# CBSE Sample Papers for Class 11 Applied Mathematics Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 3 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 11 Applied Mathematics Set 3 with Solutions

Time Allowed: 3 Hours

Maximum Marks: 80

General Instructions:

- This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there is some internal choice in some questions.
- Section A has 18 MCQs and 2 Assertion Reason-based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) questions of 2 marks each.
- Section C has 6 Short Answer (SA) questions of 3 marks each.
- Section D has 4 Long Answer (LA) questions of 5 marks each.
- Section E has 3 source-based/case-based/passage-based/integrated units of assessment (4 marks each) with sub-parts.
- Internal Choice is provided in 2 questions in Section B, 2 questions in Section C, and 2 Questions in Section D. You have to attempt only one alternative in all such questions.

Section-A

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

Question 1.

The binary equivalent of the decimal number 10 is

(a) 0010

(b) 10

(c) 1010

(d) 010

Answer:

(c) 1010

Explanation:

To get the binary equivalent of any number, we need to divide the number by 2 and obtain the remainder as:

Now, we write the remainder in the reverse order as 1010.

Question 2.

Which is true?

(a) \(2^0>\left(\frac{1}{2}\right)^0\)

(b) \(2^0<\left(\frac{1}{2}\right)^0\)

(c) \(2^0=\left(\frac{1}{2}\right)^0\)

(d) none of these

Answer:

(c) \(2^0=\left(\frac{1}{2}\right)^0\)

Explanation:

According to zero index rule,

2^{0} = 1 and \(\left(\frac{1}{2}\right)^0\) = 1

∴ \(2^0=\left(\frac{1}{2}\right)^0\)

Question 3.

The average of all the numbers between 6 and 34, which are divisible by 5 is

(a) 18

(b) 20

(c) 24

(d) 30

Answer:

(b) 20

Explanation:

The numbers between 6 and 34, which are divisible by 5 are 10, 15, 20, 25, 30.

∴ Average = \(\frac{10+15+20+25+30}{5}\) = 20

Question 4.

If A = {0, {0, 1}}, then cardinal number set of P(A) is

(a) 2

(b) 3

(c) 4

(d) 1

Answer:

(c) 4

Explanation:

P(A) = {φ, {0}, {{0, 1}}, {0, {0, 1}}}

Question 5.

If the sum of n terms of an A.P. is given by S_{n} = 3n + 2n^{2}, then the common difference of the A.P. is

(a) 3

(b) 2

(c) 6

(d) 4

Answer:

(d) 4

Explanation:

Given, S_{n} =3n + 2n^{2}

S_{1} = 3(1) + 2(1)^{2} = 5

S_{2} = 3(2) + 2(2)^{2} = 14

Since, S_{1} = a_{1} = 5 and S_{2} – S_{1} = a_{2}

∴ a_{2} = 14 – 5 = 9

Hence, common difference, d = a_{2} – a_{1} = 9 – 5 = 4.

Note: If S_{n} = an^{2} + bn where a and b are constant then the common difference of an A.P. = 2 × Coefficient of n^{2}.

Question 6.

If ^{n}C_{12} = ^{n}C_{8}, then n is equal to

(a) 20

(b) 12

(c) 6

(d) 30

Answer:

(a) 20

Explanation:

Given, ^{n}C_{12} = ^{n}C_{8}

⇒ ^{n}C_{12} = ^{n}C_{n-8} [∵ ^{n}C_{r} = ^{n}C_{n-r}]⇒ 12 = n – 8

⇒ n = 20

Question 7.

Introducing Asha to guests, Gopal said, “Her father is the only son of my father.” How is Asha related to Gopal?

(a) Daughter

(b) Mother

(c) Sister

(d) Niece

Answer:

(a) Daughter

Explanation:

The only son of Gopal’s father is Gopal himself. This means that Gopal is the father of Asha. Hence, Asha is the daughter of Gopal.

Question 8.

If [x]^{2} – 5[x] + 6 = 0, where [.] denote the greatest integer function, then

(a) x ∈ [3, 4](b) x ∈ (2, 3](c) x ∈ [2, 3](d) x ∈ [2, 4)

Answer:

(c) x ∈ [2, 3]Explanation:

Given, [x]^{2} – 5[x] + 6 = 0

⇒ [x]^{2} – 2[x] – 3[x] + 6 = 0

⇒ [x] ([x] – 2) – 3([x] – 2) = 0

⇒ ([x] – 3)([x] – 2) = 0

⇒ [x] = 2, 3

Hence, x ∈ [2, 3]

Question 9.

If f(x) = 2x and g(x) = \(\frac{x^2}{2}+1\) then which of the following can be a discontinuous function at x = 0?

(a) f(x) + g(x)

(b) f(x) – g(x)

(c) f(x) . g(x)

(d) \(\frac{g(x)}{f(x)}\)

Answer:

(d) \(\frac{g(x)}{f(x)}\)

Explanation:

Since f(x) = 2x and g(x) = \(\frac{x^2}{2}+1\) are continuous functions, then by using the algebra of continuous function, the functions f(x) + g(x), f(x) – g(x), f(x).g(x) are also continuous functions but \(\frac{g(x)}{f(x)}\) is discontinuous function at x = 0.

Question 10.

If the probability for A to fail in an examination is 0.2 and that of B is 0.3, then the probability that either A or B fails is

(a) > 0.5

(b) 0.5

(c) < 0.5

(d) 0

Answer:

(c) < 0.5

Explanation:

Given, P(A fails) = 0.2 and P(B fails) = 0.3

∴ P(either A fails or B fails) ≤ P(A fails) + P(B fails)

≤ 0.2 + 0.3

≤ 0.5

Question 11.

If P(\(\frac{A}{B}\)) > P(A), then which of the following is correct?

(a) P(\(\frac{B}{A}\)) < P(B)

(b) P(A ∩ B) < P(A).P(B)

(c) P(\(\frac{B}{A}\)) > P(B)

(d) P(\(\frac{B}{A}\)) = P(B)

Answer:

(c) P(\(\frac{B}{A}\)) > P(B)

Explanation:

Given, P(\(\frac{A}{B}\)) > P(A)

⇒ \(\frac{P(A \cap B)}{P(B)}\) > P(A)

⇒ P(A ∩ B) > P(A).P(B)

⇒ \(\frac{P(A \cap B)}{P(A)}\) > P(B)

⇒ P(\(\frac{B}{A}\)) > P(B)

Question 12.

Third Quartile is

(a) a measure used to describe the distribution of data

(b) also called upper quartile

(c) also called lower quartile

(d) the sum of numbers in a set of data divided by the number of pieces of data.

Answer:

(b) also called upper quartile

Explanation:

The third Quartile, Q_{3} is the upper quartile whereas the first quartile, Q_{1} is the lower quartile.

Question 13.

The coefficient of _________ is a measure of the shape of a distribution.

(a) Skewness

(b) Kurtosis

(c) Both (a) and (b)

(d) None

Answer:

(c) Both (a) and (b)

Explanation:

Kurtosis and Skewness both are the measures of shape of a distribution.

Question 14.

The process of calculating the future value of money from the present value is classified as _________

(a) Compounding

(b) Discounting

(c) Annuity

(d) None of these

Answer:

(a) Compounding

Explanation:

Compounding is the process of finding out the future value of present money.

Question 15.

Interest obtained on a sum of ₹ 5000 for 3 years ₹ 1500. Find the rate percent.

(a) 5%

(b) 8%

(c) 10%

(d) 12.5%

Answer:

(c) 10%

Explanation:

Let the rate be i%

We have, I = Pit

Here, I = 1500, t = 3, P = 5000 and i% = \(\frac{i}{100}\)

1500 = 5000 × \(\frac{i}{100}\) × 3

⇒ i = \(\frac{1500 \times 100}{5000 \times 3}\) = 10%

Thus, the rate of interest is 10%.

Question 16.

What does the P stands for in this formula?

Present value = \(P\left[\frac{1-(1+i)^n}{i}\right]\)

(a) The fixed payment amount

(b) The number of payments

(c) The future value

(d) The present value

Answer:

(a) The fixed payment amount

Explanation:

Present value = cash flow × \(\frac{\left[(1+i)^n-1\right]}{i(1+i)^n}\) = fixed payment amount × \(\left[\frac{1-(1+i)^n}{i}\right]\)

Question 17.

The length of the latus-rectum of the parabola y^{2} + 8x – 2y + 17 = 0 is

(a) 2

(b) 4

(c) 8

(d) 16

Answer:

(c) 8

Explanation:

Given equation of parabola y^{2} + 8x – 2y + 17 = 0

⇒ (y – 1)^{2} + 8x + 16 = 0

⇒ (y – 1)^{2} = -8(x + 2)

Let X = x + 2 and Y = y – 1

∴ Y^{2} = -8X

Compared with Y^{2} = -4aX, we get 4a = 8, which is the length of the latus-rectum.

Question 18.

The equation of the line perpendicular to line x – 7y + 5 = 0 and having X-intercept 3 is

(a) 7x – y – 21 = 0

(b) 7x + y – 21 = 0

(c) 7x + y + 21 = 0

(d) 7x – y + 21 = 0

Answer:

(b) 7x + y – 21 = 0

Explanation:

Any line perpendicular to the line x – 7y + 5 = 0 is 7x + y + λ = 0

Since, it passes through (3, 0), so

7 × 3 + 0 + λ = 0

⇒ λ = -21

Thus, 7x + y – 21 = 0 is the required line equation.

**Direction (Q.19 & Q.20): In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.**

(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): The hands of a clock coincide 22 times in a day.

Reason (R): The hands of a clock coincide 24 times in a day.

Answer:

(c) (A) is true but (R) is false.

Explanation:

The hands of a clock coincide 11 times every 12 hours (since between 11 and 1, they coincide only once, i.e., at 12 O’clock).

∴ The hands coincide 22 times in a day.

Question 20.

Assertion (A): The value of \(\frac{d y}{d x}\) for the following functions which are expressed in the parametric form: x = 3t^{3}, y = 3t^{4} + 5 is 4t/3.

Reason (R): To find \(\frac{d y}{d x}\) in case of parametric functions, we use following formula \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)

Answer:

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:

Given, x = 3t^{3}, y = 3t^{4} + 5

\(\frac{d x}{d t}\) = 9t2 and \(\frac{d y}{d t}\) = 12t^{3}

We know that, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)

\(\frac{d y}{d x}=\frac{12 t^3}{9 t^2}=\frac{4 t}{3}\)

Section-B

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

Question 21.

The reporting time of an employee is given below:

Day |
Mon | Tue | Wed | Thu | Fri | Sat |

Time (a.m.) |
10:35 | 10:20 | 10:22 | 10:27 | 10:25 | 10:40 |

If the reporting time is 10:30 a.m., then find the probability of his coming late.

OR

A die is rolled. If E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}, find P[(E ∪ F)/G].

Answer:

Let ‘S’ be the sample space and ‘E’ be the event that the employee is coming late.

∴ n(S) = 6

E = {10 : 35, 10 : 40}

∴ n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

Thus, the probability of the employee coming late is \(\frac{1}{3}\).

OR

Given, E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}

p(E) = \(\frac{3}{6}=\frac{1}{2}\)

P(F) = \(\frac{3}{6}=\frac{1}{2}\)

P(G) = \(\frac{4}{6}=\frac{2}{3}\)

Also, P(E ∩ F) = \(\frac{1}{6}\)

and P(E ∪ F) = P(E) + P(F) – P(E ∩ F) = \(\frac{2}{3}\)

Now, (E ∪ F) ∩ G = {2, 3, 5}

∴ P[(E ∪ F) ∩ G] = \(\frac{3}{6}=\frac{1}{2}\)

∴ \(P\left[\frac{(E \cup F)}{G}\right]=\frac{p[(E \cup F) \cap G]}{P(G)}\)

= \(\frac{\frac{1}{2}}{\frac{2}{3}}\)

= \(\frac{3}{4}\)

Question 22.

Find the unknown value in the following table.

Class Interval |
Frequency |
Cumulative Frequency |

0-10 | 5 | 5 |

10-20 | 7 | x_{1} |

20-30 | x_{2} |
18 |

30-40 | 5 | x_{3} |

40-50 | x_{4} |
30 |

OR

Find the mean deviation about the mean of the distribution.

Size |
20 | 21 | 22 | 23 | 24 |

Frequency |
6 | 4 | 5 | 1 | 4 |

Answer:

x_{1} = 5 + 7 = 12

x_{2} = 18 – x_{1} = 18 – 12 = 6

x_{3} = 18 + 5 = 23

x_{4} = 30 – x_{3} = 30 – 23 = 7

OR

Question 23.

Mr. Roy needs to borrow money. His neighborhood bank charges 8% interest compounded semi-annually. An internet bank charges 7.9% interest compounded monthly. At which bank will Mr. Roy pay a lesser amount of interest?

Answer:

Compare effective rates:

Neighbourhood bank, E = \(\left(1+\frac{0.08}{2}\right)^2-1\)

= 0.0816

≅ 8.16%

Internet bank, E = \(\left(1+\frac{0.79}{12}\right)^{12}-1\)

= 0.08192

≅ 8.19%

The neighborhood bank has a lower effective rate although it has a higher nominal (actual) rate.

Question 24.

Mr. Kohli, a citizen of India, has been an export manager of Arjun Overseas Limited, an Indian Company, since 1.5.2016. He has been regularly going to the USA for export promotion. He spent the following days in the U.S.A. for the last five years:

Previous Year Ended |
No. of Days Spent in the USA |

31.3.2017 | 317 days |

31.3.2018 | 150 days |

31.3.2019 | 271 days |

31.3.2020 | 311 days |

31.3.2021 | 294 days |

Determine his residential status for the assessment year 2021-22 assuming that before 1.5.2016 he had never traveled abroad.

Answer:

Total stays in India

During the previous year 2020-21, his stay in India is 71 days, and in the four preceding years = 48 + 216 + 94 + 54 = 412 days.

Question 25.

Arrange each of the following words in meaningful logical order.

(i) 1. Probation, 2. Interview, 3. Selection, 4. Appointment, 5. Advertisement, 6. Application

(ii) 1. Gold, 2. Iron, 3. Sand, 4. Platinum, 5. Diamond

Answer:

(i) As for a job, a person first sees an advertisement, then fills application form and goes for an interview. If he/she is selected, got an appointment letter and the final stage is probation. Hence, the Correct arrangement of given words is 5, 6, 2, 3, 4, 1.

(ii) All the given words represent substances that can be arranged in the increasing order of their cost. The least costly is sand after which comes the cost of iron, followed by gold, diamond, and costliest among all is platinum. Hence, the Correct arrangement of given words is 3, 2, 1, 5, 4.

Section-C

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

Question 26.

₹ 5000 is paid every year for three years to pay off a loan. What is the loan amount if the interest rate is 14% per annum compounded annually?

Answer:

Question 27.

Read the following information carefully to answer the questions given below.

‘A + B’ means ‘A is the father of B’

‘A – B’ means ‘A is the wife of B’

‘A × B’ means ‘A is the brother of B’

‘A ÷ B’ means ‘A is the daughter of B’

(i) If P ÷ R + S + Q, which of the following is true?

(a) P is the daughter of Q

(b) Q is the aunt of P

(c) P is the aunt of Q

(d) P is the mother of Q

(ii) If P – R + Q, which of the following statements is true?

(a) P is the mother of Q

(b) Q is the daughter of P

(c) P is the aunt of Q

(d) P is the sister of Q

Answer:

(i) ‘S + Q’ means S is the father of Q and ‘R + S’ means R is the father of S. This means that R is the granddaughter of Q. Now, P ÷ R means P is the daughter of R. This means P is the aunt (father’s sister) of Q. Hence, statement (c) is true.

(ii) From the relation given in the question, P – R + Q, R is the father of Q, and P is the wife of R. It is clear that P is the mother of Q. Hence, statement (a) is true.

Question 28.

The product of the first three terms of G.P. is 1000. If 6 is added to its second term and 7 is added to its third term, the terms become in A.P. Find G.P.

Answer:

Let the numbers in G.P be \(\frac{a}{r}\), a, ar

Product \(\frac{1}{2}\).a.ar = 1000

⇒ a^{3} = 1000

⇒ a = 10

According to question

a_{1} = \(\frac{a}{r}=\frac{10}{r}\)

a_{2} = a + 6 = 10 + 6 = 16

a_{3} = ar + 7 = 10r + 7

Also, a_{3} = a_{1} + 2(a_{2} – a_{1}) [∵ a_{1}, a_{2}, a_{3} are in A.P.]⇒ 10r + 7 = \(\frac{10}{r}+2\left[16-\frac{10}{r}\right]\)

⇒ 10r^{2} + 7r = 10 + 32r – 20

⇒ 10r^{2} – 25r + 10 = 0

⇒ (r – 2)(10r – 5) = 0

⇒ r = 2 or r = \(\frac{1}{2}\)

Substituting the value of a and r in equation (i), we get G.P. 5, 10, 20,…… when r = 2 and G.P. 20, 10, 5,……. when r = \(\frac{1}{2}\)

Question 29.

Find the value of ‘k’ if \(\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}\)

Answer:

Question 30.

The heights (to the nearest cm) of 60 students of a certain school are given in the following frequency distribution table.

Heights (in cm) |
151 | 152 | 153 | 154 | 155 | 156 | 157 |

No. of Students |
6 | 4 | 11 | 9 | 16 | 12 | 2 |

Find the

(i) median

(ii) lower quartile

(iii) upper quartile

(iv) inter quartile range.

OR

The coefficient of variation of the two distributions are 50 and 60 and their arithmetic means are 30 and 25 respectively. Find the difference in their standard deviation.

Answer:

The given variates are already in ascending order. We construct the cumulative frequency table:

Question 31.

Find the equation of a straight line that makes an acute angle with the positive direction of the X-axis, passes through the point (-5, 0), and is at a perpendicular distance of 3 units from the origin.

OR

Find the length of the line segment joining the vertex of the parabola y^{2} = 4ax and a point on the parabola, where the line segment makes an angle θ with the X-axis.

Answer:

Let ‘α’ be the acute angle made by the line with a positive X-axis.

∴ Equation of the line is x cos α + y sin α = 3 ……(i)

Since the line passes through (-5, 0), So

-5 . cos α + 0 . sin α = 3

cos α = \(-\frac{3}{5}\)

sin α = \(\sqrt{1-\left(-\frac{3}{5}\right)^2}\)

= \(\sqrt{1-\frac{9}{25}}\)

= \(\frac{4}{5}\)

From (i), \(x\left(-\frac{3}{5}\right)+y\left(\frac{4}{5}\right)\) = 3

⇒ 3x – 4y + 15 = 0 is the required equation of a line.

OR

The given equation of the parabola is y^{2} = 4ax.

Let the coordinates of any point on the parabola be (at^{2}, 2at).

Section-D

(This section comprises long answer type questions (LA) of 5 marks each)

Question 32.

A person opened an account in April 2019 with a deposit of ₹ 800. The account paid 6% interest compounded quarterly. On October 1, 2019, he closed the account and added enough additional money to invest in a 6-month time deposit for ₹ 1000, earning 6% compounded monthly.

(a) How much additional amount did the person invest on October 1?

(b) What was the maturity value of his time deposit on April 1, 2020?

(c) How much total interest was earned?

Given that (1 + i)^{n} is 1.03022500 for i = 1\(\frac{1}{2}\), n = 2 and (1 + i)^{n} is 1.03037751 for i = \(\frac{1}{2}\)%, n = 6.

Answer:

(a) The initial investment earned for April-June and July-September quarter i.e., two quarters.

In this case, i = \(\frac{6}{4}=1 \frac{1}{2}\)% = 0.015, n = \(\frac{6}{12}\) × 4 = 2

compounded amount = 800(1 + 0.015)^{2}

= 800 × 1.03022500

= ₹ 824.18

The additional amount invested = ₹ (1000 – 824.18) = ₹ 175.82

(b) In this case the time-deposit earned interest compounded monthly for six-month

Here, i = \(\frac{6}{12}=\frac{1}{2}\)% = 0.005, n = 6 and P = ₹ 1000

Maturity Value = 1000(1 + 0.005)6

= 1000 × 1.03037751

= ₹ 1030.38

(c) Total interest earned = ₹ (824.18 – 800) + (1030.38 – 1000)

= ₹ (24.18 + 30.38)

= ₹ 54.56

Question 33.

Find the correlation coefficient between the heights of husbands and wives based on the following data (given in inches) and interpret the result.

Couple |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

Height of Husband |
76 | 75 | 75 | 72 | 72 | 71 | 71 | 70 | 68 | 68 | 68 | 68 | 67 | 67 | 62 |

Height of Wife |
71 | 70 | 70 | 67 | 71 | 65 | 65 | 67 | 64 | 65 | 65 | 66 | 63 | 65 | 61 |

Answer:

We use assumed mean A = 70, B = 66

and use the formula \(r=\frac{\Sigma u v-\frac{1}{N} \Sigma u \cdot \Sigma v}{\sqrt{\Sigma u^2-\frac{(\Sigma u)^2}{N}} \sqrt{\Sigma v^2-\frac{(\Sigma v)^2}{N}}}\) …….(i)

Interpretation: As r = 0.90, which is a strong positive correlation. This shows that tall men usually marry tall women and short men marry short women (called assertive matching).

Question 34.

Simplify: \(\frac{1}{1+a^{m-n}+a^{m-p}}+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}}\)

OR

(i) Prove that the last day of a century cannot be Tuesday, Thursday, or Saturday.

(ii) How many times does the 29th day of the month occur in 400 consecutive years?

Answer:

OR

(i) 100 years contain 5 odd days.

∴ The last day of the 1st century is Friday.

200 years contain (5 × 2) = 10 days = 3 odd days.

∴ The last day of the 2nd century is Wednesday.

300 years contain (5 × 3) = 15 days = 1 odd day

∴ The last day of the 3rd century is Monday.

400 years contain 0 odd days.

∴ The last day of the 4th century is Sunday.

This cycle is repeated.

∴ The last day of a century cannot be Tuesday Thursday or Saturday.

Hence Proved.

(ii) In 400 consecutive years, there are 97 leap years.

Hence, in 400 consecutive years, February has the 29th day 97 times

and the remaining 11 months have the 29th day = 400 × 11 = 4400 times.

Hence, the 29th day of the month occurs (4400 + 97) = 4497 times.

Question 35.

If all the letters of the word “MOTHER” are written in all possible orders and the word so formed are arranged in a dictionary order, then find the rank of the word ‘MOTHER’?

OR

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if

(i) They can be of any colour.

(ii) two must be white and two red.

(iii) they must all be of the same colour.

Answer:

The given word is MOTHER. In the dictionary, the words are arranged in alphabetical order. The words beginning with E, H, M, O, R, and T are written in order.

Number of words starting with E = 5! = 120

Number of words starting with H = 5! = 120

Number of words starting with M = 5! = 120

But one of these words is MOTHER.

Now, the number of words with M E = 4! = 24

Number of words with M H = 4! = 24

Number of words with M O = 4! = 24

But one of these words is MOTHER.

Now, the number of words with M O E = 3! = 6

Number of words with M O H = 3! = 6

Number of words with MOT = 3! = 6

But one of these words is MOTHER.

The first word beginning with MOT is MOTHER.

The second word is MOTERH.

The third word is MOTHER.

Therefore, rank of MOTHER = 2(120) + 2(24) + 3(6) + 3

= 240 + 48 + 18 + 3

= 309

OR

Total number of marbles = 6 white + 5 red = 11 marbles

(i) If they can be of any colour, we must select 4 marbles out of 11.

∴ Required Number of ways = ^{11}C_{4}

= \(\frac{11 !}{7 ! 4 !}\)

= 330

(ii) If two must be white, then the selection will be ^{6}C_{2}, and if two must be red, then the selection will be ^{5}C_{2}.

∴ Required number of ways = ^{6}C_{2} × ^{3}C_{2}

= 15 × 10

= 150

(iii) If they all must be the same colour, then the selection of 4 white marbles out of 6 = ^{6}C_{4}

and selection of 4 red marble out of 5 = ^{5}C_{4}

∴ Required number of ways = ^{6}C_{4} + ^{5}C_{4}

= 15 + 5

= 20

Section-E

(This section comprises 3 source-based questions (Case Studies) of 4 marks each)

Question 36.

Rahul and Ravi planned to play a Business Board game with two dice. In which they will get the turn one by one, they roll the dice and continue the game.

Based on the given information answer the following questions:

(i) Ravi got the first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?

OR

Rahul got the next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13?

(ii) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12?

(iii) Rahul got the next change. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7? Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8?

Answer:

(i) Favourable cases (2, 6), (6, 2), (4, 4), (5, 3), (3, 5)

Probability of getting the sum as eight = \(\frac{5}{36}\)

OR

As the sum of numbers on two dice is 13 will not be possible, therefore zero probability.

(ii) As the sum on the two dice will always be less than or equal to 12, therefore probability = 1

(iii) Favourable events of getting sum = 7

i.e., (1, 6), (6, 1), (5, 2), (4, 3), (3, 4), (2, 5)

Required probability = \(\frac{6}{36}=\frac{1}{6}\)

Favorable events = (6, 3), (3, 6), (5, 4), (4, 5), (5, 5), (5, 6), (6, 5), (6, 6), (4, 6), (6, 4)

Required probability = \(\frac{10}{36}=\frac{5}{18}\)

Question 37.

A manufacturer produces 600 computers in third year and 700 computers in the seventh year. Assuming that the production increases uniformly by a constant number every year.

Based on this information answer the following questions:

(i) Find the number of production of computers which increase in every year.

(ii) Find the production in the first year.

(iii) Find total production in 8 years.

OR

Find the difference in members of computers manufactured in the 4th year and 8th year.

Answer:

(i) Here, a_{3} = a + (3 – 1)d

⇒ a_{3} = a + 2d

⇒ a_{3} = 600 ……(i)

a_{7} = a + (7 – 1)d

⇒ a_{7} = a + 6d

⇒ a_{7} = 700 …..(ii)

Solving (i) and (ii), we get d = 25

(ii) Substituting the value of d in (i), we get a = 550

(iii) We know that S_{n} = \(\frac{n}{2}\) [2a + (n – 1)d]Here, n = 8, a = 550 and d = 25

S_{8} = \(\frac{8}{2}\) [2 × 550 + (7)25]S_{8} = 4(1100 + 175) = 5100

OR

S_{4} = \(\frac{4}{2}\) [2 × 550 + (3)25]= 2(1100 + 75)

= 2350

and S_{8} = 5100

Therefore, difference = 5100 – 2350 = 2750

Question 38.

Algebra of Real Functions

Let f: D_{1} → R and g: D_{2} → R be two real functions with domains D_{1} and D_{2}, respectively. Then, algebraic operations such as addition, subtraction, multiplication, and division of two real functions are given below.

(I) Addition of two real functions: The sum function (f + g) is defined by

(f + g)(x) = f(x) + g(x), ∀ x ∈ D_{1} ∩ D_{2}

The domain of (f + g) is D_{1} ∩ D_{2}.

(II) Subtraction of two real functions: The difference function (f – g) is defined by

(f – g)(x) = f(x) – g(x), ∀ x ∈ D_{1} ∩ D_{2}

The domain of (f – g) is D_{1} ∩ D_{2}.

(III) Multiplication of two real functions: The product function (fg) is defined by

(fg)(x) = f(x).g(x), ∀ x ∈ D_{1} ∩ D_{2}

The domain of (fg) is D_{1} ∩ D_{2}.

(IV) Quotient of two real functions: The quotient function is defined by

\(\frac{f}{g}(x)=\frac{f(x)}{g(x)}\), ∀ x ∈ D_{1} ∩ D_{2} – [x: g(x) ≠ 0]The domain of (\(\frac{f}{g}\)) is D_{1} ∩ D_{2} – [x: g(x) ≠ 0](V) Multiplication of a real function by a scalar: The scalar multiple function cf is defined by

(cf)(x) = c.f(x), ∀ x ∈ D_{1} where c is scalar (real number).

The domain of cf is D_{1}.

Based on the given information answer the following questions:

(i) If f and g are real functions defined by f(x) = x^{2} + 7 and g(x) = 3x + 5, find each of the following:

(a) f(3) + g(-5)

(b) f(\(\frac{1}{2}\)) × g(14)

(c) f(t) – f(-2)

(d) \(\frac{f(t)-f(5)}{t-5}\), if t ≠ 5

OR

(ii) If f(x) = x – \(\frac{1}{x}\), prove that [f(x)]3 = f(x3) + 3f(\(\frac{1}{x}\)).

Answer:

Given, f and g are real functions defined by

f(x) = x^{2} + 7 and g(x) = 3x + 5

(a) f(3) = (3)^{2} + 7 = 9 + 7 = 16

and g(-5) = 3(-5) + 5 = -15 + 5 = -10

∴ f(3) + g(-5) = 16 – 10 = 6

(b) \(f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+7\)

= \(\frac{1}{4}\) + 7

= \(\frac{29}{4}\)

and g(14) = 3(14) + 5 = 42 + 5 = 47

∴ f(\(\frac{1}{2}\)) × g(14) = \(\frac{29}{4}\) × 47 = \(\frac{1363}{4}\)

(c) f(t) = (t)^{2} + 7 = t2 + 7

and f(-2) = (-2)^{2} + 7 = 4 + 7 = 11

∴ f(t) – f(-2) = t^{2} + 7 – 11 = t^{2} – 4

(d) f(t) = t^{2} + 7

and f(5) = (5)^{2} + 7 = 25 + 7 = 32

∴ \(\frac{f(t)-f(5)}{t-5}=\frac{t^2+7-25-7}{t-5}\)

= \(\frac{t^2-25}{t-5}\)

= t + 5

OR