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

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

## CBSE Sample Papers for Class 12 Applied Maths Set 12 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 carhes 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 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 inequality |x – 7| < 2 is

(a) (3, 7)

(b) (5, 9)

(c) (8, 9)

(d)( -5, -9)

Solution:

(b) – 2 < (x – 7) < 2

⇒ -2 + 7 < x < 2 + 7

∴ 5 < x < 9

Hence, solution set is (5, 9).

Question 2.

If A is a non-singular 3 × 3 matrix and B is its adjoint such that |B| = 100, then |A| is equal to

(a) 81

(b) +9

(c) +10

(d) 15

Solution:

(c) Given, A is a square matrix of order 3 and 6 is its adjoint such that 161 = 100

⇒ |adj A| = |B|

∵ |adj A| = |A|^{n-1} for any non-singular square matrix A of order n.

For given matrix A of order 3 × 3, we have

|adj A| = |A|^{3-1} = |A|²

⇒ |A|² = 100

⇒ |A| = ±10

Question 3.

Two pipes A and B can fill a tank in 24 min and 32 min respectively. If both the pipes are opened simultaneously, after x min should B be closed so that the tank can be filled in 18 min. The value of x is

(a) 2

(b) 4

(c) 6

(d) 8

Solution:

(d) Let x be the time after which pipe B should be closed so that the tank can be filled in 18 min. Part of tank filled by pipe A in 1 min = \(\frac{1}{24}\)

Part of tank filled by pipe B in 1 min = \(\frac{1}{32}\)

Pipe A is opened for the whole time i.e. for 18 min and B is opened for x min.

According to the question,

Hence, pipe B should be closed after 8 min.

Question 4.

A person can row with the stream at 8 km/h and against the stream at 6 km/h. The speed of the current is

(a) 1 km/h

(b) 2 km/h

(c) 4 km/h

(d) 5 km/h

Solution:

(a) Given, upstream speed of person = 6 km/h

Downstream speed of person = 8 km/h

Speed of stream or current

= \(\frac{1}{2}\)(Downstream speed – Upstream speed)

= \(\frac{1}{2}\)(8 – 6) = \(\frac{1}{2}\) × 2 = 1 km/h

Question 5.

If a coin is tossed 20 times and the coin falls on head after any toss, it is success. Suppose the probability of success is 0.5, the probability that the number of success is less than or equal to 12 is

(a) 0.542

(b) 0.749

(c) 0.8133

(d) 0.6431

Solution:

(c) Given, µ = np =20 × 0.5 = 10

Question 6.

The general solution of \(\frac{dy}{dx}\) = 2xe^{x² – y}

(a) e^{x² – y} = C

(b) e^{-y}

(c) e^{y} = e^{x²} + C

(d) e^{x² + y} = C

Solution:

Question 7.

Between the hours 3 and 5 pm, the average number of phone calls per minute coming into the switchboard of a company is 2. The probability that during one particular minute there will be no phone call at all is (given e^{-2} = 0.13435)

(a) 0.12536

(b) 0.86565

(c) 0.13435

(d) None of these

Solution:

(c) Let X denotes the number of phone cails received during a minute.

Question 8.

A shopkeeper purchases 40 kg of sugar at the rate of ₹ 35 per kg and 50 kg of sugar at the rate of ₹ 40 per kg. He mixed these two types of sugar. He sell the mixture to make a 20% profit in the transaction at the price of

(a) ₹45.33 per kg

(b) ₹47.67 per kg

(c) ₹35.39 per kg

(d) ₹37.30 per kg

Solution:

(a) Let CP (Cost Price) of mixed sugar be ₹x per kg.

Using alligation rule,

Question 9.

The interval in which the function f(x) = 10 – 6x – 2x² is decreasing is

Solution:

(c) We have, f(x) = 10 – 6x – 2x²

⇒ f'(x) = -6 – 4x

For f(x)to be decreasing, we must have

f'(x) < 0

⇒ -6 – 4x < 0

⇒ -(6 + 4x) < 0

⇒ 6 + 4x > 0

⇒ x > –\(\frac{3}{2}\) ⇒ x ∈ (-\(\frac{3}{2}\), ∞)

Thus, f(x) is increasing on the interval (\(\frac{-3}{2}\), ∞).

Question 10.

If A and B are square matrices such that B = -A^{-1} BA, then (A + B)² = is equal to

(a) 0

(b) A² + B²

(c) A² + 2AB + B²

(d) A + B

Solution:

(b) Given, B = -A^{-1} BA

⇒ AB = -AA^{-1}BA

⇒ AB = -BA ⇒ AB + BA = 0

Now, (A+ B)² = (A + B)(A + B)

⇒ A² + AB + BA + B²

= A² + 0 + B² [∵ AB + BA = 0]= A² + B²

Question 11.

If in a binomial distribution n = 4, P(X = 0) = \(\frac{16}{81}\), then P(X = 4) equals

(a) \(\frac{1}{16}\)

(b) \(\frac{1}{81}\)

(c) \(\frac{1}{27}\)

(d) \(\frac{1}{8}\)

Solution:

Question 12.

If x = at2 and y = 2 at, then value of \(\frac{d^2y}{dx^2}\) is

(a) –\(\frac{1}{2at^3}\)

(b) –\(\frac{1}{2at^2}\)

(c) \(\frac{1}{t^2}\)

(d) \(\frac{-2a}{t}\)

Solution:

Question 13.

For the purpose of t-test of significance, a random sample of size (n) 48 is drawn from a normal population, then the degree of freedom (ν) is

(a) 49

(b) \(\frac{1}{48}\)

(c) 47

(d) 48

Solution:

(c) Degree of freedom = n – 1

Here, n = 48

∴ Degree of freedom = 48 – 1 = 47

Question 14.

The effective rate that is equivalent to a nominal rate of 8% compounded quarterly is

(a) 8.16%

(b) 8.5%

(c) 8.75%

(d) None of these

Solution:

Question 15.

Prosperity, recession and depression in a business is an example of

(a) seasonal trend

(b) cyclical trend

(c) irregular trend

(d) secular trend

Solution:

(b) Time series show rise and fall of the values in a short period time. The variations in a time series which operate themselves over a span of more than one year are called cyclic variation.

Question 16.

In the trend line equation y = a + bx, b is the

(a) slope

(b) mean

(c) x-intercept

(d) y-intercept

Solution:

(a) ‘b’ represents the slope of trend line.

Question 17.

A company intends to create a sinking fund to replace at the end of 20th year assets costing ₹ 50000. Calculate the amount to be retained out of profit every year, if the interest rate is 5%

[given (1.05)^{20} = 2.65 32](a) ₹ 15122.18

(b) ₹ 15322.18

(c) ₹ 15422.18

(d) ₹ 15022.18

Solution:

(a) Here,S = ₹50000, i = 5% = 0.05 and n = 20yr

Question 18.

The feasible region (shaded) for a LPP is shown in following figure.

Then, the maximum value of Z = 3x + 4y is

(a) 156

(b) 196

(c) 152

(d) 180

Solution:

(b) As clear from the graph, corner points are O, A E and D with coordinates (0, 0), (52,0), (44,16) and (0, 38), respectively. Also, given region is bounded.

Here, Z = 3x + 4y

∵ 2x + y = 104 and 2x + 4y = 152

⇒ -3y = -48 ⇒ y = 16 and x = 44

Comer points | Corresponding vaue of Z =3x + 4y |

O(0, 0) | 0 |

A(52, 0) | 156 |

E(44, 16) | 196 (Maximum) |

D(0, 38) | 152 |

Hence, Z at E(44, 16) is maximum and its maximum value is 196.

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) The ratio of copper and zinc in brass is 13 : 7. In 100 kg of brass, there is 35 kg zinc.

Solution:

(b) Assertion Let the quantity of copper and zinc in brass is 13x and 7x respectively. Total quantity of brass = 13x + 7x = 20x.

According to the question,

20x = 100 ⇒ x = 5

Quantity of zinc = 7 × 5 = 35 kg

Hence, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 20.

Assertion (A) A machine costing ₹ 30000 is expected to have a useful life of 13 yr and a final scrap value of ₹ 4000. The annual depreciation charge using straight line method is ₹ 2000.

Solution:

(a) Reason (R) is true.

Assertion (A) is true

Hence, both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

Section B

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

Question 21.

A sales promotion company sells tickets for ₹ 100 each to win a prize of ₹ 5 lakhs. If a person buys one of the 10000 tickets sold, then find his expected gain or loss (in ₹).

Solution:

Let X denotes the gaining amount.

X can take the value either 0 or 500000.

The probability distribution of X is

Question 22.

Solution:

Question 23.

Find the present value of perpetuity of ₹ 800 at the end of each quarter, if money is worth 10% compounded quarterly.

Solution:

Given, R = ₹800 and r = 10% = 0.10

Question 24.

The marginal revenue function of a commodity is given by MR = 6 – 5x + x².

Find the demand function.

Or

Evaluate \(\int_0^4\) [|x|+|x – 2|+|x – 4|]dx.

Solution:

We have, MR = 6 – 5x + x²

Question 25.

At what rate of interest will the present value of a perpetuity of ₹ 300 payable at the end of each quarter be ₹ 24000?

Solution:

Let the rate of interest be r% per annum, then

Hence, rate of interest is 5%.

Section C

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

Question 26.

Tap A can fill a tank 3 times faster than tap B and takes 28 min less than tap B to fill the tank. If both the taps are opened simultaneously, then find the time taken to fill the tank.

Solution:

Let the time taken by tap A to fill the tank be x min. Then, time taken by tap 6 to fill the tank be 3x min.

Then, according to the question,

3x – x = 28 ⇒ 2x = 28

⇒ x = \(\frac{28}{2}\) = 14min

Time taken by tap A = 14 min;

Part filled by tap A in 1 min = \(\frac{1}{14}\)

Time taken by tap B = 3 × 14 = 42 min

Part filled by tap 6 in 1 min = \(\frac{1}{42}\)

∴ Part filled by both taps in 1 min

= \(\frac{1}{14}+\frac{1}{42}=\frac{3+1}{42}=\frac{4}{42}=\frac{2}{21}\)

So, time taken by both taps to fill the tank0 working together = \(\frac{21}{2}\) min

Question 27.

A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is ₹ 50 and that on typeB circuit is ₹ 60, formulate this problem as an LPP, so that the manufacturer can maximise his profit.

Solution:

Let the manufacturer produces x units of type A circuits and y units of type B circuits.

From the given information, we have following corresponding constraint table.

Thus, we see that total profit Z = 50x + 60y (in ₹).

Now, we have the following mathematical model for the given problem.

Maximise Z = 50x + 60y ….(i)

Subject to the constraints are

20x + 10y ≤ 200 [resistors constraint]⇒ 2x+ y ≤ 20 …(ii)

and 10x + 20y ≤ 120 [transistor constraint]⇒ x + 2y ≤ 12 …(iii)

and 10x + 30y ≤ 150 [capacitor constraint]⇒ x + 3y ≤ 15 ,..(iv)

and x ≥ 0, y ≥ 0 [non-negative constraint] …(v)

So, maximise Z = 50x + 60y,

Subject to constraints

2x + y ≤ 20, x + 2y ≤ 12, x + 3y ≤ 15 x ≥ 0, y ≥ 0.

Question 28.

Find the interval for which the function f(x) = [x(x – 2)]² is a strictly increasing function.

Or

Find the absolute maximum value of the function f given by

Solution:

Given function is

Therefore, y is strictly increasing in (0, 1) and (2, ∞)

Hence, absolute maximum value of function f is 18 at x = -1

Question 29.

Solution:

Question 30.

Naman borrowed ₹ 1000000 from a bank to purchase a house and decided to repay the loan by equal monthly installments in 10 yr. If bank charges interest at 9% per annum compounded monthly, calculate the EMI. [given(1.0075)^{120} = 2.4514]Or

Has set up a sinking fund in order to have ₹ 50000 after 12 yr for his children’s equation. How much should be set aside bi-annually into on account paying 5% per annum compounded half-yearly?

[given (1.025)^{24} = 1.809]Solution:

Question 31.

A trader mixes 14 kg rice of variety A which costs ₹ 60 per kg with 18 kg of quantity of type B rice. He sells the mixture at ₹ 65 per kg and earns a profit of \(\frac{100}{3}\)%. Then, find the cost price of type B rice.

Solution:

Let cost price of mixture be ₹ y per kg.

Section D

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

Question 32.

There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.

Solution:

Here.S ={(1, 3), (1, 5), (1, 7), (3, 1), (3, 5), (3, 7), (5, 1), (5, 3), (5, 7), (7, 1), (7, 3), (7, 5)} ⇒ n(S) = 12.

Let random variable X denotes the sum of the numbers on two cards drawn. So, the random variables X may have values 4, 6, 8,10 and 12.

Question 33.

Consider the following data

Dates in the month of April | Number of units sold |

12 | 2 |

13 | 5 |

14 | 0 |

15 | 12 |

16 | 13 |

17 | 25 |

18 | 45 |

19 | 13 |

20 | 31 |

21 | 18 |

22 | 11 |

23 | 2 |

24 | 3 |

25 | 1 |

Calculate 3-days moving average and display these and the original figures on the same graph.

Or

The production of a soft drink company in thousand of litres during each month of a year is as follows

Calcuate the 5-monthly moving average and show these moving averages on a graph.

Solution:

According to the question,

Dates in the month of April | Number of units sold |

12 | 2 |

13 | 5 |

14 | 0 |

15 | 12 |

16 | 13 |

17 | 25 |

18 | 45 |

19 | 13 |

20 | 31 |

21 | 18 |

22 | 11 |

23 | 2 |

24 | 3 |

25 | 1 |

Let the 3-days moving average is m.

On the basis of above data, we can draw the following graph.

Or

According to the question,

Question 34.

The marginal cost function for a product is given by MC = \(\frac{5}{\sqrt{2x+9}}\) and fixed cost is ₹ 20 find the average cost for 10 units of output.

Or

The demand function of a product is p = 10 e^{-x}. Find the consumer’s surplus when the market price is p = L

(given log_{10} e = 0.4343).

Solution:

When x = 10, we get

Hence, average cost for 10 units of output is ₹ 3.195.

Or

Given, the demand function is

Question 35.

Rajesh purchased a house from a company for ₹ 2500000 and made a down payment of ₹ 500000. He repays the balance in 25 yr by monthly instalments at the rate of 9% per annum compounded monthly

(i) What are the monthly payment?

(ii) What is the total interest payment? [given (1.0075)^{-300} = 0.1062]Solution:

Cost of house = ₹ 2500000

Down payment = ₹ 500000

∴ Principal amount = ₹ (2500000 – 500000)

Hence, total interest is ₹ 3034681.

Section E

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

Question 36.

Consider the following hypothesis test

H_{0} : p ≤ 16

H_{a} : p > 16

A sample of 30 provided mean \(\overline{\mathrm{x}}\) = 18 and a sample standard deviation, S = 5.14.

Based on the above information, answer the following questions.

(i) Find the degree of freedom.

(ii) Find the value of test statistics.

(iii) Find the range of p-value.

Or

Find the sum of the minimum value and maximum value of p.

Solution:

(iii) Since, t = 2.13 > 0

Therefore, p-value of 2.13

= Area under the f-distribution curve to the right off. Since, t = 2.13 lies between 2.045 and 2.462 for which area lies between 0.01 and 0.025.

∴ 0.01 < p-value < 0.025

Or

Sum of minimum value and maximum value of

p = 0.01 + 0.025 = 0.035

Question 37.

A manufacturer has three machine I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 h, whereas machine III must be operated for atleast 5 h a day. She produces only two items M and N each requiring the use of all the three machines.

The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table.

She makes a profit of ₹ 600 and ₹ 400 on items M and N, respectively.

Let x and y be the number of items M and N, respectively and assume that she can sell all the items that she produced.

On the basis of above information, answer the following questions.

(i) Find the objective functions.

(ii) Find the number of corner points of feasible regions.

(iii) How many units of item M should she produce so as to maximise her profit?

Or

How many units of item N should she produce so as to maximise her profit?

Solution:

(i) Objective function is given by maximise

Z = 600x + 400y

(ii) Mathematical formulation of the given problem is as follows

Maximise Z = 600x + 400y

Subject to the constraints are

x + 2y ≤ 12 … (i)

2x + y ≤ 12 …(ii)

x + \(\frac{5}{4}\) ≥ 5 …(iii)

x ≥ 0, y ≥ 0 …(iv)

Let us draw the graph of constraints (i) to (iv). ABODE is the feasible region (shaded) as shown below. Observe that the feasible region is bounded, coordinates of corner points A, B,C,D and E are (5,0), (6, 0), (4, 4), (0, 6) and (0, 4), respectively.

∵ Corner points of the feasible region are (5, 0), (6, 0), (4, 4), (0, 6) and (0, 4).

∴ Required answer is 5.

(iii) Let us evaluate Z = 600x + 400y at these corner points.

Corner points | Value of Z = 600x + 400y |

(5, 0) | 600 × 5 + 400 × 0 = 3000 |

(6, 0) | 600 × 6 + 400 × 0 = 3600 |

(4, 4) | 600 × 4 + 400 × 4 = 4000 (Maximum) |

(0, 6) | 600 × 0+ 400 × 6 = 2400 |

(0, 4) | 600 × 0 + 400 × 4 = 1600 |

We see that the point (4, 4) is giving the maximum value of Z. Hence, 4 units of item M should be produced to maximise the profit.

Or

∵ The point (4, 4) is giving the maximum value of Z.

∴ A units of item N should be produced to maximise the profit.

Question 38.

A factory produces three items every day. Their production on certain day is 45 tonnes. It is found that the production of third items exceeds the production of first item by 8 tonnes while the total production of first and third item is twice the production of second item.

On the basis of above information, answer the following questions.

Find the production level of each product by using matrix method.

Or

Determine the production level of each product using Cramer’s rule.

Solution:

Let the production level of first, second and third product on a certain day be x tonnes, y tonnes and z tonnes, respectively.

According to the question,

x + y + z = 45 …(i)

⇒ x + 8= z

⇒ x + 0 . y – z = -8 …(ii)

and x + z = 2y

⇒ x – 2y + z = 0 …..(iii)

Matrix from of equation

Or

According to question

Hence the company products 11 tonnes of first product, 15 tonnes of second product and 19 tonnes third product everyday.