ISC Class 12 Sample Paper 2027 Applied Mathematics
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Page 1
FOR ISC CLASS 12 EXAM PREPARATION
ISC Class 12 2027
Sample Paper · Applied
Mathematics
EXAM YEAR TYPE SUBJECT
ISC Class 12 2027 Sample Paper Applied Mathematics
Notes · Sample Papers · Previous Year Papers · Mock Tests
ISC Class 12 2027
Sample Paper · Applied
Mathematics
EXAM YEAR TYPE SUBJECT
ISC Class 12 2027 Sample Paper Applied Mathematics
Notes · Sample Papers · Previous Year Papers · Mock Tests
Page 2
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APPLIED MATHEMATICS
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Maximum Marks: 80
Time Allotted: Three Hours
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Reading Time: Additional Fifteen Minutes
e m
m
eCandidates l as
as
Instructions to
ag
lare allowed an additional fifteen minutes for only reading the paper.
a
1. Youg
2. You must NOT start writing during reading time.
3. The question paper has 14 printed pages.
4. It consists of 20 questions and four sections: A, B, C and D. All questions
are compulsory.
5. Section A comprises very short answer questions of 1 mark each.
6. Section B consists of short answer questions of 2 marks each.
m
7. Section C consists of moderately long answer questions of 3 marks each.
.co
8. Section D consists of long answer questions of 5 marks each.
m
e
9. Internal choices have been provided in three questions, each in Sections B,
s
C and D.
l a
ag
10. While attempting Multiple Choice Questions in Section A, you are
required to write only ONE option as the answer.
11. The intended marks for questions or parts of questions are given in the
brackets [].
12. All workings, including rough work, should be done on the same page as,
and adjacent to, the rest of the answer.
13. Mathematical tables and graph papers are provided.
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.co
Instruction to Supervising Examiner
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Kindly read aloud the instructions given above to all the candidates present in the
s e m
s e examination hall.
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g la a
a
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1 ISC SPECIMEN QUESTION PAPER 2027
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Page 1 of 34
.co s e m
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APPLIED MATHEMATICS
m
Maximum Marks: 80
Time Allotted: Three Hours
om . co
. c
Reading Time: Additional Fifteen Minutes
e m
m
eCandidates l as
as
Instructions to
ag
lare allowed an additional fifteen minutes for only reading the paper.
a
1. Youg
2. You must NOT start writing during reading time.
3. The question paper has 14 printed pages.
4. It consists of 20 questions and four sections: A, B, C and D. All questions
are compulsory.
5. Section A comprises very short answer questions of 1 mark each.
6. Section B consists of short answer questions of 2 marks each.
m
7. Section C consists of moderately long answer questions of 3 marks each.
.co
8. Section D consists of long answer questions of 5 marks each.
m
e
9. Internal choices have been provided in three questions, each in Sections B,
s
C and D.
l a
ag
10. While attempting Multiple Choice Questions in Section A, you are
required to write only ONE option as the answer.
11. The intended marks for questions or parts of questions are given in the
brackets [].
12. All workings, including rough work, should be done on the same page as,
and adjacent to, the rest of the answer.
13. Mathematical tables and graph papers are provided.
m
.co
Instruction to Supervising Examiner
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m .co
Kindly read aloud the instructions given above to all the candidates present in the
s e m
s e examination hall.
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g la a
a
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1 ISC SPECIMEN QUESTION PAPER 2027
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Page 1 of 34
Page 3
Please note that Specimen Question Paper in the subject provides a realistic format of the
Board Question Paper and should be used as a practice tool. However, the Board Question
Paper may not necessarily replicate the Specimen Question Paper as individual questions may
vary. The weightage allocated to various topics, as given in the syllabus, will be strictly
adhered to.
SECTION A – 20 MARKS
Question 1
In subparts (i) to (xvii) choose the correct options and in subparts (xviii) to (xx), answer
the questions as instructed.
(i) A relation 𝑅 is defined on the set of students of a class such that [1]
𝑅 = {(𝑎, 𝑏): student 𝑎 and student 𝑏 have at least one common friend}
Which of the following statements is correct? (Understand)
(a) 𝑅 is an equivalence relation.
(b) 𝑅 is neither symmetric nor transitive.
(c) 𝑅 is symmetric but not necessarily transitive.
(d) 𝑅 is reflexive and transitive but not symmetric.
(ii) A bank offers two investment schemes: [1]
• Scheme A: 10% compounded annually
• Scheme B: 10% compounded quarterly
Which scheme gives a higher effective annual return? (Understand)
(a) Scheme A
(b) Scheme B
(c) Cannot be determined
(d) Both give the same return
(iii) 𝑥 0 [1]
If 𝐴 = ( ) is both symmetric and skew-symmetric, then 𝑥 + 𝑦 is: (Apply)
0 𝑦
(a) 0
(b) 3
(c) −3
(d) 6
2 ISC SPECIMEN QUESTION PAPER 2027
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Board Question Paper and should be used as a practice tool. However, the Board Question
Paper may not necessarily replicate the Specimen Question Paper as individual questions may
vary. The weightage allocated to various topics, as given in the syllabus, will be strictly
adhered to.
SECTION A – 20 MARKS
Question 1
In subparts (i) to (xvii) choose the correct options and in subparts (xviii) to (xx), answer
the questions as instructed.
(i) A relation 𝑅 is defined on the set of students of a class such that [1]
𝑅 = {(𝑎, 𝑏): student 𝑎 and student 𝑏 have at least one common friend}
Which of the following statements is correct? (Understand)
(a) 𝑅 is an equivalence relation.
(b) 𝑅 is neither symmetric nor transitive.
(c) 𝑅 is symmetric but not necessarily transitive.
(d) 𝑅 is reflexive and transitive but not symmetric.
(ii) A bank offers two investment schemes: [1]
• Scheme A: 10% compounded annually
• Scheme B: 10% compounded quarterly
Which scheme gives a higher effective annual return? (Understand)
(a) Scheme A
(b) Scheme B
(c) Cannot be determined
(d) Both give the same return
(iii) 𝑥 0 [1]
If 𝐴 = ( ) is both symmetric and skew-symmetric, then 𝑥 + 𝑦 is: (Apply)
0 𝑦
(a) 0
(b) 3
(c) −3
(d) 6
2 ISC SPECIMEN QUESTION PAPER 2027
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Page 4
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(iv) The function 𝑦 = −𝑥 3 + 𝑎𝑥 2 + 2𝑥 − 27 attains its maximum at 𝑥 = 1. Then the [1]
value of 𝑎 is: (Apply)
−1
(a)
2
1
m
(b)
co
2
(c) 2
. c om em .
(d) −2
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(v) a s
l (P) The number of trials is finite.
Trials of a random experiment are called Bernoulli trials if:
ag[1]
ag
(Q) The trials are dependent on each other.
(R) The probability of success or failure remains the same in each trial.
Choose the correct answer. (Recall)
(a) Only (P) and (Q)
(b) Only (Q) and (R)
m
(c) Only (P) and (R)
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(d) (P), (Q) and (R)
s e
l a
ag
(vi) [1]
𝑦 = 3𝑥 − 1
m
m The open half plane shaded in the above graph formed by the line 𝑦 = 3𝑥 − 1 is
.co
m .co represented as: (Understand)
s e m
s e (a) 3𝑥 − 𝑦 ≥ 1
g la
g la (b) 3𝑥 − 𝑦 ≤ 1 a
a (c) 3𝑥 − 𝑦 > 1
(d) 3𝑥 − 𝑦 < 1
m . c
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3 ISC SPECIMEN QUESTION PAPER 2027
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Page 3 of 34
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(iv) The function 𝑦 = −𝑥 3 + 𝑎𝑥 2 + 2𝑥 − 27 attains its maximum at 𝑥 = 1. Then the [1]
value of 𝑎 is: (Apply)
−1
(a)
2
1
m
(b)
co
2
(c) 2
. c om em .
(d) −2
e m l as
(v) a s
l (P) The number of trials is finite.
Trials of a random experiment are called Bernoulli trials if:
ag[1]
ag
(Q) The trials are dependent on each other.
(R) The probability of success or failure remains the same in each trial.
Choose the correct answer. (Recall)
(a) Only (P) and (Q)
(b) Only (Q) and (R)
m
(c) Only (P) and (R)
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(d) (P), (Q) and (R)
s e
l a
ag
(vi) [1]
𝑦 = 3𝑥 − 1
m
m The open half plane shaded in the above graph formed by the line 𝑦 = 3𝑥 − 1 is
.co
m .co represented as: (Understand)
s e m
s e (a) 3𝑥 − 𝑦 ≥ 1
g la
g la (b) 3𝑥 − 𝑦 ≤ 1 a
a (c) 3𝑥 − 𝑦 > 1
(d) 3𝑥 − 𝑦 < 1
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3 ISC SPECIMEN QUESTION PAPER 2027
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Page 5
(vii) Find the present value of perpetuity of ₹1000 payable at the end of each year, if [1]
money is worth 5% per annum. (Apply)
(a) ₹18,000
(b) ₹28,000
(c) ₹10,000
(d) ₹20,000
(viii) The production rate of a manufacturing company is modelled by the function: [1]
𝑃(𝑡) = 500 + 20𝑡, where 𝑃(𝑡) denotes the number of units produced per day
after 𝑡 days.
Assertion: The rate of change of production with respect to time is constant.
Reason: The production function is a linear function of 𝑡.
Choose the correct answer from the following options. (Understand)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
for Assertion
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(ix) Which of the following is an advantage of weighted index numbers in real life [1]
situations? (Recall)
(a) Base year is not needed.
(b) All items are treated equally.
(c) Calculations are always simple.
(d) Commodities are assigned weights according to their relative importance.
(x) Let 𝐴 and 𝐵 be square matrices of the same order. [1]
2 2 2
Statement I: If 𝐴𝐵 = 𝐵𝐴, then (𝐴 + 𝐵) = 𝐴 + 2𝐴𝐵 + 𝐵 .
Statement II: If 𝐴𝐵 ≠ 𝐵𝐴, then (𝐴 + 𝐵)2 = 𝐴2 + 𝐵 2 .
(Understand)
(a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.
4 ISC SPECIMEN QUESTION PAPER 2027
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money is worth 5% per annum. (Apply)
(a) ₹18,000
(b) ₹28,000
(c) ₹10,000
(d) ₹20,000
(viii) The production rate of a manufacturing company is modelled by the function: [1]
𝑃(𝑡) = 500 + 20𝑡, where 𝑃(𝑡) denotes the number of units produced per day
after 𝑡 days.
Assertion: The rate of change of production with respect to time is constant.
Reason: The production function is a linear function of 𝑡.
Choose the correct answer from the following options. (Understand)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
for Assertion
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(ix) Which of the following is an advantage of weighted index numbers in real life [1]
situations? (Recall)
(a) Base year is not needed.
(b) All items are treated equally.
(c) Calculations are always simple.
(d) Commodities are assigned weights according to their relative importance.
(x) Let 𝐴 and 𝐵 be square matrices of the same order. [1]
2 2 2
Statement I: If 𝐴𝐵 = 𝐵𝐴, then (𝐴 + 𝐵) = 𝐴 + 2𝐴𝐵 + 𝐵 .
Statement II: If 𝐴𝐵 ≠ 𝐵𝐴, then (𝐴 + 𝐵)2 = 𝐴2 + 𝐵 2 .
(Understand)
(a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.
4 ISC SPECIMEN QUESTION PAPER 2027
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Page 6
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4 [1]
(xi) If the mean and variance of a binomial distribution are 2 and 3 respectively, then
the number of trials, n is: (Apply)
(a) 6
(b) 1
6
m
(c) 5
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6
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(d)
a
l the graphs given below:
36 g
a [1]
ag
(xii) Observe
x1 x2 x1 x2
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e
Figure 1 Figure 2
l as
ag
x1 x2 x1 x2
Figure 3 Figure 4
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.co
Statement I: Figure 1 represents an increasing function and Figure 4 represents
m
.co em
a strictly decreasing function in the defined interval.
e m Statement II: Figure 2 represents a strictly increasing function and Figure 3
las
las represents a decreasing function in the defined interval.
g
a (Understand)
ag (a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.
m . c
c. o
5 ISC SPECIMEN QUESTION PAPER 2027
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Page 5 of 34
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4 [1]
(xi) If the mean and variance of a binomial distribution are 2 and 3 respectively, then
the number of trials, n is: (Apply)
(a) 6
(b) 1
6
m
(c) 5
om . co
6
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(d)
a
l the graphs given below:
36 g
a [1]
ag
(xii) Observe
x1 x2 x1 x2
m
m .co
e
Figure 1 Figure 2
l as
ag
x1 x2 x1 x2
Figure 3 Figure 4
m
.co
Statement I: Figure 1 represents an increasing function and Figure 4 represents
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a strictly decreasing function in the defined interval.
e m Statement II: Figure 2 represents a strictly increasing function and Figure 3
las
las represents a decreasing function in the defined interval.
g
a (Understand)
ag (a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.
m . c
c. o
5 ISC SPECIMEN QUESTION PAPER 2027
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Page 5 of 34
Page 7
(xiii) If the feasible region of a Linear Programming Problem (LPP) is unbounded, [1]
then the maximum or minimum value of the objective function: (Recall)
(a) always exists.
(b) never exists.
(c) may or may not exist.
(d) can never be determined as the feasible region has no corner point.
(xiv) In which EMI method, does the interest decrease with every installment paid? [1]
(Recall)
(a) Flat rate method
(b) Discount method
(c) Simple interest method
(d) Reducing balance method
(xv) Assertion: The mean of a probability distribution is always non-negative. [1]
Reason: Probability of an event is always non-negative.
(Understand)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(xvi) The system of equation 3𝑥 − 2𝑦 = 4 and 6𝑥 − 4𝑦 = 9 is: (Understand) [1]
(a) consistent
(b) dependent
(c) inconsistent
(d) homogeneous
(xvii) If the rate of return on an investment increases, then the time taken for the [1]
investment to double will generally: (Understand)
(a) increase
(b) decrease
(c) become zero
(d) remain unchanged
2 1/2
(xviii) Find the degree of the differential equation (1+ (𝑑𝑦) ) = 𝑥 + 𝑦. (Recall)
𝑑𝑥
6 ISC SPECIMEN QUESTION PAPER 2027
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then the maximum or minimum value of the objective function: (Recall)
(a) always exists.
(b) never exists.
(c) may or may not exist.
(d) can never be determined as the feasible region has no corner point.
(xiv) In which EMI method, does the interest decrease with every installment paid? [1]
(Recall)
(a) Flat rate method
(b) Discount method
(c) Simple interest method
(d) Reducing balance method
(xv) Assertion: The mean of a probability distribution is always non-negative. [1]
Reason: Probability of an event is always non-negative.
(Understand)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(xvi) The system of equation 3𝑥 − 2𝑦 = 4 and 6𝑥 − 4𝑦 = 9 is: (Understand) [1]
(a) consistent
(b) dependent
(c) inconsistent
(d) homogeneous
(xvii) If the rate of return on an investment increases, then the time taken for the [1]
investment to double will generally: (Understand)
(a) increase
(b) decrease
(c) become zero
(d) remain unchanged
2 1/2
(xviii) Find the degree of the differential equation (1+ (𝑑𝑦) ) = 𝑥 + 𝑦. (Recall)
𝑑𝑥
6 ISC SPECIMEN QUESTION PAPER 2027
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Page 8
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(xix) In a batch of 25 students at a dance academy, there are 12 boys and 13 girls. [1]
Among them, 5 boys and 7 girls are members of the Indian classical dance team.
A student is selected at random to lead the team at an international dance festival.
If it is known that the selected student is from the Indian classical dance team,
find the probability that the student is a girl. (Apply)
(xx) Evaluate ∫ 𝑒 𝑥 (𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔𝑠𝑖𝑛𝑥)𝑑𝑥 (Evaluate) [1]
m
om SECTION B – 14 MARKS . co
. c e m
m as
Question 2
ase a triangular plot of land with vertices: A(10,20), B(40,25), C(30,60). agl
A landownerlowns
agplans to utilise the plot for starting a new business based on its area.
The owner
•
If the area of the plot is greater than 100 square units, the owner wants to construct
a grocery shop.
• Otherwise, the owner wants to open a tea shop.
Based on the above information, answer the following questions:
(i) Using determinants, find the area of the triangular plot. (Analyse) [1]
om
(ii) Determine the type of shop the owner should construct. (Analyse) [1]
. c
e m
Question 3
l as [2]
ag OR ∫
3 √4−𝑥
(i) Using properties of definite integral, evaluate: 1 √4−𝑥+ √𝑥
𝑑𝑥 (Evaluate)
3
(ii) Evaluate: ∫−2|𝑥 − 1| 𝑑𝑥 (Evaluate)
Question 4 [2]
A group of friends is bowling at an entertainment mall. Each player gets 10 attempts.
5
m
.co
If the probability of a specific player hitting the pins on any given throw is 6, what is the
m
.co
probability that he will hit the pins at least 9 times? (Apply)
e m
e m las
las ag
ag
m . c
c. o
7 ISC SPECIMEN QUESTION PAPER 2027
e m
em as
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For more Question Papers, Sample Papers, Notes & Syllabus visit
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Page 7 of 34
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s em l a
a ag
(xix) In a batch of 25 students at a dance academy, there are 12 boys and 13 girls. [1]
Among them, 5 boys and 7 girls are members of the Indian classical dance team.
A student is selected at random to lead the team at an international dance festival.
If it is known that the selected student is from the Indian classical dance team,
find the probability that the student is a girl. (Apply)
(xx) Evaluate ∫ 𝑒 𝑥 (𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔𝑠𝑖𝑛𝑥)𝑑𝑥 (Evaluate) [1]
m
om SECTION B – 14 MARKS . co
. c e m
m as
Question 2
ase a triangular plot of land with vertices: A(10,20), B(40,25), C(30,60). agl
A landownerlowns
agplans to utilise the plot for starting a new business based on its area.
The owner
•
If the area of the plot is greater than 100 square units, the owner wants to construct
a grocery shop.
• Otherwise, the owner wants to open a tea shop.
Based on the above information, answer the following questions:
(i) Using determinants, find the area of the triangular plot. (Analyse) [1]
om
(ii) Determine the type of shop the owner should construct. (Analyse) [1]
. c
e m
Question 3
l as [2]
ag OR ∫
3 √4−𝑥
(i) Using properties of definite integral, evaluate: 1 √4−𝑥+ √𝑥
𝑑𝑥 (Evaluate)
3
(ii) Evaluate: ∫−2|𝑥 − 1| 𝑑𝑥 (Evaluate)
Question 4 [2]
A group of friends is bowling at an entertainment mall. Each player gets 10 attempts.
5
m
.co
If the probability of a specific player hitting the pins on any given throw is 6, what is the
m
.co
probability that he will hit the pins at least 9 times? (Apply)
e m
e m las
las ag
ag
m . c
c. o
7 ISC SPECIMEN QUESTION PAPER 2027
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Page 7 of 34
Page 9
Question 5
(i) The graph below shows the Employee Efficiency Index 𝑆(𝑥) of a company with
respect to the number of training sessions conducted (𝑥).
12
Employee Efficiency Index S(x)
10
8
6
4
2
0
0 1 2 3 4 5 6 7
Number of Training Sessions (x)
Based on the graph, answer the following questions:
(a) State the domain and range of the function represented by the graph. [1]
(Understand)
(b) Interpret the graph in the context of training sessions and Employee [1]
Efficiency Index. (Analyse)
OR
(ii) An online learning platform first doubles the course fee 𝑥 and adds a registration
charge of ₹200. This is represented by 𝑓(𝑥) = 2𝑥 + 200. A maintenance charge of
₹50 is then added, represented by 𝑔(𝑥) = 𝑥 + 50. Find:
(a) (𝑔 ∘ 𝑓)(𝑥) (Understand) [1]
(b) the final amount payable if the course fee is ₹1500. (Analyse) [1]
Question 6 [2]
1 𝑑2 𝑦
Given that 𝑥 = log 𝑡 and 𝑦 = for 𝑡 > 0 , find . (Apply)
𝑡 𝑑𝑥 2
8 ISC SPECIMEN QUESTION PAPER 2027
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(i) The graph below shows the Employee Efficiency Index 𝑆(𝑥) of a company with
respect to the number of training sessions conducted (𝑥).
12
Employee Efficiency Index S(x)
10
8
6
4
2
0
0 1 2 3 4 5 6 7
Number of Training Sessions (x)
Based on the graph, answer the following questions:
(a) State the domain and range of the function represented by the graph. [1]
(Understand)
(b) Interpret the graph in the context of training sessions and Employee [1]
Efficiency Index. (Analyse)
OR
(ii) An online learning platform first doubles the course fee 𝑥 and adds a registration
charge of ₹200. This is represented by 𝑓(𝑥) = 2𝑥 + 200. A maintenance charge of
₹50 is then added, represented by 𝑔(𝑥) = 𝑥 + 50. Find:
(a) (𝑔 ∘ 𝑓)(𝑥) (Understand) [1]
(b) the final amount payable if the course fee is ₹1500. (Analyse) [1]
Question 6 [2]
1 𝑑2 𝑦
Given that 𝑥 = log 𝑡 and 𝑦 = for 𝑡 > 0 , find . (Apply)
𝑡 𝑑𝑥 2
8 ISC SPECIMEN QUESTION PAPER 2027
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Page 10
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Question 7 [2]
The prices of the following commodities in the base year and current year are given
below:
Commodity Base Year Price (₹) (𝒑𝟎 ) Current Year Price (₹) (𝒑𝟏 )
Rice 40 50
m
om . co
Sugar
. c 30 36
e m
e m l as
s
Tea 20 24
a
l Aggregative Price Index Number. ag
a g
Find the Simple (Understand)
Question 8 [2]
(i) An investment doubled itself in 8 years under compound growth. Find the annual
rate of return correct to two decimal places. (Understand)
OR
(ii) The population of a town increased from 80,000 to 1,00,000 in 5 years. Find the
m
annual growth rate. (Understand)
SECTION C – 21 c o
. MARKS
e m
l as
Question 9
g
a by 𝑝 = √9 − 𝑥 for 0 ≤ 𝑥 ≤ 9.
The demand function of a product is given
[3]
Find the level of output, x at which the total revenue will be the maximum. (Analyse)
Question 10 [3]
The probability distribution of a random variable X is graphically represented below.
m
.co
P(X)
m
.co
1
2
e m
e m 1
las
las 4
ag
ag 1
8
1 2 3 4 X
Find the mean and variance of the probability distribution. (Understand)
m . c
c. o
9 ISC SPECIMEN QUESTION PAPER 2027
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em as
s
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.co s e m
s em l a
a ag
Question 7 [2]
The prices of the following commodities in the base year and current year are given
below:
Commodity Base Year Price (₹) (𝒑𝟎 ) Current Year Price (₹) (𝒑𝟏 )
Rice 40 50
m
om . co
Sugar
. c 30 36
e m
e m l as
s
Tea 20 24
a
l Aggregative Price Index Number. ag
a g
Find the Simple (Understand)
Question 8 [2]
(i) An investment doubled itself in 8 years under compound growth. Find the annual
rate of return correct to two decimal places. (Understand)
OR
(ii) The population of a town increased from 80,000 to 1,00,000 in 5 years. Find the
m
annual growth rate. (Understand)
SECTION C – 21 c o
. MARKS
e m
l as
Question 9
g
a by 𝑝 = √9 − 𝑥 for 0 ≤ 𝑥 ≤ 9.
The demand function of a product is given
[3]
Find the level of output, x at which the total revenue will be the maximum. (Analyse)
Question 10 [3]
The probability distribution of a random variable X is graphically represented below.
m
.co
P(X)
m
.co
1
2
e m
e m 1
las
las 4
ag
ag 1
8
1 2 3 4 X
Find the mean and variance of the probability distribution. (Understand)
m . c
c. o
9 ISC SPECIMEN QUESTION PAPER 2027
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s
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Page 11
Question 11
(i) Solve the following Linear Programming Problem graphically: [3]
Maximise Z = 3𝑥 + 2𝑦 subject to the constraints
𝑥 + 𝑦 ≤ 6, 𝑥 ≥ 2, 𝑦 ≤ 3 , 𝑥, 𝑦 ≥ 0 (Apply)
OR
(ii) The feasible region for a Linear Programming Problem is shown in the following
graph:
Y
B (0, 4)
D (0, 3)
P
O A(2, 0) C (3,0) X
Based on the given graph, answer the following questions: (Apply)
(a) Write the constraints for the Linear Programming Problem. [1]
(b) Find the coordinate of point P. [1]
(c) Calculate the least value of the objective function 𝑍 = 3𝑥 + 4𝑦. [1]
Question 12 [3]
𝑥2
(i) Evaluate: ∫ 2 𝑑𝑥 (Evaluate)
𝑥 −𝑎2
OR
𝜋
(ii) Find the value of: ∫02 𝑥 2 𝑠𝑖𝑛𝑥 3 𝑑𝑥 (Evaluate)
Question 13
A loan of ₹25,00,00 at the interest rate of 6% p.a. compounded monthly is to be amortised
by equal payments at the end of each month for 5 years. Find:
(i) the size of each monthly payment. (Apply) [2]
(ii) the principal outstanding at the beginning of 40th month. (Apply) [1]
10 ISC SPECIMEN QUESTION PAPER 2027
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(i) Solve the following Linear Programming Problem graphically: [3]
Maximise Z = 3𝑥 + 2𝑦 subject to the constraints
𝑥 + 𝑦 ≤ 6, 𝑥 ≥ 2, 𝑦 ≤ 3 , 𝑥, 𝑦 ≥ 0 (Apply)
OR
(ii) The feasible region for a Linear Programming Problem is shown in the following
graph:
Y
B (0, 4)
D (0, 3)
P
O A(2, 0) C (3,0) X
Based on the given graph, answer the following questions: (Apply)
(a) Write the constraints for the Linear Programming Problem. [1]
(b) Find the coordinate of point P. [1]
(c) Calculate the least value of the objective function 𝑍 = 3𝑥 + 4𝑦. [1]
Question 12 [3]
𝑥2
(i) Evaluate: ∫ 2 𝑑𝑥 (Evaluate)
𝑥 −𝑎2
OR
𝜋
(ii) Find the value of: ∫02 𝑥 2 𝑠𝑖𝑛𝑥 3 𝑑𝑥 (Evaluate)
Question 13
A loan of ₹25,00,00 at the interest rate of 6% p.a. compounded monthly is to be amortised
by equal payments at the end of each month for 5 years. Find:
(i) the size of each monthly payment. (Apply) [2]
(ii) the principal outstanding at the beginning of 40th month. (Apply) [1]
10 ISC SPECIMEN QUESTION PAPER 2027
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Question 14 [3]
(i) Solve for 𝑥 ∶ sin−1 𝑥 − cos −1 𝑥 = sin−1(3𝑥 − 2) (Apply)
OR
√1+𝑥 2 +√1−𝑥2 𝜋 1
(ii) Prove that: tan−1 (√1+𝑥 2 ) = + cos −1 𝑥 2 (Apply)
−√1−𝑥2 4 2
m
om . co
. c e m
s
Question 15 [3]
e m l a
s
Given below is the marketing and advertising expenditure in lakhs of rupees, of an
a ag
gl Mar Apr May Jun
apparel manufacturing company during each month of a particular year:
Jan aFeb Jul Aug Sep Oct Nov Dec
1·3 0·7 1·5 1·7 2 2 3·7 4·9 3·5 1·8 0·8 1·3
Calculate the 3-monthly moving averages and illustrate graphically. (Apply)
SECTION D – 25 MARKS
Question 16
m
.co
A supermarket owner wants to analyse the contribution of three departments —
m
Groceries, Clothing, and Electronics, towards the weekly profit of the store.
e
as
During a financial review, following observations were made:
l
ag
• The combined contribution of 2 Grocery sections, 1 Clothing section, and
1 Electronics section resulted in a weekly profit of ₹17,000.
• The combined contribution of 1 Grocery section, 2 Clothing sections, and 1
Electronics section resulted in a weekly profit of ₹20,000.
• The combined contribution of 1 Grocery section, 1 Clothing section, and 2
Electronics sections resulted in a weekly profit of ₹19,000.
The owner wants to determine the individual profit contribution of each department to
make future business decisions.
Based on the above information, answer the following:
o m
c
m the matrix method to determine the weekly profit contribution ofmGroceries,
.
c. (i)o Use s e (Apply)
[4]
m Clothing and Electronics.
a
e (ii) If the owner plans to expand only one department due to limitedglinvestment
l as a
funds, [1]
ag
which department should be given priority? Justify your answer using the obtained
values. (Evaluate)
m . c
c. o
11 ISC SPECIMEN QUESTION PAPER 2027
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s
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.co s e m
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Question 14 [3]
(i) Solve for 𝑥 ∶ sin−1 𝑥 − cos −1 𝑥 = sin−1(3𝑥 − 2) (Apply)
OR
√1+𝑥 2 +√1−𝑥2 𝜋 1
(ii) Prove that: tan−1 (√1+𝑥 2 ) = + cos −1 𝑥 2 (Apply)
−√1−𝑥2 4 2
m
om . co
. c e m
s
Question 15 [3]
e m l a
s
Given below is the marketing and advertising expenditure in lakhs of rupees, of an
a ag
gl Mar Apr May Jun
apparel manufacturing company during each month of a particular year:
Jan aFeb Jul Aug Sep Oct Nov Dec
1·3 0·7 1·5 1·7 2 2 3·7 4·9 3·5 1·8 0·8 1·3
Calculate the 3-monthly moving averages and illustrate graphically. (Apply)
SECTION D – 25 MARKS
Question 16
m
.co
A supermarket owner wants to analyse the contribution of three departments —
m
Groceries, Clothing, and Electronics, towards the weekly profit of the store.
e
as
During a financial review, following observations were made:
l
ag
• The combined contribution of 2 Grocery sections, 1 Clothing section, and
1 Electronics section resulted in a weekly profit of ₹17,000.
• The combined contribution of 1 Grocery section, 2 Clothing sections, and 1
Electronics section resulted in a weekly profit of ₹20,000.
• The combined contribution of 1 Grocery section, 1 Clothing section, and 2
Electronics sections resulted in a weekly profit of ₹19,000.
The owner wants to determine the individual profit contribution of each department to
make future business decisions.
Based on the above information, answer the following:
o m
c
m the matrix method to determine the weekly profit contribution ofmGroceries,
.
c. (i)o Use s e (Apply)
[4]
m Clothing and Electronics.
a
e (ii) If the owner plans to expand only one department due to limitedglinvestment
l as a
funds, [1]
ag
which department should be given priority? Justify your answer using the obtained
values. (Evaluate)
m . c
c. o
11 ISC SPECIMEN QUESTION PAPER 2027
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Page 13
Question 17
A company measures the effectiveness of its customer support system using a service
effectiveness index 𝑥. The customer satisfaction score 𝑦 is approximately modelled by the
function 𝑦 = (tan−1 𝑥)2 .
(i) Determine the marginal increase in the customer satisfaction score for a small [2]
increase in the service effectiveness index. Evaluate this marginal increase when
𝑥 = 1. (Analyse)
𝑑2 𝑦 𝑑𝑦 [3]
(ii) Hence, prove that (𝑥 2 + 1)2 + 2𝑥(𝑥 2 + 1) =2 (Apply)
𝑑𝑥 2 𝑑𝑥
Question 18
(i) In 2021, Rohan invested ₹80,000 in a mutual fund. After 4 years, the value of his
investment became ₹1,20,000.
At the same time, his sister Meera deposited ₹50,000 in a bank fixed deposit
scheme offering 8% per annum compounded quarterly.
Based on the above information, answer the following questions:
(a) Find the total Rate of Return earned by Rohan on his mutual fund investment. [1]
(Apply)
(b) Calculate the Compound Annual Growth Rate (CAGR) of Rohan’s [2]
investment. (Apply)
(c) Find the Nominal Rate of Return per annum for Meera’s fixed deposit if the [1]
interest is compounded quarterly. (Apply)
(d) Which investment appears to grow faster annually: Rohan’s mutual fund or [1]
Meera’s fixed deposit? Give a reason based on your calculations. (Evaluate)
OR
(ii) A school management plans to replace its old buses after 8 years. The estimated
cost of replacement is ₹24,00,000. The school decides to invest equal yearly
deposits into a sinking fund earning 6% interest compounded annually.
Based on the above information, answer the following questions:
(a) Determine the yearly deposit required to accumulate ₹24,00,000 in 8 years. [2]
(Apply)
(b) Find the total interest earned by the sinking fund at the end of 8 years. [1]
(Apply)
(c) The school management decides to collect the yearly sinking fund [2]
contribution only from Classes XI and XII students. If there are 80 students
in Class XI and 120 students in Class XII, and the contribution per student of
Classes XI and XII is in the ratio 1:2 respectively, find the annual
contribution per student of each class. (Evaluate)
12 ISC SPECIMEN QUESTION PAPER 2027
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A company measures the effectiveness of its customer support system using a service
effectiveness index 𝑥. The customer satisfaction score 𝑦 is approximately modelled by the
function 𝑦 = (tan−1 𝑥)2 .
(i) Determine the marginal increase in the customer satisfaction score for a small [2]
increase in the service effectiveness index. Evaluate this marginal increase when
𝑥 = 1. (Analyse)
𝑑2 𝑦 𝑑𝑦 [3]
(ii) Hence, prove that (𝑥 2 + 1)2 + 2𝑥(𝑥 2 + 1) =2 (Apply)
𝑑𝑥 2 𝑑𝑥
Question 18
(i) In 2021, Rohan invested ₹80,000 in a mutual fund. After 4 years, the value of his
investment became ₹1,20,000.
At the same time, his sister Meera deposited ₹50,000 in a bank fixed deposit
scheme offering 8% per annum compounded quarterly.
Based on the above information, answer the following questions:
(a) Find the total Rate of Return earned by Rohan on his mutual fund investment. [1]
(Apply)
(b) Calculate the Compound Annual Growth Rate (CAGR) of Rohan’s [2]
investment. (Apply)
(c) Find the Nominal Rate of Return per annum for Meera’s fixed deposit if the [1]
interest is compounded quarterly. (Apply)
(d) Which investment appears to grow faster annually: Rohan’s mutual fund or [1]
Meera’s fixed deposit? Give a reason based on your calculations. (Evaluate)
OR
(ii) A school management plans to replace its old buses after 8 years. The estimated
cost of replacement is ₹24,00,000. The school decides to invest equal yearly
deposits into a sinking fund earning 6% interest compounded annually.
Based on the above information, answer the following questions:
(a) Determine the yearly deposit required to accumulate ₹24,00,000 in 8 years. [2]
(Apply)
(b) Find the total interest earned by the sinking fund at the end of 8 years. [1]
(Apply)
(c) The school management decides to collect the yearly sinking fund [2]
contribution only from Classes XI and XII students. If there are 80 students
in Class XI and 120 students in Class XII, and the contribution per student of
Classes XI and XII is in the ratio 1:2 respectively, find the annual
contribution per student of each class. (Evaluate)
12 ISC SPECIMEN QUESTION PAPER 2027
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Question 19
(i) A company has three departments - Manufacturing (M), Marketing (K), and
Research & Development (R). The projects handled by the company are
distributed among the departments as follows:
Department M K R
m
co
Percentage of Projects 50% 30% 20%
. c om e m .
Past records show the following data of a project being a high-success project:
m
e Department l as
l as M K R
ag
ag Percentage of High Success Projects 80% 60% 90%
A project is selected at random.
Based on the above information, answer the following:
(a) Find the probability that the selected project is a high-success project. [1]
(Apply)
(b) What is the probability that the high-success project belongs to the [1]
Manufacturing Department? (Apply)
(c) Which department is most likely responsible for the selected high-profit [1]
m
.co
project? (Analyse)
em
(d) Suppose the proportion of projects handled by the Research & Development [2]
l as
Department increases from 20% to 30%, while the proportion handled by the
ag
Marketing Department decreases from 30% to 20%.
Find the new probability that a randomly selected project becomes a high-
success project.
Hence comment on whether the overall probability of obtaining a high-
success project increase or decreases. (Evaluate)
OR
(ii) During the Parents’ Appreciation Day, students at a school prepare greeting cards
and gifts for their parents.
m
Past records show that 3% of the items prepared by students contain defects such
.co
as improper folding, damaged decoration or incomplete finishing.
m
.co e m
The Headmaster decides that a batch will be approved for exhibition only if the
e m number of defective items is less than 4.
las
las A batch of 100 handmade gifts is submitted for approval.
ag
ag Using Poisson distribution, answer the following:
(a) Find the probability that the batch contains no defective items. (Apply) [2]
(b) What is the probability that the batch will be approved by the Headmaster? [2]
(Apply)
(c) Based on your answer, comment whether the quality of students’ work is [1]
satisfactory. (Evaluate)
m . c
c. o
13 ISC SPECIMEN QUESTION PAPER 2027
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Question 19
(i) A company has three departments - Manufacturing (M), Marketing (K), and
Research & Development (R). The projects handled by the company are
distributed among the departments as follows:
Department M K R
m
co
Percentage of Projects 50% 30% 20%
. c om e m .
Past records show the following data of a project being a high-success project:
m
e Department l as
l as M K R
ag
ag Percentage of High Success Projects 80% 60% 90%
A project is selected at random.
Based on the above information, answer the following:
(a) Find the probability that the selected project is a high-success project. [1]
(Apply)
(b) What is the probability that the high-success project belongs to the [1]
Manufacturing Department? (Apply)
(c) Which department is most likely responsible for the selected high-profit [1]
m
.co
project? (Analyse)
em
(d) Suppose the proportion of projects handled by the Research & Development [2]
l as
Department increases from 20% to 30%, while the proportion handled by the
ag
Marketing Department decreases from 30% to 20%.
Find the new probability that a randomly selected project becomes a high-
success project.
Hence comment on whether the overall probability of obtaining a high-
success project increase or decreases. (Evaluate)
OR
(ii) During the Parents’ Appreciation Day, students at a school prepare greeting cards
and gifts for their parents.
m
Past records show that 3% of the items prepared by students contain defects such
.co
as improper folding, damaged decoration or incomplete finishing.
m
.co e m
The Headmaster decides that a batch will be approved for exhibition only if the
e m number of defective items is less than 4.
las
las A batch of 100 handmade gifts is submitted for approval.
ag
ag Using Poisson distribution, answer the following:
(a) Find the probability that the batch contains no defective items. (Apply) [2]
(b) What is the probability that the batch will be approved by the Headmaster? [2]
(Apply)
(c) Based on your answer, comment whether the quality of students’ work is [1]
satisfactory. (Evaluate)
m . c
c. o
13 ISC SPECIMEN QUESTION PAPER 2027
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Page 15
Question 20
(i) A health research centre studies the sleeping pattern of elderly people participating
in a wellness programme. The researchers observe that awareness programmes
and medical support help improve healthy sleeping habits among elderly
participants.
Let 𝑥 represent the number of elderly people achieving healthy sleep patterns after
𝑦 weeks.
𝑑𝑥
The rate of improvement is modelled by: + 𝑥 = 100
𝑑𝑦
(a) Solve the differential equation and obtain the expression for the number of [3]
elderly participants achieving healthy sleep patterns after 𝑦 weeks. Use the
initial condition x = 40 when y = 0. (Apply)
(b) An old age home with 100 elderly residents plans to introduce a monthly
late-night entertainment programme (music, cultural events, movie
screening, etc.) to improve social interaction.
The management decides that the programme will be introduced only if at
least 80 residents achieve healthy sleep patterns.
(1) Using the expression obtained in (a), determine the number of [1]
residents achieving healthy sleep patterns after 2 weeks. (Analyse)
(2) Hence, suggest whether the old age home management should [1]
introduce the monthly late-night entertainment programme or not.
(Evaluate)
OR
(ii) A company manufacturing smart devices observes that its marginal revenue
function (in thousand rupees) for selling 𝑥 hundred units of the product is given
by:
𝑀𝑅(𝑥) = 𝑥 2 𝑒 𝑥
The company also knows that when no units are sold, the revenue is zero, i.e.,
𝑅(0) = 0
Based on the above information, answer the following questions:
(a) Find the revenue function 𝑅(𝑥). (Evaluate) [3]
(b) Another company earns revenue according to the model: [2]
𝑅1 (𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
Compare the revenues of the two companies at 𝑥 = 1 and determine which
company earns more revenue. (Analyse)
14 ISC SPECIMEN QUESTION PAPER 2027
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(i) A health research centre studies the sleeping pattern of elderly people participating
in a wellness programme. The researchers observe that awareness programmes
and medical support help improve healthy sleeping habits among elderly
participants.
Let 𝑥 represent the number of elderly people achieving healthy sleep patterns after
𝑦 weeks.
𝑑𝑥
The rate of improvement is modelled by: + 𝑥 = 100
𝑑𝑦
(a) Solve the differential equation and obtain the expression for the number of [3]
elderly participants achieving healthy sleep patterns after 𝑦 weeks. Use the
initial condition x = 40 when y = 0. (Apply)
(b) An old age home with 100 elderly residents plans to introduce a monthly
late-night entertainment programme (music, cultural events, movie
screening, etc.) to improve social interaction.
The management decides that the programme will be introduced only if at
least 80 residents achieve healthy sleep patterns.
(1) Using the expression obtained in (a), determine the number of [1]
residents achieving healthy sleep patterns after 2 weeks. (Analyse)
(2) Hence, suggest whether the old age home management should [1]
introduce the monthly late-night entertainment programme or not.
(Evaluate)
OR
(ii) A company manufacturing smart devices observes that its marginal revenue
function (in thousand rupees) for selling 𝑥 hundred units of the product is given
by:
𝑀𝑅(𝑥) = 𝑥 2 𝑒 𝑥
The company also knows that when no units are sold, the revenue is zero, i.e.,
𝑅(0) = 0
Based on the above information, answer the following questions:
(a) Find the revenue function 𝑅(𝑥). (Evaluate) [3]
(b) Another company earns revenue according to the model: [2]
𝑅1 (𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
Compare the revenues of the two companies at 𝑥 = 1 and determine which
company earns more revenue. (Analyse)
14 ISC SPECIMEN QUESTION PAPER 2027
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APPLIED MATHEMATICS
ANSWER KEY
SECTION A – 20 MARKS
Question 1
m
om
In answering Multiple Choice Questions, candidates have to write either the correct
. co
. c
option number or the explanation against it. Please note that only ONE correct
e m
e m
answer should be written.
a
l[1]s
as ag
glScheme B
(i) (c) 𝑅 is symmetric but not necessarily transitive.
(ii) a(b) [1]
(iii) (a) 0 [1]
(iv) 1 [1]
(b) 2
(v) (c) Only (P) and (R) [1]
(vi) (c) 3𝑥 − 𝑦 > 1 [1]
m
(vii) (d) ₹ 20,000
m .co [1]
(viii)
s e
(a) Both Assertion and Reason are true, and Reason is the correct explanation for [1]
Assertion.
l a
(ix) agweights according to their relative importance.
(d) Commodities are assigned [1]
(x) (a) Statement I is true and Statement II is false. [1]
(xi) (a) 6 [1]
(xii) (d) Both the statements are false. [1]
(xiii) (c) may or may not exist. [1]
m
c. o
(xiv) (d) Reducing balance method [1]
m
c. o(xv) (d) Assertion is false and Reason is true. s em
[1]
e m (xvi) (c) inconsistent la
las ag
[1]
ag (xvii) (b) decrease [1]
(xviii) Degree is 2 [1]
m . c
c. o
1 ISC SPECIMEN ANSWER KEY 2027
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Page 15 of 34
.co s e m
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a ag
APPLIED MATHEMATICS
ANSWER KEY
SECTION A – 20 MARKS
Question 1
m
om
In answering Multiple Choice Questions, candidates have to write either the correct
. co
. c
option number or the explanation against it. Please note that only ONE correct
e m
e m
answer should be written.
a
l[1]s
as ag
glScheme B
(i) (c) 𝑅 is symmetric but not necessarily transitive.
(ii) a(b) [1]
(iii) (a) 0 [1]
(iv) 1 [1]
(b) 2
(v) (c) Only (P) and (R) [1]
(vi) (c) 3𝑥 − 𝑦 > 1 [1]
m
(vii) (d) ₹ 20,000
m .co [1]
(viii)
s e
(a) Both Assertion and Reason are true, and Reason is the correct explanation for [1]
Assertion.
l a
(ix) agweights according to their relative importance.
(d) Commodities are assigned [1]
(x) (a) Statement I is true and Statement II is false. [1]
(xi) (a) 6 [1]
(xii) (d) Both the statements are false. [1]
(xiii) (c) may or may not exist. [1]
m
c. o
(xiv) (d) Reducing balance method [1]
m
c. o(xv) (d) Assertion is false and Reason is true. s em
[1]
e m (xvi) (c) inconsistent la
las ag
[1]
ag (xvii) (b) decrease [1]
(xviii) Degree is 2 [1]
m . c
c. o
1 ISC SPECIMEN ANSWER KEY 2027
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Page 17
(xix) Let [1]
𝐺: Selected student is a girl.
•
• 𝐶: Selected student is from the Indian classical dance team.
There are:
• Girls in the classical dance team = 7
• Total students in the classical dance team = 5 + 7 = 12
Given that the selected student is from the Indian classical dance team,
𝑃(𝐺∩𝐶)
𝑃(𝐺 ∣ 𝐶) =
𝑃(𝐶)
7
7
= 25
12 =
12
25
(xx) We know, [1]
∫ 𝑒 𝑥 (𝑓(𝑥) + 𝑓 ′ (𝑥))𝑑𝑥 = 𝑒 𝑥 𝑓(𝑥) + 𝐶
Assuming:
𝑓(𝑥) = 𝑙𝑜𝑔𝑠𝑖𝑛𝑥
⇒ 𝑓 ′ (𝑥) = 𝑐𝑜𝑡𝑥
⸫ Given Integral: ∫ 𝑒 𝑥 (𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔𝑠𝑖𝑛𝑥)𝑑𝑥 = 𝑒 𝑥 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 + 𝐶
SECTION B – 14 MARKS
Question 2
(i) Given the vertices of the triangular plot: [1]
𝐴(10,20), 𝐵(40,25), 𝐶(30,60)
Using determinant for the area of a triangle:
1 𝑥1 𝑦1 1
Area = |𝑥2 𝑦2 1|
2 𝑥 𝑦 1
3 3
Substituting the given points:
1 10 20 1
Area = |40 25 1|
2
30 60 1
Expanding the determinant:
1
= ∣ 10(25 − 60) − 20(40 − 30) + 1(40 × 60 − 25 × 30) ∣
2
1
= ∣ 10(−35) − 20(10) + (2400 − 750) ∣
2
1
= ∣ −350 − 200 + 1650 ∣
2
1
= ∣ 1100 ∣
2
1100
=
2
= 550
Area of the triangular plot: 550 square units
2 ISC SPECIMEN ANSWER KEY 2027
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𝐺: Selected student is a girl.
•
• 𝐶: Selected student is from the Indian classical dance team.
There are:
• Girls in the classical dance team = 7
• Total students in the classical dance team = 5 + 7 = 12
Given that the selected student is from the Indian classical dance team,
𝑃(𝐺∩𝐶)
𝑃(𝐺 ∣ 𝐶) =
𝑃(𝐶)
7
7
= 25
12 =
12
25
(xx) We know, [1]
∫ 𝑒 𝑥 (𝑓(𝑥) + 𝑓 ′ (𝑥))𝑑𝑥 = 𝑒 𝑥 𝑓(𝑥) + 𝐶
Assuming:
𝑓(𝑥) = 𝑙𝑜𝑔𝑠𝑖𝑛𝑥
⇒ 𝑓 ′ (𝑥) = 𝑐𝑜𝑡𝑥
⸫ Given Integral: ∫ 𝑒 𝑥 (𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔𝑠𝑖𝑛𝑥)𝑑𝑥 = 𝑒 𝑥 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 + 𝐶
SECTION B – 14 MARKS
Question 2
(i) Given the vertices of the triangular plot: [1]
𝐴(10,20), 𝐵(40,25), 𝐶(30,60)
Using determinant for the area of a triangle:
1 𝑥1 𝑦1 1
Area = |𝑥2 𝑦2 1|
2 𝑥 𝑦 1
3 3
Substituting the given points:
1 10 20 1
Area = |40 25 1|
2
30 60 1
Expanding the determinant:
1
= ∣ 10(25 − 60) − 20(40 − 30) + 1(40 × 60 − 25 × 30) ∣
2
1
= ∣ 10(−35) − 20(10) + (2400 − 750) ∣
2
1
= ∣ −350 − 200 + 1650 ∣
2
1
= ∣ 1100 ∣
2
1100
=
2
= 550
Area of the triangular plot: 550 square units
2 ISC SPECIMEN ANSWER KEY 2027
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Page 18
.
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(ii) Since the area of the plot is greater than 100 square units, 550 > 100 the owner [1]
should construct a Grocery Shop.
Question 3 [2]
3
m
(i) √4−𝑥
𝑑𝑥 …………(1)
co
Let 𝐼 = ∫
m .
√4−𝑥+√𝑥
1
c. o
Using the property:
s e m
∫ 𝑓(𝑥)e𝑑𝑥
𝑏
m= ∫ 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥
𝑏
l a
𝑎
s
la𝑎 = 1and 𝑏 = 3, so 𝑎 + 𝑏 = 4
𝑎
ag
g
Here,
aConsider,
√4−𝑥
𝑓(𝑥) = 4−𝑥+
√ √𝑥
√𝑥
𝑓(4 − 𝑥) =
√𝑥 + √4 − 𝑥
Therefore,
3
√𝑥
𝐼=∫ 𝑑𝑥 ………(2)
m
.co
√𝑥+√4−𝑥
1
m
Adding (1) & (2):
3
s e
2𝐼 = ∫
√4−𝑥 √𝑥
[ 4−𝑥+ 𝑥 + 𝑥+ 4−𝑥] 𝑑𝑥
l a
ag
√ √ √ √
1
3
√4−𝑥+√𝑥
2𝐼 = ∫ 𝑑𝑥
√4−𝑥+√𝑥
1
3
2𝐼 = ∫1 1 𝑑𝑥
2𝐼 = [𝑥]13
2𝐼 = 3 − 1
2𝐼 = 2
𝐼=1
m
.co
OR
m
c. o(ii)
3 1 3
em
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = ∫−2(1 − 𝑥) 𝑑𝑥 + ∫1 (𝑥 − 1) 𝑑𝑥
e m 3 𝑥2
1
𝑥2
3
las
las ∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [𝑥− 2 ] + [ 2 −𝑥]
−2 1
ag
ag 3 1
∫ ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [(1 − ) − (−2 − )] + [( − 3) − ( − 1) ]
−2 2
4
2 2
9 1
2
3 1 3 −1
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [(2) − (−4)] + [( ) − ( ) ]
2 2
3 9
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = 2 + 2
3 13
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = 2
m . c
c. o
3 ISC SPECIMEN ANSWER KEY 2027
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(ii) Since the area of the plot is greater than 100 square units, 550 > 100 the owner [1]
should construct a Grocery Shop.
Question 3 [2]
3
m
(i) √4−𝑥
𝑑𝑥 …………(1)
co
Let 𝐼 = ∫
m .
√4−𝑥+√𝑥
1
c. o
Using the property:
s e m
∫ 𝑓(𝑥)e𝑑𝑥
𝑏
m= ∫ 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥
𝑏
l a
𝑎
s
la𝑎 = 1and 𝑏 = 3, so 𝑎 + 𝑏 = 4
𝑎
ag
g
Here,
aConsider,
√4−𝑥
𝑓(𝑥) = 4−𝑥+
√ √𝑥
√𝑥
𝑓(4 − 𝑥) =
√𝑥 + √4 − 𝑥
Therefore,
3
√𝑥
𝐼=∫ 𝑑𝑥 ………(2)
m
.co
√𝑥+√4−𝑥
1
m
Adding (1) & (2):
3
s e
2𝐼 = ∫
√4−𝑥 √𝑥
[ 4−𝑥+ 𝑥 + 𝑥+ 4−𝑥] 𝑑𝑥
l a
ag
√ √ √ √
1
3
√4−𝑥+√𝑥
2𝐼 = ∫ 𝑑𝑥
√4−𝑥+√𝑥
1
3
2𝐼 = ∫1 1 𝑑𝑥
2𝐼 = [𝑥]13
2𝐼 = 3 − 1
2𝐼 = 2
𝐼=1
m
.co
OR
m
c. o(ii)
3 1 3
em
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = ∫−2(1 − 𝑥) 𝑑𝑥 + ∫1 (𝑥 − 1) 𝑑𝑥
e m 3 𝑥2
1
𝑥2
3
las
las ∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [𝑥− 2 ] + [ 2 −𝑥]
−2 1
ag
ag 3 1
∫ ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [(1 − ) − (−2 − )] + [( − 3) − ( − 1) ]
−2 2
4
2 2
9 1
2
3 1 3 −1
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = [(2) − (−4)] + [( ) − ( ) ]
2 2
3 9
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = 2 + 2
3 13
∫−2 ∣ 𝑥 − 1 ∣ 𝑑𝑥 = 2
m . c
c. o
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Page 19
Question 4 [2]
Let the random variable 𝑋 be the number of times the player hits the pins.
Given:
• Number of attempts, 𝑛 = 10
5
• Probability of hitting the pins, 𝑝 =
6
Since this is a binomial distribution,
5
𝑋 ∼ 𝐵(10, )
6
We need to find the probability that the player hits the pins at least 9 times:
𝑃(𝑋 ≥ 9) = 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10)
Using the binomial probability formula:
𝑃(𝑋 = 𝑟) = (𝑛𝑟)𝑝𝑟 (1 − 𝑝)𝑛−𝑟
5 9 1
𝑃(𝑋 = 9) = (10
9
)( ) ( )
6 6
5 9 1
= 10 ( ) (6)
6
10⋅59
= 610
5 10
𝑃(𝑋 = 10) = (10
10
)( )
6
5 10
=( )
6
510
= 610
10⋅59 510
𝑃(𝑋 ≥ 9) = 610 + 610
59 (10+5)
=
610
15⋅59
= 610
So,
15×1953125
𝑃(𝑋 ≥ 9) =
60466176
29296875
=
60466176
≈ 0 · 4845
Therefore, the probability that the player hits the pins at least 9 times is approximately
0 · 4845 or 48 · 45%
Question 5
(i) (a) From the graph, 𝑥 varies from 1 to 6. [1]
Domain = [1,6]
From the graph, the minimum value of 𝑆(𝑥)is 4 and the maximum value is
10.
Range = [4,10]
4 ISC SPECIMEN ANSWER KEY 2027
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Let the random variable 𝑋 be the number of times the player hits the pins.
Given:
• Number of attempts, 𝑛 = 10
5
• Probability of hitting the pins, 𝑝 =
6
Since this is a binomial distribution,
5
𝑋 ∼ 𝐵(10, )
6
We need to find the probability that the player hits the pins at least 9 times:
𝑃(𝑋 ≥ 9) = 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10)
Using the binomial probability formula:
𝑃(𝑋 = 𝑟) = (𝑛𝑟)𝑝𝑟 (1 − 𝑝)𝑛−𝑟
5 9 1
𝑃(𝑋 = 9) = (10
9
)( ) ( )
6 6
5 9 1
= 10 ( ) (6)
6
10⋅59
= 610
5 10
𝑃(𝑋 = 10) = (10
10
)( )
6
5 10
=( )
6
510
= 610
10⋅59 510
𝑃(𝑋 ≥ 9) = 610 + 610
59 (10+5)
=
610
15⋅59
= 610
So,
15×1953125
𝑃(𝑋 ≥ 9) =
60466176
29296875
=
60466176
≈ 0 · 4845
Therefore, the probability that the player hits the pins at least 9 times is approximately
0 · 4845 or 48 · 45%
Question 5
(i) (a) From the graph, 𝑥 varies from 1 to 6. [1]
Domain = [1,6]
From the graph, the minimum value of 𝑆(𝑥)is 4 and the maximum value is
10.
Range = [4,10]
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(b) The graph suggests that conducting around 4 training sessions is optimal for [1]
achieving the maximum employee efficiency.
OR
(ii) (a) Given: 𝑓(𝑥) = 2𝑥 + 200 [1]
𝑔(𝑥) = 𝑥 + 50
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))
m
co
𝑔(𝑓(𝑥)) = (2𝑥 + 200) + 50
om .
= 2𝑥 + 250
. c e m
Hence, (𝑔 ∘ 𝑓)(𝑥) = 2𝑥 + 250
e m l as
(b)
l asSubstitute 𝑥 = 1500 into (𝑔 ∘ 𝑓)(𝑥):
(𝑔 ∘ 𝑓)(𝑥) = 2(1500) + 250 ag
[1]
ag = 3000 + 250
= 3250
Hence, the final amount payable is: ₹3250
Question 6 [2]
1
Given: 𝑥 = log 𝑡, 𝑦 = , 𝑡 > 0
𝑡
1
𝑦= 𝑥 = log 𝑡
𝑡
m
.co
𝑑𝑦 1 𝑑𝑥 1
=− 2 =
𝑑𝑡 𝑡 𝑑𝑡 𝑡
e
𝑑𝑦
m
s
𝑑𝑦
= 𝑑𝑡
a
Using parametric differentiation:
l
𝑑𝑥 𝑑𝑥
ag
𝑑𝑡
𝑑𝑦 −1/𝑡 2
=
𝑑𝑥 1/𝑡
𝑑𝑦 1
=−
𝑑𝑥 𝑡
𝑑 𝑑𝑦
𝑑2 𝑦 ( )
Using: 2
= 𝑑𝑡𝑑𝑥𝑑𝑥
𝑑𝑥
𝑑𝑡
𝑑2 𝑦 1/𝑡 2
= 1/𝑡
𝑑𝑥 2
𝑑2 𝑦 1
=
𝑑𝑥 2
m
𝑡
om 7 .co
c. Question em
[2]
e m Simple Aggregative Price Index Number is: 𝑃 = ∑𝑝1
× 100
las
l as ∑𝑝 = 40 + 30 + 20 01 ∑𝑝0
ag
ag
0
∑𝑝 = 90 0
∑𝑝1 = 50 + 36 + 24
∑𝑝1 = 110
110
𝑃01 = × 100
90
𝑃01 = 12222 × 100
𝑃01 = 122 · 22
Therefore, the Simple Aggregative Price Index Number is: 122·22
m . c
c. o
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(b) The graph suggests that conducting around 4 training sessions is optimal for [1]
achieving the maximum employee efficiency.
OR
(ii) (a) Given: 𝑓(𝑥) = 2𝑥 + 200 [1]
𝑔(𝑥) = 𝑥 + 50
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))
m
co
𝑔(𝑓(𝑥)) = (2𝑥 + 200) + 50
om .
= 2𝑥 + 250
. c e m
Hence, (𝑔 ∘ 𝑓)(𝑥) = 2𝑥 + 250
e m l as
(b)
l asSubstitute 𝑥 = 1500 into (𝑔 ∘ 𝑓)(𝑥):
(𝑔 ∘ 𝑓)(𝑥) = 2(1500) + 250 ag
[1]
ag = 3000 + 250
= 3250
Hence, the final amount payable is: ₹3250
Question 6 [2]
1
Given: 𝑥 = log 𝑡, 𝑦 = , 𝑡 > 0
𝑡
1
𝑦= 𝑥 = log 𝑡
𝑡
m
.co
𝑑𝑦 1 𝑑𝑥 1
=− 2 =
𝑑𝑡 𝑡 𝑑𝑡 𝑡
e
𝑑𝑦
m
s
𝑑𝑦
= 𝑑𝑡
a
Using parametric differentiation:
l
𝑑𝑥 𝑑𝑥
ag
𝑑𝑡
𝑑𝑦 −1/𝑡 2
=
𝑑𝑥 1/𝑡
𝑑𝑦 1
=−
𝑑𝑥 𝑡
𝑑 𝑑𝑦
𝑑2 𝑦 ( )
Using: 2
= 𝑑𝑡𝑑𝑥𝑑𝑥
𝑑𝑥
𝑑𝑡
𝑑2 𝑦 1/𝑡 2
= 1/𝑡
𝑑𝑥 2
𝑑2 𝑦 1
=
𝑑𝑥 2
m
𝑡
om 7 .co
c. Question em
[2]
e m Simple Aggregative Price Index Number is: 𝑃 = ∑𝑝1
× 100
las
l as ∑𝑝 = 40 + 30 + 20 01 ∑𝑝0
ag
ag
0
∑𝑝 = 90 0
∑𝑝1 = 50 + 36 + 24
∑𝑝1 = 110
110
𝑃01 = × 100
90
𝑃01 = 12222 × 100
𝑃01 = 122 · 22
Therefore, the Simple Aggregative Price Index Number is: 122·22
m . c
c. o
5 ISC SPECIMEN ANSWER KEY 2027
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Page 21
Question 8 [2]
(i) Let the principal amount be 𝑃.
Since the investment doubles in 8 years,
𝐴 = 2𝑃
Using the compound growth formula:
𝐴 = 𝑃(1 + 𝑟)𝑛
where:
• 𝐴 = 2𝑃
• 𝑛=8
• 𝑟 = annual rate of return
2𝑃 = 𝑃(1 + 𝑟)8
2 = (1 + 𝑟)8
1 + 𝑟 = 21/8
𝑟 = 21/8 − 1
Now 21/8 ≈ 1 · 0905
𝑟 ≈ 1 · 0905 − 1
𝑟 ≈ 0 · 0905
𝑟 ≈ 9 · 05%
Therefore, the annual rate of return is approximately 9·05% per annum.
OR
(ii) Let the annual growth rate be 𝑟
Initial population: 𝑃 = 80,000
Final population after 5 years: 𝐴 = 1,00,000
Using the compound growth formula: 𝐴 = 𝑃(1 + 𝑟)𝑛
where 𝑛 = 5.
Substituting the values:
100000 = 80000(1 + 𝑟)5
100000
= (1 + 𝑟)5
80000
1 · 25 = (1 + 𝑟)5
1 + 𝑟 = (1 · 25)1/5
1 + 𝑟 ≈ 1 · 0456
𝑟 ≈ 1 · 0456 − 1
𝑟 ≈ 0 · 0456
𝑟 ≈ 4 · 56%
Therefore, the annual growth rate of the population is approximately 4·56% per
annum.
6 ISC SPECIMEN ANSWER KEY 2027
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(i) Let the principal amount be 𝑃.
Since the investment doubles in 8 years,
𝐴 = 2𝑃
Using the compound growth formula:
𝐴 = 𝑃(1 + 𝑟)𝑛
where:
• 𝐴 = 2𝑃
• 𝑛=8
• 𝑟 = annual rate of return
2𝑃 = 𝑃(1 + 𝑟)8
2 = (1 + 𝑟)8
1 + 𝑟 = 21/8
𝑟 = 21/8 − 1
Now 21/8 ≈ 1 · 0905
𝑟 ≈ 1 · 0905 − 1
𝑟 ≈ 0 · 0905
𝑟 ≈ 9 · 05%
Therefore, the annual rate of return is approximately 9·05% per annum.
OR
(ii) Let the annual growth rate be 𝑟
Initial population: 𝑃 = 80,000
Final population after 5 years: 𝐴 = 1,00,000
Using the compound growth formula: 𝐴 = 𝑃(1 + 𝑟)𝑛
where 𝑛 = 5.
Substituting the values:
100000 = 80000(1 + 𝑟)5
100000
= (1 + 𝑟)5
80000
1 · 25 = (1 + 𝑟)5
1 + 𝑟 = (1 · 25)1/5
1 + 𝑟 ≈ 1 · 0456
𝑟 ≈ 1 · 0456 − 1
𝑟 ≈ 0 · 0456
𝑟 ≈ 4 · 56%
Therefore, the annual growth rate of the population is approximately 4·56% per
annum.
6 ISC SPECIMEN ANSWER KEY 2027
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Page 22
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SECTION C – 21 MARKS
Question 9 [3]
Given 𝑝 = √9 − 𝑥 for 0 ≤ 𝑥 ≤ 9
Revenue function 𝑅(𝑥) = 𝑥𝑝 = 𝑥√9 − 𝑥
1/2
𝑅(𝑥) = 𝑥(9 − 𝑥)
𝑑𝑅 𝑥
m
co
= √9 − 𝑥 −
om .
𝑑𝑥 2√9−𝑥
𝑑𝑅 2(9−𝑥)−𝑥 18−3𝑥
𝑑𝑥
=
2√9−𝑥
=
.
2√9−𝑥c e m
e
For maximum revenue, m 𝑑𝑅
= 0 ⟹ 18 − 3𝑥 = 0 ⟹ 𝑥 = 6
l as
𝑑2 𝑅 as
= [ l ]
3 𝑥−12
𝑑𝑥
ag
𝑑𝑥 2
𝑑2 𝑅
ag 4 (9−𝑥)32
< 0 at 𝑥 = 6, hence the revenue is maximum.
𝑑𝑥 2
Question 10 [3]
From the graph, the probability distribution is:
1 𝑋2 3 4
1 1
𝑃(𝑋) m
1 1
8 2
m .co 4 8
The mean of a probability distribution is:
s e
𝜇 = ∑𝑥𝑃(𝑥)
l a
1 1 1
𝜇 = 1 (8) + 2 (2) + 3 (4) + 4 (8)
1 3 1
1
ag
= +1+ +
8 4 2
1+8+6+4
=
8
19
=
8
𝜇 = 2 · 375
𝐸(𝑋 2 ) = ∑𝑥 2 𝑃(𝑥)
m
1 1 1 1
= 12 ( ) + 22 ( ) + 32 ( ) + 42 ( )
.co
8 2 4 8
m = + 4 ( ) + 9 ( ) + 16 ( )
.co
1 1 1 1
8 2 4 8
e m
e m
1
= +2+ +2
9
las
s
8 4
la =
1+16+18+16
ag
ag =
51
8
8
Variance = 𝜎 2 = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2
51 19 2
= −( )
8 8
51 361 408−361
= − =
8 64 64
47
= = 0 · 734375
64
m . c
c. o
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SECTION C – 21 MARKS
Question 9 [3]
Given 𝑝 = √9 − 𝑥 for 0 ≤ 𝑥 ≤ 9
Revenue function 𝑅(𝑥) = 𝑥𝑝 = 𝑥√9 − 𝑥
1/2
𝑅(𝑥) = 𝑥(9 − 𝑥)
𝑑𝑅 𝑥
m
co
= √9 − 𝑥 −
om .
𝑑𝑥 2√9−𝑥
𝑑𝑅 2(9−𝑥)−𝑥 18−3𝑥
𝑑𝑥
=
2√9−𝑥
=
.
2√9−𝑥c e m
e
For maximum revenue, m 𝑑𝑅
= 0 ⟹ 18 − 3𝑥 = 0 ⟹ 𝑥 = 6
l as
𝑑2 𝑅 as
= [ l ]
3 𝑥−12
𝑑𝑥
ag
𝑑𝑥 2
𝑑2 𝑅
ag 4 (9−𝑥)32
< 0 at 𝑥 = 6, hence the revenue is maximum.
𝑑𝑥 2
Question 10 [3]
From the graph, the probability distribution is:
1 𝑋2 3 4
1 1
𝑃(𝑋) m
1 1
8 2
m .co 4 8
The mean of a probability distribution is:
s e
𝜇 = ∑𝑥𝑃(𝑥)
l a
1 1 1
𝜇 = 1 (8) + 2 (2) + 3 (4) + 4 (8)
1 3 1
1
ag
= +1+ +
8 4 2
1+8+6+4
=
8
19
=
8
𝜇 = 2 · 375
𝐸(𝑋 2 ) = ∑𝑥 2 𝑃(𝑥)
m
1 1 1 1
= 12 ( ) + 22 ( ) + 32 ( ) + 42 ( )
.co
8 2 4 8
m = + 4 ( ) + 9 ( ) + 16 ( )
.co
1 1 1 1
8 2 4 8
e m
e m
1
= +2+ +2
9
las
s
8 4
la =
1+16+18+16
ag
ag =
51
8
8
Variance = 𝜎 2 = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2
51 19 2
= −( )
8 8
51 361 408−361
= − =
8 64 64
47
= = 0 · 734375
64
m . c
c. o
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Page 23
Question 11
(i) [3]
𝑥, 𝑦 ≥ 0
OR
(ii) (a) 2𝑥 + 𝑦 ≥ 4 and 𝑥 + 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 [1]
(b) Solving 2𝑥 + 𝑦 = 4 and 𝑥 + 𝑦 = 3 [1]
𝑥 = 1, 𝑦 = 2
P (1,2)
8 ISC SPECIMEN ANSWER KEY 2027
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(i) [3]
𝑥, 𝑦 ≥ 0
OR
(ii) (a) 2𝑥 + 𝑦 ≥ 4 and 𝑥 + 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 [1]
(b) Solving 2𝑥 + 𝑦 = 4 and 𝑥 + 𝑦 = 3 [1]
𝑥 = 1, 𝑦 = 2
P (1,2)
8 ISC SPECIMEN ANSWER KEY 2027
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Page 24
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(c) For minimum value: [1]
Corner Point Value of Z = 3𝑥 + 4𝑦
B(0, 4) Z = 3(0) + 4(4) = 16
P(1, 2) Z = 3(1) + 4(2) = 11
C(3, 0) Z = 3(3) + 4(0) = 9
Minimum value of Z = 9
Now the open half plane 3𝑥 + 4𝑦 < 9 has no point in common with the
m
om
feasible region as shown in the graph below. Thus Minimum Value of the
. co
. c
objective function exists and is equal to 9.
e m
e m l as
l as ag
ag B (0, 4)
D (0, 3)
P
m
.co
O A(2, 0) C (3,0) X
e m
l as
ag
Question 12 [3]
(i) 𝑥2 (𝑥 2 −𝑎2 )+𝑎2
∫ 𝑥 2 −𝑎2 𝑑𝑥 = ∫ 𝑥 2 −𝑎2
𝑑𝑥
𝑎2
= ∫(1 + 𝑥 2−𝑎2 ) 𝑑𝑥
Thus,
𝑥2 1
m
.co
∫ 𝑑𝑥 = ∫ 1 𝑑𝑥 + 𝑎2 ∫ 2 𝑑𝑥
m 2
𝑥 −𝑎 2 𝑥 − 𝑎2
m .co ∫ 2
𝑥2
𝑑𝑥 = 𝑥 + 𝑎2 ∫
1
𝑑𝑥
s e m
e (𝑥 − 𝑎)(𝑥 + 𝑎)
la
𝑥 −𝑎 2
las 𝑥2 1 1
ag
g
1
∫ 𝑑𝑥 = 𝑥 + 𝑎2 ∫ ( − ) 𝑑𝑥
a 2
𝑥 −𝑎
𝑥2
2
𝑎
2𝑎 𝑥−𝑎 𝑥+𝑎
∫ 2 2
𝑑𝑥 = 𝑥 + [log ∣ 𝑥 − 𝑎 ∣ − log ∣ 𝑥 + 𝑎 ∣] + 𝐶
𝑥 −𝑎 2
𝑥2 𝑎 𝑥−𝑎
∫ 𝑥 2−𝑎2 𝑑𝑥 = 𝑥 + log | |+𝐶
2 𝑥+𝑎
OR
m . c
c. o
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(c) For minimum value: [1]
Corner Point Value of Z = 3𝑥 + 4𝑦
B(0, 4) Z = 3(0) + 4(4) = 16
P(1, 2) Z = 3(1) + 4(2) = 11
C(3, 0) Z = 3(3) + 4(0) = 9
Minimum value of Z = 9
Now the open half plane 3𝑥 + 4𝑦 < 9 has no point in common with the
m
om
feasible region as shown in the graph below. Thus Minimum Value of the
. co
. c
objective function exists and is equal to 9.
e m
e m l as
l as ag
ag B (0, 4)
D (0, 3)
P
m
.co
O A(2, 0) C (3,0) X
e m
l as
ag
Question 12 [3]
(i) 𝑥2 (𝑥 2 −𝑎2 )+𝑎2
∫ 𝑥 2 −𝑎2 𝑑𝑥 = ∫ 𝑥 2 −𝑎2
𝑑𝑥
𝑎2
= ∫(1 + 𝑥 2−𝑎2 ) 𝑑𝑥
Thus,
𝑥2 1
m
.co
∫ 𝑑𝑥 = ∫ 1 𝑑𝑥 + 𝑎2 ∫ 2 𝑑𝑥
m 2
𝑥 −𝑎 2 𝑥 − 𝑎2
m .co ∫ 2
𝑥2
𝑑𝑥 = 𝑥 + 𝑎2 ∫
1
𝑑𝑥
s e m
e (𝑥 − 𝑎)(𝑥 + 𝑎)
la
𝑥 −𝑎 2
las 𝑥2 1 1
ag
g
1
∫ 𝑑𝑥 = 𝑥 + 𝑎2 ∫ ( − ) 𝑑𝑥
a 2
𝑥 −𝑎
𝑥2
2
𝑎
2𝑎 𝑥−𝑎 𝑥+𝑎
∫ 2 2
𝑑𝑥 = 𝑥 + [log ∣ 𝑥 − 𝑎 ∣ − log ∣ 𝑥 + 𝑎 ∣] + 𝐶
𝑥 −𝑎 2
𝑥2 𝑎 𝑥−𝑎
∫ 𝑥 2−𝑎2 𝑑𝑥 = 𝑥 + log | |+𝐶
2 𝑥+𝑎
OR
m . c
c. o
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Page 25
𝜋
(ii) Let, 𝐼 = ∫0 𝑥 2 𝑠𝑖𝑛𝑥 3 𝑑𝑥
2
Using the substitution,
𝑢 = 𝑥3
𝑑𝑢 = 3𝑥 2 𝑑𝑥
𝑑𝑢
𝑥 2 𝑑𝑥 =
3
Changing the limits:
When 𝑥 = 0, 𝑢 = 03 = 0
𝜋 𝜋 3 𝜋3
When 𝑥 = , 𝑢 = ( ) = 8
2 2
𝜋3
1
So, 𝐼 = ∫0 sin 𝑢 𝑑𝑢8
3
𝜋3
1
Therefore, 𝐼 = [−cos 𝑢]0 8
3
1 𝜋3
= [−cos ( 8 ) + cos 0]
3
1 𝜋3
𝐼 = (1 − cos ( 8 ))
3
Question 13
Given:
• Loan amount 𝑃 = ₹250000
• Rate of interest = 6% p.a. compounded monthly
• Monthly interest rate:
6
𝑖= = 0 · 005
12×100
Time = 5 years = 60 months
(i) Size of each monthly payment [2]
𝑃𝑖
𝐴=
1−(1+𝑖)−𝑛
Substituting:
250000(0.005)
𝐴=
1−(1·005)−60
1250
𝐴=
1−(1·005)−60
Now,
( 1 · 005)−60 ≈ 0 · 7414
So,
1250
𝐴=
1−0·7414
1250
𝐴=
0·2586
𝐴 ≈ 4833 · 57
Monthly payment ₹4833 · 57
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(ii) Let, 𝐼 = ∫0 𝑥 2 𝑠𝑖𝑛𝑥 3 𝑑𝑥
2
Using the substitution,
𝑢 = 𝑥3
𝑑𝑢 = 3𝑥 2 𝑑𝑥
𝑑𝑢
𝑥 2 𝑑𝑥 =
3
Changing the limits:
When 𝑥 = 0, 𝑢 = 03 = 0
𝜋 𝜋 3 𝜋3
When 𝑥 = , 𝑢 = ( ) = 8
2 2
𝜋3
1
So, 𝐼 = ∫0 sin 𝑢 𝑑𝑢8
3
𝜋3
1
Therefore, 𝐼 = [−cos 𝑢]0 8
3
1 𝜋3
= [−cos ( 8 ) + cos 0]
3
1 𝜋3
𝐼 = (1 − cos ( 8 ))
3
Question 13
Given:
• Loan amount 𝑃 = ₹250000
• Rate of interest = 6% p.a. compounded monthly
• Monthly interest rate:
6
𝑖= = 0 · 005
12×100
Time = 5 years = 60 months
(i) Size of each monthly payment [2]
𝑃𝑖
𝐴=
1−(1+𝑖)−𝑛
Substituting:
250000(0.005)
𝐴=
1−(1·005)−60
1250
𝐴=
1−(1·005)−60
Now,
( 1 · 005)−60 ≈ 0 · 7414
So,
1250
𝐴=
1−0·7414
1250
𝐴=
0·2586
𝐴 ≈ 4833 · 57
Monthly payment ₹4833 · 57
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Page 26
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(ii) Principal outstanding at the beginning of the 40th month [1]
At the beginning of the 40th month, 39 payments have already been made.
Remaining payments:
60 − 39 = 21
Outstanding balance equals the present value of remaining payments:
1−(1+𝑖)−21
𝐵 = 𝐴[ ]
m
co
𝑖
om .
Substitute the values:
. c 1−(1·005)−21
e m
𝐵 = 4833 · 57 [
e m 0·005
]
l as
Now,
l as ag
ag
( 1 · 005)−21 ≈ 0.9006
1−0·9006
𝐵 = 4833 · 57 [ ]
0·005
0·0994
𝐵 = 4833 · 57 [ ]
0·005
𝐵 = 4833 · 57(19 · 88)
𝐵 ≈ 96090 · 35
Principal outstanding at the beginning of the 40th month ₹ 96090 · 35
m
c. o
Question 14 [3]
(i) Given:
e m
l as
sin−1 𝑥 − cos−1 𝑥 = sin−1 (3𝑥 − 2)
g 2)
−1 𝜋 −1 −1
sin 𝑥 − ( −sin 𝑥) = sin (3𝑥 −
Simplify:
2
a
𝜋
2sin−1 𝑥 − = sin−1 (3𝑥 − 2)
2
Let 𝜃 = sin−1 𝑥 then, 𝑥 = sin 𝜃
𝜋
2𝜃 − = sin−1 (3𝑥 − 2)
2
𝜋
sin (2𝜃− ) = 3𝑥 − 2
2
m
.co
−cos 2𝜃 = 3𝑥 − 2
m
.co m
Now,
m cos 2𝜃 = 1 − 2sin2 𝜃
s e
s e Since sin 𝜃 = 𝑥,
g la
g la −(1 − 2𝑥 2 ) = 3𝑥 − 2 a
a −1 + 2𝑥 2 = 3𝑥 − 2
2𝑥 2 − 3𝑥 + 1 = 0
(2𝑥 − 1)(𝑥 − 1) = 0
1
Thus, 𝑥 = or 𝑥 = 1
2
OR
m . c
c. o
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(ii) Principal outstanding at the beginning of the 40th month [1]
At the beginning of the 40th month, 39 payments have already been made.
Remaining payments:
60 − 39 = 21
Outstanding balance equals the present value of remaining payments:
1−(1+𝑖)−21
𝐵 = 𝐴[ ]
m
co
𝑖
om .
Substitute the values:
. c 1−(1·005)−21
e m
𝐵 = 4833 · 57 [
e m 0·005
]
l as
Now,
l as ag
ag
( 1 · 005)−21 ≈ 0.9006
1−0·9006
𝐵 = 4833 · 57 [ ]
0·005
0·0994
𝐵 = 4833 · 57 [ ]
0·005
𝐵 = 4833 · 57(19 · 88)
𝐵 ≈ 96090 · 35
Principal outstanding at the beginning of the 40th month ₹ 96090 · 35
m
c. o
Question 14 [3]
(i) Given:
e m
l as
sin−1 𝑥 − cos−1 𝑥 = sin−1 (3𝑥 − 2)
g 2)
−1 𝜋 −1 −1
sin 𝑥 − ( −sin 𝑥) = sin (3𝑥 −
Simplify:
2
a
𝜋
2sin−1 𝑥 − = sin−1 (3𝑥 − 2)
2
Let 𝜃 = sin−1 𝑥 then, 𝑥 = sin 𝜃
𝜋
2𝜃 − = sin−1 (3𝑥 − 2)
2
𝜋
sin (2𝜃− ) = 3𝑥 − 2
2
m
.co
−cos 2𝜃 = 3𝑥 − 2
m
.co m
Now,
m cos 2𝜃 = 1 − 2sin2 𝜃
s e
s e Since sin 𝜃 = 𝑥,
g la
g la −(1 − 2𝑥 2 ) = 3𝑥 − 2 a
a −1 + 2𝑥 2 = 3𝑥 − 2
2𝑥 2 − 3𝑥 + 1 = 0
(2𝑥 − 1)(𝑥 − 1) = 0
1
Thus, 𝑥 = or 𝑥 = 1
2
OR
m . c
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Page 27
(ii) Let 𝜃 = 𝑐𝑜𝑠 −1 (𝑥 2 ), then cos 𝜃 = 𝑥 2
𝜃 𝜃
√1 + 𝑥 2 = √1 + cos 𝜃 = √2cos 2, and √1 − 𝑥 2 = √1 − cos 𝜃 = √2sin 2
𝜃 𝜃
√1+𝑥 2 +√1−𝑥 2 cos +sin
Hence, √1+𝑥 2 −√1−𝑥 2
= 2
𝜃
2
𝜃
cos −sin
2 2
𝜋 1+tan 𝐴 𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝐴
Using, tan ( +𝐴) = =
4 1−tan 𝐴 𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴
𝜃 𝜃
cos +sin 𝜋 𝜃
2
𝜃 𝜃
2
= tan ( + )
cos −sin 4 2
2 2
√1+𝑥 2 +√1−𝑥2 𝜋 𝜃
Therefore, tan−1 (√1+𝑥 2 )= +
−√1−𝑥2 4 2
−1 2
Since 𝜃 = cos (𝑥 ),
√1+𝑥 2 +√1−𝑥 2 𝜋 1
we get tan−1 (√1+𝑥 2 ) = + cos−1 (𝑥 2 )
−√1−𝑥 2 4 2
Question 15 [3]
Months Actual Expenditure 3 Monthly Moving Average
Jan 1·3
Feb 0·7 1·17
Mar 1·5 1·30
Apr 1·7 1·73
May 2 1·90
Jun 2 2·57
Jul 3·7 3·53
Aug 4·9 4·03
Sep 3·5 3·40
Oct 1·8 2·03
Nov 0·8 1·30
Dec 1·3
3 Monthly Moving Average
6
Expenditure (in lakhs of Rupees)
5
4
3
2
1
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Months
Actual Expenditure 3 Monthly Moving Average
12 ISC SPECIMEN ANSWER KEY 2027
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𝜃 𝜃
√1 + 𝑥 2 = √1 + cos 𝜃 = √2cos 2, and √1 − 𝑥 2 = √1 − cos 𝜃 = √2sin 2
𝜃 𝜃
√1+𝑥 2 +√1−𝑥 2 cos +sin
Hence, √1+𝑥 2 −√1−𝑥 2
= 2
𝜃
2
𝜃
cos −sin
2 2
𝜋 1+tan 𝐴 𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝐴
Using, tan ( +𝐴) = =
4 1−tan 𝐴 𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴
𝜃 𝜃
cos +sin 𝜋 𝜃
2
𝜃 𝜃
2
= tan ( + )
cos −sin 4 2
2 2
√1+𝑥 2 +√1−𝑥2 𝜋 𝜃
Therefore, tan−1 (√1+𝑥 2 )= +
−√1−𝑥2 4 2
−1 2
Since 𝜃 = cos (𝑥 ),
√1+𝑥 2 +√1−𝑥 2 𝜋 1
we get tan−1 (√1+𝑥 2 ) = + cos−1 (𝑥 2 )
−√1−𝑥 2 4 2
Question 15 [3]
Months Actual Expenditure 3 Monthly Moving Average
Jan 1·3
Feb 0·7 1·17
Mar 1·5 1·30
Apr 1·7 1·73
May 2 1·90
Jun 2 2·57
Jul 3·7 3·53
Aug 4·9 4·03
Sep 3·5 3·40
Oct 1·8 2·03
Nov 0·8 1·30
Dec 1·3
3 Monthly Moving Average
6
Expenditure (in lakhs of Rupees)
5
4
3
2
1
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Months
Actual Expenditure 3 Monthly Moving Average
12 ISC SPECIMEN ANSWER KEY 2027
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SECTION D – 25 MARKS
Question 16
(i) Let the weekly profit contribution of: [4]
• Groceries be x
• Clothing be y
m
co
• Electronics be z
. c
2x + 1y + 1z = 17,000
om e m .
e m
1x + 2y + 1z = 20,000
l as
as
1x + 1y + 2z = 19,000
lsystem can be written in the matrix form 𝐴𝑋 = 𝐵 ag
a g
This
2 1 1 𝑥 17,000
[1 2 1] [𝑦] = [20,000]
1 1 2 𝑧 19,000
1
X = A−1B, where A−1 = |𝐴| adj(A)
|𝐴| = 2(3) − 1(1) + 1(−1) = 6 -1-1 = 4
3 −1 −1
𝑎𝑑𝑗(𝐴) = [−1 3 −1]
m
.co
−1 −1 3
m
−1
X=A B
1 3 −1 −1 17,000 s e
l a
ag
𝑋 = [−1 3 −1] [20,000]
4 19,000
−1 −1 3
3000
= [6000]
5000
The individual weekly profit contributions are:
• Groceries (x): ₹3,000
• Clothing (y): ₹6,000
• Electronics (z): ₹5,000
m
c. o
(ii)m If the owner can only expand one department, Clothing should be given priority.
m.co [1]
m Based on the calculated values, the Clothing department yields the highest
s e
s e individual weekly profit contribution (₹6,000), followed by Electronics (₹5,000)
g la
g la and Groceries (₹3,000). Allocating the limited investment funds to Clothing
a
a maximises the return on investment per section.
m . c
c. o
13 ISC SPECIMEN ANSWER KEY 2027
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s
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.co s e m
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a ag
SECTION D – 25 MARKS
Question 16
(i) Let the weekly profit contribution of: [4]
• Groceries be x
• Clothing be y
m
co
• Electronics be z
. c
2x + 1y + 1z = 17,000
om e m .
e m
1x + 2y + 1z = 20,000
l as
as
1x + 1y + 2z = 19,000
lsystem can be written in the matrix form 𝐴𝑋 = 𝐵 ag
a g
This
2 1 1 𝑥 17,000
[1 2 1] [𝑦] = [20,000]
1 1 2 𝑧 19,000
1
X = A−1B, where A−1 = |𝐴| adj(A)
|𝐴| = 2(3) − 1(1) + 1(−1) = 6 -1-1 = 4
3 −1 −1
𝑎𝑑𝑗(𝐴) = [−1 3 −1]
m
.co
−1 −1 3
m
−1
X=A B
1 3 −1 −1 17,000 s e
l a
ag
𝑋 = [−1 3 −1] [20,000]
4 19,000
−1 −1 3
3000
= [6000]
5000
The individual weekly profit contributions are:
• Groceries (x): ₹3,000
• Clothing (y): ₹6,000
• Electronics (z): ₹5,000
m
c. o
(ii)m If the owner can only expand one department, Clothing should be given priority.
m.co [1]
m Based on the calculated values, the Clothing department yields the highest
s e
s e individual weekly profit contribution (₹6,000), followed by Electronics (₹5,000)
g la
g la and Groceries (₹3,000). Allocating the limited investment funds to Clothing
a
a maximises the return on investment per section.
m . c
c. o
13 ISC SPECIMEN ANSWER KEY 2027
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Question 17
(i) The marginal increase in the customer satisfaction score is, [2]
𝑑𝑦 −1 1 𝑑𝑦 2tan−1 𝑥
= 2tan 𝑥⋅ ⟹ =
𝑑𝑥 1+𝑥 2 𝑑𝑥 1+𝑥 2
−1 𝜋
At 𝑥 = 1, tan (1) = .
4
𝜋
𝑑𝑦 𝜋 2( )
4
∣ = 1+1 = .
𝑑𝑥 𝑥=1 4
(ii) 𝑑𝑦 2tan−1 𝑥 [3]
=
𝑑𝑥 1 + 𝑥2
𝑑𝑦
(1 + 𝑥 2 ) = 2tan−1 𝑥
𝑑𝑥
2
𝑑 𝑦 𝑑𝑦 1
(1 + 𝑥 2 ) + 2𝑥 = 2
𝑑𝑥 2 𝑑𝑥 1 + 𝑥2
2
𝑑 𝑦 𝑑𝑦
(1 + 𝑥 2 )2 + 2𝑥 (1 + 𝑥 2)
= 2
𝑑𝑥 2 𝑑𝑥
Question 18
(i) (a) Total Rate of Return (RoR) for Rohan [1]
Final Value−Initial Value
Total RoR = × 100
Initial Value
• Initial Value (Principal): ₹80,000
• Final Value: ₹1,20,000
1,20,000−80,000 40,000
Total RoR = × 100 = × 100 = 50%
80,000 80,000
(b) Compound Annual Growth Rate (CAGR) for Rohan [2]
1
Final Value 𝑛
CAGR = ( ) −1
Initial Value
• Number of years (n): 4
1
1,20,000 4
CAGR = ( ) − 1 = (1 · 5)0.25 − 1
80,000
( 1 · 5)0.25 ≈ 1 · 10668
CAGR = 1 · 10668 − 1 = 0 · 10668 or 10 · 67%
Answer: The CAGR of Rohan’s investment is approximately 10·67%.
(c) Nominal Rate of Return per Annum for Meera [1]
Given that Meera’s bank fixed deposit scheme offers 8% per annum
compounded quarterly.
The Nominal Rate of Return per annum for Meera’s fixed deposit is 8%.
14 ISC SPECIMEN ANSWER KEY 2027
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(i) The marginal increase in the customer satisfaction score is, [2]
𝑑𝑦 −1 1 𝑑𝑦 2tan−1 𝑥
= 2tan 𝑥⋅ ⟹ =
𝑑𝑥 1+𝑥 2 𝑑𝑥 1+𝑥 2
−1 𝜋
At 𝑥 = 1, tan (1) = .
4
𝜋
𝑑𝑦 𝜋 2( )
4
∣ = 1+1 = .
𝑑𝑥 𝑥=1 4
(ii) 𝑑𝑦 2tan−1 𝑥 [3]
=
𝑑𝑥 1 + 𝑥2
𝑑𝑦
(1 + 𝑥 2 ) = 2tan−1 𝑥
𝑑𝑥
2
𝑑 𝑦 𝑑𝑦 1
(1 + 𝑥 2 ) + 2𝑥 = 2
𝑑𝑥 2 𝑑𝑥 1 + 𝑥2
2
𝑑 𝑦 𝑑𝑦
(1 + 𝑥 2 )2 + 2𝑥 (1 + 𝑥 2)
= 2
𝑑𝑥 2 𝑑𝑥
Question 18
(i) (a) Total Rate of Return (RoR) for Rohan [1]
Final Value−Initial Value
Total RoR = × 100
Initial Value
• Initial Value (Principal): ₹80,000
• Final Value: ₹1,20,000
1,20,000−80,000 40,000
Total RoR = × 100 = × 100 = 50%
80,000 80,000
(b) Compound Annual Growth Rate (CAGR) for Rohan [2]
1
Final Value 𝑛
CAGR = ( ) −1
Initial Value
• Number of years (n): 4
1
1,20,000 4
CAGR = ( ) − 1 = (1 · 5)0.25 − 1
80,000
( 1 · 5)0.25 ≈ 1 · 10668
CAGR = 1 · 10668 − 1 = 0 · 10668 or 10 · 67%
Answer: The CAGR of Rohan’s investment is approximately 10·67%.
(c) Nominal Rate of Return per Annum for Meera [1]
Given that Meera’s bank fixed deposit scheme offers 8% per annum
compounded quarterly.
The Nominal Rate of Return per annum for Meera’s fixed deposit is 8%.
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(d) Growth Comparison and Reason [1]
Rohan’s mutual fund CAGR = 10.67% p.a.
Meera’s fixed deposit nominal rate = 8% p.a.
Since 10.67% > 8%, Rohan’s mutual fund appears to grow faster
annually.
Rohan’s mutual fund grows faster because its annual growth rate (10·67%) is
higher than Meera’s fixed deposit (8%).
m
om
OR
. co
. c e m
s
(ii) Given:
• m a
s e Future value required = ₹24,00,000
g l
l a • Rate of interest = 6% p.a. compounded annually
a
agFor •a sinking
Time = 8 years
fund, the yearly deposit 𝑃is given by:
𝐴×𝑟
𝑃 = (1+𝑟)𝑛−1
where
𝐴 = 24,00,000, 𝑟 = 0 · 06, 𝑛 = 8
(a) Yearly deposit required: [2]
24,00,000×0.06
𝑃=
m
8
(1.06) −1
.co
8
( 1.06) ≈ 1.5938
𝑃=
1,44,000
e m
𝑃=
1.5938−1
1,44,000
l as
ag
0.5938
𝑃 ≈ 2,42,505
Hence, the yearly deposit required is: ₹2,42,505 (approx.)
(b) Total interest earned [1]
Total amount accumulated after 8 years:
= ₹24,00,000
Total deposits made:
= 8 × 2,42,505
m
m = ₹19,40,040
.co
m .co Interest earned:
s e m
s e = 24,00,000 − 19,40,040
g la
la
= ₹4,59,960
g a
a (c)
Hence, the total interest earned is: ₹4,59,960
Contribution per student of Classes XI and XII [2]
Students in Class XI = 80
Students in Class XII = 120
Contribution ratio per student 1: 2
Let Class XI contribution per student = ₹x
m . c
c. o
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(d) Growth Comparison and Reason [1]
Rohan’s mutual fund CAGR = 10.67% p.a.
Meera’s fixed deposit nominal rate = 8% p.a.
Since 10.67% > 8%, Rohan’s mutual fund appears to grow faster
annually.
Rohan’s mutual fund grows faster because its annual growth rate (10·67%) is
higher than Meera’s fixed deposit (8%).
m
om
OR
. co
. c e m
s
(ii) Given:
• m a
s e Future value required = ₹24,00,000
g l
l a • Rate of interest = 6% p.a. compounded annually
a
agFor •a sinking
Time = 8 years
fund, the yearly deposit 𝑃is given by:
𝐴×𝑟
𝑃 = (1+𝑟)𝑛−1
where
𝐴 = 24,00,000, 𝑟 = 0 · 06, 𝑛 = 8
(a) Yearly deposit required: [2]
24,00,000×0.06
𝑃=
m
8
(1.06) −1
.co
8
( 1.06) ≈ 1.5938
𝑃=
1,44,000
e m
𝑃=
1.5938−1
1,44,000
l as
ag
0.5938
𝑃 ≈ 2,42,505
Hence, the yearly deposit required is: ₹2,42,505 (approx.)
(b) Total interest earned [1]
Total amount accumulated after 8 years:
= ₹24,00,000
Total deposits made:
= 8 × 2,42,505
m
m = ₹19,40,040
.co
m .co Interest earned:
s e m
s e = 24,00,000 − 19,40,040
g la
la
= ₹4,59,960
g a
a (c)
Hence, the total interest earned is: ₹4,59,960
Contribution per student of Classes XI and XII [2]
Students in Class XI = 80
Students in Class XII = 120
Contribution ratio per student 1: 2
Let Class XI contribution per student = ₹x
m . c
c. o
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Then Class XII contribution per student = ₹2x
Total contribution:
80(𝑥) + 120(2𝑥) = 2,42,505
80𝑥 + 240𝑥 = 2,42,505
320𝑥 = 2,42,505
𝑥 ≈ 757 · 83
So,
Class XI contribution per student: ₹758 (approx.)
Class XII contribution per student:
= 2𝑥 ≈ 1515 · 66
=₹1,516 (approx.)
Question 19
(i) Let us define the events representing the selection of a project from each
department:
• 𝐸1 : The project is from the Manufacturing department (M).
• 𝐸2 : The project is from the Marketing department (K).
𝐸3 : The project is from the Research & Development department (R).
•
Let 𝐴 be the event that the selected project is a high success project.
Given Probabilities:
𝑃(𝐸1 ) = 50% = 0 · 50
𝑃(𝐸2 ) = 30% = 0 · 30
𝑃(𝐸3 ) = 20% = 0 · 20
𝑃(𝐴 ∣ 𝐸1 ) = 80% = 0 · 80
𝑃(𝐴 ∣ 𝐸2 ) = 60% = 0 · 60
𝑃(𝐴 ∣ 𝐸3 ) = 90% = 0 · 90
(a) By the Law of Total Probability, the overall probability 𝑃(𝐴) that a [1]
randomly chosen project is a high success project is:
𝑃(𝐴) = 𝑃(𝐸1 ) ⋅ 𝑃(𝐴 ∣ 𝐸1 ) + 𝑃(𝐸2 ) ⋅ 𝑃(𝐴 ∣ 𝐸2 ) + 𝑃(𝐸3 ) ⋅ 𝑃(𝐴 ∣ 𝐸3 )
Substitute the given values into the formula:
𝑃(𝐴) = (0 · 50 × 0 · 80) + (0 · 30 × 0 · 60) + (0 · 20 × 0 · 90)
𝑃(𝐴) = 0 · 40 + 0 · 18 + 0 · 18
𝑃(𝐴) = 0 · 76
Answer: The probability that the selected project is a high success project
0·76 (or 76%).
(b) Probability of selected high success project is from Manufacturing [1]
department:
𝑃(𝐸 )⋅𝑃( 𝐴∣𝐸 ) 0·40 40
𝑃( 𝐸1 ∣ 𝐴 ) = 1 𝑃(𝐴) ∣ 1 = = ≈ 0 · 5263 (52 · 63%)
0·76 76
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Total contribution:
80(𝑥) + 120(2𝑥) = 2,42,505
80𝑥 + 240𝑥 = 2,42,505
320𝑥 = 2,42,505
𝑥 ≈ 757 · 83
So,
Class XI contribution per student: ₹758 (approx.)
Class XII contribution per student:
= 2𝑥 ≈ 1515 · 66
=₹1,516 (approx.)
Question 19
(i) Let us define the events representing the selection of a project from each
department:
• 𝐸1 : The project is from the Manufacturing department (M).
• 𝐸2 : The project is from the Marketing department (K).
𝐸3 : The project is from the Research & Development department (R).
•
Let 𝐴 be the event that the selected project is a high success project.
Given Probabilities:
𝑃(𝐸1 ) = 50% = 0 · 50
𝑃(𝐸2 ) = 30% = 0 · 30
𝑃(𝐸3 ) = 20% = 0 · 20
𝑃(𝐴 ∣ 𝐸1 ) = 80% = 0 · 80
𝑃(𝐴 ∣ 𝐸2 ) = 60% = 0 · 60
𝑃(𝐴 ∣ 𝐸3 ) = 90% = 0 · 90
(a) By the Law of Total Probability, the overall probability 𝑃(𝐴) that a [1]
randomly chosen project is a high success project is:
𝑃(𝐴) = 𝑃(𝐸1 ) ⋅ 𝑃(𝐴 ∣ 𝐸1 ) + 𝑃(𝐸2 ) ⋅ 𝑃(𝐴 ∣ 𝐸2 ) + 𝑃(𝐸3 ) ⋅ 𝑃(𝐴 ∣ 𝐸3 )
Substitute the given values into the formula:
𝑃(𝐴) = (0 · 50 × 0 · 80) + (0 · 30 × 0 · 60) + (0 · 20 × 0 · 90)
𝑃(𝐴) = 0 · 40 + 0 · 18 + 0 · 18
𝑃(𝐴) = 0 · 76
Answer: The probability that the selected project is a high success project
0·76 (or 76%).
(b) Probability of selected high success project is from Manufacturing [1]
department:
𝑃(𝐸 )⋅𝑃( 𝐴∣𝐸 ) 0·40 40
𝑃( 𝐸1 ∣ 𝐴 ) = 1 𝑃(𝐴) ∣ 1 = = ≈ 0 · 5263 (52 · 63%)
0·76 76
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(c) To find out which department is most likely responsible for this high success [1]
project, calculate the posterior conditional probabilities 𝑃( 𝐸𝑖 ∣ 𝐴 )using
Bayes' Theorem:
𝑃(𝐸𝑖 )⋅𝑃( 𝐴∣∣𝐸𝑖 )
𝑃( 𝐸𝑖 ∣ 𝐴 ) = 𝑃(𝐴)
1≤𝑖≤3
1. Manufacturing (𝐸1 ):
m
0·40 40
𝑃(𝐸1 ∣ 𝐴) = = ≈ 0 · 5263 (52 · 63%)
co
0·76 76
. c om
2. Marketing (𝐸2 ):
e m .
m
𝑃(𝐸2 ∣ 𝐴) =
e
0·18
=
18
≈ 0 · 2368 (23 · 68%)
l as
s
0·76 76
a
l𝑃(𝐸 ∣ 𝐴) =
3. Research & Development (𝐸3 ):
ag
ag 3
0·18
0·76
=
18
76
≈ 0 · 2368 (23 · 68%)
Answer: The Manufacturing Department is most likely responsible for the
selected high-success project, with a conditional probability of approximately
52·63%.
(d) • Manufacturing (𝐸1 ) remains unchanged: 𝑃(𝐸1 ) = 0 · 50 [2]
• Marketing (𝐸2 ) decreases to 20%: 𝑃𝑛𝑒𝑤 (𝐸2 ) = 0 · 20
• Research & Development (𝐸3 ) increases to 30%: 𝑃𝑛𝑒𝑤 (𝐸3 ) = 0 · 30
m
.co
The probabilities of a project being high-success from each individual
department
e m
𝑃(𝐴 ∣ 𝐸𝑖 ) 1 ≤ 𝑖 ≤ 3 remains the same.
la(0s · 20 × 0 · 60) + (0 · 30 × 0 · 90)
𝑃𝑛𝑒𝑤 (𝐴) = 𝑃(𝐸1 ) ⋅ 𝑃(𝐴 ∣ 𝐸1 ) + 𝑃𝑛𝑒𝑤 (𝐸2 ) ⋅ 𝑃(𝐴 ∣ 𝐸2 ) + 𝑃𝑛𝑒𝑤 (𝐸3 ) ⋅ 𝑃(𝐴 ∣ 𝐸3 )
𝑃 (𝐴) = (0 · 50 × 0 · 80)g+
𝑛𝑒𝑤
a
𝑃 (𝐴) = 0 · 40 + 0 · 12 + 0 · 27
𝑛𝑒𝑤
𝑃𝑛𝑒𝑤 (𝐴) = 0 · 79
• Old overall probability 𝑃(𝐴) = 0 · 76
• New overall probability 𝑃𝑛𝑒𝑤 (𝐴) = 0 · 79
The overall probability of getting a high-success project increases from 76%
to 79%.
m
.co
OR
m
c. (ii)o Given: s em
e m Total number of handmade items (n) = 100
la
las Probability of a defective item (p) = 3% = 0·03
g
Since 𝑛 is large (𝑛 ≥ 20) and 𝑝 is small (𝑝 ≤ 0 · 05), wea can approximate the
ag Binomial distribution using a Poisson distribution with parameter 𝜆(mean):
𝜆 = 𝑛 ⋅ 𝑝 = 100 × 0 · 03 = 3
The probability mass function of a Poisson distribution is given by:
𝑒 −𝜆 ⋅𝜆𝑥
𝑃(𝑋 = 𝑥) = 𝑥!
For 𝜆 = 3:
𝑒 −3 ⋅3𝑥
𝑃(𝑋 = 𝑥) = 𝑥!
m . c
c. o
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(c) To find out which department is most likely responsible for this high success [1]
project, calculate the posterior conditional probabilities 𝑃( 𝐸𝑖 ∣ 𝐴 )using
Bayes' Theorem:
𝑃(𝐸𝑖 )⋅𝑃( 𝐴∣∣𝐸𝑖 )
𝑃( 𝐸𝑖 ∣ 𝐴 ) = 𝑃(𝐴)
1≤𝑖≤3
1. Manufacturing (𝐸1 ):
m
0·40 40
𝑃(𝐸1 ∣ 𝐴) = = ≈ 0 · 5263 (52 · 63%)
co
0·76 76
. c om
2. Marketing (𝐸2 ):
e m .
m
𝑃(𝐸2 ∣ 𝐴) =
e
0·18
=
18
≈ 0 · 2368 (23 · 68%)
l as
s
0·76 76
a
l𝑃(𝐸 ∣ 𝐴) =
3. Research & Development (𝐸3 ):
ag
ag 3
0·18
0·76
=
18
76
≈ 0 · 2368 (23 · 68%)
Answer: The Manufacturing Department is most likely responsible for the
selected high-success project, with a conditional probability of approximately
52·63%.
(d) • Manufacturing (𝐸1 ) remains unchanged: 𝑃(𝐸1 ) = 0 · 50 [2]
• Marketing (𝐸2 ) decreases to 20%: 𝑃𝑛𝑒𝑤 (𝐸2 ) = 0 · 20
• Research & Development (𝐸3 ) increases to 30%: 𝑃𝑛𝑒𝑤 (𝐸3 ) = 0 · 30
m
.co
The probabilities of a project being high-success from each individual
department
e m
𝑃(𝐴 ∣ 𝐸𝑖 ) 1 ≤ 𝑖 ≤ 3 remains the same.
la(0s · 20 × 0 · 60) + (0 · 30 × 0 · 90)
𝑃𝑛𝑒𝑤 (𝐴) = 𝑃(𝐸1 ) ⋅ 𝑃(𝐴 ∣ 𝐸1 ) + 𝑃𝑛𝑒𝑤 (𝐸2 ) ⋅ 𝑃(𝐴 ∣ 𝐸2 ) + 𝑃𝑛𝑒𝑤 (𝐸3 ) ⋅ 𝑃(𝐴 ∣ 𝐸3 )
𝑃 (𝐴) = (0 · 50 × 0 · 80)g+
𝑛𝑒𝑤
a
𝑃 (𝐴) = 0 · 40 + 0 · 12 + 0 · 27
𝑛𝑒𝑤
𝑃𝑛𝑒𝑤 (𝐴) = 0 · 79
• Old overall probability 𝑃(𝐴) = 0 · 76
• New overall probability 𝑃𝑛𝑒𝑤 (𝐴) = 0 · 79
The overall probability of getting a high-success project increases from 76%
to 79%.
m
.co
OR
m
c. (ii)o Given: s em
e m Total number of handmade items (n) = 100
la
las Probability of a defective item (p) = 3% = 0·03
g
Since 𝑛 is large (𝑛 ≥ 20) and 𝑝 is small (𝑝 ≤ 0 · 05), wea can approximate the
ag Binomial distribution using a Poisson distribution with parameter 𝜆(mean):
𝜆 = 𝑛 ⋅ 𝑝 = 100 × 0 · 03 = 3
The probability mass function of a Poisson distribution is given by:
𝑒 −𝜆 ⋅𝜆𝑥
𝑃(𝑋 = 𝑥) = 𝑥!
For 𝜆 = 3:
𝑒 −3 ⋅3𝑥
𝑃(𝑋 = 𝑥) = 𝑥!
m . c
c. o
17 ISC SPECIMEN ANSWER KEY 2027
e m
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s
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Page 31 of 34
Page 33
(a) Probability that the batch contains no defective items [2]
We need to find 𝑃(𝑋 = 0):
𝑒 −3 ⋅30 𝑒 −3 ⋅1
𝑃(𝑋 = 0) = = = 𝑒 −3
0! 1
𝑃(𝑋 = 0) ≈ 0 · 0498
Answer: The probability that the batch contains no defective items is
approximately 0·0498 (or 4·98%).
(b) The headmaster approves the batch only if the number of defective items is [2]
less than 4 (𝑋 < 4). This means we need to find the cumulative probability
for 𝑋 = 0,1,2 and 3:
𝑃(𝑋 < 4) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)
Let’s calculate each term:
• 𝑃(𝑋 = 0) = 𝑒 −3 ≈ 0.04979
𝑒 −3 ⋅31
• 𝑃(𝑋 = 1) = = 3 ⋅ 𝑒 −3 ≈ 3 × 0 · 04979 = 0 · 14937
1!
𝑒 −3 ⋅32 9
• 𝑃(𝑋 = 2) = = ⋅ 𝑒 −3 = 4 · 5 × 0 · 04979 = 0 · 22406
2! 2
𝑒 −3 ⋅33 27
• 𝑃(𝑋 = 3) = = ⋅ 𝑒 −3 = 4 · 5 × 0 · 04979 = 0 · 2240
3! 6
𝑃(𝑋 < 4) = 𝑒 −3 ⋅ (1 + 3 + 4 · 5 + 4 · 5)
𝑃(𝑋 < 4) = 𝑒 −3 ⋅ (13)
𝑃(𝑋 < 4) = 13 × 0 · 04979 = 0 · 64727
Answer: The probability that the batch will be approved by the headmaster is
approximately 0·6473 (or 64·73%)
(c) The probability of approval is approximately 64·72%, which means the batch [1]
has a reasonably good chance of meeting the quality requirement.
Hence, the quality of students’ work may be considered satisfactory, though
there is still scope for improvement since approval is not very high.
Question 20
(i) (a) Let 𝑥 represent the number of elderly people achieving healthy sleep patterns [3]
after 𝑦 weeks. The rate of improvement is modeled by the first-order
differential equation:
𝑑𝑥
+ 𝑥 = 100
𝑑𝑦
𝑑𝑥
= 100 − 𝑥
𝑑𝑦
1
𝑑𝑥 = 𝑑𝑦
100 − 𝑥
Integrating both sides with respect to their corresponding variables:
1
∫ 𝑑𝑥 = ∫ 𝑑𝑦
100 − 𝑥
18 ISC SPECIMEN ANSWER KEY 2027
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We need to find 𝑃(𝑋 = 0):
𝑒 −3 ⋅30 𝑒 −3 ⋅1
𝑃(𝑋 = 0) = = = 𝑒 −3
0! 1
𝑃(𝑋 = 0) ≈ 0 · 0498
Answer: The probability that the batch contains no defective items is
approximately 0·0498 (or 4·98%).
(b) The headmaster approves the batch only if the number of defective items is [2]
less than 4 (𝑋 < 4). This means we need to find the cumulative probability
for 𝑋 = 0,1,2 and 3:
𝑃(𝑋 < 4) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)
Let’s calculate each term:
• 𝑃(𝑋 = 0) = 𝑒 −3 ≈ 0.04979
𝑒 −3 ⋅31
• 𝑃(𝑋 = 1) = = 3 ⋅ 𝑒 −3 ≈ 3 × 0 · 04979 = 0 · 14937
1!
𝑒 −3 ⋅32 9
• 𝑃(𝑋 = 2) = = ⋅ 𝑒 −3 = 4 · 5 × 0 · 04979 = 0 · 22406
2! 2
𝑒 −3 ⋅33 27
• 𝑃(𝑋 = 3) = = ⋅ 𝑒 −3 = 4 · 5 × 0 · 04979 = 0 · 2240
3! 6
𝑃(𝑋 < 4) = 𝑒 −3 ⋅ (1 + 3 + 4 · 5 + 4 · 5)
𝑃(𝑋 < 4) = 𝑒 −3 ⋅ (13)
𝑃(𝑋 < 4) = 13 × 0 · 04979 = 0 · 64727
Answer: The probability that the batch will be approved by the headmaster is
approximately 0·6473 (or 64·73%)
(c) The probability of approval is approximately 64·72%, which means the batch [1]
has a reasonably good chance of meeting the quality requirement.
Hence, the quality of students’ work may be considered satisfactory, though
there is still scope for improvement since approval is not very high.
Question 20
(i) (a) Let 𝑥 represent the number of elderly people achieving healthy sleep patterns [3]
after 𝑦 weeks. The rate of improvement is modeled by the first-order
differential equation:
𝑑𝑥
+ 𝑥 = 100
𝑑𝑦
𝑑𝑥
= 100 − 𝑥
𝑑𝑦
1
𝑑𝑥 = 𝑑𝑦
100 − 𝑥
Integrating both sides with respect to their corresponding variables:
1
∫ 𝑑𝑥 = ∫ 𝑑𝑦
100 − 𝑥
18 ISC SPECIMEN ANSWER KEY 2027
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Page 34
.
.co s e m
s em l a
a ag
−log ∣ 100 − 𝑥 ∣= 𝑦 + 𝐶1
log ∣ 100 − 𝑥 ∣= − 𝑦 − 𝐶1
100 − 𝑥 = 𝑒 −𝑦−𝐶1
100 − 𝑥 = 𝑒 −𝑦 ⋅ 𝑒 −𝐶1
Let 𝑒 −𝐶1 be a new constant, 𝐴:
100 − 𝑥 = 𝐴𝑒 −𝑦
𝑥 = 100 − 𝐴𝑒 −𝑦 m
om
We are given the initial condition: 𝑥 = 40 when 𝑦 = 0. . co
. c
40 = 100 − 𝐴𝑒 −0 e m
e m
40 = 100 − 𝐴(1)
l as
l as
𝐴 = 100 − 40 = 60 ag
agSubstituting 𝐴 = 60 gives the complete expression for the number of
participants achieving healthy sleep patterns after 𝑦 weeks:
𝑥 = 100 − 60𝑒 −𝑦
(b) (1) Substituting 𝑦 = 2 into the expression derived in Part a: [1]
𝑥 = 100 − 60𝑒 −2
60
𝑥 = 100 − 2
𝑒
Using the standard mathematical value 𝑒 ≈ 2 · 71828, we find 𝑒 2 :
m
.co
𝑒 2 ≈ 7 · 389
m
Now, substituting this back into the equation:
e
𝑥 ≈ 100 −
60
l as
ag
7 · 389
𝑥 ≈ 100 − 8 · 12
𝑥 ≈ 91 · 88
Since x is the number of individual human beings, rounding the answer to the
nearest whole number, gives 92 residents.
(2) [1]
• Approximately 92 residents will achieve healthy sleep patterns after 2
weeks.
• Target Requirement: At least 80 residents.
m
m The old age home management should introduce the monthly late-night
.co
m .co entertainment program.
s e m
s e OR
g la
g la (ii) (a) R(x) = ∫MR(x) dx =∫x2ex dx a [3]
a ∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − ∫ 2𝑥𝑒 𝑥 𝑑𝑥
∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑅(𝑥) = 𝑥 2 𝑒 𝑥 − 2(𝑥𝑒 𝑥 − 𝑒 𝑥 ) + 𝐶
𝑅(𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) + 𝐶
Given that when no units are sold, the revenue is zero (𝑅(0) = 0):
0 = 𝑒 0 (02 − 2(0) + 2) + 𝐶
m . c
c. o
19 ISC SPECIMEN ANSWER KEY 2027
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s
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.co s e m
s em l a
a ag
−log ∣ 100 − 𝑥 ∣= 𝑦 + 𝐶1
log ∣ 100 − 𝑥 ∣= − 𝑦 − 𝐶1
100 − 𝑥 = 𝑒 −𝑦−𝐶1
100 − 𝑥 = 𝑒 −𝑦 ⋅ 𝑒 −𝐶1
Let 𝑒 −𝐶1 be a new constant, 𝐴:
100 − 𝑥 = 𝐴𝑒 −𝑦
𝑥 = 100 − 𝐴𝑒 −𝑦 m
om
We are given the initial condition: 𝑥 = 40 when 𝑦 = 0. . co
. c
40 = 100 − 𝐴𝑒 −0 e m
e m
40 = 100 − 𝐴(1)
l as
l as
𝐴 = 100 − 40 = 60 ag
agSubstituting 𝐴 = 60 gives the complete expression for the number of
participants achieving healthy sleep patterns after 𝑦 weeks:
𝑥 = 100 − 60𝑒 −𝑦
(b) (1) Substituting 𝑦 = 2 into the expression derived in Part a: [1]
𝑥 = 100 − 60𝑒 −2
60
𝑥 = 100 − 2
𝑒
Using the standard mathematical value 𝑒 ≈ 2 · 71828, we find 𝑒 2 :
m
.co
𝑒 2 ≈ 7 · 389
m
Now, substituting this back into the equation:
e
𝑥 ≈ 100 −
60
l as
ag
7 · 389
𝑥 ≈ 100 − 8 · 12
𝑥 ≈ 91 · 88
Since x is the number of individual human beings, rounding the answer to the
nearest whole number, gives 92 residents.
(2) [1]
• Approximately 92 residents will achieve healthy sleep patterns after 2
weeks.
• Target Requirement: At least 80 residents.
m
m The old age home management should introduce the monthly late-night
.co
m .co entertainment program.
s e m
s e OR
g la
g la (ii) (a) R(x) = ∫MR(x) dx =∫x2ex dx a [3]
a ∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − ∫ 2𝑥𝑒 𝑥 𝑑𝑥
∫ 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑅(𝑥) = 𝑥 2 𝑒 𝑥 − 2(𝑥𝑒 𝑥 − 𝑒 𝑥 ) + 𝐶
𝑅(𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) + 𝐶
Given that when no units are sold, the revenue is zero (𝑅(0) = 0):
0 = 𝑒 0 (02 − 2(0) + 2) + 𝐶
m . c
c. o
19 ISC SPECIMEN ANSWER KEY 2027
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s
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Page 35
0 = 1(2) + 𝐶
𝐶 = −2
R(x) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
(b) Comparing the revenues at 𝑥 = 1 [2]
For the first company,
𝑅(1) = 𝑒 1 (12 − 2(1) + 2) − 2
= 𝑒(1 − 2 + 2) − 2
=𝑒−2
For the second company,
𝑅1 (𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
Substituting 𝑥 = 1,
𝑅1 (1) = 𝑒(1 − 2 + 2) − 2
=𝑒−2
Thus, 𝑅(1) = 𝑅1 (1) = 𝑒 − 2
Therefore, both companies earn the same revenue at 𝑥 = 1.
20 ISC SPECIMEN ANSWER KEY 2027
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𝐶 = −2
R(x) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
(b) Comparing the revenues at 𝑥 = 1 [2]
For the first company,
𝑅(1) = 𝑒 1 (12 − 2(1) + 2) − 2
= 𝑒(1 − 2 + 2) − 2
=𝑒−2
For the second company,
𝑅1 (𝑥) = 𝑒 𝑥 (𝑥 2 − 2𝑥 + 2) − 2
Substituting 𝑥 = 1,
𝑅1 (1) = 𝑒(1 − 2 + 2) − 2
=𝑒−2
Thus, 𝑅(1) = 𝑅1 (1) = 𝑒 − 2
Therefore, both companies earn the same revenue at 𝑥 = 1.
20 ISC SPECIMEN ANSWER KEY 2027
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Document Details
| Board / Org | CISCE |
| Exam | Class 12 |
| Type | Sample Paper |
| Pages | 35 |
| Language | english |
| Updated | 24 Sep 2026 |