# CBSE Sample Papers for Class 12 Applied Maths Set 6 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 6 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths Set 6 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.
- Section-A carries 20 marks weightage, Section-B carries 10 marks weightage, Section-C carries 18 marks weightage, Section-D carries 20 marks weightage and Section-E carries 3 case-based with total weightage of 12 marks.
- Sections-A comprises of 20 MCQs of 1 marks each.
- Section-B comprise of 5 VSA Type Questions of 2 marks each.
- Section-C comprises of 6 SA Type Questions of 3 marks each.
- Section-D comprises of 4 LA Type Questions of 5 marks each.
- Section-E has 3 Case Studies. Each case study comprises of 3 case-based questions, where 2 VSA Type Questions of 1 marks each and 1 SA Type Questions is of 2 marks. Internal choice is provided in 2 marks questions in each case-study.
- Internal choice is provided in 2 questions in Section-B, 2 questions in Section-C, 2 questions in Section-D. You have to attempt only one of the alternatives in all such questions.

Section A

All questions are compulsory. No Internal Choice is provided in this section

Question 1.

The solution set of the inequalities 6 ≤ – 3(2x – 4) ≤ 12 is

(a) (-∞, 1]

(b) (0, 1]

(c) (0, 1] ∪ [1, ∞)

(d) [1, ∞)

Solution:

(b) The given inequality 6 ≤ – 3(2x – 4) < 12

⇒ 6 ≤ -6x + 12<12

Adding (- 12) to each term,

6 – 12 ≤ -6x + 12 – 12 < 12 – 12

⇒ -6 ≤ – 6x < 0

Dividing by (-6) to each term,

∴ Solution set is (0, 1]

Question 2.

The function f(x) = x³ is

(a) decreasing on R

(b) increasing on R

(c) Both (a) and (b)

(d) None of these

Solution:

(b) We have, f(x) = x³

⇒ f'(x) = 3x²

Since 3x² > 0 ∀ x ∈ R

∴ f'(x) > 0

So, f(x) is increasing on R.

Question 3.

(a) scalar matrix

(b) diagonal matrix

(c) identity matrix

(d) None of these

Solution:

In a square matrix, if a_{ij} = 0 for i ≠ j, then the matrix is said to be a diagonal matrix.

Hence, the given matrix A is a diagonal matrix.

Question 4.

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The minimum value of F occurs at

(a) only (0, 2)

(b) only (3, 0)

(c) the mid-point of the line segment joining the points (0, 2) and (3, 0)

(d) any point on the line segment joining the points (0, 2) and (3, 0)

Solution:

Corner points | Value 0f F = 4x + 6y |

(0, 2) | 4 × 0 + 6 × 2 = 12 (Minimum) |

(3, 0) | 4 × 3 + 6 × 0 = 12 (Minimum) |

(6, 0) | 4 × 6 + 6 × 0 = 24 |

(6, 8) | 4 × 6 + 6 × 8 = 72 (Maximum) |

(0, 5) | 4 × 0 + 6 × 5 = 30 |

Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2) and (3, 0).

Question 5.

The least non-negative remainder when 612 is divided by 7 is

(a) 1

(b) 3

(c) 4

(d) 6

Solution:

(a) We find that remainder when 6^{12} is divided by 7

∴ 6 = -1(mod 7)

⇒ (6)^{12} = (-1)^{12} (mod 7)

⇒ 6^{12} = 1 mod 7

Hence, the remainder when 6^{12} is divided by 7 is 1.

Question 6.

The probability distribution of a discrete random variable X is given below

The value of k is

(a) 8

(b) 16

(c) 32

(d) 48

Solution:

(c) We know that ∑(X) = 1

⇒ \(\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}\) = 1

⇒ \(\frac{32}{k}\)

∴ k = 32

Question 7.

Multiplicative model for time series O is equal to

(a) T + S × C × I

(b) T × S × C + I

(c) T × S × C × I

(d) TS + S × I

Solution:

(c) In this model, we represents a particular observations in a time series as the product of these four components.

Question 8.

The time series analysis helps

(a) to make predictions

(b) to compare the two or more series

(c) to know the behaviour of business

(d) All of the above

Solution:

(d) The time series analysis helps to make predictions, to compare the two or more series and to know the behaviour of business.

Question 9.

Let X is a normal variable with mean is 10 and standard deviation is 2. If P(12 < X < a) = 0.1465, then the value of a is

(a) 12

(b) 22

(c) 14.5

(d) 18.5

Solution:

(c) Let µ be the mean and σ be the deviation.

Given, µ = 10 and σ = 2

Question 10.

The annual depreciation of a car is ₹30000, if the scrap value of the car after 15 yr is ₹50000, then original cost of the car, when depreciation is linear.

(a) ₹500000

(b) ₹450000

(c) ₹600000

(d) None of these

Solution:

(a) Here, annual depreciation = ₹ 30000

Scrap value of car = ₹ 50000

Useful life of car = 15yr

Let the original cost of car is ₹ x.

Hence, the original cost of car was ₹ 500000.

Question 11.

Random sampling is useful as it is

(a) reasonably more accurate as compared to other methods

(b) economical in nature

(c) free from personal biases of the investigator

(d) All of the above

Solution:

(d) All the given options are correct.

Question 12.

Solution:

Question 13.

The effective rate of interest which is equivalent to a nominal rate of 8% compounded semi-annually is

(a) 8.24%

(b) 8.33%

(c) 8.16%

(d) 8.06%

Solution:

∴ Effective rate of interest = (0.0816 × 100)%

= 8.16%

Question 14.

A cover’s a distance of 1 km in 4 min 54 s and B in 5 min. In a race of 1000 km at what distance B should stand ahead of A, so that both reach last point together

(a) 20 m

(b) 25 m

(c) 15 m

(d)10m

Solution:

(a) A covers distance of 1000m in 294 s.

B covers distance of 1000 m in 300 s.

Start distance between A and B is

300 s → 1000 m

6 s → 20 m

Means B should stand 20 m ahead of A.

Question 15.

For the system of equations x – 2y = 4 and 3x – 5y = 7, the values of x and y are respectively

(a) – 6 and – 5

(b) 6 and 5

(c) – 6 and 5

(d) 6 and – 5

Solution:

(a) The given system can be written as AX = B, where

Question 16.

One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is

(a) 12\(\frac{1}{7}\)%

(b) 24\(\frac{2}{7}\)%

(c) 37%

(d) 46%

Solution:

(b) Given, a glass is filled with 10 parts of first liquid and 4 parts of second liquid.

∴ Total volume of glass = 14

Question 17.

If X is a Poisson variable such that P(X = K) = P(X = K + 1), then variance of X is

(a) K – 1

(b) K

(c) K + 1

(d) K + 2

Solution:

(c) Given, P(X = K) = P(X = K + 1)

Question 18.

Pipes A and B can fill a tank in 15 and 12 min, respectively. Pipe C empties the tank at the rate of 2 L/min. If all the pipes are opened at the same time, then the tank is filled in 8\(\frac{4}{7}\) min. The capacity of tank is

(a) 80 L

(b) 75 L

(c) 60 L

(d) 90 L

Solution:

(c) Part of tank filled by pipe A in 1 min = \(\frac{1}{15}\)

Part of tank filled by pipe 8 In 1 min = \(\frac{1}{12}\)

Let pipe C can empty the tank in ‘a’ min.

Then, part of tank emptied by pipe C in 1 min = \(\frac{1}{a}\)

According to the question,

Hence, pipe C will empty the tank in 30 min. Rate of flow of pipe C = 2 LVmin

∴ Capacity of tank = (30 × 2)L = 60 L

Directions In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

(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) (186 × 93) (mod 7) = 2

Reason (R) (a . b)(mod n) = a(mod n) . b(mod n)

Solution:

(d) Reason We know that,

(a . b) (mod n) = a (mod n) . b (mod n)

Assertion (186 × 93) (mod 7)

= (186) (mod 7) × (93) (mod 7)

= 4 (mod 7) × 2 (mod 7)

= 4 × 2 (mod7)

= 8mod7 31 mod 7

∴ (186 × 93) mod 7 = 1

Hence, Assertion is false but Reason is true.

Question 20.

Assertion (A) The points

A(a, b + c), B(b, c + a), C(c, a + b) are collinear

Reason (R) Area of triangle with three collinear points are zero.

Solution:

(a) Assertion We know that if area of triangle is zero, then the points are collinear.

Since, area of ∆ABC = 0

Hence, points are collinear.

Hence, both Assertion and Reason are true and Reason is correct explanation of Assertion.

Section B

All questions are compulsory. In case of Internal choice, attempt any one question only

Question 21.

The population of a town grows at the rate of 10% per year. Using differential equation, find how long it will take for the population to grows 3 times.

Or

The marginal cost of producing x pairs of tennis shoes is given by MC = 60 + \(\frac{400}{x+1}\). If the fixed cost is ₹ 3000, find the total cost function.

Solution:

Let P_{0} be the population initially and P be the population after t yr,

C = 60x + 400log|(x +1)| + K …(i)

If x = 0, C = ₹3000

∴ 3000 = 0 + 400 log (1) + K

⇒ K = 3000

∴ C = 60x + 400 log |(x + 1)| + 3000 [from Eq. (i)]

Question 22.

Find the present value of perpetuity of ₹ 900 at the end of each quarter, if money is worth 6% compounded quarterly.

Or

To what sum will ₹ 16000 accumulate in 8 yr, if invested at an effective rate of 10%. [given (1.08)^{8} = 1.85]

Solution:

Given, R = ₹900

Hence, the required sum is ₹ 29600.

Question 23.

The corner points of the feasible region determined by the following systems of linear inequalities

2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0 and y ≥ 0 are (-0, 0), (5, 0), (3, 4) and (0, 5).

Let Z = px + qy, when p, q > 0, then find the relation between p and q, so that the maximum of Z occurs at both points (3, 4) and (0, 5).

Solution:

The values of Z = px + qy at the points (3, 4) and (0, 5) are 3p + 4q and 5q, respectively.

As, Z has maximum value of both points (3, 4) and (0, 5), we get

3p + 4g = 5 q

⇒ 3p = q

or 3p – q = 0

which is the required relation between p and q.

Question 24.

Suppose 2 × 3 matrix A = [a_{ij}], elements are given by a_{ij} = \(\frac{i+2j}{2}\) construct the matrix A.

Solution:

In general, the matrix A of order 2 × 3 is given by

Question 25.

A random sample of 36 workers is drawn from a factory to test, if their average monthly wages is ₹ 1500 or not. The population wages distribution is assumed to be normal with standard deviation of ₹ 300. The average monthly wages based on the 36 observations comes out to be ₹ 1800. Perform a test at 5% level of significance. Calculate test statistics.

Solution:

Section C

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

Question 26.

Sanchit borrowed a home loan amount of ₹ 5000000 from a bank at interest rate of 12% annum for 30 yr. Find the EMI. Sanchit has to pay the bank.

[use [101]^{-360} = 0.02781]

Solution:

Question 27.

Find the value of ∫\(\frac{1}{\sqrt{x^2+2x+2}}\) dx.

Or

Evaluate ∫(x² + 1)log x dx.

Solution:

Question 28.

Consider the probability distribution of a random variable X

Calculate the value of Var(X).

Solution:

Given, probability distribution table

Question 29.

Calculate the quarterly trend values by method of least square for quarterly data for last 5 yr given below.

Or

Apply the method of least squares to obtain the trend values from the following data.

Year | Sales (in lakh tonne) |

2006 | 100 |

2007 | 120 |

2008 | 110 |

2009 | 140 |

2010 | 80 |

Also, predict the sales for the year 2007.

Solution:

Here, n = 5 (odd)

So, middle year i.e, 1966 is taken as origin, we will fit linear trend equation between average quarterly values (y) anc! time variable x (in year).

∴ Linear trend equation is y, = 112 + 23x

From Eqs. (I) and (ii), yearly increment in trend value

b = 23

Quarterly increment = \(\frac{23}{4}\) = 5.75

Or

Here, n = 5 (odd)

So, middle year i.e. 2008 is taken as origin.

The required equation is,

y_{t} = 110 – 2x

For x = -2, y_{t} = 110 -2(-2) = 114

Thus, for 2007 the trend value will be 114 – 2 = 112.

Question 30.

If Mr Nirav deposits ₹250 at the beginning interest of 6% per annum compounded monthly, how many months will be required for the deposit to amount to at least ₹6390?

[given log(1.127) = 0.5190 and log (1.005) = 0.021]

Solution:

Let Mr Nirav deposit amount per month is a = ₹250

The total amount of annuity is (M) = ₹6390

Rate of interest, r = 6% = \(\frac{6}{600}\) per annum

Let the number of months be n.

We know that,

Taking log on both sides of Eq. (i), we get

∴ Number of months required = 24 months

Question 31.

Consider a hypothetical population comprising only four values 4, 6, 8 and 10. Find point estimation of population variance (σ²). Also, find standard error of sample variance (S²).

Solution:

Section D

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

Question 32.

A propeller costs ₹ 180000 and its effective life is estimated to be 10 yr. A sinking fund is created for replacing the propeller by a new model at the end of its lifetime, when its scrap realises a sum of ₹ 34000 only. The price of the new model is estimated to be 30% more than the price of the present one. What amount should be put into the sinking fund at the end of each year, if it accumulates at 4% per annum compound interest? [given, (1.04)^{10} = 1.480]

Solution:

According to the question,

The present value of the machine is ₹ 180000.

Since, the price of new model is 30% more than the price of present machine.

∴ Price of new model = \(\frac{130}{100}\) × 180000 = ₹234000

Scrap value of the present machine is ₹ 34000.

The net amount (A) to be paid after 10 yr for new model = 234000 – 34000

= ₹ 200000

Let the value of each sinking fund is a.

The rate of interest for 10 yr is

Hence, the value of each sinking fund is ₹ 16666.67 to be paid at the end of each year.

Question 33.

Determine the product of

solve the system of equations

x – y + z = 4, x – 2y – 2z = 9 and 2x + y + 3z = 1.

Solution:

On comparing corresponding elements, we get

x = 3, y = -2 and z = -1

Question 34.

There are two types of fertilisers F_{1} and F_{2}, F_{1} consists of 10% nitrogen and 6% phosphoric acid and F_{2} consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F_{1} costs ₹ 6 per kg and F_{2} costs ₹ 5 per kg. Determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost? What is the minimum cost?

Or

Two godowns A and B have a grain storage capacity of 100 quintals and 50 quintals, respectively. They supply it to 3 ration shops D, E and F, whose requirements are 60, 50 and 40 quintals, respectively. The cost of transportation per quintal from godowns to the shops are given in the following table

How should the supplies be transported in order that the transportation cost is minimum?

Solution:

Suppose, the farmer uses x kg of F_{1}, and y kg of F_{2}.

We have to construct the following table

Total cost of fertilisers,

Z = 6x + 5y

So, our problem is to minimise Z = 6x + 5y

Subject to constraints are

On putting (0, 0) in the inequality

2x + y ≥ 280, we get

2 × 0 + 0 ≥ 280

⇒ 0 ≥ 280, which is not true.

So, the half plane is away from the origin.

Table for line 3x + 5y = 700 is

So, line passes through the points (0, 140) and (\(\frac{700}{3}\), 0)

On putting (0, 0) in the inequality

3x + 5y ≥ 700, we get

3 × 0 + 5x 0 ≥ 700

⇒ 0 ≥ 700, which is not true.

So, the half plane is away from the origin.

Also, x, y ≥ 0, so the region lies in the I quadrant.

On solving the equations

2x + y = 280 and 3x + 5y = 700, we get B(100, 80).

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are

A(\(\frac{700}{3}\), 0) B(100, 80) and C(0, 280).

The values of Z at the corner points are given below

As, the feasible region is unbounded, therefore 1000 may or may not be the minimum value of Z. For this, we draw a graph of the inequality 6x + 5y < 1000 and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with 6x + 5y < 1000. Therefore, 100 kg of fertiliser F_{1}, and 80 kg of fertiliser F_{2} should be used to minimise the cost and the minimum cost is ₹1000.

Or

Let godown A supplies x and y quintals of grain to the shops D and E, respectively.

Then, (100 – x – y) will be supplied to shop F.

The requirement at shop D is 60 quintals, since x quintals are transported from godown A.

Therefore, the remaining (60 – x) quintals are transported from godown B Similarly,

(50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shops E and F, respectively. The given problem can be represented diagrammatically as follows

Let Z be the total cost of transportation, then

Z = 6x + 3y + 2.50(100 – x – y) + 4(60 – x) + 2(50 – y) + 3[(x + y) – 60]

= 6x + 3y + 250 – 2.50x – 2.50y + 240 – 4x + 100 – 2y + 3x + 3y – 180

= 2.50 x + 1.50y+410 …(i)

Subject to constraints,

60 – x ≥ 0

or x ≤ 60 …(ii)

50 – y ≥ 0

or y ≤ 50 …(iii)

100 – (x + y) ≥ 0

or x + y ≤ 100 …..(iv)

x + y – 60 ≥ 0

or x + y ≥ 60 …(v)

and x, y ≥ 0 …(vi)

Table for line x + y = 100 is

So, line passes through the points (100, 0) and (0, 100).

On putting (0, 0) in the inequality x + y ≤ 100, we get

0 + 0 ≤ 100

⇒ 0 ≤ 100, which is true.

So, the half plane is towards the origin.

Table for line x + y = 60 is

So, line passes through the points (0, 60) and (60, 0).

On putting (0, 0) in the inequality x + y ≥ 60, we get 0 + 0 ≥ 60

⇒ 0 ≥ 60, which is not true.

So, the half plane is away from the origin.

Now, draw the graph of lines x = 60 and y = 50, which is perpendicular to X and Y-axes, respectively.

Clearly, the half planes x ≤ 60 and y ≤ 50 is towards Y and X-axes, respectively.

Also, x, y ≥ 0, so the region lies in the I quadrant.

The points of intersection of lines corresponding to Eqs. (ii), (iii), (iv) and (v) are

A(60, 0), B(60, 40), C(50, 50) and D(10, 50).

The graphical representation of these lines is given below

The shaded region in the graph represents the feasible region and its corner points are A (60, 0), e (60, 40), C (50, 50) and D (10, 50).

The values of Z at the corner points are given below

Corner points | Value of Z = 2.5x + 1.5y + 410 |

A(60, 0) | Z = 2.5(60) + 1.5(0) + 410 = 560 |

B(60, 40) | Z = 2.5(60) + 1.5(40) + 410 = 620 |

C(50, 50) | Z = 2.5(50) + 1.5(50) + 410 = 610 |

D(10, 50) | Z = 2.5(10) +1.5 (50) + 410 = 510(Minimum) |

The minimum value of Z is 510 at D (10, 50).

Thus, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals and 40 quintals, respectively and from B to D, E and F is 50 quintals, 0_{r} quintal and 0 quintal, respectively.

The minimum cost is ₹ 510.

Question 35.

In a bank principal increases at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it be worth after 10 yr (given, e^{05} = 1.648)?

Or

Suppose, it is given that the rate at which some bacteria multiply is proportional to the instantaneous number present. If the original number of bacteria triples in three hours. In how many hours will it be four times.

Solution:

Let P and f be the principal and time, respectively.

It is given that the principal increases continuously at the rate of 5% per year.

Hence, after 10 yr, the amount will worth ₹ 1648.

Or

Let the original count of bacteria be N0 and at any time the count of bacteria be N.

Section E

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

Question 36.

A, B, C, D are the four contestant in a 1 km race. A can run 2.16 km in 9 min, B can run 1.8 km in 7.5 min C can run 1.5 km in 5 min and D can run 9 km/h.

Based on above information answer the following question

(i) Find the ratio of speed A : D.

(ii) Which contestant win the race without any start?

(iii) If C gives a start of 40 m to B and still beat him by how much seconds?

Or

Find the B’ time over the course.

Solution:

Here, C win by 40 s

Or

Speed of B = 4 m/s

B’s time over the course = \(\frac{1}{4}\) × 1000

= 250 s

Question 37.

Suman was doing a project on a school survey in the average number of hours spent on study by student selected at random. At the end of survey suman prepared the following report relating the data.

Let X denotes the average number of hours spent to the study by students. The probability that X can take the value x has the following, where k is some unknown constant.

Based on above information answer the following question.

(i) Find the value of k.

(ii) What is the probability that the average study time of student is exactly 2 h.

(iii) What is the probability that the average study time of student is atleast 3 h.

Or

What is the probability that the average study time of student is atleast 1 h?

Solution:

(i) We know, ∑P_{1} = 1

Then, 0.2 + k + 2k + 3k + 2k + 0 = 1

⇒ 8k = 1 – 0.2 = 0.8

⇒ k = \(\frac{0.8}{8}\) = 0.1

(ii) P(average study time is exactly 2 h)

= P(X = 2) = 2k = 2(0.1) = 02

(iii) P (average study time is at least 3 h)

= P(X ≥ 3) = P(X = 3) + P(X = 4)

= 0.3 + 0.2 = 0.5

Or

P(average study time is at least 1 h)

P(X ≥ 1) = 1 – P(X = 0)

= 1 – 0.2 = 0.8

Question 38.

Suppose, C represents the total cost function.

Then, the marginal cost function is given by

MC = \(\frac{dC}{dx}\)

⇒ dC = MC dx

⇒ ∫dC = ∫MC dx + C

⇒ C = ∫MC dx + k

where k is a constant of integration if we divided the cost function by x, we get the average cost function i.e. AC = \(\frac{C}{x}\)

The marginal cost function of manufacturing x units of a commodity is 6 + 10x – 6x². The total cost of producing one unit of the commodity is ₹ 12. Find the total cost function and average cost function.

Or

The marginal cost function MC for a product is given by MC = \(\frac{5}{\sqrt{2x+9}}\) and fixed cost is ₹ 20 find the average cost for 8 units of outputs.

Solution”

Given, marginal cost = 6 + 10x – 6x²

C = ∫MC dx + k

⇒ C = ∫(6 + 10x – 6x²)dx + k

⇒ C = 6x + 5x² – 2x³ + k ,..(i)

It is given that x = 1, C = 12

12 = 6+ 5-2 + k

⇒ k = 3

On putting k = 3 in Eq. (i), we get

C = 6x + 5x² – 2x³ + 3

Hence, total cost function is

C = 6x + 5x² – 2x³ + 3