# CBSE Sample Papers for Class 12 Applied Maths Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 3 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths 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.
- 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 inequality |3x – 2| ≤ \(\frac{1}{2}\) is

Solution:

Question 2.

A fire in a factor delaying production for some weeks is

(a) secular trend

(b) irregular trend

(c) cyclical trend

(d) seasonal trend

Solution:

(b) Since, the irregular component of the time series is the factor which causes the variations that are random.

Question 3.

The best fitting trend is one in which the sum of squares of residuals is

(a) maximum

(b) zero

(c) minimal

(d) None of these

Solution:

(c) The sum of the square of deviations y must be minimal.

Question 4.

The equation of normal at (1, 2) to the curve y² = 4x is

(a) 4x – y = 2

(b) x + y – 3 = 0

(c) y – x = 1

(d) None of these

Solution:

(b) We have,

Question 5.

Evaluate (57 – 42) mod 11 is

(a) 3

(b) 5

(c) 9

(d) 4

Solution:

(d) To find (57 – 42) mod 11, let us divide

So, (57 – 42)modi 1 = 4

Question 6.

If ∫|x|dx = kx |x| + C x ≠ 0, then the value of k is

(a) 2

(b) -2

(c) –\(\frac{1}{2}\)

(d) \(\frac{1}{2}\)

Solution:

(d) Let l = ∫|x|dx

l = x|x| – l + C

2l = x + |x| + C

l = \(\frac{1}{2}\)x|x| + C

∴ k = \(\frac{1}{2}\)

Question 7.

In a binomial distribution, the probability of getting a success is \(\frac{1}{4}\) and the standard deviation is 3, then its mean is

(a) 6

(b) 8

(c) 12

(d) 48

Solution:

Question 8.

A, B and C are three contestants in 800 m race. If A can give B a start of 20 m and A can given C a start of 32 m, then B gives C a start of

(a) 12.310 m

(b) 11.307 m

(c) 13.307 m

(d) 12.703 m

Solution:

(a) A cover’s 800 m.

B cover’s 800 – 20 = 780

C cover’s 800 – 32 = 768

So, when 6 covers 780 m, then C covers 768 m.

∴ When B covers 1 m, then C covers \(\frac{768}{780}\) m

⇒ When B cover 800m, then C cover (\(\frac{768}{780}\) × 800) = 787.69

B can give C a start of 800 – 787.69= 12.310 m

Question 9.

The present value of a perpetuity of ₹ 5000 payable at the end of each year, if money is worth 5% compounded annually is

(a) ₹ 20000

(b) ₹100000

(c) ₹ 10000

(d) ₹ 25000

Solution:

(b) We know that

∴ Present value = ₹100000

Question 10.

The process of using sample data to estimate the values of unknown population parameters is called

(a) estimation

(b) population

(c) sampling

(d) interval estimation

Solution:

(a) Clearly, to estimate the values of unknown population parameters by using sample data is called estimation.

Question 11.

Manish takes a loan of ₹300000 at an interest of 10% compounded annually for a period of 3 yr, then EMI by using flat rate method is

(a) ₹ 1083.33

(b) ₹ 1073.33

(c) ₹ 1093.33

(d) ₹ 1063.33

Solution:

(a) We have, P = Principal = ₹ 30000

Question 12.

A company intends to create a sinking fund to replace at the end of 20th year assets costing ₹500000. Calculate the amount to be retained out of profit every year, if the interest rate is 5%

(given (105)^{20} = 2.6532)

(a) ₹15122.18

(b) ₹15322.18

(c) ₹15422.18

(d) ₹15022.18

Solution:

(a) Here, S = ₹500000, i = 5% = 0.05 and n = 20 yr

Question 13.

A man can row with a speed of 8 km/h in still water. If the speed of stream is 4 km/h, then the upstream speed of boat is

(a) 4 km/h

(b) 6 km/h

(c) 5 km/h

(d) 12 km/h

Solution:

(a) Given, speed of man, x = 8 km/h

and speed of stream, y = 4 km/h

∴ Upstream speed = (x – y) km/h

= 8 – 4

= 4 km/h

Question 14.

If A² + A + I = O, then A^{-1} is equal to

(a) A – I

(b) I – A

(c) -(A + I)

(d) None of the above

Solution:

(c) We have, A² + A + I = 0

On pre multiplying by A^{-1} both sides, we get

A^{-1}(A² + A + l) = A^{-1}O

⇒ A^{-1}A² + A^{-1}A + A^{-1}l = 0

⇒ lA + l + A^{-1} = 0

⇒ A + l + A^{-1} = 0

⇒ A^{-1} = -(A + l)

Question 15.

A 95% confidence interval for a population was reported to be 152 to 160. If σ = 15, then sample size is

(a) 60

(b) 54

(c) 72

(d) 65

Solution:

(b) Let the sample mean be \(\overline{\mathrm{x}}\) and margin of error be E.

Then, \(\overline{\mathrm{x}}\) – E = 152 ….(i)

and \(\overline{\mathrm{x}}\) + E = 160 ….(ii)

On solving Eqs. (i) and (ii), we get

Question 16.

The comer points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Then, the condition on p and q, so that the maximum of Z occurs at both points (15, 15) and (0, 20) is

(a) p = q

(b) p = 2q

(c) q = 2p

(d) q = 3p

Solution:

(d) The corner point of feasible region are (0, 10), (5, 5), (15, 15), (0, 20).

Maximise Z = px + qy

Maximum value of Z at (15, 15) and (0, 20)

∴ 15p + 15q = 0 + 20g

⇒ 15 p = 5g

⇒ 3p = q

Question 17.

A machine costing ₹150000 has scrap value of ₹22500. If annual depreciation charge is ₹8500, then useful life of the machine is

(a) 10 yr

(b) 12 yr

(c) 15 yr

(d) 16 yr

Solution:

(c) Here, original cost of machine = ₹150000

Scrap value of machine = ₹22500

Annual depreciation = ₹8500

∴ Annual depreciation

Question 18.

If the probability that an individual suffers a bad reaction from injection of a given serum is 0.002, then the probability that out of 1000 individuals exactly 2 individuals will suffer from a bad reaction is (use e^{-2} = 0.1353)

(a) 0.7936

(b) 0.7206

(c) 0.2706

(d) 0.2936

Solution:

(c) Let X be the number of individuals suffering a bad reaction.

Then, X has a poisson distribution with

n = 1000 and p = 0.002

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) Rohan can row with a speed of 16 km/h in still water and the speed of stream is 12 km/h. Then, speed of Rohan downstream will be 26 km/h.

Reason (R) If the speed of a boat in still water is x km/h and speed of stream is y km/h, then speed of the down stream is (x + y) km/h.

Solution:

(d) Assertion Given, speed of Rohan in still water x = 16 km/h.

and speed of stream, y = 12 km/h

∴ Speed in down stream = (x + y) km/h

= (16 + 12)

= 28 km/h

Hence, Assertion is false but Reason is true.

Question 20.

singular.

Reason (R) A square matrix A, is said to be singular, if |A| = 0

Solution:

Hence, A is singular matrix.

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

Section B

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

Question 21.

The feasible region of the LPP

Minimise Z =3x + 2y

Subject to constraints are

2x + y ≥ 6, x – y ≥ 0, x ≥ 0, y ≥ 0 is given below

Determine the optimal solution. Justify your answer.

Solution:

We have, Minimise Z = 3x + 2y

Subject to constraints are

2x + y ≥ 6, x – y ≥ 0, x ≥ 0, y ≥ 0.

The feasible region of the constraints is given below.

Corner point | Value of Z = 3x + 2y |

P(2, 2) | 3 × 2 + 2 × 2 = 6 + 4 = 10 |

Q(3, 0) | 3 × 3 + 2 × 0 = 9 + 0 = 9 |

The smallest value of Z is 9. Since, the feasible region is unbounded, we draw the graph of 3x + 2y < 9. The resulting open half plane has common point with feasible region.

∴ Z = 9 is not the minimum value of Z, Hence, the optimal solution does not exist.

Question 22.

A mixture contains milk and water in the ratio 2 : 3. A certain quantity of milk is added to the mixture such that the ratio of milk and water becomes 2 : 1. Find the percentage of milk added in the original mixture with respect to original quantity of solution.

Or

Three pipes A, B and C can fill a tank in 30 min, 20 min and 10 min, respectively. When the tank is empty, all the three pipes are opened. If A, B and C discharge chemical solutions P,Q and R respectively, then find the part of solution R in the liquid in the tank after 3 min.

Solution:

Question 23.

Find the present value of perpetuity of ₹600 at end of each quarter, if money is worth 8% compounded quarterly.

Solution:

Given, R = ₹600

Let P be the present value of the perpetuity Also, given rate of interest is quarterly, then

Hence, P is ₹ 30000.

Question 24.

A person invested ₹15000 in a mutual fund and the value of investment at the time of redemption was ₹25000. If CAGR for this investment is 8.88%, calculate the time period for which the amount was invested? [given log(1.667) = 0.2219 and log(1.089) = 0.037]Solution:

Given, Begining value (BV) = ₹15000

Ending value (EV) = ₹ 25000

Let number of years = n and CAGR = 8.88%

We know that,

n = \(\frac{0.2219}{0.037}\)

n = 5.99 ≈ 6

Hence, time period is 6 yr.

Question 25.

The marginal revenue function for a commodity is given by MR = 9 + 2x – 6x². Find the demand function.

Or

The marginal cost of producing x pairs of tennis shoes is given by MC = 50 + \(\frac{300}{x+1}\) If the fixed cost is ₹ 2000, find the total cost function.

Solution:

We have, MR = 9 + 2x – 6x²

We know that, \(\frac{dy}{dx}\) = MR dx

⇒ R = ∫MRdx + k

⇒ R = ∫(9 + 2x – 6x²)dx + k

⇒ R = 9x + x² – 2x³ + k

When, x = 0, R = 0, we get

C = 50x + 300 log(x + 1)+ K ,..(i)

When x = 0, C = 2000, we get

K = 2000

∴ C = 50x + 300 log {x + 1) + 2000 [from Eq. (i)]Hence, total cost function is

50x+ 300 log (x + 1) + 2000.

Section C

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

Question 26.

A machine produces washers of thickness 0.50 mm. To determine whether the machine is in proper working order, a sample of 10 washers is chosen for which the mean thickness is 0.53 mm and the standard deviation is 0.03 mm. Test the hypothesis at 5% level of significance that the machine is working in proper order.

[given t_{0.025} = 2.262 at ν = 9]Solution:

It is given that, µ = Population mean = 0.50 mm

\(\overline{\mathrm{X}}\) = Sample mean = 0.53 mm

n = Sample size = 10

and S = Sample standard deviation = 0.03 mm

We define null hypothesis H_{0} and alternative hypothesis H_{1}, as follows

H_{0} : µ = 0.50 mm

H_{1} : µ ≠ 0.50 mm

Thus, a two-tailed test is applied under hypothesis H 0, we have

Since, the calculated value of t = 3 does not lie in the internal -t_{0.025} to t_{0.025} i.e. -2.262 to 2.262 for 10 – 1 = 9 degree of freedom. So, we reject H_{0} at 0.05 level. Hence, we conclude that machine is not working properly.

Question 27.

The following table shows the quarterly sales (₹ in crore) of a real estate company. Compute the trend by quarterly moving averages.

Or

Fit a straight line trend by the method of least squares and estimate the trend for the year 2023.

Solution:

The trend value are given by 4-quarterly centered moving average.

Or

Here, n = 7 (odd)

So, middle year i.e 2017 taken as origin, Construct the table as under

Question 28.

Find the maximum value of \(\frac{log x}{x}\), x > 0

Solution:

Question 29.

A machine costing ₹ 50000 is to be replaced at the end of 10 yr, when it will have a salvage value of ₹ 5000. In order to provide money at that time for a machine costing the same amount, a sinking fund is set up. If equal payments are placed in the fund at the end of each quarter and the fund earns 8% compounded quarterly, then what should each payment be? [given (1.02)^{40} = 2. 208]Solution:

We have, cost of machine = ₹ 50000

Salvage value of machine = ₹ 5000

Hence, the money required for new machine after 10 yr

= ₹ (50000 – 5000) = ₹ 45000

So, we have S = ₹ 45000, i = \(\frac{8}{400}\) = 0.02

Hence, ₹ 745.03 are set aside each year to purchase new machine.

Question 30.

Evaluate ∫\(\frac{1}{x-x^3}\)dx.

Or

Evaluate ∫x² e^{x}dx.

Solution:

Question 31.

Solution:

Section D

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

Question 32.

Solve the system of equations x + 2 y – 3z = 6, 3x + 2y – 2z = 3 and 2x – y + z – 2 by using matrix method.

Solution:

The given system of equations can be written as AX = B

Question 33.

A company produces TV’s. The probability that any one TV is defective is \(\frac{1}{50}\) and they are packed in 10 big boxes. From a single big box, find the probability that more than 8 TV’s work properly.

Or

A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, then in the probability distribution of number of tails, find the maximum probability term.

Solution:

Let X is the random variable which denotes that a TV is defective.

Or

Let X be the random variable which denotes the number of tails when a biased coin is tossed twice.

So, X may have values 0, 1 or 2.

Since, the coin is biased in which head is 3 times as likely to occur as a tail.

Question 34.

The demand function for a commodity is p = 22 – x- x². Find the consumer’s surplus at equilibrium price p_{0} = 10.

Solution:

Given, the demand function is

p = 22 – x – x²

At the equilibrium price p_{0} = 10

We have,

Since, x_{0} ≠ -4 ⇒ x_{0} = 3

∴ Consumer’s surplus (CS) at p_{0} = 10

Hence, consumer’s surplus is 22.5.

Question 35.

A decorative item dealer deals in two items A and B. He has ₹ 15000 to invest and a space to store at the most 80 pieces. Item A costs him ₹ 300 and item B costs him ₹ 150. He can sell items A and B at respective, profits of ₹ 50 and ₹ 28. Assuming, he can sell all he buys, formulate the linear programming problem in order to maximise his profit and solve it graphically.

Or

An aeroplane can carry a maximum of 200 passengers. A profit of ₹ m is made on each executive class ticket and a profit of ₹ n is made on each economy class ticket, where m and n are obtained on multiplying first term and common ratio by 100 of the geometric series 10, 60, 360, … times respectively. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine, how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

Solution:

Let the number of items of the type A and B be x and y, respectively. Then, the required LPP is

Maximise Z = 50x + 28y

Subject to the constraints,

x + y ≤ 80, 300x + 150y ≤ 15000

or 2x + y ≤ 100 and x, y ≥ 0

Let us consider the inequalities as equations, we get

x + y = 80 … (i)

2x+y = 100 …..(ii)

Table for line x + y = 80 is

So, it passes through the points (0, 80) and (80, 0). On putting (0, 0) in the inequality x + y ≤ 80, we get

0 + 0 ≤ 80

⇒ 0 ≤ 80 (which is true)

So, the half plane is towards the origin.

Table for line 2x + y = 100 is

So, it passes through the points (0, 100) and (50, 0). On putting (0, 0) in the inequality 2x + y < 100, we get

2 (0) + 0 ≤ 100

⇒ 0 ≤ 100 (which is true)

So, the half plane is towards the origin.

On solving Eqs. (i) and (ii), we get

x = 20 and y = 60

So, the point of intersection is (20, 60).

The feasible region is OBPC whose corner points are 0(0,0), 6(50,0), 6(20,60) and C(0, 80).

Comer points | Value of Z = 50x + 28y |

0(0,0) | 0 |

C (0,80) | 2240 |

P(20, 60) | 2680 (Maximum) |

B (50, 0) | 2500 |

From the above table, the maximum value of Z is 2680 at P(20, 60), i.e. when 20 items of type A and 60 items of type B are purchased and sold. He get maximum profit.

Or

Given geometric series is 10, 60, 360,…

Here, first term (a) = 10 and common ratio

r = \(\frac{60}{10}\) = 6

According to the given condition,

m = a × 100 and n = r × 100

∴ m = 10 × 100 and n = 6 × 100

⇒ m = 1000 and n = 600

Let x passengers travel by executive class and y passengers travel by economy class.

We construct the following table

Class | Number of tickets | Profit (in ₹) |

Executive | x | 1000x |

Economy | y | 600y |

Total | x + y | 1000x + 600y |

So, our problem is to maximise

Z = 1000x + 600y …(i)

Subject to constraints

x + y ≤ 200 …(ii)

x ≥ 20 …(iii)

y – 4x ≥ 0 or y ≥ 4x …(iv)

and x ≥ 0, y ≥ 0 …(v)

Table for line x + y = 200 is

So, line passes through the points (0, 200) and (200, 0).

On putting (0, 0) in the inequality x + y ≤ 200, we get 0 + 0 ≤ 200

⇒ 0 ≤ 200, which is true.

So, the half plane is towards the origin. Table for line y = 4x is

So, line passes through the points (0, 0) and (20, 80).

On putting (10, 0) in the inequality y ≥ 4x, we get

0 ≥ 4 × 10

⇒ 0 ≥ 40, which is not true.

So, the half plane is towards Y-axis.

Now, draw the graph of the line x = 20.

On putting (0, 0) in the inequality x ≥ 20, we get 0 ≥ 20, 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 Eqs. (ii), (iii) and (iv), we get

A(20, 80), B(40, 160) and C (20, 180).

∴ Feasible region is ABCA.

The corner points of the feasible region are A(20, 80), B (40, 140) and C (20, 180).

The value of Z at the corner points are given below

Comer points | Value of Z = 1000x + 600y |

A (20, 80) | Z = 1000 × 20 + 600 × 80 = 68000 |

B(40, 160) | Z = 1000 × 40+ 600 × 160 = 136000 (Maximum) |

C (20, 180) | Z = 1000 × 20 + 600 × 180 = 128000 |

Thus, the maximum value of Z is 136000 at B(40, 160).

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit and the maximum profit is ₹ 136000.

Section E

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

Question 36.

Three taps A, B and C can fill a tank in 12 h, 15 h and 20 h respectively. If A is open all the time B and C are open for one hour each alternately.

Based on the above information, answer the following questions.

(i) Find the part of the tank filled in 2 h.

(ii) Find the part of the tank filled in 6 h.

(iii) Find the part of the tank filled by A and B in one hour.

Or

Find the time taken by all the three pipes to fill the tank.

Solution:

(i) Tap A always remains open and taps 6 and C are opened alternately for 1 h each.

So, after 2 h, tap A has filled the tank for 2 h, S for 1 h and C for 1 h.

∴ Part of the tank filled after 2 h

(\(\frac{2}{12}+\frac{1}{15}+\frac{1}{20}=\frac{17}{60}\))

(ii) So, after 3 rounds of 2 h each the tank will be partially unfilled and in 4th round it overflow.

(\(\frac{1}{4}<\frac{17}{60}<\frac{1}{3}\))

Part of the tank filled in 6 h (2 × 3)h

Hence, the time taken by all the three pipes to fill the tank is (6 + 1) = 7 h

Question 37.

Varun and Isha decided to play with dice to keep themselves busy at home as their schools are closed due to covid pandemic Varun throw a dice, until a six is obtained. He denote the number of throws required by X.

Based on above information answer the following question.

(i) Find the probability that X = 2.

(ii) Find the probability that X = 4.

(iii) Find the probability that X > 3.

Or

Find the value of P (X ≥ 6).

Solution:

(i) P(X = 2) = Probability of not getting x at first throw x probability of getting second throw.

Question 38.

In year 2018, Mr Verma took a loan of ₹250000 at the interest of 6% per annum compounded monthly is to be amortized by equal payment at the end of each of 5 yr. [given (1.005)^{60} = 1.3489 and (1.005)^{21} = 1.1104]Based on the above information, answer the following questions.

Find the size of each monthly payment.

Or

Find the interest paid in 40th payment.

Solution:

Given, P = ₹ 250000,