# CBSE Sample Papers for Class 11 Applied Mathematics Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 4 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 11 Applied Mathematics Set 4 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 value of \(8^{1 / 3}\) is

(a) \(\sqrt[3]{2}\)

(b) 4

(c) 2

(d) None of these

Answer:

(c) 2

Explanation:

\(8^{1 / 3}\) = \(\left(2^3\right)^{1 / 3}\) = 2^{1} = 2.

Question 2.

The average of 20 numbers is zero. Of them, at most, how many may be greater than zero?

(a) 0

(b) 1

(c) 10

(d) 19

Answer:

(d) 19

Explanation:

An average of 20 numbers = 0

∴ Sum of 20 numbers = (0 × 20) = 0

It is quite possible that 19 of these numbers may be positive and if their sum is a, then the 20th number is (-a).

Question 3.

A car moves at a speed of 80 km/h. What is the speed of the car in meters per second?

(a) 8 m/s

(b) 20\(\frac{1}{9}\) m/s

(c) 22\(\frac{2}{9}\) m/s

(d) None of these

Answer:

(c) 22\(\frac{2}{9}\) m/s

Explanation:

Question 4.

The power set of A = φ is

(a) {φ}

(b) {0}

(c) {φ, {0}}

(d) None of these

Answer:

(a) {φ}

Explanation:

As only a subset of φ is φ, thus P(A) = {φ}.

Question 5.

The minimum value of the expression \(3^x+3^{1-x}\), x ∈ R, is

(a) 0

(b) \(\frac{1}{3}\)

(c) 3

(d) 2√3

Answer:

(d) 2√3

Explanation:

We know that, A.M. > G.M. for positive numbers.

Question 6.

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5, and 6 taken all at a time is

(a) 432

(b) 108

(c) 36

(d) 18

Answer:

(b) 108

Explanation:

Suppose, we fix 3 at the unit place, then the total number of possible arrangements for the remaining numbers = 3!

Similarly, if we fix 4, 5, and 6 at unit place, then in each case total possible arrangements are 3!

Hence, the required number of arrangements in which the sum of unit digits of all such numbers is

= 3 × 3! + 4 × 3! + 5 × 3! + 6 × 3!

= (3 + 4 + 5 + 6) × 3!

= 18 × 6

= 108

Question 7.

The logical and meaningful arrangement of the following words is _________

1. Leaves

2. Branch

3. Flower

4. Tree

5. Fruit

(a) 4, 2, 1, 3, 5

(b) 5, 3, 1, 2, 4

(c) 1, 2, 3, 4, 5

(d) 5, 4, 3, 2, 1

Answer:

(a) 4, 2, 1, 3, 5

Explanation:

From the above words, it is deducted that in a tree, first come branches, then leaves, then flowers, then from flowers we get fruit. Therefore the correct arrangement is

Question 8.

If A and B are finite sets such that n(A) = 5 and n(B) = 7, then the number of functions from A to B is _________

(a) 5^{7}

(b) 7^{5}

(c) 5^{5}

(d) 7^{7}

Answer:

(b) 7^{5}

Explanation:

Given, n(A) = 5 and n(B) = 7

We know that, if n(A) = p and n(B) = q,

then number of functions from A to B = q^{p}.

∴ Number of functions from A to B = 7^{5}

Question 9.

\(\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}\) is equal to:

(a) 1

(b) 2

(c) -1

(d) -2

Answer:

(c) -1

Explanation:

Question 10.

The probability that atleast one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then P(\(\bar{A}\)) + P(\(\bar{B}\)) is equal to:

(a) 0.4

(b) 0.8

(c) 1.2

(d) 1.6

Answer:

(c) 1.2

Explanation:

Given, P(A ∪ B) = 0.6 and P(A ∩ B) = 0.2

We know that,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 0.6 = P(A) + P(B) – 0.2

⇒ P(A) + P(B) = 0.8

⇒ [1 – P(\(\bar{A}\))] + [1 – P(\(\bar{B}\))] = 0.8

⇒ 2 – [P(\(\bar{A}\)) + P(\(\bar{B}\))] = 0.8

⇒ P(\(\bar{A}\)) + P(\(\bar{B}\)) = 2 – 0.8 = 1.2

Question 11.

Let P(A) = \(\frac{7}{13}\), P(B) = \(\frac{9}{13}\) and P(A ∩ B) = \(\frac{4}{13}\). Then P(\(\frac{A^{\prime}}{B}\)) is equal to _________

(a) \(\frac{9}{5}\)

(b) \(\frac{5}{9}\)

(c) \(\frac{1}{5}\)

(d) \(\frac{1}{9}\)

Answer:

(b) \(\frac{5}{9}\)

Explanation:

\(P\left(\frac{A^{\prime}}{B}\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}\)

Question 12.

The appropriate measure of dispersion for open-end classification is

(a) Standard deviation

(b) Quartile Deviation

(c) Mean deviation

(d) All these measures

Answer:

(b) Quartile Deviation

Explanation:

Quartile deviation provides the best measure of dispersion for open-end classification. It is also less affected due to sampling fluctuations.

Question 13.

The degree of peakedness or flatness of an unimodal distribution is called

(a) Skewness

(b) Symmetry

(c) Dispersion

(d) Kurtosis

Answer:

(d) Kurtosis

Explanation:

Kurtosis is a parameter that describes the weakness at the center and flatness at the tails of a distribution.

Question 14.

Which of the following is not true regarding an annuity due?

(a) It is a series of equal cash flow

(b) It is also known as deferred annuity

(c) Cash flows occur for a specific period

(d) Payments are made at the start of each period

Answer:

(a) It is a series of equal cash flow

Explanation:

An annuity is a series of equal payments made at equal intervals. Examples of annuities are regular deposits to a savings account, monthly home mortgage payments, monthly insurance payments pension payments, etc.

Question 15.

What is the full form of GST?

(a) Goods and Supply Tax

(b) Goods and Services Tax

(c) General Sales Tax

(d) Government Sales Tax

Answer:

(b) Goods and Services Tax

Question 16.

The salary, remuneration, or compensation received by the partners is taxable under the head _________

(a) Capital gains

(b) Income from Salaries

(c) Income from Business

(d) Income from other sources

Answer:

(c) Income from Business

Explanation:

The salary, remuneration, or compensation received by the partners is taxable under the head Income from Business.

Question 17.

A line cutting off intercept – 3 from the Y-axis and tangent at an angle to the X-axis is \(\frac{3}{5}\), its equation is

(a) 5y – 3x + 15 = 0

(b) 3y – 5x + 15 = 0

(c) 5y – 3x – 15 = 0

(d) None of these

Answer:

(a) 5y – 3x + 15 = 0

Explanation:

Since the lines cut off intercept – 3 on the Y-axis, then the line is passing through the point (0, -3).

Also, given that tan θ = \(\frac{3}{5}\)

⇒ m = \(\frac{3}{5}\) (slope of the line)

So, the equation of the required line is (y – y_{1}) = m(x – x_{1})

⇒ [y – (-3)] = \(\frac{3}{5}\)(x – 0)

⇒ 5y + 15 = 3x

⇒ 5y – 3x + 15 = 0

Question 18.

The equation of the circle which passes through the origin and whose center is (3, 4) will be:

(a) x^{2} + y^{2} – 3x – 4y = 0

(b) x^{2} + y^{2} + 3x + 4y = 0

(c) x^{2} + y^{2} + 6x + 8y = 0

(d) x^{2} + y^{2} – 6x – 8y = 0

Answer:

(d) x^{2} + y^{2} – 6x – 8y = 0

Explanation:

If the origin is on the circumference of the circle and the coordinates of the center C be (3, 4), then from the adjoining figure, it is clear that

OC^{2} = OM^{2} + CM^{2}

⇒ r^{2} = 3^{2} + 4^{2}

⇒ r^{2} = 9 + 16

⇒ r^{2} = 25

Thus, the equation of a required circle with center (3, 4) and radius 25 is given by

(x – 3)^{2} + (y – 4)^{2} = 25

⇒ (x^{2} – 6x + 9) + (y^{2} – 8y + 16) = 25

⇒ x^{2} + y^{2} – 6x – 8y = 0

**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): If the area of the trapezium, whose parallel sides are 6 cm and 10 cm is 32 sq. cm, then the distance between the parallel sides is 4 cm.

Reason (R): Area of trapezium = \(\frac{1}{2}\)(a + b) × h

Answer:

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

Explanation:

Area of trapezium = \(\frac{1}{2}\)(a + b) × h

where a and b are parallel sides and h is altitude (distance)

⇒ 32 = \(\frac{1}{2}\)(6 + 10) × h

⇒ h = 4 cm

Therefore, the distance between parallel sides is 4 cm.

Question 20.

Assertion (A): If y = \(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !} \ldots\), then \(\frac{d y}{d x}\) = y

Reason (R): If y = \(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !} \ldots\), then \(\frac{d y}{d x}\) = 0

Answer:

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

Explanation:

Given that,

Section-B

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

Question 21.

Out of the following, find the odd man out and give a reason for your answer.

(i) DEHG

(ii) RSVU

(iii) XYBA

(iv) LMQP

(v) JKNM

Answer:

Hence, in all the given parts except (iv), the placement of the alphabet is in the order (+1, +3, -1). So, LMQP is the odd one out.

Question 22.

Three identical dice are rolled. Find the probability that the same number appears on each of them.

OR

If A and B are two events such that P(\(\frac{A}{B}\)) = p, P(A) = p, P(B) = \(\frac{1}{3}\) and P(A ∪ B) = \(\frac{5}{9}\), then find the value of p.

Answer:

Since three identical dice are rolled, so number of elements in the sample space S is

n(S) = 63 = 216

Let E be the event of getting the same number on each of them

i.e., E = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}

∴ n(E) = 6

P(E) = \(\frac{n(E)}{n(S)}\)

= \(\frac{6}{216}\)

= \(\frac{1}{36}\)

∴ The probability that the same number appears on all three dice is \(\frac{1}{36}\).

OR

Question 23.

Define direct and indirect taxes.

Answer:

Direct Tax:

- A person who pays the tax to the Government directly bears the incidence of the tax.
- Progressive in Nature-high rate of taxes for people having a higher ability to pay.
- The burden of tax is borne by the person himself.

Indirect Tax:

- The person paying the tax to the Government collects the same from the ultimate consumer.
- Regressive in Nature- All consumers equally bear the burden, irrespective of their ability to pay.
- The burden of tax shifted to another person.

Question 24.

₹ 1000 is invested every 3 months at 4.8% compounded quarterly. How much will the annuity be worth in 2 years? [Given (1.012)^{8} = 1.1001]Answer:

Hence, in two years the annuity worth will be ₹ 8344.18.

Question 25.

The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110 not 100. Find the true mean and median.

OR

Find the mean deviation from the median and coefficient of deviation for the following series:

Marks: 17, 18, 25, 20, 26, 28, 38, 30, 22

Answer:

Mean = \(\frac{\Sigma f x}{\Sigma f}\)

⇒ 50 = \(\frac{\Sigma f x}{100}\)

⇒ Σfx = 5000

True mean = 5000 – 100 + 110 = 5010

∴ True mean = \(\frac{5010}{100}\) = 50.1

The median will remain the same i.e., median = 52.

OR

Arranging the marks in ascending order, the series obtained is 17, 18, 20, 22, 25, 26, 28, 30, 38

Here, number of terms n = 9 is odd, so that

Median, Md = Value of \(\left(\frac{9+1}{2}\right)^{\mathrm{th}}\) term

= Value of 5th term

= 25

Hence, the Mean deviation

Also, coefficient of deviation = \(\frac{\text { Mean deviation }}{\text { Median }} \times 100\)

= \(\frac{5}{25}\) × 100

= 20

Section-C

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

Question 26.

Study the following statements A, B, C, D, and E, and answer the following questions:

(A) 1, 5, 9 means ‘you better go’

(B) 1, 6, 7 means ‘better come here’

(C) 5, 6, 7 means ‘you come here’

(D) 1, 5, 6 means ‘better you here’

(E) 3, 7, 9 means ‘come and go’

(i) How many minimum number of statements are necessary to find the code number of ‘better’?

(ii) Which numeral means ‘go’?

(iii) Which numeral means ‘you’?

Answer:

(i) In statements A and B, A and D & B and D common word is ‘better’ and in all the combinations common code is 1.

Hence, code no. of better is 1 and a minimum number of required statements is 2 (two).

(ii) In statements A and E common word is ‘go’ and the common code no. is ‘9’. Hence, the numeral ‘9’ means ‘go’.

(iii) In statements A and C common words is ‘you’ and common code no. is ‘5’. Hence, the numeral ‘5’ means ‘you’.

Question 27.

In what ratio the line joining (-1, 1) and (5, 7) is divided by the line x + y = 4?

OR

A double ordinate of parabola y^{2} = 4ax is of length 8a. Prove that the lines from the vertex to its ends are at right angles.

Answer:

Let the given points be A(x_{1}, y_{1}) = (-1, 1) and B(x_{2}, y_{2}) = (5, 7) and P(x_{3}, y_{3}) be the point which

divides AB in the ratio m : n.

∴ Co-ordinates of P are \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\) = \(\left(\frac{5 m-n}{m+n}, \frac{7 m+n}{m+n}\right)\)

Since, the point P lies on line x + y = 4.

∴ \(\frac{5 m-n}{m+n}+\frac{7 m+n}{m+n}\) = 4

⇒ \(\frac{12 m}{m+n}\) = 4

⇒ 8m = 4n

⇒ m : n = 1 : 2

OR

Let PQ be the double ordinate of length 8a of the parabola, y^{2} = 4ax. Then, PR = QR = 4a.

Let AR = x_{1}, then coordinates of point P are (x_{1}, 4a) and coordinates of point R are (x_{1}, -4a)

Since P lies on y^{2} = 4ax

∴ (4a)^{2} = 4ax_{1}

⇒ x_{1} = 4a

So, the coordinates of P and Q are (4a, 4a) and (4a, -4a), respectively.

Also, the coordinates of the vertex A are (0, 0)

∴ m_{1} = slope of AP = \(\frac{4 a-0}{4 a-0}\) = 1

and m_{2} = slope of AQ = \(\frac{-4 a-0}{4 a-0}\) = -1

Clearly, m_{1}m_{2} = -1

Hence, AP ⊥ AQ.

Question 28.

The marked price of the goods is ₹ 5,000 and the rate of GST on it is 18%. A shopkeeper buys the goods at a reduced price and sells it at its marked price. If the shopkeeper paid ₹ 144 as CGST to the government, find the amount (inclusive of GST) paid by the shopkeeper.

Answer:

Let the shopkeeper buy the goods at ₹ x.

∴ Cost price = ₹ x

and Selling price = ₹ 5,000

According to the question,

CGST paid = ₹ 144

and SGST paid = ₹ 144

Total GST = ₹ (144 + 144) = ₹ 288

i.e., GST on Selling Price – GST on Cost Price = ₹ 288

⇒ \(\frac{18}{100}\) × 5000 – \(\frac{18}{100}\) × x = 288

⇒ \(\frac{18}{100}\) × x = 900 – 288

⇒ x = \(\frac{612 \times 100}{18}\)

⇒ x = ₹ 3,400

∴ The shopkeeper buys the goods for ₹ 3400.

Total amount paid by the shopkeeper = Cost price of the goods + GST

= ₹ (3,400 + \(\frac{18}{100}\) × 3400)

= ₹ 4012

Question 29.

Find the mean, variance, and standard deviation of the first 10 multiples of 4.

Answer:

Here, the variables are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40.

Here, N = 10

So, mean, \(\bar{x}=\frac{4+8+\ldots+40}{10}\)

∴ Mean = 22, Variance = 132 (Approx.), and Standard deviation = 11.49 (Approx.)

Question 30.

If y = \(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\), prove that (2xy)\(\frac{d y}{d x}=\frac{x}{a}-\frac{a}{x}\).

Answer:

Question 31.

Determine n, if ^{2n}C_{3} : ^{n}C_{2} = 12 : 1.

Answer:

Section-D

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

Question 32.

X, who is a person with a disability, submits the following information.

Particulars |
Amount (₹) |

(i) Salary (per annum) | 3,05,000 |

(ii) Rent received (per month) | 2,000 |

(iii) Dividend from Co-operative Society | 3,000 |

(iv) Interest on Bank Deposits | 7,000 |

(v) Interest on Government Securities | 5,000 |

(vi) Winnings from Lotteries | 2,000 |

(vii) NSC (VIII Issue) purchased during the year | 15,000 |

(viii) Deposit under the PPF Scheme | 35,000 |

He earned a long-term capital gain of ₹ 78,000 on the sale of shares during the year. Compute

(a) the taxable income

(b) the tax payable for the assessment year 2021-22.

Answer:

(a) Computation of Total Income

Note: Income from winning lotteries cannot be shifted to other income to claim a full exemption of ₹ 2,50,000.

Question 33.

Examine whether there is any correlation between age and blindness based on the following data:

Age in Years |
0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

No. of Persons: (in thousands) |
90 | 120 | 140 | 100 | 80 | 60 | 40 | 20 |

No. of Blind Persons: |
10 | 15 | 18 | 20 | 15 | 12 | 10 | 6 |

OR

For a certain distribution, Σx_{i} = 20, \(\Sigma x_i^2\) = 100 and \(\Sigma x_i^3\) = 1500, and n = 10, find coefficient of skewness.

Answer:

Let us denote the mid-value of age in years as x and the number of blind persons per lakh as y. Then, to compute the coefficient of correlation between x and y, we construct the following table:

The correlation coefficient between age and blindness is given by

= \(\frac{1090}{1136.268}\)

= 0.959

≅ 0.96

Which exhibits a very high degree of positive correlation between age and blindness.

OR

Question 34.

In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls sit together in a back row on adjacent seats?

OR

Find the sum of the first ‘n’ terms of the series 0.7 + 0.77 + 0.777 + ……..

Answer:

We have 14 seats on two vans and there are 9 boys and 3 girls, i.e., a total of 12 people.

The number of ways of arranging 12 people on 14 seats without restriction is ^{14}P_{12}.

Now, ^{14}P_{12} = \(\frac{14 !}{2 !}\)

= \(\frac{14.13 !}{2 !}\)

= 7 × 13! ways

Three girls can be seated together in the back row on adjacent seats in the following ways:

1, 2, 3 or 2, 3, 4 of the first van

1, 2, 3 or 2, 3, 4 of the second van

In each way, the three girls can interchange among themselves in 3! ways.

∴ Total number of ways in which 3 girls sit together in a back row = 4 × 3! = 24 ways

9 boys are to seated on 11 seats = ^{11}P_{9} = \(\frac{11 !}{2 !}\)

Hence, the total number of ways = \(\frac{24 \times 11 !}{2 !}\) = 12 × 11!

OR

We have, 0.7 + 0.77 + 0.777 + ….. to n terms

= 7 × 0.1 + 7 × 0.11 + 7 × 0.111 + …. to n terms

= 7{0.1 + 0.11 + 0.111 + ….. to n terms}

= \(\frac{7}{9}\){0.9 + 0.99 + 0.999 + ….. to n terms}

Question 35.

(i) A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder to have its vertex at the center of one end, and the other end as its base. Find the ratio of the volume of the cylinder, hemisphere, and cone.

(ii) A regular hexagonal prism has a perimeter of its base of 600 cm and a height equal to 200 cm. How many liters of petrol can it hold? Find the weight of the petrol if the density is 0.8 gram cc.

Answer:

(i) We have, radius of the hemisphere = radius of the cone = height of the cone = r (say)

Then, ratio of volumes of cylinder, hemisphere and cone = πr^{3} : \(\frac{2}{3}\)πr^{3} : \(\frac{1}{3}\)πr^{3}

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

= 3 : 2 : 1

(ii) Side of hexagon = \(\frac{\text { Perimeter }}{\text { No. of sides }}\)

= \(\frac{600}{6}\)

= 100 cm

Area of base of regular hexagon = \(\frac{3 \sqrt{3}}{2}\) × 100 × 100 = 25980 sq. cm

[∵ Area of regular hexagon = \(\frac{3 \sqrt{3}}{2}\) × (side)^{2}]Volume = Base Area × height

= 25980 × 200

= 5196,000 cu. cm.

= 5.19 cu. m.

= 5196 Litres

Weight of Petrol = Volume × Density

= 519600 × 0.8 gm/cc

= 4152000 gm

= 4152 kg

Section-E

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

Question 36.

In a library, 25 students are reading books. It was found that 15 students are reading Mathematics, 12 reading Physics, and 11 students are reading Chemistry. If 5 students are reading both Mathematics and Chemistry, 9 students reading Physics and Mathematics 4 students reading Physics and Chemistry and 3 students are reading all three subjects.

Based on the above information answer the following questions:

(i) The number of students reading only chemistry books?

OR

The number of students reading only one subject?

(ii) The number of students reading atleast one subject?

(iii) If the total time taken to read a chemistry book is 14 hrs, then how many hours were devoted to reading only chemistry books by those students who are reading only chemistry books?

Answer:

(i) We can make the following Venn diagram to understand the problem easily.

Here, we have n(C ∩ M ∩ P) = 3 (given),

where C represents Chemistry, P represents Physics and M represents Mathematics.

n(U) = Total students = 25

By using the Venn diagram, we can easily see that students who read only chemistry are 5.

OR

The total students who read only one subject

n(c) + n(p) + n(m) = 5 + 2 + 4 = 11

(ii) Here we need to find:

n(P ∪ C ∪ M) = n(M) + n(P) + n(C) – n(P ∩ C) – n(P ∩ M) – n(M ∩ C) + n(P ∩ C ∩ M) = 23

(adding all the numbers from the Venn diagram will give us the result)

(iii) Since the total number of students who read only Chemistry books is 5,

therefore total time devoted to reading only chemistry books by these students is 5 × 14 = 70 hours.

Question 37.

Varun, Abhay, and Rohit were shooting. Varun hits the target 4 times in 5 shots, Abhay hits the target 3 times in 4 shots and Rohit hits the target 2 times in 3 shots.

Now based on this information answer the following questions:

(i) Find the probability that Varun, Abhay, and Rohit all may hit the target.

OR

What is the probability that Abhay and Rohit may hit but Varun may not hit?

(ii) What is the probability that none of them will hit the target?

(iii) Find the probability that any two among Varun, Abhay, and Rohit may hit the target.

Answer:

(i) Probability of Varun hitting the target = \(\frac{4}{5}\)

Probability of Abhay hitting the target = \(\frac{3}{4}\)

Probability of Rohit to hit the target = \(\frac{2}{3}\)

If all may hit the target, then required probability = \(\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{2}{5}\)

OR

Required probability = \(\left(1-\frac{4}{5}\right) \times \frac{3}{4} \times \frac{2}{3}=\frac{1}{10}\)

(ii) For none of them will hit the target, we have to find when they will not hit the target individually = \(\left(1-\frac{4}{5}\right) \times\left(1-\frac{3}{4}\right) \times\left(1-\frac{2}{3}\right)\)

= \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3}\)

= \(\frac{1}{60}\)

(iii) Required probability

Question 38.

The derivative tells us the rate of change of one quantity compared to another at a particular instant or point (So, we call it the “instantaneous rate of change”). This concept has many applications in electricity, economics, fluid flow, population modeling, queueing theory, and so on.

Let y = f(x) be a function of x. Let Δy be the change corresponding to the small change Δx in x. Then, \(\frac{\Delta y}{\Delta x}\) represents the change in y due to a unit change in x. In other words, \(\frac{\Delta y}{\Delta x}\) represents the average rate of change of y concerning x as x changes from x to x + Δx.

As Δx → 0, the limiting value of this average rate of change of y concerning x in the interval [x, x + Δx] becomes the instantaneous rate of change of y concerning x.

Thus, \(\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}\) = Instantaneous rate of change of y with respect to x

⇒ \(\frac{d y}{d x}\) = Rate of change of y with respect to x. [∵ \(\lim _{\Delta \dot{x} \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{d y}{d x}\)

(i) The radius r of a right circular cone is decreasing at the rate of 3 cm/min and the height h is increasing at the rate of 2 cm/min. When r = 9 and h = 6 cm, find the rate of change of its volume.

OR

(ii) Find the point on the curve y^{2} = 8x + 3 for which the y-coordinate changes 8 times more than the coordinate of x.

Answer:

(i) Given, \(\frac{d r}{d t}\) = -3 cm/min

and \(\frac{d h}{d t}\) = 2 cm/min

We know that,

Volume of cone, V = \(\frac{1}{3} \pi r^2 h\)

\(\frac{d V}{d t}=\frac{\pi}{3}\left[r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right]\)

At r = 9 and h = 6,

\(\frac{d V}{d t}\) = \(\frac{\pi}{3}\) [(9)^{2} × (2) + 2 × 9 × 6 × (-3)]= \(\frac{\pi}{3}\) (81 × 2 – 324)

= \(-\frac{\pi}{3}\) × 162

= -54π cm^{3}/min

Hence, volume is decreasing at the rate = 54π cm^{3}/min.

OR

(ii) Let the required point be P(x, y).

Given that,

Rate of change of y coordinate = 8(Rate of change of x-coordinate)