(i) (b): The upper limit of a class and the lower class of its succeeding class differ by 1.
(ii) (d) : Here, class intervals are in inclusive form. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows:

\(\text { Here, } \Sigma f_{i} \text { i.e., } N=60 \)
\(\Rightarrow \frac{N}{2}=30\)

Now, the class interval whose cumulative frequency is
just greater than 30 is 219.5 - 229.5.
\(\therefore\) Median class is 219.5 - 229.5.
(iii) (b): Clearly, the cumulative frequency of the class preceding the median class is 18
(iv) (d): Median \(=l+\left[\frac{\frac{N}{2}-c . f .}{f}\right] \times h\)
\(=219.5+\left(\frac{30-18}{26}\right) \times 10 \)
\(=219.5+\frac{12 \times 10}{26}=219.5+4.62=224.12\)
\(\therefore\) Median of the distance travelled is 224.12 km
(v) (d): We know, Mode = 3 Median - 2 Mean
\(\therefore \quad \text { Mean }=\frac{1}{2}(3 \text { Median }-\text { Mode }) \)
\(=\frac{1}{2}(672.36-223.78)=224.29 \mathrm{~km}\)

Given frequency distribution table can be drawn as:

(i) (d): Clearly, average mileage
\(=\frac{7240}{50}=144.8 \mathrm{~km} / \text { charge }\)
(ii) (b) : Since, highest frequency is IS, therefore,
modal class is 140-160.
Here, l = 140,f₁ = 18,f₀ = 12,f₂ = 13, h = 20
\(\therefore \quad \text { Mode }=140+\frac{18-12}{36-12-13} \times 20=140+\frac{6}{11} \times 20 \)
\(=140+\frac{120}{11}=140+10.91=150.91\)
(iii) (b) : Here \(\frac{N}{2}=\frac{50}{2}=25\) and the corresponding class whose cumulative frequency is just greater than
25 is 140-160.
Here, l = 140, c.f = 19, h = 20 and f= 18
\(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h\)
\(=140+\frac{25-19}{18} \times 20=140+\frac{60}{9}=146.67\)
(iv) (a) : Assumed mean method is useful in determining the mean.
(v) (a): Since, Mean = 144.S, Mode = 150.91 and Median = 146.67 and minimum of which is 144 approx, therefore manufacturer can claim the mileage for his scooter 144 km/charge.

(i) (d): Required number of persons = 9 + 6 = 15
(ii) (c): Required number of persons = 6 + 8 + 2 = 16
(iii) (a) : 50-60 is the modal class as the maximum frequency is 9.
(iv) (b) : The cumulative frequency distribution table for the given data can be drawn as :

\(\text { Here, } \frac{N}{2}=\frac{25}{2}=12.5\)
The cumulative frequency just greater than 12.5 lies in the interval 60-70.
Hence, the median class is 60-70.
(v) (a): We know, Mode = 3 Median - 2 Mean
\(\therefore\) 3 Median = Mode + 2 Mean.

(i) (a): If f_i and x_i are very small, then direct method is appropriate method for calculating mean.
(ii) (a) : The frequency distribution table from the given data can be drawn as :

\(\therefore \quad \text { Mean }=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{326}{40}=8.15 \text { litres }\)

(iii) (b) : If 2000 vehicles comes daily and average quantity of diesel required for a vehicle is 8.15 litres, then total quantity of diesel required = 2000 x 8.15
= 16300 litres
(iv) (c) : Here \(N=40 \text { and } \frac{N}{2}=20\) c.f for the distribution are 5, 15,25,32,40
Now, c.f just greater than 20 is 25 which is corresponding to the class interval 7-9.
So median class is 7-9.
\(\therefore\) Required sum of upper limit and lower limit
= 7 + 9 = 16
(v) (b): We know, Mode = 3 Median -2 Mean
= 3(8) - 2(8.15) = 24 - 16.3 = 7.7

Total number of coins = 50 + 48 + 36 + 28 + 8 = 170
(i) (c): Number of \(\begin{equation} ₹ \end{equation} \) 5 coins = 36
Required probability = \(\begin{equation} \frac{36}{70}=\frac{18}{85} \end{equation}\)
(ii) (b): Number of \(\begin{equation} ₹ \end{equation} \)20 coins = 8
Required probability = \(\begin{equation} \frac{8}{170}=\frac{4}{85} \end{equation}\)
(iii) (d): Number of \(\begin{equation} ₹ \end{equation} \)10 coins = 28
Probability (coin is of \(\begin{equation} ₹ \end{equation} \)10) = \(\begin{equation} \frac{28}{170} \end{equation}\)
Required probability = 1 - P(coin is of 10)
\(\begin{equation} =1-\frac{28}{170}=\frac{142}{170}=\frac{71}{85} \end{equation}\)
(iv) (a) : Total number of coins of \(\begin{equation} ₹ \end{equation} \)10 and \(\begin{equation} ₹ \end{equation} \) 20
= 28 + 8 = 36
Required probability = \(\begin{equation} \frac{36}{170}=\frac{18}{85} \end{equation}\)
(v) (a): Total number of coins of \(\begin{equation} ₹ \end{equation} \)1, \(\begin{equation} ₹ \end{equation} \)2 and \(\begin{equation} ₹ \end{equation} \) 5
 = 50 + 48 + 36 = 134
Required probability = \(\begin{equation} \frac{134}{170}=\frac{67}{85} \end{equation}\)

Since, every student get one chocolate. So, number of chocolates Rohit has is equal to the number of students in the class.
(i) (a): Let number of milk chocolates Rohit has = x
Probability of distributing milk chocolates = \(\begin{equation} \frac{1}{3} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{x}{54}=\frac{1}{3} \Rightarrow x=\frac{54}{3}=18 \end{equation}\)
(ii) (c): Let number of dark chocolates Rohit has = y
Probability of distributing dark chocolates = \(\begin{equation} \frac{4}{9} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{y}{54}=\frac{4}{9} \Rightarrow y=\frac{4 \times 54}{9}=24 \end{equation}\)
(iii) (d) : Number of white chocolates Rohit has = 54 -(18 + 24) = 12
Required probability = \(\begin{equation} \frac{12}{54}=\frac{2}{9} \end{equation}\)
(iv) (b) : Total number of milk and white chocolates = 18 + 12 = 30
Required probability = \(\begin{equation} \frac{30}{54}=\frac{5}{9} \end{equation}\)
(v) (b): Since all students gets one chocolate. So, total number of chocolates distributed = 54
\(\begin{equation} \text { Required probability }=\frac{54}{54}=1 \end{equation}\)

Total number of blocks in the kit = 120
Number of red blocks = 40
Number of blue blocks = 25
Number of green blocks = 30
Number of yellow blocks = 120 - (40 + 25 + 30)
= 120 - 95 = 25
(i) (d): P(block is red) \(\begin{equation} =\frac{40}{120}=\frac{1}{3} \end{equation}\)
(ii) (c): P(block is not yellow) = 1 - P(block is yellow)
\(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\)
(iii) (c) : P(block is green) = \(\begin{equation} \frac{30}{120}=\frac{1}{4} \end{equation}\)
(iv) (b) : P(block is yellow) = \(\begin{equation} \frac{5}{24} \end{equation}\)
(v) (b): P(block is not blue) = 1 - P(block is blue)
\(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\)

Total number of fish in the aquarium = 13 + 18 + 12 + 11 = 54
(i) (b): Number of male fish in the aquarium = 36
Number of female fish in the aquarium = 54 - 36 = 18
So, required probability = \(\begin{equation} \frac{18}{54}=\frac{1}{3} \end{equation}\)
(ii) (b): Required probability = \(\begin{equation} \frac{18}{54}=\frac{1}{3} \end{equation}\)
(iii) (d) : P(selecting a koi fish) = \(\begin{equation} \frac{12}{54}=\frac{2}{9} \end{equation}\)
P(not selecting a koi fish) = 1 - P(selecting a koi fish) \(\begin{equation} =1-\frac{2}{9}=\frac{7}{9} \end{equation}\)
(iv) (b) : Total number of guppy fish and koi fish = 13 + 12 = 25
P(selecting neither angel fish nor flowerhorn fish =25/54.
(v) (c): P(selecting a guppy fish) = 13/54.

(i) (b):Total number of possible outcomes on throwing two dice simultaneously = 6 x 6 = 36
(ii) (c): As we know, that maximum sum of numbers on two dice = 6 + 6 = 12
It is an impossible event.
So, required probability = 0
(iii) (c) : Let A be the event that both the numbers on dice are prime.
 A = {(2, 2), (2, 3), (2, 5), (3,2), (3, 3), (3, 5), (5, 2), (5,3), (5, 5)}
\(\Rightarrow\)n(A) = 9
\(\begin{equation} P(A)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(iv) (c) : Let B the event that product of two numbers is odd.
B = {(1, 1), (1, 3), (1, 5), (3,1), (3, 3), (3, 5), (5,1), (5,3), (5, 5)}
\(\Rightarrow\)n(B) = 9
\(\begin{equation} P(B)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(v) (c) : Let C be the event that difference of two numbers is zero.
C = {(1, 1), (2,2), (3, 3), (4,4), (5, 5), (6, 6)}
\(\Rightarrow\)n(C) = 6
\(\begin{equation} P(C)=\frac{6}{36}=\frac{1}{6} \end{equation}\)

Total number of cards left = 52 - (4 + 4 + 4) = 52 - 12 = 40
There are four 2's, four 5's and four Jacks in a deck
(i) (c): P(getting 6 of spade) = \(\begin{equation} \frac{1}{40} \end{equation}\)
(ii) (a): As there is no black diamond in the deck of cards.
P(black diamond) = 0
(iii) (b) : Total number of face cards in a deck = 12
But all Jacks are removed
Number offace cards left = 12 - 4 = 8
So, P(getting a face card) = \(\begin{equation} \frac{8}{40}=\frac{1}{5} \end{equation}\)
(iv) (d) : Number of cards of club left = 13 - 3 = 10
3 cards are removed of each kind
P(getting a club) = \(\begin{equation} \frac{10}{40}=\frac{1}{4} \end{equation}\)
(v) (c): Number of red cards left = 26 - 6 = 20
P(getting a red card) = \(\begin{equation} \frac{20}{40}=\frac{1}{2} \end{equation}\)

Total number of possible outcomes = 10
(i) (b): P(getting 100% discount) = \(\begin{equation} \frac{1}{10} \end{equation}\)
(ii) (c): P(getting no prize) = \(\begin{equation} \frac{3}{10} \end{equation}\)
(iii) (b) : P(getting a soccer ball) = \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\)
(iv) (b) : P(getting 15% discount) = \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\)
(v) (a): P(getting a free spin) = \(\begin{equation} \frac{1}{10} \end{equation}\)

Total number of footballs produced in one day = 220000
(i) (c): P(selecting a defective football) = \(\begin{equation} \frac{15}{100}=\frac{3}{20} \end{equation}\)
(ii) (d): P(selecting a non-defective football) = 1- P(selecting a defective football) = \(\begin{equation} 1-\frac{3}{20}=\frac{17}{20} \end{equation}\)
(iii) (b) : Total number of defective footballs produced in 1 day = \(\begin{equation} \frac{3}{20} \times 22000=3300 \end{equation}\)
(iv) (a) : Total number of non-defective footballs produced in 1 day = 22000 - 3300 = 18700
(v) (d): P(selecting defective football) \(\begin{equation} =\frac{8}{25} \end{equation}\)
P (selecting non-defective football) = \(\begin{equation} 1-\frac{8}{25}=\frac{17}{25} \end{equation}\)
Number of non-defective footballs produced \(\begin{equation} =\frac{17}{25} \times 22000=14960 \end{equation}\)

Let Band G denotes boy and girl respectively.
Then,
Possible outcomes for Gupta's family having two kids is {BB, BG, GG}
Similarly, possible outcomes for Singhal's family having 3 kids is {BBB, BBG, BGG, GGG}
(i) (b): Required probability = \(\begin{equation} \frac{1}{4} \end{equation}\)
(ii) (b): Required probability = \(\begin{equation} \frac{2}{3} \end{equation}\)
(iii) (b): Required probability = \(\begin{equation} \frac{2}{3} \end{equation}\)
(iv) (b) : Required probability = \(\begin{equation} \frac{1}{4} \end{equation}\)
(v) (c): Required probability = \(\begin{equation} \frac{1}{3}+\frac{1}{4}=\frac{7}{12} \end{equation}\)

Total number of cartons in the store = 80 + 90 + 38 + 42 = 250
(i) (d) :P(choosing an apple juice carton) = \(\begin{equation} \frac{90}{250}=\frac{9}{25} \end{equation}\)
(ii) (c) : P( choosing an orange juice carton ) = \(\begin{equation} \frac{80}{250}=\frac{8}{25} \end{equation}\)
P(choosing not an orange juice carton) = \(\begin{equation} =1-\frac{8}{25}=\frac{17}{25} \end{equation}\)
(Iii) (d):P(choosing a guava juice carton) = \(\begin{equation} \frac{42}{250}=\frac{21}{125} \end{equation}\)
(iv) (c) : Total number of cartons Vishal bought =4+3+3=10
Number of tetrapacks in 1 carton = 10
Total number of tetrapacks Vishal has = 100
So, P( customer picks a guava juice tetrapack) \(\begin{equation} =\frac{3 \times 10}{100}=\frac{3}{10} \end{equation}\)
(v) (b): Number of cartons left with storekeeper = 250 - 10 = 240
Number of cartons he bought = 14
Total number of cartons are with storekeeper now = 240 + 14 = 254
So, P(selecting a tetrapack of apple juice from store) \(\begin{equation} =\frac{(90-4+14) \times 10}{254 \times 10}=\frac{100}{254}=\frac{50}{127} \end{equation}\)

Total number of days in June = 30
(i) (c): Number of sunny days = PCsunny day) x 30 = \(\begin{equation} \frac{1}{2} \times 30=15 \end{equation}\)
(ii) (b): Number of cloudy days in June = 5
\(\begin{equation} x=\frac{5}{30}=\frac{1}{6} \end{equation}\)
(iii) (a): Required probability = \(\begin{equation} \frac{1}{2}+\frac{1}{6}+\frac{1}{5}=\frac{13}{15} \end{equation}\)
(iv) (d) : We have, \(\begin{equation} x+y=\frac{3}{10} \Rightarrow y=\frac{3}{10}-\frac{1}{6}=\frac{2}{15} \end{equation}\)
So, number of rainy days = \(\begin{equation} \frac{2}{15} \times 30=4 \end{equation}\)
(v) (c): Number of partially cloudy days = P(partially cloudy days) x 30 \(\begin{equation} =\frac{1}{5} \times 30=6 \end{equation}\)