# CBSE Sample Papers for Class 12 Applied Maths Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 7 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths Set 7 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.

following is defined?

(a) A + B

(b) B + C

(c) C + D

(d) B + D

Solution:

(d) Only B + D is defined because matrices of the same order can only be added.

Question 2.

The rate of interest will the present value of a perpetuity of ₹500 payable at the end of every 6 months be ₹10000 is

(a) 5%

(b) 8%

(c) 10%

(d) 4%

Solution:

(c) Here, P = ₹ 10000 and R = ₹500

Let rate of interest be r% per annum

Question 3.

Population value is called

(a) variable

(b) parameters

(c) data

(d) statistics

Solution:

(b) The values (measurable characteristics) obtained from the study of population such as the population mean (µ), population variance (σ²), population standard deviation (a) and etc. are called parameters.

Question 4.

(19 x 121) mod 6 is

(a) 2

(b) 1

(c) 3

(d) 4

Solution:

(b) To determine (19 × 121) mod 6, lets divide 19 by 6 and 121 by 6.

Using property,

(ad)mod n=[(a mod n)(b mod n)]mod n

Thus, (19 × 121) mod 6 = (1 × 1) mod 6 = 1.

Question 5.

In a container, milk and water are present in the ratio 7 : 5. If 15 L water is added to this mixture, the ratio of milk and water becomes 7:8. The quantity of water in the new mixture is

(a) 40 L

(b) 30 L

(c) 45 L

(d) 35 L

Solution:

(a) Let the quantity of milk and water in initial mixture be 7x L and 5x L.

Then, according to the question,

\(\frac{7x}{5x+15}=\frac{7}{8}\)

⇒ 7x × 8 = 7 (5x + 15)

⇒ 56x = 35x + 105

⇒ 56x – 35x = 105

⇒ x = \(\frac{105}{21}\) = 5

∴ Quantity of water in initial mixture = 5 × 5 = 25 L

and quantity of water in new mixture = 25 + 15 = 40 L

Question 6.

A six faced die is thrown until 1 comes, the probability that 1 comes in even number of trials, is

Solution:

(a) Clearly, in a single throw of a dice, probability of getting 1 is \(\frac{1}{6}\) and probability of not getting 1 is \(\frac{1}{6}\).

Then, getting 1 in even number of trials is getting 1 i.e. in 2nd trial or in 4th or in 6th trial and so on.

Required probability

Question 7.

A boat goes 48 km downstream in 20 h. it takes 4 h more to cover the same distance against the stream. The speed of the boat in still water is

(a) 2.2 km/h

(b) 2 km/h

(c) 4 km/h

(d) 4.2 km/h

Solution:

Question 8.

The equation of tangent to curve y² = 9x at the point (1, 1) is

(a) 2x – 9y = 7

(b) 9x – 2y = 7

(c) 9x + 2y = 7

(d) 2x + 9y = 7

Solution:

(b) The equation of the given curve is y² = 9x

Differentiating w.r.t. x, we get

Question 9.

Sampling which provides for a known non-zero equal chance of selection is

(a) probability sampling

(b) non-probability sampling

(c) snowball sampling

(d) convenience sampling

Solution:

(a) When selection of objects from the population is random, so objects of the population has an equal.

Question 10.

The feasible region for an LPP is shown in the following figure.

Let F = 3x – 4y be the objective function. Maximum value of F is

(a) 0

(b) 8

(c) 12

(d) -18

Solution:

(c) The feasible region as shown in the figure, has objective function F = 3x – 4y.

Corner points | Corresponding value of F = 3x – 4y |

(0, 0) | 3 × 0 – 4 × 0 = 0 |

(12, 6) | 3 × 12 – 4 × 6= 12 (Maximum) |

(0, 4) | 3 × 0 – 4 × 4 = -16 (Minimum) |

Hence, the maximum value of F is 12 at (12, 6).

Question 11.

Moving average method is used for measurement of trend when

(a) trend is non-linear

(b) trend is linear

(c) trend is curvilinear

(d) trend is parabolic

Solution:

(b) Moving average method is used for measurement of trend, when trend is linear.

Question 12.

If X is a Binomial variate with the range {0, 1, 2, 3, 4, 5, 6} and P(X = 2) = 4P(X = 4), then the parameter P of X is

Solution:

(a) Here, n = 6

According to the question,^{6}C_{2}p²g^{4} = ^{6}C_{4}p^{4}q²

⇒ q² = 4p²

⇒ (1 – p)² = 4p²

⇒ 3p² + 2p – 1 = 0

⇒ (q + 1)(3p – 1) = 0

⇒ p = \(\frac{1}{3}\) [since, p cannot be negative]

Question 13.

(a) 0

(b) 1

(c) -3

(d) 3

Solution:

Question 14.

If y = e^{ax}, then \(\frac{d^2y}{dx^2}\) is equal to

(a) a²e^{ax}

(b) e^{ax}

(c) ae^{ax}

(d) a²e^{x}

Solution:

(a) We have, y = e^{ax}

Differentiating on both sides w.r.t. x, we get

\(\frac{dy}{dx}\) = ae^{ax}

Again, differentiating on both sides w.r.t. x, we get

\(\frac{d^2y}{dx^2}\) = a²e^{ax}

Question 15.

The sum of order and degree of differential equation

1 + (\(\frac{dy}{dx}\))^{4} = 7(\(\frac{d^2y}{dx^2}\))³ is

(a) 5

(b) 4

(c) 3

(d) 2

Solution:

So, order = 2 and degree = 3

Sum of order and degree = 2 + 3 = 5

Question 16.

The function f(x)-= x2 is strictly decreasing in the interval

(a) (-∞, 0)

(b) (0, ∞)

(c) (1, 4)

(d) None of these

Solution:

(a) Given, f(x) = x²

∵ f'(x) = 2x < 0 ∀ x ∈ (-∞, 0)

∴ f is strictly decreasing function in (-∞, 0)

Question 17.

(a) 10

(b) 12

(c) 7

(d) 15

Solution:

On comparing the corresponding elements, we get

8 + y = 0 and 2x + 1 = 5

⇒ y = – 8 and x = \(\frac{5 – 1}{2}\) = 2

∴ x – y = 2 – (-8) = 10

Question 18.

If |2x + 3| < 7, x ∈ R, then

(a) x ∈ (- 5, 2](b) x ∈ (- 5, 2)

(c) x ∈ (-∞, -5) ∪ (2, ∞)

(d) x ∈ (-∞, -5) ∪ (2, ∞)

Solution:

(b) Given, |2x + 3| < 7

∴ -7 < (2x + 3) < 7

⇒ -7 – 3 < 2x + 3 – 3 < 7 – 3

⇒ -10 < 2x < 4

⇒ \(\frac{10}{2}<\frac{2x}{2}<\frac{4}{2}\)

⇒ -5 < x < 2

∴ x ∈ (-5, 2)

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) A can run a kilometer in 4 min 50 s and B in 5 min. Then, A must give 10 s start to B in a kilometer race, so that the race may end in a dead heat.

Reason (R) A can give B a start of t min means that A will start ‘t’ min after B starts from the starting point and both A and B reach the finishing point at the same time.

Solution:

(a) Assertion A runs 1000 m in 290 s and B in 300 s. For the race to end in a dead heat, A and S must reach the goal at the same time i.e. in 290 s. So, A must give 10 s start to B. Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

Question 20.

Assertion (A) X is normally distributed with mean 20 and standard deviation 4. Then, standard normal variable Z corresponding to X = 21 is 0.25.

Reason (R) P(Z > z) = 1 – F(z)

Solution:

Reason P(Z > z) = 1 – F(z)

Hence, Assertion and Reason both are true and Reason is not the correct explanation of Assertion.

Section B

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

Question 21.

A Lefievo computer, whose cost is ₹ 600000 will depreciate to a scrap value of ₹ 60000 in 5 yr. Using linear method of depreciation, find the book value of the computer at the end of third year.

Solution:

We know, C = Original cost of computer = ₹ 600000

S = Scrap value of computer = ₹ 60000

n = useful life of computer = 5 yr

∴ Annual depreciation

The book value at the end of third year

= 600000 – 3 × 108000

= ₹ 276000

Question 22.

A furniture dealer deals in only two items-tables and chairs. He has ₹ 50000 to invest and has storage space of at most 60 pieces. A table costs ₹ 2500 and a chair ₹ 500. Then, find the constraints of the above problem (where, x is number of tables and y is number of chairs).

Solution:

The constraints in the given problem are

x ≥ 0, y ≥ 0

2500x + 500y ≤ 50000

or 5x + y ≤100

and x + y ≤ 60

Question 23.

A bag contains 2 white and 4 black balls. A ball is drawn 5 times with replacement. Find the probability that atleast 4 of the balls drawn are white.

Or

There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1. Find the probability that all the lines are busy.

Solution:

Question 24.

A pipe can fill a cistern in 6 h. Due to a leak in its bottom, it is filled in 7 h. If the cistern is full, then find the time will it be emptied by the leak.

Solution:

In 1 h, \(\frac{1}{6}\) of the cistern can be filled.

In 1 h, only \(\frac{1}{7}\) of the cistern can be filled due to leak in its bottom.

∴ In 1 h, \(\frac{1}{6}-\frac{1}{7}=\frac{1}{42}\) of the cistern is empty.

Hence, the whole cistern will be emptied in 42 h.

Question 25.

satisfy the equation AX = B, then the matrix A is equal to

Solution:

Section C

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

Question 26.

Deepa invested ₹ 30000 in a mutual fund in 2015. The value of mutual fund increased to ₹ 48000 in year 2020. Calculate the compound annual growth rate of his investment.

[given, (1.6)^{1/5} = 1.099]Solution:

Given, beginning value of the investment (BV) = ₹ 30000

Number of years = 5

Ending value of the investment (EV) = ₹ 48000

∴ Compound annual growth rate (CAGR)

Hence, CAGR% is 0.099 × 100% = 9.9%

Question 27.

Solution:

Question 28.

If y = (x² + x + 3)², find \(\frac{d^2y}{dx^2}\) at x = -1.

Or

Find the maximum value of (\({(\frac{1}{x})}^x\))

Solution:

Given, y = (x² + x + 3)²

Question 29.

A random sample of size 16 has 53 mean. The sum of the squares of the deviations taken from mean is 150. Compute the value of the test statistics.

Solution:

Question 30.

If AT is normally distributed with mean and standard deviation 4. Find x such that the probability of X between 10 and x is 0.4772. [given, F (2) = 0.9772]Solution:

Given, µ = 10 and σ = 4

Question 31.

A sinking fund is created for the redemption of debentures of ₹ 100000 at the end of 25 yr. How much money should be provided out of profits each year for the sinking fund, if the investment can earn interest 4% per annum.

[given (1.04)^{25} = 2.6658]Solution:

Here, S = ₹100000, n = 25yr and i= 4% =0.04

Section D

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

Question 32.

A machine is bought for ₹ 320000. Its effective life is 8 yr, after which its salvage value would be ₹ 25000. It is decided to create a sinking fund to replace this machine at the end of its effective life by making half yearly payments that will earn an interest of 8% per annum compounded half-yearly. If it is known that the cost of machine increases by 5% per annum. Calculate the amount of , each payment to the sinking fund.

[given(1.04)^{16} = 1.8730 and (1.05)^{8} = 1.4774]Or

Which gives better yields.

(i) 8.4% per annum compounded monthly.

(ii) 8.5% per annum compounded semi-annually.

[given, (1.007)^{12} = 1.0873 and (1.0425)² = 1.0868]Solution:

Let each semi-annually deposit in the sinking fund of ₹ R. Since, the cost of new machine is increases by 5% per annum the cost of present.

Cost of machine at present = ₹ 320000

Cost of machine after increasing 5% per annum after 8yr

= 320000(1 + \(\frac{5}{100}\))^{8}

= 32000 (1.05)^{8}

= 320000 × 1.4774

= ₹ 472768

Salvage value of present machine = ₹ 25000

So, net amount required at the end of 8 yr to purchase the new model is ₹ (472768 – 25000) = ₹447768

We know that

R = \(\frac{i\timesS}{(1+i)^n-1}\)

Here, S = ₹ 447768, n = 8 × 2 = 16yr

Thus, × 20516.28 deposited half yearly out of the profit to purchase the new model of the machine.

Or

(i) The effective rate corresponding to 8.4% compounded monthly is

Hence, the option (i) given better yields.

Question 33.

If the interest is compounded continuously at 6% per annum, how much worth ₹ 1000 will be 10 yr? [given e^{0.6} = 1.822]Solution:

Let P be the principal amount at any time t and the rate of interest be r% per annum compounded continuously.

Hence, in 10 yr, the worth of ₹ 1000 will be come ₹ 1822.

Question 34.

A dietician wishes to mix two types of foods in such a way that the vitamin contents of mixture contains atleast 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C, while food II contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs ₹ 5 per kg to purchase food I and ₹ 7 per kg to purchase food II. Find the minimum cost of such a mixture. Formulate above as an LPP and solve it graphically.

Or

Solve the following LPP graphically. Maximise Z = 3x + 2 y, subject to constraints are x + 2 y ≤ 10, 3x + y ≤ 15 and x, y ≥ 0.

Also, determine the area of the feasible region.

Solution:

The given data can be put in the tabular form as follows

Suppose the diet contains x kg of food I and y kg of food II.

Then, the required LPP is minimise

Z = 5x + 7y

Subject to constraints

2x + y ≥ 8,

x + 2y ≥ 10

and x ≥ 0, y ≥ 0

On considering the inequalities as equations, we get

2x + y = 8 ,..(i)

and x + 2y = 10 …(ii)

Table for line 2x + y = 8 is

So, it passes through the points (0, 8) and (4, 0), [1] On putting (0,0) in the inequality

2x + y ≥ 8, we get

0 ≥ 8 [which is false]So, the half plane is away from the origin.

Table for line x + 2y = 10 is

So, it passes through the points (10, 0) and (0, 5). On putting (0,0) in the inequality x + 2y ≥ 10, we get

0 ≥ 10 [which is false]So, the half plane is away from the origin.

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

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

x = 2 and y = 4

So, these lines intersect at P(2, 4).

The graphical representation of the system of inequations given below

From the graph, the feasible region is BPC which is unbounded.

The values of Z at corner points are as follows

Corner points | Value of Z = 5x + 7y |

C(10, 0) | Z = 5(10) + 7(0) = 50 |

P(2, 4) | Z = 5(2) + 7(4) = 10 + 28 = 38 (Minimum) |

B(0, 8) | Z = 5(0) + 7(8) = 0 + 56 = 56 |

From table, the minimum value of Z is 38 at P (2, 4). As the feasible region is unbounded, therefore 38 may or may not be the minimum value of Z. For this, we draw a dotted graph of the inequality 5x + 7y < 38 and check whether the resulting half plane has point in common with the feasible region or not,

It can be seen that the feasible region has no common point with 5x + 7y < 38.

Hence, the minimum cost is ₹ 38, when x =2 and y = 4.

Or

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

Subject to constraints,

x + 2y ≤ 10 ….(ii)

3x + y ≤ 15 ….(iii)

x, 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 x 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 first 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 (Maximum) |

C(0, 5) | Z = 3 × 0 + 2 × 5 = 10 |

Therefore, the maximum value of Z is 18 at the point 6(4, 3).

∴ Area of feasible region

Question 35.

Write a trend line equation to the following time series data.

Solution:

Section E

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

Question 36.

Three schools DPS, CVC and KVS decided to organize a fair for collecting money for helping the flood victims.

They sold handmade fans, mats and plates from recycled material at a cost of ₹25, ₹100 and ₹50 each respectively. The numbers of articles sold are given as

Based on the information given above, answer the following questions.

(i) How many articles (in total) are sold by three schools?

(ii) What is the total money (in ₹) collected by the school DPS?

(iii) What is the total money (in ₹) collected by the school CVC?

Or

What is the total money (in ₹) collected by the school KVS?

Solution:

(i) Total number of articles are sold by three schools

= (40 + 50 + 20)+ (25 + 40 + 30) + (35 + 50 + 40)

= 330

(ii) Total money collected by the school DPS

Hence, total money collected by the school CVC is ₹6125.

Hence, total total money collected by the school KVS is ₹ 7875.

Question 37.

A pipe is connected to a tank or cistern. It is used to fill or empty the cistern. The amount of work done by a pipe is a part of the tank filled or emptied in unit time.

Three pipes A, B and C are connected to a tank. A and B fill the tank in 5 h and 9 h, respectively, when operated independently. Pipe C empty the full tank in 12 h when opened alone.

Based on the above information, answer the following questions.

(i) If both pipes A and B are opened together, then find the time of the tank can be filled.

(ii) If pipes A and C are opened together, then find the time of the tank can be filled.

(iii) If all three pipes A, B and C are opened together, then find the time of the tank can be filled.

Or

If all the pipes are opened, then find the part of the tank which remains unfilled after 3 h.

Solution:

Question 38.

Suppose a function f(x, y) can be expressed as the product of two functions x and y i.e. F(x, y) = g(x) . g(y), such that

\(\frac{dy}{dx}\) = F(x, y) = g(x) . g(y), g(y) ≠ 0

In which, we separate the variables and integrate it, to get the solution of differential equation.

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

Suppose that \(\frac{dy}{dx}=\frac{x+1}{y+1}\) and y = 2, when x = 1.

Find the positive value of x, when y = 5.

Or

Find the solution of \(\frac{dy}{dx}=\frac{xy+y}{xy+x}\)

Solution: