CBSE Sample Papers for Class 12 Applied Mathematics Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 1 with Solutions

Maximum Marks: 80 Marks

Time Allowed : 3 Hours

General Instructions:

1. This question paper contains five sections A, B, C, D and E. Each section is compulsory.
2. Section – A carries 20 marks weightage, Section – B carries 10 marks weightage. Section – C carries 18 marks weightage. Section – D carries 20 marks weightage and Section – E carries 3 case-based with total weightage of 12 marks.
3. Section A: It comprises of 20 MCQs of 1 mark each.
4. Section B: It comprises of 5 VSA type questions of 2 marks each.
5. Section C: It comprises of 6 SA type of questions of 3 marks each.
6. Section D: It comprises of 4 LA type of questions of 5 marks each.
7. Section E: It has 3 case studies. Each case study comprises of 3 case-based questions, where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
8. Internal choice is provided in 2 questions in Section – B, 2 questions in Section – C, 2 questions in Section – D. You have to attempt only one of the alternatives in all such questions.

Section – A [20 marks]

(All questions are compulsory. No internal choice is provided in this section)

Question 1.
What is the least value of ‘x’ that satisfies x = 27 (mod 4), when 27 < x ≥ 36?
(a) 27
(b) 30
(c) 31
(d) 35
Answer:
(c) 31
x ≡ 27 (mod 4)
⇒ x – 27 = 4 K, for some integer
⇒ x = 31 as 27 < x ≤ 36

Explanation:
x ≡ 27(mod 4)
⇒ x – 27 = 4 K, for some integer
But 27 < x ≤ 36 x = 31.

Question 2.
Let p > 0 and q < 0 and p, q, ∈ Z then choose the correct inequality from the given below options to complete the statement:
(a) >
(b) ≥
(c) ≤
(d) <
Answer:
(d) <

Explanation:
Here p > 0 and q < 0
p + q = p – q
and p – q = p – (-q) = p + q
∴ p + q < p – q

Question 3.
A machine makes car wheels and in a random sample of 26 wheels, the test statistic is found to be 3.07. As per the t-distribution test (of 5% level of significance), what can you say about the quality of wheels produced by the machine? (Use t₂₅ (0.05) = 2.06)
(a) Superior quality
(b) Inferior quality
(c) Same quality
(d) Cannot say [1]
Answer:
(b) inferior quality
n = 26
⇒ |t| = 3.07 > t₂₅(0.05) = 2.06

Explanation:
Here, n = 26
|t| = 3.077 t²⁵ (0.05) = 2.06, 5% level of significance
Hence, the quality of types are inferior.

Question 4.
For the purpose of t-test of significance, a random sample of size (n) 34 is drawn from a normal population, then the degree of freedom (v) is:
(a) 134
(b) 33
(c) 34
(d) 35 [1]
Answer:
(b) 33
n = 34
⇒ v = 34 – 1 = 33

Explanation:
Sample size, n = 34
There for t-test of significance, degree of freedom, v = n – 1
= 34 – 1 = 33.

Question 5.
A person can row a boat along the stream of the river at 10 km/h and against the stream in 6 km/h. What is the speed of the stream flow?
(a) 1 km/h
(b) 2 km/h
(c) 4 km/h
(d) 5 km/h [1]
Answer:
(b) 2 km/h
Speed of boat downstream = u = 10 km/h
And, speed of boat upstream = v = 6 km/h
Speed of stream = \(\frac{1}{2}\)(u – v) = 2km/h

Explanation: Let, the speed to boat in still water be u km/hr and speed of stream be v km/hr
∴ Speed in downstream = u + v = 10 …. (i)
Speed in upstream = u — v = 6 … (ii)
on showing (i) and (ii), we get
u = 8 km/h
v = 2 km/h

Question 6.
Standard deviation of a sample from a population is called a:
(a) Standard error
(b) Parameter
(c) Statistic
(d) Central limit [1]
Answer:
(c) Statistic

Explanation: A sample standard deviation is a statistic this means that it is calculated from only some of the individuals in a population.

Question 7.
Two water supplying trucks – A and B supply water to remote areas. Truck A is carrying 100 litres of water to a village 1.5 km away and truck B is delivering 80 litres of water to another village, 1 km away. Due to bad road conditions, each truck loses 20 ml water while travelling each metre distance. Which truck is able to deliver more water and by how much more?
(a) Truck A, 20 litres
(b) Truck B, 20 litres
(c) Truck A, 10 litres
(d) Truck B, 10 litres [1]
Answer:
(c) Truck A, 10 Litres
Truck A carries water = 100 – \(\left(\frac{20 \times 1500}{1000}\right)\)
= 70l
Truck B carries water = 80 – \(\left(\frac{20 \times 1000}{1000}\right)\)
= 60l

Explanation:
Truck A, 10 liters
Quantity of water deviled by Truck A
= 100 – \(\frac{20 \times 1500}{1000}\)
= 100 – 50
= 70l
Quantity of water delivered by Track B
= 80 – \(\left(\frac{20 \times 1000}{1000}\right)\)
= 60l.

Question 8.
What is the face value of a sinking fund that yields a dividend of ₹1800 at 10% semi-annually?
(a) ₹ 3600
(b) ₹ 18000
(c) ₹ 24000
(d) ₹ 36000 [1]
Answer:
(d) 136000
Let the face value of the bond = x
Then, \(\frac{10}{200}\) × =1800 ⇒ x = 36000

Question 9.
In the given figure, the area bounded by the curve x = f(y), y-axis and abscissa y = a and y = b is equal to:

(a) \(\int_a^b f(y) d y\)
(b) \(\int_a^b f(x) d y\)
(c) \(\int_a^b|f(y)| d y\)
(d) \(\int_a^b|f(x)| d y\) [1]
Answer:
(c) \(\int_a^b|f(y)| d y\)

Explanation: \(\int_a^b|f(y)| d y\)
‘Since, curve is x = f(y)
Therefore, required area = \(\int_a^b|f(y)| d y\)

Question 10.
A factory production is delayed for three weeks due to breakdown of a machine and unavailability of spare parts. Under which trend oscillation does this situation fall under?
(a) Seasonal
(b) Cyclical
(c) Secular
(d) irregular [1]
Answer:
(d) Irregular

Explanation: Irregular variations are of irregular and do not repeat in a definite pattern. It is not definite and can occur at any point in irregular manner.

Question 11.
A newspaper printing machine costs ₹ 4,80,000 and estimated scrap value of ₹ 25,000 at the end of its useful life of 10 years. What is its annual depredation as per linear method?
(a) ₹ 4,550
(b) ₹ 45,500
(c) ₹ 50,500
(d) ₹ 61,500 [1]
Answer:
(b) ₹ 45,500
D = \(\frac{\mathrm{C}-\mathrm{S}}{n}\) = \(\frac{480000-25000}{10}\) = 45500

Explanation:
Annual depreciation

Question 12.
In the given figure, what is the LPP shaded region known as?

(a) Feasible region
(b) Feasible solution
(c) Optimal region
(d) Objective region [1]
Answer:
(a) Feasible region

Explanation: The common region determined by all the constraints including non-negative constraints x, y ≥ 0 of a linear programming problem is called feasible region.

Question 13.
General solution of differential equation: ylogy dx – xdy = 0 is:
(a) y = log |Cx|
(b) y = e^|Cx|
(c) y = e^-Cx
(d) log y = |C + x| [1]
Answer:
(b) y = e^|Cx|
\(\int \frac{d y}{y \log y}\) = \(\int \frac{d x}{x}\)
⇒ log(log y) = Log|x| + log|C|
⇒ log(log y) = log|Cx|
⇒ y = e^|Cx|

Question 14.
An investment of ₹ 10,000 becomes ₹ 60,000 in 4 years, then the CAGR (compound annual growth rate) is given by:
(a) \(\frac{\sqrt[4]{6}-1}{100}\)
(b) \(\frac{\sqrt[4]{6}+1}{100}\)
(C) \([\sqrt[4]{6}-1] \times 100\)
(d) \([\sqrt[4]{6}+1] \times 100\) [1]
Answer:
(C) \([\sqrt[4]{6}-1] \times 100\)

Question 15.
In what ratio shall I add water to the liquid detergent costing ₹ 480 per litre to get resulting mixture worth ₹ 300 per litre?
(a) 5 : 3
(b) 3 : 8
(c) 3 : 5
(d) 5 : 8 [1]
Answer:
(c) 3 : 5
⇒ 180 : 300 = 3 : 5

Explanation:

∴ Required ratio = 180 : 300
= 3 : 5

Question 16.
A grain whole-seller visits the granary market. While going around to make a good purchase, he takes a handful of rice from random sacks of rice, in order to inspect the quality of farmers produce. The handful of rice taken from a sack of rice for quality inspection is a:
(a) statistic
(b) population
(c) parameter
(d) sample [1]
Answer:
(d) sample

Explanation: Sample is a part, portion or fraction of the population. It considered as the representative of the whole data.

Question 17.
For predicting the straight-line trend ¡n the sales of scooters (in thousands) on the basis of 6 consecutive years data, the company makes use of 4-year moving averages method, if the sales of scooters for respective years are a, b, c, d, e and f respectively, then which of the following average will not be computed?
(a) \(\frac{a+b+c+d}{4}\)
(b) \(\frac{b+c+d+e}{4}\)
(c) \(\frac{a+c+d+e}{4}\)
(d) \(\frac{c+d+e+f}{4}\) [1]
Answer:
(c) \(\frac{a+c+d+e}{4}\)

Explanation:
Rest we will be computed while evaluating a 4 gear-moving average.

Question 18.
In a school, a random sample of 145 students is taken to check whether a student’s average calory intake is 1500 or not. The collected data of average calories intake of sample students is presented in a frequency distribution, which is called a:
(a) Statistics
(b) Sampling distribution
(c) Parameter
(d) Population sampling [1]
Answer:
(b) Sampling distribution

Explanation: When a collected sample from populates is represented in the form of a frequency distribution, it is called sampling distribution.

Assertion-Reason Questions

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 19.
Assertion (A): Kuhu and Beena are two equally capable badminton players. Probability that Beena will beat Kuhu in 3 games out of 4 is 25%.
Reason (R): C The probability of successes in trials, denoted by P(X = r) is given by = P(X = r) = ⁿC_rp_r q_n-r, r = 0, 1,…, n where denotes success and denotes failure in each trial. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
P(Win in one game) = P(Lose in one game)
= 1/2
⇒ P(Beena to win in 3 out of 4 games) = ⁴C₃
\(\left(\frac{1}{2}\right)^4\) = \(\frac{1}{4}\) = 25%
Assertion is correct and Reason is the correct explanation for it.

Question 20.
Assertion (A): If the nominal rate of interest is 12.5% and the inflation is 2%, then the effective rate of interest is 10.5%
Reason (R): If the interest is calculated only at the end of an year, then the effective rate of interest is same as the nominal rate of interest. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of the (A)
Effective rate of interest = Nominal rate – inflation rate = 12.5 – 2 = 10.5%
Assertion is correct
Reason is true but not supportive of assertion

Section – B

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 21.
₹ 2,50,000 cash is equivalent to a perpetuity of ₹ 7,500 payable at the end of each quarter. What is the rate of interest convertible quarterly? [2]
Answer:
P = 250000, R = 7500, i = \(\frac{r}{400}\)
⇒ 250000 = \(\frac{7500 \times 400}{r}\)
⇒ r = 12

Question 22.
Find value of 2a + 3b – c, if A
= \(\left[\begin{array}{ccc}
0 & -1 & 28 \\
a-8 & 0 & 3 b \\
-c+2 & 2 & 0
\end{array}\right]\) is a skew-symmetric matrix
OR
There are two real value(s) of x, for which the value of the determinant Δ = \(\left[\begin{array}{ccc}
1 & -2 & 5 \\
2 & x & -1 \\
0 & 4 & 2 x
\end{array}\right]\) is 86. Find the value(s) of x.
Answer:
⇒ a – 8 = 1
⇒ a = 9
3b = -2
⇒ b = \(-\frac{2}{3}\)
-c + 2 = -28
c = 30
⇒ 2a + 3b – c = -14
OR
Expanding C₁, we get A = 1(2x² + 4) – 2 (-4x – 20) = 86
⇒ x² + 4x – 21 = 0
∴ x = 3, – 7

Question 23.
A book publisher sells a hard cover edition of a book for ₹ 72 and a paperback edition for ₹ 40. In addition to a fixed weekly cost of ₹ 9,600, the cost of printing hardcover and paperback editions are ₹ 56 and ₹ 28 per book respectively. Each edition requires 5 minutes on the printing machine whereas hardcover binding takes 10 minutes and paperback takes 2 minutes on the binding machine. The printing machine and the binding machine are available for 80 hours each week. Formulate the linear programming problem to maximise the publisher’s profit. [2]
Answer:
Let the number of hardcopy and paperback copies be x and y respectively
⇒ Maximum profit Z = (72x + 40y) – (9600 + 56x + 28y)
= 16x+ 12y – 9600
Subject to constraints:
x + y ≤ 960
5x + y ≤ 2400
x, y ≥ 0

Question 24.
A boatman takes half as much time in rowing his boat for a certain distance downstream than upstream. What is the ratio between his speed of rowing the boat in still water and speed of current?
OR
In a 200-metre race, Anuj can beat Param by 5 metre or 3 seconds. How much time did Anuj take to complete the race? [2]
Answer:
Speed of boat in still waters = x km/h
Speed of stream = x km/h
Distance travelled = x km
Time taken to travel downstream = \(\frac{d}{x+y}\)
Time taken to travel upstream = \(\frac{d}{x-y}\)
Then, \(\frac{2 d}{x+y}\) = \(\frac{d}{x-y}\)
⇒ x : y = 3 : 1
OR
Param runs 5 m in 3 seconds
⇒ time taken to run 200 m = \(\frac{3}{5}\) × 200
= 120 seconds
Anuj’s time = 120 – 3 = 117 seconds

Question 25.
Mitul invested ₹ 3,50,000 in a fund. At the end of the year the value of the fund is ₹ 4,37,500. What is the nominal rate of interest, if the market price is same at the end of the year? [2]
Answer:
V_f = 437500, V_i = 350000
Nominal rate = \(\frac{V_f-V_i}{V_i}\) × 100 = 25%

Section – C

(All questions are compulsory. In case of internal choice. attempt any one question only)

Question 26.
Find the interval(s) in which the function f(x) = \(\frac{x^4}{4}\) – 2^x3 + \(\frac{11 x^2}{2}\) – 6x, is strictly increasing and strictly decreasing. [3]
Answer:
P(x) = x³ – 6x² + 11x – 6 = (x – 1)(x – 2)(x – 3)
⇒ x = 1, 2, 3
Strictly increasing in (1, 2) ∪ (3, ∞)
Strictly decreasing in (-∞, 1) ∪ (2, 3)

Question 27.
Two badminton teams A and B are staying in the same hotel Team A has 2 male and 3 female players accompanied by 1 coach. Team B comprises of 1 male, 2 female players and 2 coaches. The daily diet requirement (calories and protein) for each person is as given below:

	Calories	Protein
Male Player	2500	65 g
Female Player	1900	50 g
Coach	2000	54 g

Use matrix algebra to calculate the total diet requirement of calories and protein for each team. [3]
Answer:
Daily diet of team A = [2 3 1]\(\left[\begin{array}{ll}
2500 & 65 \\
1900 & 50 \\
2000 & 54
\end{array}\right]\)
= \(\left[\begin{array}{c}
12700 \\
334
\end{array}\right]\)
Team A consumes 12700 calories and 334 g vitamin
Daily diet of team B = [1 2 2]\(\left[\begin{array}{ll}
2500 & 65 \\
1900 & 50 \\
2000 & 54
\end{array}\right]\)
= \(\left[\begin{array}{c}
10300 \\
273
\end{array}\right]\)
Team B consumes 10300 calories and 273g vitamin

Question 28.
Evaluate \(\int \frac{d x}{\left(1+e^x\right)\left(1+e^{-x}\right)}\)
OR
Evaluate \(\int x \log \left(1+x^2\right) d x\) [3]
Answer:

⇒ 3000 = \(\frac{(4000000-x) \times 0.0075}{1-0.1062}\)
⇒ x = 424800
Down payment = ₹ 4,24,800

Question 29.
Under the pure market competition scenario, the demand function p_d and the supply function p_s = for a certain commodity are given as p_d = \(\frac{8}{x+1}\) – 2 and p_s = \(\frac{x+3}{2}\) respectively, where p is the price and x is the quantity of the commodity. Using integrals, find the producer’s surplus.
OR
The demand function p for maximising a profit monopolist is given by p = 274 – x² while the marginal cost is 4 + 3x, for x units of the commodity. Using integrals, find the consumer surplus. [3]
Answer:
Under pure competition, p_d = p_s
⇒ \(\frac{8}{x+1}\) – 2 = \(\frac{x+3}{2}\)
⇒ x² + 8x – 9 = 0
⇒ x = -9, 1
⇒ x = 1
When x₀ = 1 ⇒ p₀ = 2
∴ Produce surplus = 2 – \(\int_0^1 \frac{x+3}{2} d x\)
= 2\(\left[\frac{x^2}{4}+\frac{3 x}{2}\right]_0^1\) = \(\frac{1}{4}\)

OR

p = 274 – x²
⇒ R = px = 274x – x³
\(\frac{d R}{d x}\) = 274 – 3x².
Given MR = 4 + 3x
In profit monopolist market,
MR = \(\frac{d R}{d x}\)
⇒ 4 + 3x = 274 – 3x²
⇒ x² + x – 90 = 0
⇒ x = -10, 9
∴ x = 9
When x₀ = 9 ⇒ p₀ = 193
∴ Consumer surplus
= \(\int_0^9\)(274 – 0) – 193 × 9
= \(\left[274 x-\frac{x^3}{0}\right]_0^9\)
= 486

Question 30.
Surjeet purchased a new house, costing ₹ 40,00,000 and made a certain amount of down payment so that he can pay the balance by taking a home loan from XYZ Bank. If his equated monthly installment is ₹ 30,000, at 9% interest compounded monthly (reducing balance method) and payable for 25 years, then what is the initial down payment made by him? [3] [Use (1.0075) – 300 = 0.1062]
Answer:
Purchase = ₹ 40,00,000
Down payment = x
Balance = 40,00,000 – x
i = \(\frac{9}{1200}\) = 0.0075
n = 25 × 12 = 300
E = ₹ 30,000
⇒ 30000 = \(\frac{(4000000-x) \times 0.0075}{1-(1.0075)^{-300}}\)

Question 31.
10 years ago, Mr Mehra set up a sinking fund to save for his daughter’s higher studies. At the end of 10 years, he has received an amount of ₹ 10,21,760. What amount did he put in the sinking fund at the end of every 6 months for the tenure, which paid him 5% p.a. compounded semi-annually? [3]
[Use (1.025)20 = 1.6386]
Answer:
n = 10 × 2 = 20,
S = 10, 21, 760,
i = \(\frac{5}{200}\) = 0.025, R = ?

Mr Mehra set aside an amount of ₹ 40,000 at the end of every six months.

Section – D [20 Marks]

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 32.
A. It is known that 3% of plastic buckets manufactured in a factory are defective. Using the Poisson distribution on a sample of 100 buckets, find the probability of:
(A) Zero defective buckets
(B) At most one bucket is defective
[Use e^-3 = 0.049]
OR
In a math aptitude test, student scores are found to be normally distributed having mean as 45 and standard deviation 5. What percentage of students have scores:
(A) more than the mean score?
(B) between 30 and 50? [5]
Answer:
Probability of defective bucket = 0.03
n = 100
m = np = 100 × 0.03 = 3
Let X = number of defective buckets in a sample of 100
P(X = r) = \(\frac{m^r e^{-m}}{r !}\), r = 0, 1, 2, 3,…..
= \(\frac{3^0 e^{-m}}{0 !}\) = 0.049
(B) P (at most one defective bucket) = P(r = 0, 1)
= \(\frac{3^0 e^{-m}}{0 !}\) + \(\frac{3^1 e^{-3}}{1 !}\)
= 0.049 + 0.147 = 0.196

OR

X = scores of students, µ = 45, σ = 5
∴ Z = \(\frac{X-\mu}{\sigma}\) = \(\frac{X-45}{5}\)

(A) When X = 45, Z = 0
P (X > 45) = P (Z > 0) = 0.5
⇒ 50% students scored more than the mean score

(B) When X = 30, Z = -3 and when X = 50, Z = 1
P (30 < X < 50) = P (-3 < Z < 1) = P (-3 < Z ≤ 1)
= P(-3 < Z ≤ 0) + P(0 ≤ Z < 1)
= P (0 ≤ Z < 3) + X (0 ≤ Z < 1)
= 0.4987 + 0.3413 = 0.84
84% students scored between 30 and 50 marks.

Question 33.
An event management company charges ₹ 4,800 per guest, for a bulk booking for 100 guests. In addition, it offers a discount of ₹ 200 for each group of 10 guests over and above 100 guest booking. What is the number of guests that will maximise the amount of money the company receives on a booking? What is the maximum profit on such booking?
OR
To manufacture ‘x’ number of dolls, a company’s total cost function C(x) is given by C(x) = 100 + 0.025x² and the total revenue function R(x) is described as R(x) = 5x. Given that C(x) and R(x) are in thousand rupees, what number of dolls shall be manufactured to maximise the profit of the company? What is the maximum profit? [5]
Answer:
Let x be the number of guests for the booking
Clearly, x >100 to avail discount
Profit, P = [4800 – \(\frac{200}{10}\)(x – 100)]x = 6800
x – 20x²
⇒ \(\frac{d P}{d x}\) = 6800 – 40x ⇒ x = 170
As \(\frac{d^2 P}{d x^2}\) = -40 < 0, ∀ x
∴ Manufacturing loo dolls will maximise the profit of the company and, profit = ₹ 1,50,000.

Question 34.
Rahul is at the whole sale market to purchase folding tables and chairs, to later sell them at his furniture shop. He has only ₹ 5,760 to spend and his van has space to carry at the most 20 items. A table costs him ₹ 360 and a chair costs ₹ 240. Back at his shop, he plans to sell a table at a profit of ₹ 22 and a chair at a profit of 18. Given that he can sell all the items that he purchases, how many tables and chairs shall he purchase In order to maximise his profit? [5]
Answer:
Let the number of tables and chairs be x and y respectively
(Max profit) Z = 22x+ 18y
Subject to constraints:
x + y ≤ 20
3x + 2y ≤ 48
x, y ≥ 0

The feasible region OABCA is closed (bounded)

Corner points	Z = 22x + 18y
O(0, 0)	0
A(0, 20)	360
B(8, 12)	392
B(8, 12)	352

Buying 8 tables and 12 chairs will, maximise the profit

Question 35.
The equilibrium conditions for three competitive markets are described as given below, where p₁, p₂ and p₃ are the equilibrium price for each market respectively.
p₁ + P₂ + p₃ = 85
3p₁ + 2p₂ + 2p₃ = 105
2p₁ + 3p₂ + 2p₃ = 110
Using matrix method, find the values of respective equilibrium prices. [5]
Answer:

Section – E

(All questions are compulsory. In case of internal choice, attempt any one question only)

Question 36.
Case Study-1
O An overhead water tank has three pipes A, B and C attached to it (as shown in figure (II)). The inlet pipes A and B can fill the empty tank independently in 15 hours and 12 hours respectively. The outlet pipe C alone can empty a full tank in 20 hours.
Based on the above information, answer the following questions. Show steps to support your answers.

(A) For a routine cleaning of the tank, the tank needs to be emptied. If pipes A and B are closed at the time when the tank is filled to two-fifth of its total capacity, how long will pipe C take to empty the tank completely? [1]
(B) How long will it take for the empty tank to fill completely, if all the three pipes are opened simultaneously? [1]
(C) On a given day, pipes A, B and C are opened (in order) at 5 am, 8 am and 9 am respectively, to fill the empty tank. In how many hours will the tank be filled completely?
OR
Given that the tank is half-full, only pipe C is opened at 6 AM, to empty the tank. After closing the pipe C and an hour’s cleaning time, tank is filled completely by pipe A and B together. What is the total time taken in the whole process? [2]
Answer:
(A) Pipe C empties 1 tank in 20 h ⇒ \(\frac{2}{5} \text { th }\) tank in \(\frac{2}{5}\) × 20 = 8 hours

(B) Part of tank filled in 1 hour = \(\frac{1}{15}\) + \(\frac{1}{12}\) + \(\frac{1}{20}\)
= \(\frac{1}{10} \text { th }\) time taken to fill tank completely
= 10 hours

(C) At 5 am,
Let the tank be completely filled in ‘t’ hours
⇒ pipe A is opened for ‘t’ hours
pipe B is opened for ‘t – 3’ hours
And, pipe C is opened for ‘t – 4’ hours
⇒ In one hour,
part of tank filled by pipe A = \(\frac{t}{15}\)^th
part oftankfilled by pipe B = \(\frac{t-3^{\text {th }}}{15}\)
and, part of tank emptied by pipe C = \(\frac{t-4^t}{15}\)
Therefore \(\frac{t}{15}\) + \(\frac{t-3}{12}\) – \(\frac{t-4}{20}\) = 1
⇒ t = 10.5
Total time to fill the tank = 10 hours 30 minutes

OR

6 am, pipe C is opened to empty 1/2 filled tank
Time to empty = 10 hours
Time for cleaning = 1 hour
Part of tank filled by pipes A and B in 1 hour
= \(\frac{1}{15}\) + \(\frac{1}{12}\) = \(\frac{3}{20}\)^th tank
⇒ time taken to fill the tank completely = \(\frac{20}{3}\) hours
Total time taken in the process = 10 + 1 + \(\frac{20}{3}\) = 17 hour 40 minutes

Question 37.
Case Study-II
O When observed over a long period of time, a time series data can predict trend that can forecast increase or decrease or stagnation of a variable under consideration. Such analytical studies can benefit a business for forecasting or prediction of future estimated sales or production.

Mathematically, for finding a line of best-fit to represent a trend, many methods are available. Methods like moving-averages and least- squares squares are some of the techniques to predict such trends.

Mrs. Shamita runs a bread factory and the record of her sales of bakery items for the period of 2015 – 2019 is as follows:

Based on the above information, answer the following questions. Show steps to support your answers.
(A) By taking year 2017 as origin, use method of least-squares to find the best-fit trend line equation for Mrs. Shamita’s business. Show the steps of your working.
OR
Demonstrate the technique to fit the best-suited straight-line trend by the method of 3-years moving averages. Also draw the trend line. [2]
(B) What are the estimated sales for Mrs. Shamita’s business for year 2022? [1]
(C) Mrs Shamita wishes to grow her business to yearly sale of ₹ 67400. In which year will she be able to reach her target? [1]
Answer:

(OR)

(A)

Year	Y	3-year moving average
2015	35	–
2016	42	41
2017	46	43
2018	41	45
2019	48	–

OR

(B) For year 2022,
Y₂₀₂₂ = 42.4 + 2.5(2022 – 2017) = 54.9
⇒ the estimated sales for year 2022 = ₹ 54,900
(C) Y_C = 42.4 + 2.5
⇒ 67.4 = 42.4 + 2.5X
⇒ X = 10
Sales will be ₹ 67,400 in year (2017 + 10) = year 2027

Question 38.
Case Study-III
According to an educational board survey, it was observed that class XII students apply at least one to four weeks ahead of colleges application deadline. Let X represent the week when an average student applies ahead of a college’s application deadline and the probability of student to get admission in the college P(X = x) is given as follows:
when x = 0, 1 or 2 when x = 3 when x = 4 when x > 4

Where k is a real number.
Based on the above information, answer the following questions. Show steps to support your answers.
(A) Find the value of k. [1]
(B) What is the probability that Sonali will get admission in the college, given that she applied at least 2 weeks ahead of application deadline? [1]
(C) Calculate the mathematical expectation of number of weeks taken by a student to apply ahead of a college’s application deadline.
OR
To promote early admissions, the college is offering scholarships to the students for applying ahead of deadline as follows:
₹ 50000 for applying 4 weeks early,
₹ 20000 for applying 3 weeks early,
₹ 12000 for applying 2 weeks early, and ₹ 9600 for applying 1 Week yearly
What is the expected scholarship offered by the college? [2]
Answer:
(A) \(\frac{k}{6}\) + \(\frac{2 k}{6}\) + \(\frac{3(1-k)}{6}\) + \(\frac{4 k}{2}\) = 1
⇒ k = \(\frac{1}{4}\)

(B) P (getting admission on applying at least 2 weeks ahead of application deadline)
= P (X = 2, 3, 4)
\(\frac{1}{12}\) + \(\frac{3}{8}\) + \(\frac{1}{12}\) = \(\frac{23}{24}\)
[alternate method: 1 – P] (X = 1)
= 1 – \(\frac{1}{24}\) = \(\frac{23}{24}\)] [1]

(C) X = week applied ahead of application deadline

X	1	2	3	4
P(X)	\(\frac{1}{24}\)	\(\frac{1}{12}\)	\(\frac{3}{8}\)	\(\frac{1}{2}\)
XP(X)	\(\frac{1}{24}\)	\(\frac{1}{6}\)	\(\frac{9}{8}\)	2

∴ E(X) = \(\frac{80}{24}\) = 3\(\frac{1}{3}\) weeks
OR
X = Scholarship money awarded for the week applied in before the deadline

Week applied in in	1	2	3	4
X	9600	12000	20000	50000
P(X)	\(\frac{1}{24}\)	\(\frac{1}{12}\)	\(\frac{3}{8}\)	\(\frac{1}{2}\)
XP(X)	\(\frac{9600}{24}\)	\(\frac{12000}{12}\)	\(\frac{60000}{8}\)	\(\frac{5000}{2}\)

∴ E(X) = ₹33.900.