# CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 11 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths Set 11 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 smallest non-negative integer congruent to 2796 (mod 7) is

(a) 4

(b) 2

(c) 5

(d) 3

Solution:

(d) Let x be a non-negative integer congruent to 2796 (mod 7)

x ≡ 2796 (mod 7)

x = 3 (mod 7)

Question 2.

If -3 ≤ 4 – \(\frac{7x}{2}\) ≤ 18, then

(a) x ∈ (-∞, -4](b) x ∈ [-4, ∞)

(c) x ∈ [-4, 2](d) x ∈ [2, ∞)

Solution:

∴ Solution set is [- 4,2].

Question 3.

The declared rate of return compounded semi-annually which is equivalent to 10.25% effective rate of return, is

(a) 10.13%

(b) 10.05%

(c) 10%

(d) 9.89%

Solution:

So, the declared rate of interest

= (0.10 × 100)% = 10%

Question 4.

A mixture of 150 L of wine and water contains 20% water. The amount of water which should be added so that the water becomes 25% of the new mixture is

(a) 8 L

(b) 10 L

(c) 12 L

(d) 14 L

Solution:

(b) In 150 L of mixture, quantity of water

= 20% of 150

= 30

∴ Quantity of wine = (150 – 30) L

= 120 L

Let the water added in the mixture be x.

According to the question,

30 + x = 25% of (150 + x)

⇒ 30 + x = \(\frac{150+x}{4}\)

⇒ 120 + 4x = 150 + x

⇒ 3x = 30

⇒ x = 10

Hence, 10 L of water is added to the mixture.

Question 5.

what is the value of x?

(a) 2

(b) 3

(c) -1

(d) 5

Solution:

(d) ∵ A = A’

⇒ 2x – 3 = x + 2

⇒ x = 5

Question 6.

The area of the triangle, whose vertices are (3, 8), (-4, 2) and (5, 1), is

(a) 60 sq units

(b) 61 sq units

(c) \(\frac{61}{2}\) sq units

(d) 30 sq units

Solution:

(c) The area of triangle is given by

Question 7.

A coin is tossed 4 times. The probability of getting atleast one head is

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

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

(c) \(\frac{7}{8}\)

(d) \(\frac{15}{16}\)

Solution:

(d) ∴ Required probability

Question 8.

The value of ∫\(\frac{1}{x(5+logx)}\) dx is

(a) \(\frac{1}{5}\) log |5 + logx| + C

(b) 5log|5 + logx| + C

(c) \(\sqrt{5+logx}\) + C

(d) log|5 + logx| + C

Solution:

Question 9.

A random variable ‘X’ has the following distribution:

The value of k is

(a) -1

(b) –\(\frac{1}{10}\)

(c) 1

(d) \(\frac{1}{10}\)

Solution:

(d) ∑P_{i} = 1

∴ k + k + 2k + 3k + k² + 2k² + 7k² + 2k = 1

⇒ 10k² + 9k – 1 = 0

⇒ 10k² + 10k – k – 1 = 0

⇒ (10k – 1)(k + 1)= 0

⇒ k = \(\frac{1}{10}\), k ≠ 1

Question 10.

A machine costing ₹ 50000 has a useful life of 4 yr. The estimated scrap value is ₹ 10000. The rate of depreciation per annum is

(a) 20%

(b) 30%

(c) 25%

(d) 15%

Solution:

(c) Annual depreciation amount = ₹ 10000

Original cost of machine – Salvage value of machine

= ₹ (50000 – 10000)

= ₹40000

∴ Depreciation rate percent

Question 11.

For a random variable X, E(X) = 3 and E(X²) = 11. Then, variance of X is

(a) 8

(b) 5

(c) 2

(d) 1

Solution:

(c) Variance = E(X)² – (E(X))² = 11 – (3)² = 11 – 9 = 2

Question 12.

The function f(x) = x^{4} – 4x is strictly

(a) decreasing in [1, ∞)

(b) increasing in (1, ∞)

(c) increasing in (-∞, 1)

(d) increasing in (-1, 1)

Solution:

(b) f(x) = x^{4} – 4x

⇒ f'(x) = 4x³ – 4

⇒ f'(x) = 0

⇒ 4x³ – 4 = 0

⇒ 4(x³ – 1) = 0 , x = 1

∴ f(x) is strictly increasing in (1, ∞) and decreasing in (-∞, 1).

Question 13.

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 14.

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 option are correct.

Question 15.

The degree of the differential equation

(a) 1

(b) 2

(c) 3

(d) not defined

Solution:

(d) The degree of given differential equation is not defined because when we expand sin (\(\frac{dy}{dx}\)), we get

an infinite series in the increasing powers of \(\frac{dy}{dx}\). Therefore, its degree is not defined.

Question 16.

If the null hypothesis is false, then which of the following is accepted?

(a) Alternate hypothesis

(b) Null hypothesis

(c) Negative hypothesis

(d) Positive hypothesis

Solution:

(a) If the null hypothesis is false, then alternative hypothesis is accepted, it is also called as research hypothesis.

Question 17.

The variables x and y in a linear programming problem are called

(a) decision variables

(b) linear variables

(c) optimal variables

(d) None of these

Solution:

(a) The variables x and y in a linear programming problem are called decision variables.

Question 18.

In a 100 m race, A runs at a speed of \(\frac{5}{6}\) m/s. If A gives a start of 4 m to B and still beats him by 12 s. Then, the speed of B is

(a) \(\frac{5}{4}\) m/s

(b) \(\frac{7}{5}\) m/s

(c) \(\frac{4}{3}\) m/s

(d) \(\frac{6}{5}\) m/s

Solution:

(c) Let the speed of 6 be x m/s.

According to the question,

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) In 30 L mixture of acid, the ratio of acid and water is 2 : 3. The amount of water should be added to the mixture so that the ratio of acid and water becomes 2 : 5 is 12 L.

Reason (R) Mean price is always lesser than cost price of cheaper quantity and higher than the cost price of dearer quantity.

Solution:

(c) Assertion Quantity of acid in the mixture

\(\frac{2}{5}\) × 30 = 12 L

Quantity of water in the mixture = 30 -12 = 18 L Let the amount of water added to the mixture be x. According to the question,

\(\frac{12}{18x+x}=\frac{2}{5}\)

⇒ 36 + 2x = 60

⇒ 2x =24

⇒ x = 12 L

Reason The mean price of the quantity will be always higher than the cost price of cheaper quantity and lower than the cost price of dearer quantity.

Hence, Assertion is true but the Reason is false.

Question 20.

Assertion (A) If x is real, then the minimum value of x² – 8x + 17 is 1.

Reason (R) If f”(x) > 0 at critical point, then the value of the function at critical point will be the minimum value of the function.

Solution:

(a) Assertion Let f(x) = x² – 8x + 17

∴ f'(x) = 2x – 8

So, f'(x) = 0, gives x = 4

Here, x = 4 is the critical number.

Now, f”(x) = 2 > 0, ∀ x

So, x = 4 is the point of local minima.

∴ Minimum value of f(x) at x = 4,

f (4) = 4 × 4 – 8 × 4 + 17 = 1

Hence, we can say that both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

Section B

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

Question 21.

A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then find the time taken to fill the tank.

Or

Find the last digit of 12^{12}.

Solution:

Net part filled in 1 h when both (tap and leakage) work

∴ Required time to fill the tank is 30 h.

Or

To find the last digit of 12^{12}, we find 12^{12} (mod 10)

Since, 12 ≡ 2 (mod 10)

⇒ 12^{4} = 2^{4} (mod 10) = 16 (mod 10)

⇒ 12^{4} = 6(mod 10)

⇒ (12^{4})³ = (6)³ mod 10

⇒ 12^{12} = 216 mod 10

⇒ 12^{12} = 6mod 10

Hence, the last digit of 12^{12} is 6.

Question 22.

Find the maximum value of Z = 4x + 3y, if the feasible region for an LPP is shown in following figure.

Solution:

The feasible region is bounded. Therefore, maximum of Z must occurs at the corner points of the feasible region.

The corner points of the feasible region are given below

Corner points | Value of Z = 4 x + 3y |

O(0, 0) | 4(0) + 3(0) = 0 |

A(25, 0) | 4(25) + 3(0) = 100 |

B(16, 16) | 4(16) + 3(16) = 112 ← Maximum |

C(0, 24) | 4(0) + 3 (24) = 72 |

Hence, the maximum value of Z is 112 at 6 (16, 16).

Question 23.

Consider the following hypothesis tests H_{0} : µ = 22; H_{a} : µ ≠ 22

A sample of 65 provided a sample mean, \(\overline{\mathrm{x}}\) = 20 and sample standard deviation, S = 6.4. Compute the value of the test statistic.

Solution:

Question 24.

Solution:

Question 25.

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

Solution:

Let the rate of interest be r% per annum.

Hence, rate of interest is 5% per annum.

Section C

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

Question 26.

Find the matrix X for which

Solution:

Question 27.

A person buys a house for which be agrees to pay ₹ 25000 at the end of each month for 8 yr. If money is worth 12% converted monthly, what is the cash price of house? [given (1.01)^{-96} = 0.3847],

Solution:

Given, EMI = ₹ 25000, n = 12 × 8 = 96

Hence, the cash price of house is ₹ 1538250.

Question 28.

In a binomial distribution, the sum of its mean and variance is 1.8 find the probability of two success if the events was conducted 5 times.

Or

Two biased dice are thrown together. For the first die P(6) = \(\frac{1}{2}\), the other scores being equally likely while for the second die P(1) = \(\frac{2}{5}\) and the other scores are equally likely. Find the probability distribution of ‘the number of one’s seen’.

Solution:

Here, np + npq = 1.8 and n = 5

5p + 5pq = 1.8

⇒ 5p(1 + qr) = 1.8

⇒ 5(1 – q)(1 + q) = 1.8 [∵ p + q = 1]⇒ 1 – q² = \(\frac{1.8}{5}\)

⇒ q² = 1 – 0.36 = 0.64

⇒ q = 0.8 [∵ 0 ≤ q ≤ 1]∴ p = 1 – 0.8 = 02

n = 5, p = 0.2, q = 0.8

∴ p(r = 2) = ^{5}C_{2}p²q³

= 10(0.2)² (0.8)³

= 0.2048

Or

Question 29.

A monopolistic demand function for one of its product is p(x) = ax + b. He knows that he can sell 1400 units when the price is ₹ 4 per unit and he can sell 1800 units at a price of ₹ 2 per unit. Find the marginal revenue function.

Or

A manufacturer’s marginal revenue function is given by MR = 275 – x – 0.3x². Find the increase in the manufacturer’s total increase in revenue, if the production is increased from 10 to 20 units.

Solution:

We have, p(x) = ax + b

It is given that p = ₹ 4 when x = 1400 and p = ₹ 2 when x = 1800

Substituting these values in Eq. (i), we get

∴ 4 = 1400 a+b

2 = 1800a + b

Solving these two equations, we get

We have to find the total increase in revenue, if the production is increased from 10 to 20 units

i.e. we have to find R(20) – R (10)

Hence, increase in revenue is ₹ 1900

Question 30.

Find the general solution of

log(\(\frac{dy}{dx}\)) = 3x +4y

Solution:

which is the required general solution of given differential equation.

Question 31.

A person invested of ₹ 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? [log(L667) = 0.2219 and log(1.089) = 0.037]Solution:

Given, beginning value = ₹ 15000

ending value = ₹ 25000

CAGR = 8.88%

Let number of years = x

On taking log both sides., we get

Hence, time period = 6 yrs

Section D

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

Question 32.

The table given below shows the daily attendance in thousand at a certain exhibition over a period of two weeks

Calculate seven day moving averages and illustrate these and original information on the same graph using the same scales.

Solution:

According to the question,

On the basis of above data we can draw the following moving average graph

Question 33.

The sum of three numbers is 20. If we multiply the first number by 2 and add the second number to result and subtract the third number, we get 23. By adding second and third numbers to three times the first number, we get 46. Represent the above problem algebraically and use Cramer’s rule to find the number of those equations.

Or

Find AB, use this to solve the system of equations x – y = 3, 2x + 3y + 4z – 17 and y + 2z = 7.

Solution:

Let three numbers be x, y and z.

Given, x + y + z = 20

2x + y – z = 23

3x + y + z = 46

Hence, three numbers are 13, 2, 5.

Or

On comparing corresponding elements, we get

x = 2, y = -1 and z = 4

Question 34.

A person borrows ₹ 68962 on the condition that he will repay money with compound interest at 5% per annum in 4 equal annual installment the first one being payable at the end of the first year. Find the value of each installment.

[given (1.05)^{-4} = 0.8227]Solution:

According to the question, the borrow amount (principal amount = P) is ₹ 68962.

So, the value of each installment is ₹ 19447.82.

Question 35.

Solve the following LPP graphically. Maximise Z =3x + 2y,

Subject to constraints are x + 2y ≤ 10, 3x + y ≤ 15 and x ≥ 0, y ≥ 0.

Also, determine the area of the feasible region.

Or

Find the maximum value of Z for the problem maximise Z =2x + 3y,

Subject to constraints

x + 2y ≤ 10, 2x + y ≤ 14, x,y ≥ 0.

Solution:

Our problem is to maximise Z = 3x + 2y … (i)

Subject to constraints are

x + 2y ≤ 10 …(ii)

3x + y ≤ 15 …(iii)

and x ≥ 0, y ≥ 0 …(iv)

Table for line x + 2y = 10 is

So, the line passes through the points (0, 5) and (4,3).

On putting (0, 0) in the inequality x + 2y ≤ 10, we get

0 + 2 × 0 ≤ 10 ⇒ 0 ≤ 10, which is true.

So, the half plane is towards the origin.

Table for line 3x + y = 15 is

So, the line passes through the points (4, 3) and (5, 0).

On putting (0, 0) in the inequality 3x + y ≤ 15, we get

3 × 0 + 0 ≤ 15 ⇒ 0 ≤ 15, which is true.

So, the half plane is towards the origin.

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

On solving equations x + 2y = 10 and 3x + y = 15, we get

x = 4 and y = 3

So, the intersection point is B (4, 3).

∴ Feasible region is OABCO.

The corner points of the feasible region are 0(0, 0), A(5, 0), 6(4, 3) and C(0, 5).

The values of Z at the corner points are given below

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

O(0, 0) | Z = 3 × 0 + 2 × 0 = 0 |

A(5, 0) | Z = 3 × 5 + 2 × 0 = 15 |

B(4, 3) | Z = 3 × 4 + 2 × 3 = 18 |

C(0, 5) | Z = 3 × 0 + 2 × 5 = 10 |

Therefore, the maximum value of Z is 18 at the point B(4, 3).

∴ Area of feasible region

Or

Given, maximise Z = 2x + 3y

Subject to constraints are

x + 2y ≤ 10

2x + y ≤ 14

and x ≥ 0, y ≥ 0

Shade the region to the right of Y-axis to show x ≥ 0 and above X-axis to show y ≥ 0.

Table for line x + 2y = 10 is

So, the line is passing through the points (0, 5), (6,2) and (10,0).

On putting (0, 0) in the inequality x + 2y ≤ 10, we get 0 + 0 ≤ 10, which is true.

So, the half plane is towards the origin.

Table for line 2x + y = 14 is

So, the line is passing through the points (4, 6), (6,2) and (7, 0).

On putting (0, 0) in the inequality 2x + y ≤ 14, we get

0 + 0 ≤ 14, which is true.

So, the half plane is towards the origin.

The intersection point of lines corresponding to

Eqs. (i) and (ii) is 6(6,2).

On shading the common region, we get the feasible region OABD.

The values of Z at corner points are given below

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

0(0, 0) | Z = 2 × 0 + 3 × 0 = 0 |

A(7, 0) | Z = 2 × 7 + 3 × 0 = 14 |

8(6, 2) | Z = 2 × 6 + 3 × 2 = 18 |

D(0, 5) | Z = 2 × 0 + 3 × 5 = 15 |

Hence, the maximum value of Z is 18 at the point 6(6, 2).

Section E

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

Question 36.

An urn contains 25 balls of which 10 balls bear a mark X and remaining 15 bear a mark Y. A balls is drawn at random from the urn, its mark noted down and it is replace. In this way 6 balls are drawn.

Based on the above information, answer the following questions:

(i) Find the probability that all balls will bear X marks.

(ii) Find the probability that all balls will not bear X mark.

(iii) Find the probability that at most 2 balls will bear mark Y.

Or

Find the probability that at least 2 balls will bear mark Y.

Solution:

Given total balls = 25

n(X) = 10, n(Y) = 15

Let p = probability of getting mark X ball

q = probability of getting mark Y ball

Question 37.

There are 24 h in a day. There are mainly two types of clocks used 12 h clock and 24 h clocks 12 h clock repeat itself twice in a day i.e. 24 h of a day are divided into two periods called am and pm. Each period consists of 12 h numbered 12 (acting as 0), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Then, 24 h per day cycle start at 12 mid-night (usually indicated as 12 am) runs through 12 noon (usually indicated as 12 pm) and continues before mid-night at the end of the day.

Based on above information, answer the following questions

(i) It is currently 6:00 pm in 12 h clock. What is the time after 375 h.

(ii) What time is 24 h clock in equivalent to 5:00 pm in 12 h clocks.

(iii) If the time after 490 h from now will be 7:00 pm. Then, what is current time?

Or

What time in 12h clock is equivalent to 22:00 in 24 h clock?

Solution:

(i) 375 = 24 × 15 + 15

∴ 375 h = (24 × 15 + 15) h

Time = 6 pm + 15 h = 6 am + 3h = 9 am

(ii) In 24 h clock 5:00 in 12 h equivalent = 17:00

(iii) 490 = 24 × 20 + 10

The current time = 7:00 pm – 10 h

= 9:00 am

Or

22:00 = 10:00 pm in 12 h clock.

Question 38.

An architect designs a building for a multi-national company The floor consists of a rectangular region with semi-circular ends having a perimeter of 200 m as shown below

Based on the above information answer the following

Find the maximum value of area A.

Or

The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen find the value of x.

Solution:

From the figure, it is clear that perimeter of floor = 200 m