.

Solution:
Let the tree break at point B, AB = h m, BD = 8 m (distance on ground).
Broken part AD touches ground, ∠ADB = 30°, so AD = hypotenuse = BD / cos 30°.

$A D = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}} m .$

Vertical part = AB = AD sin 30° = $\frac{16}{\sqrt{3}} \times \frac{1}{2} = \frac{8}{\sqrt{3}}$

Total height = AB + BD = $\frac{8}{\sqrt{3}} +$ (?? Wait) hold. Actually BD is ground distance, so total height of tree = AB + BC where BC = AD sin 30°, etc. Correcting:
Let unbroken part = BE and broken part = ED touches ground.
If DE makes 30° with ground and BD = 8 m = DE cos 30°.

$D E = \frac{8}{\cos 30 °} = \frac{8}{\sqrt{3} / 2} = \frac{16}{\sqrt{3}} m .$

Vertical BE = DE sin 30° = $\frac{16}{\sqrt{3}} \times \frac{1}{2} = \frac{8}{\sqrt{3}}$

Hence total height = BE + ED sin 30° = $\frac{8}{\sqrt{3}} + \frac{8}{\sqrt{3}}$ = $\frac{16}{\sqrt{3}}$ =9.24 m (approx).

Height of tree = 9.24 m (approx).

Q 3.

Question:
Two slides—one for small children (height 1.5 m, 30° incline) and one for elders (height 3 m, 60° incline). Find the length of each slide.

Solution:

(i) For small children:

$\sin 30 ° = \frac{1.5}{L_{1}} \Rightarrow \frac{1}{2} = \frac{1.5}{L_{1}} \Rightarrow L_{1} = 3 m .$

(ii) For elder children:

$\sin 60 ° = \frac{3}{L_{2}} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{L_{2}} \Rightarrow L_{2} = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2 \sqrt{3} = 3.46 m (a p p r o x) .$

✅ Lengths: 3 m and 3.46 m (approx).

Q 4.

Question:
Angle of elevation of top of tower = 30°, distance from tower = 30 m. Find height.

Solution:

$\tan 30 ° = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10 \sqrt{3} = 17.32 m .$

✅ Height of tower = 17.32 m.

Q 5.

Question:
Kite height = 60 m, inclination of string = 60°. Find string length (no slack).

Solution:

$\sin 60 ° = \frac{60}{l} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{l} \Rightarrow l = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = 40 \sqrt{3} = 69.28 m .$

✅ Length of string = 69.3 m (approx).

Q 6.

Question:
A 1.5 m tall boy is standing from a 30 m building; angle of elevation from 30° to 60° as he walks towards building. Find distance walked.

Solution:
Let initial distance = x m, final distance = y m.

At 30°:

$\tan 30 ° = \frac{30 - 1.5}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{x} \Rightarrow x = 28.5 \sqrt{3} = 49.35.$

At 60°:

$\tan 60 ° = \frac{28.5}{y} \Rightarrow \sqrt{3} = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}} = 16.46.$

Distance walked = x − y = 49.35 − 16.46 = 32.89 m.

He walked ≈ 32.9 m.

Q 7.

Question:
From a point on ground, angles of elevation of bottom and top of transmission tower fixed atop a 20 m building are 45° and 60° respectively. Find height of tower.

Solution:
Let tower = h m. Distance from building = x m.

From top:

$\tan 60 ° = \frac{20 + h}{x} \Rightarrow \sqrt{3} = \frac{20 + h}{x} \Rightarrow x = \frac{20 + h}{\sqrt{3}} .$

From bottom:

$\tan 45 ° = \frac{20}{x} \Rightarrow 1 = \frac{20}{x} \Rightarrow x = 20.$

Equate x’s:

$20 = \frac{20 + h}{\sqrt{3}} \Rightarrow 20 \sqrt{3} = 20 + h$

$\Rightarrow h = 20 (\sqrt{3} - 1) = 20 (1.732 - 1) = 14.64 m .$

Height of tower = 14.64 m.

Q 8.

Question:
A statue 1.6 m tall stands on a pedestal. From a point on ground, angle of elevation of top of statue = 60°, and of top of pedestal = 45°. Find height of pedestal.

Solution:
Let pedestal height = h m, distance from point = x m.

From top of statue:

$\tan 60 ° = \frac{h + 1.6}{x} \Rightarrow \sqrt{3} = \frac{h + 1.6}{x} .$

From top of pedestal:

$\tan 45 ° = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h .$

Substitute:

$\sqrt{3} = \frac{h + 1.6}{h} \Rightarrow \sqrt{3} h = h + 1.6 \Rightarrow h (\sqrt{3} - 1) = 1.6$

$\Rightarrow h = \frac{1.6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = 0.8 (√ 3 + 1) = 0.8 \times 2.732 = 2.19 m .$

Height of pedestal = 2.19 m (approx).

Q 9.

Question:
Angle of elevation of top of building from foot of tower = 30°, and of top of tower from foot of building = 60°. Tower height = 50 m. Find height of building.

Solution:
Let distance between tower and building = x m, height of building = h m.

From foot of tower:

$\tan 30 ° = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow x = h \sqrt{3} .$

From foot of building:

$\tan 60 ° = \frac{50}{x} \Rightarrow \sqrt{3} = \frac{50}{x} \Rightarrow x = \frac{50}{\sqrt{3}}$

Equate:

$h \sqrt{3} = \frac{50}{\sqrt{3}} \Rightarrow h = \frac{50}{3} = 16.67 m .$

Height of building = 16.67 m.

Q 10.

Question:
Two equal poles opposite each other across a road 80 m wide. From a point between them, angles of elevation are 60° and 30°. Find height and distances.

Solution:
Let distance from point to first pole = x m, so to other pole = 80 − x m, height = h m.

From 1st pole (60°):

$\tan 60 ° = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x √ 3.$

From 2nd pole (30°):

$\tan 30 ° = \frac{h}{80 - x} \Rightarrow \frac{1}{√ 3} = \frac{h}{80 - x} \Rightarrow h = \frac{80 - x}{√ 3} .$

Equate:

$x √ 3 = \frac{80 - x}{√ 3} \Rightarrow 3 x = 80 - x \Rightarrow 4 x = 80 \Rightarrow x = 20.$

So $h = 20 √ 3 = 34.64 m$

Height of each pole = 34.64 m, Distances = 20 m and 60 m.

Q 11.

Question:
A TV tower stands on one bank of a canal. From a point on the opposite bank, angle of elevation = 60°. From another point 20 m away on same line, angle of elevation = 30°. Find height of tower and width of canal.

Solution:
Let height = h m, width of canal = x m.

From near point (60°):

$\tan 60 ° = \frac{h}{x} \Rightarrow √ 3 = \frac{h}{x} \Rightarrow h = x √ 3.$

From far point (30°):

$\tan 30 ° = \frac{h}{x + 20} \Rightarrow \frac{1}{√ 3} = \frac{h}{x + 20} \Rightarrow h = \frac{x + 20}{√ 3} .$

Equate:

$x √ 3 = \frac{x + 20}{√ 3} \Rightarrow 3 x = x + 20 \Rightarrow 2 x = 20 \Rightarrow x = 10$

Then $h = 10 √ 3 = 17.32 m$

Height = 17.32 m, Width = 10 m.

Q 12.

Question:
From top of 7 m building, angle of elevation of top of tower = 60°, angle of depression of foot = 45°. Find height of tower.

Solution:
Let tower height = h m, distance = x m.

From angle of depression 45°:

$\tan 45 ° = \frac{7}{x} \Rightarrow x = 7.$

From angle of elevation 60°:

$\tan 60 ° = \frac{h - 7}{x} \Rightarrow √ 3 = \frac{h - 7}{7} \Rightarrow h - 7 = 7 √ 3 \Rightarrow h = 7 (1 + √ 3) = 7 (2.732) = 19.12 m .$

Height of tower = 19.12 m.

Q 13.

Question:
From a 75 m high lighthouse, angles of depression of two ships are 30° and 45°. Find distance between ships on same side of lighthouse.

Solution:
Let distances from lighthouse = x₁ (for 45°) and x₂ (for 30°).

$\tan 45 ° = \frac{75}{x_{1}} \Rightarrow x_{1} = 75.$

$\tan 30 ° = \frac{75}{x_{2}} \Rightarrow \frac{1}{√ 3} = \frac{75}{x_{2}} \Rightarrow x_{2} = 75 √ 3 = 129.9$

Distance between ships = 54.9 m.

Q14 (redo)

Question (restated briefly).
A 1.2 m tall girl watches a balloon flying horizontally at height 88.2 m above the ground. From her eyes the angle of elevation of the balloon is first $60^{\circ}$ and later becomes $30^{\circ}$ . Find the horizontal distance the balloon travelled during this time.

Solution.

Height of balloon above the girl’s eyes

$H = 88.2 - 1.2 = 87.0 m$

Let the horizontal distances of the balloon from the girl’s vertical line at the times when the elevation angles are $60^{\circ}$ and $30^{\circ}$ be $x_{1}$ and $x_{2}$ respectively.

Using $\tan θ = \frac{opposite}{adjacent}$

For $θ = 60^{\circ}$ (first position),

$\tan 60^{\circ} = \sqrt{3} = \frac{H}{x_{1}} \Rightarrow x_{1} = \frac{H}{\sqrt{3}} = \frac{87}{\sqrt{3}} = 29 \sqrt{3} (m) .$

For $θ = 30^{\circ}$ (later),

$\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{H}{x_{2}} \Rightarrow x_{2} = H \sqrt{3} = 87 \sqrt{3} (m) .$

Distance travelled by the balloon $= x_{2} - x_{1}$ :

$x_{2} - x_{1} = 87 \sqrt{3} - 29 \sqrt{3} = 58 \sqrt{3} m .$

Numeric approximation ( $\sqrt{3} \approx 1.732$ ):

$58 \sqrt{3} \approx 58 \times 1.732 = 100.46 m (approx).$

Answer: $58 \sqrt{3} m$ $\approx 100.46 m$

Q15 (redo)

Question (restated briefly).
From the top of a tower a man sees a car on the straight highway at an angle of depression $30^{\circ}$ . The car moves toward the foot of the tower; 6 seconds later the angle of depression becomes $60^{\circ}$ . How long (from that second observation) will the car take to reach the foot of the tower?

Solution.

Let tower height = $h$ . Let horizontal distances from the foot of the tower to the car at the two observations be $x_{1}$ (when angle $30^{\circ}$ ) and $x_{2}$ (when angle $60^{\circ}$ ).

Angle of depression equals corresponding angle of elevation downward, so:

For $30^{\circ}$ :

$\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{x_{1}} \Rightarrow x_{1} = h \sqrt{3} .$

For $60^{\circ}$ :

$\tan 60^{\circ} = \sqrt{3} = \frac{h}{x_{2}} \Rightarrow x_{2} = \frac{h}{\sqrt{3}} .$

Distance the car covers in 6 s:

$Δ x = x_{1} - x_{2} = h \sqrt{3} - \frac{h}{\sqrt{3}} = h ⁣ (\frac{3 - 1}{\sqrt{3}}) = \frac{2 h}{\sqrt{3}} .$

So its speed $v = \frac{Δ x}{6} = \frac{2 h}{6 \sqrt{3}} = \frac{h}{3 \sqrt{3}}$

Distance remaining to the foot of the tower at the second observation = $x_{2} = \frac{h}{\sqrt{3}}$

Time to reach foot from that instant:

$t = \frac{x_{2}}{v} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3 \sqrt{3}}} = 3 seconds .$

Answer: The car will reach the foot of the tower in 3 seconds from the second observation.

October 10, 2025
Exercise 8.1, Class 10th, Maths, Chapter 8, NCERT

1. In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A
(ii) sin C, cos C

Answer 1.
AC (hypotenuse) = √(AB² + BC²) = √(24² + 7²) = √(576 + 49) = √625 = 25.

(i) For angle A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = 25.
$\sin A = \frac{7}{25}, \cos A = \frac{24}{25} .$

(ii) For angle C: opposite = AB = 24, adjacent = BC = 7, hypotenuse = 25.
$\sin C = \frac{24}{25}, \cos C = \frac{7}{25}$

2. In Fig. 8.13, find $\tan P - \cot R .$

Answer 2.
If Fig. 8.13 is the standard right triangle with PQ = 4, QR = 3, PR = 5 (a common 3–4–5 right triangle), then
$\tan P = \frac{Q R}{P Q} = \frac{3}{4}$ and $\cot R = \frac{Q R}{P Q} = \frac{3}{4}$ , so $\tan P - \cot R = 0$

3. If $\sin A = \frac{3}{4}$ , calculate $\cos A$ and $\tan A$ .

Answer 3.
$\cos A = \sqrt{1 - \sin^{2} A} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

$\tan A = \frac{\sin A}{\cos A} = \frac{3 / 4}{\sqrt{7} / 4} = \frac{3}{\sqrt{7}} = \frac{3 \sqrt{7}}{7}$

4. Given $15 \cot A = 8$ , find $\sin A$ and $\sec A$ .

Answer 4.
$\cot A = \frac{8}{15} .$ Then $\csc^{2} A = 1 + \cot^{2} A = 1 + \frac{64}{225} = \frac{289}{225}$ , so $\csc A = \frac{17}{15}$ (positive for acute angle).
Thus $\sin A = \frac{1}{\csc A} = \frac{15}{17}$

Also $\cos A = \cot A \cdot \sin A = \frac{8}{15} \cdot \frac{15}{17} = \frac{8}{17}$ , so $\sec A = \frac{1}{\cos A} = \frac{17}{8}$

5. Given $\sec θ = \frac{13}{12}$ , calculate all other trigonometric ratios.

Answer 5.
$\cos θ = \frac{12}{13} .$

$\sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$

$\tan θ = \frac{\sin θ}{\cos θ} = \frac{5 / 13}{12 / 13} = \frac{5}{12} .$

Reciprocals: $\csc θ = \frac{13}{5}, \cot θ = \frac{12}{5} .$

6. If ∠A and ∠B are acute angles such that $\cos A = \cos B$ , then show that ∠A = ∠B.

Answer 6.
On $[0^{\circ}, 90^{\circ}]$ the cosine function is strictly decreasing and therefore one-to-one. Hence $\cos A = \cos B$ (with A and B acute) implies $A = B$ .
(Equivalently: construct right triangles with same cosine value; corresponding sides/hypotenuse ratios match, triangles are similar and the acute angles equal.)

7. If $\cot θ = \frac{7}{8}$ , evaluate:

(i) $\frac{1 + \sin θ}{1 - \sin θ}$ and $\frac{1 + \cos θ}{1 - \cos θ}$

(ii) $\cot^{2} θ .$

Answer 7.
Given $\cot θ = \frac{7}{8}$ . Let $\sin θ = s$ , $\cos θ = c$ . From $\cot = \frac{c}{s} = \frac{7}{8}$ we get (as usual) $s = \frac{8}{\sqrt{113}}$ , $c = \frac{7}{\sqrt{113}}$ (since $s^{2} + c^{2} = 1$ gives factor 113).

(i)

$\frac{1 + \sin θ}{1 - \sin θ} = \frac{1 + \frac{8}{\sqrt{113}}}{1 - \frac{8}{\sqrt{113}}} = \frac{\sqrt{113} + 8}{\sqrt{113} - 8} .$ $\frac{1 + \cos θ}{1 - \cos θ} = \frac{1 + \frac{7}{\sqrt{113}}}{1 - \frac{7}{\sqrt{113}}} = \frac{\sqrt{113} + 7}{\sqrt{113} - 7} .$

(You can rationalize these if you prefer a form without roots in denominators.)

(ii) $\cot^{2} θ = (\frac{7}{8})^{2} = \frac{49}{64} .$

8. If $3 \cot A = 4$ , check whether $\frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \cos^{2} A - \sin^{2} A$

Answer 8.
From $3 \cot A = 4$ we get $\cot A = \frac{4}{3}$ so $\tan A = \frac{3}{4}$ . Compute LHS:

$\frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \frac{1 - (3 / 4)^{2}}{1 + (3 / 4)^{2}} = \frac{1 - 9 / 16}{1 + 9 / 16} = \frac{7 / 16}{25 / 16} = \frac{7}{25} .$

RHS: $\cos^{2} A - \sin^{2} A = \cos 2 A$ . Using triangle with legs 3 and 4 (hypotenuse 5): $\sin A = \frac{3}{5}, \cos A = \frac{4}{5}$ Then $\cos^{2} A - \sin^{2} A = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$

So both sides equal $\frac{7}{25}$ . The equality holds (this is a standard identity: $\cos 2 A = \frac{1 - \tan^{2} A}{1 + \tan^{2} A}$

9. In triangle ABC, right-angled at B, if $\tan A = \frac{1}{3}$ find the value of:

(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C - \sin A \sin C$

Answer 9.
In a right triangle with right angle at B, $A + C = 90^{\circ}$

(i) $\sin A \cos C + \cos A \sin C = \sin (A + C) = \sin 90^{\circ} = 1.$

(ii) $\cos A \cos C - \sin A \sin C = \cos (A + C) = \cos 90^{\circ} = 0$

10. In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

Answer 10.
Let $Q R = x$ . Then $P R = 25 - x$ . Right triangle gives $P Q^{2} + Q R^{2} = P R^{2}$
$5^{2} + x^{2} = (25 - x)^{2} = 625 - 50 x + x^{2} .$ Cancel $x^{2}$ : $25 = 625 - 50 x$ ⇒ $50 x = 600$ ⇒ $x = 12.$

So $Q R = 12, P R = 13.$ For angle P: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13.

$\sin P = \frac{12}{13}, \cos P = \frac{5}{13}, \tan P = \frac{12}{5} .$

11. State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A = \frac{12}{5}$ for some value of angle A.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
(iv) $\cot A$ is the product of cot and A.
(v) $\sin θ = \frac{4}{3}$ for some angle $θ$ .

Answer 11.
(i) False. $\tan A$ can be greater than 1 (e.g. $A = 60^{\circ}$ gives $\tan 60^{\circ} = \sqrt{3} > 1$ .
(ii) True. $\sec A = \frac{12}{5}$ means $\cos A = \frac{5}{12}$ , which is a valid cosine value for some acute angle (so such an angle exists).
(iii) False. $\cos A$ denotes cosine of A. Cosecant is written $\csc A$
(iv) False. $\cot A$ means cotangent of A, not the product “cot × A.”
(v) False. For real angles $∣ \sin θ ∣ \leq 1.$

$\frac{4}{3} > 1$ , so $\sin θ = \frac{4}{3}$ is impossible for a real angle.

October 7, 2025
Exercise-4.3, Class 10th, Maths, Chapter 4, NCERT

Q1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

(i) $2 x^{2} - 3 x + 5 = 0$
Discriminant $Δ = b^{2} - 4 a c = (- 3)^{2} - 4 (2) (5) = 9 - 40 = - 31 < 0$ .
Nature: $Δ < 0$ ⇒ no real roots (two complex conjugate roots).

(ii) $3 x^{2} - 4 \sqrt{3} x + 4 = 0$
$Δ = (- 4 \sqrt{3})^{2} - 4 (3) (4) = 48 - 48 = 0.$
Nature: $Δ = 0$ ⇒ two equal real roots (a repeated root).
Root: $x = \frac{- b}{2 a} = \frac{4 \sqrt{3}}{6} = \frac{2 \sqrt{3}}{3} .$

(iii) $2 x^{2} - 6 x + 3 = 0$
$Δ = (- 6)^{2} - 4 (2) (3) = 36 - 24 = 12 > 0$

Nature: $Δ > 0$ ⇒ two distinct real roots.
Roots: $x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2 \sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} .$

Q2. Find the values of $k$ so that the quadratic has two equal roots.

(i) $2 x^{2} + k x + 3 = 0$ .
For equal roots $Δ = 0 \Rightarrow k^{2} - 4 (2) (3) = 0 \Rightarrow k^{2} - 24 = 0.$
So $k = \pm \sqrt{24} = \pm 2 \sqrt{6}$

(ii) $k x (x - 2) + 6 = 0$ .
Expand: $k x^{2} - 2 k x + 6 = 0$ . Here $a = k, b = - 2 k, c = 6$ . For equal roots $Δ = 0$

$(- 2 k)^{2} - 4 (k) (6) = 0 \Rightarrow 4 k^{2} - 24 k = 4 k (k - 6) = 0.$

This gives $k = 0$ or $k = 6$ .
But $k = 0$ makes the equation $6 = 0$ (not a quadratic), so discard $k = 0$ .
Valid value: $k = 6$ (gives a quadratic with a repeated root).

Q3. Mango grove: length = twice breadth, area $= 800 m^{2}$ . Find length and breadth.

Let breadth $= b$ . Length $= 2 b$ . Area: $2 b^{2} = 800 \Rightarrow b^{2} = 400 \Rightarrow b = 20$ (positive).
Length $= 2 b = 40$
Answer: Breadth $20 m$ , length $40 m$ .

Q4. Ages problem: Sum of ages $= 20$ . Four years ago product of ages was $48$ . Is this possible? If so, find present ages.

Let present ages be $x$ and $20 - x$ . Four years ago their ages were $x - 4$ and $16 - x$ . Given:

$(x - 4) (16 - x) = 48.$ $Δ < 0$ ⇒ no real solution.
Conclusion: The situation is not possible (no pair of real ages satisfies the conditions).

Q5. Park: perimeter $= 80$ m and area $= 400 m^{2}$ . Is this possible? If so, find length and breadth.

Let length $= l$ , breadth $= b$ . From perimeter $l + b = 40$ . Area gives $l (40 - l) = 400$ .
So $- l^{2} + 40 l - 400 = 0 \Rightarrow l^{2} - 40 l + 400 = 0.$
Discriminant $Δ = 40^{2} - 4 \cdot 1 \cdot 400 = 1600 - 1600 = 0.$
$Δ = 0$ ⇒ equal roots: $l = \frac{40}{2} = 20$ . Then $b = 40 - l = 20$
Answer: Yes — the park is $20 m \times 20 m$ (a square).

October 4, 2025

Question	Mean	Mode
1	35.38	36.82
2	—	65.63
3	2662.50	1847.83
4	29.21	30.63
5	—	4608.7
6	—	44.71

Tag: Class 10th Maths NCERT Solutions

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Q1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

Q2. Find the values of kk so that the quadratic has two equal roots.

Q3. Mango grove: length = twice breadth, area =800 m2=800 m2. Find length and breadth.

Q4. Ages problem: Sum of ages =20=20. Four years ago product of ages was 4848. Is this possible? If so, find present ages.

Q5. Park: perimeter =80=80 m and area =400 m2=400 m2. Is this possible? If so, find length and breadth.

Q2. Find the values of $k$ so that the quadratic has two equal roots.

Q3. Mango grove: length = twice breadth, area $= 800 m^{2}$ . Find length and breadth.

Q4. Ages problem: Sum of ages $= 20$ . Four years ago product of ages was $48$ . Is this possible? If so, find present ages.

Q5. Park: perimeter $= 80$ m and area $= 400 m^{2}$ . Is this possible? If so, find length and breadth.