# CBSE Sample Papers for Class 12 Applied Maths Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 9 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths Set 9 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 37 – (3x + 5) ≥ 9x – 8(x – 3) is

(a) (-∞, 2)

(b) (-∞, -2)

(c) (-∞, 2]

(d) (-∞, -2]

Solution:

(c) We have,

37 – (3x + 5) ≥ 9x – 8 (x – 3)

⇒ (37 – 3x – 5) ≥ 9x – 8x + 24

⇒ 32 — 3x ≥ x + 24

Transferring the term 24 to LHS and the term (-3x) to RHS,

32 – 24 ≥ x + 3x

⇒ 8 ≥ 4x ⇒ 4x < 8

Dividing both sides by 4, we get

∴ Solution set is (-∞, 2]

Question 2.

In a 600 m race, A can give a start of 6 m to B and A start of 14 m to C. In the same race B give the start to C

(a) 8.09 m

(b) 9 m

(c) 9.08 m

(d) 7.08 m

Solution:

(a) In the same time A covers 600 m.

B covers (600 – 6) = 594

and C covers (600 -14) = 586

B runs 594 than C runs 586

If B runs 1 unit of distance than C runs = \(\frac{586}{594}\)

and in 600 m = \(\frac{586}{594}\) × 600 = 591.91

B give C a start of (600 – 591.91) = 8.09 m.

Question 3.

If A and B are square matrices of the same order and BA = 5I, then B^{-1} is equal to

(a) 5A

(b) \(\frac{1}{5}\)A

(c) 5A^{-1}

(d) \(\frac{1}{5}\)A^{-1}

Solution:

Question 4.

If, A is a non singular matrix of order 3 and |A| = -8, then the value of |adj A| is

(a) -8

(b) 64

(c) -64

(d) 24

Solution:

(b) We know that, |adj A| = | A|^{n-1}, where n is the order of the matrix.

Since, we have matrix A of order 3

|adj A| = |A|^{3-1} = |A|² = (-8)² = 64

Question 5.

Solution:

Question 6.

If X follows Binomial distribution with parameters n = 5, p and P(X = 2) = 9P (X = 3), then p is equal to

(a) \(\frac{1}{20}\)

(b) \(\frac{1}{40}\)

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

(d) \(\frac{1}{5}\)

Solution:

Question 7.

The income of a group of 10000 persons was found to be normally distributed with mean ₹ 1750 per month and standard deviation ₹ 50. The percentage of persons having income exceeding ₹ 1668 is

(a) 90%

(b) 92%

(c) 89%

(d) 95%

Solution:

(d) Let X denotes the income of a person

P(X >1668) = 1-P(X <1668)

= 1 – P(Z ≤ \(\frac{1668-1750}{50}\))

= 1 – P(Z < – 1.64)

= 1 – F(-1.64)

= 1 – 0.0505 = 0.9495

Hence, the number of persons having income more than 1668 is 0.9495 × 100

= 94.95% ≈ 95%

Question 8.

The supply function of a producer is given by P = \(\frac{3}{5}\)e^{3x}, where x denotes thousands units. If sales are 4000 units, then producer’s surplus (PS) is

Solution:

Question 9.

If’the demand is 150 during April 2017, 250 in may, 350 in June, 450 in July, then the 4 month simple moving average for August 2017 is

(a) 400

(b) 300

(c) 500

(d) 600

Solution:

(b) 4-month simple moving average for August 2017

Question 10.

Seasonal variations are

(a) short term

(b) long term

(c) sudden

(d) None of these

Solution:

(a) Short term are seasonal variation.

Question 11.

The general pattern of increase or decrease in economics or social phenomena is known by

(a) irregular trend

(b) secular trend

(c) seasonal trend

(d) cyclic trend

Solution:

(b) Secular trend movements are considered as long term movements.

Question 12.

At what rate of interest will the present value of a perpetuity of ₹ 500 payable at the end of each quarter be ₹ 40000

(a) 1.25%

(b) 2.5%

(c) 5%

(d) 6%

Solution:

(c) Let the rate of interest r% per annum.

Question 13.

Assume that the probability of a bomb dropped from an aeroplane will strike a certain target is \(\frac{1}{5}\). If 6 bombs are dropped, the probability that atleast 2 will strike the target is

(a) 0.301

(b) 0.3612

(c) 0.3762

(d) 0.3378

Solution:

Question 14.

The interval in which f(x) = x² + 2x – 5 is increasing is

(a) (-∞, -1)

(b) (-1, ∞)

(c) (-1, 1)

(d) R

Solution:

(b) We have, f(x) = x² + 2x – 5

⇒ f'(x) = 2x + 2

For increasing, f'(x) > 0

⇒ 2x + 2 > 0

⇒ x > -1

⇒ x ∈ (-1, ∞)

Question 15.

In an LPP, if the objective function Z = ax + by has same maximum value on two comer points of the feasible region, then the number of points at which maximum value of Z occurs is

(a) 0

(b) 2

(c) finite

(d) infinite

Solution:

(d) If same maximum value occur at two corner points of the feasible region, then all points of the line joining thse two points will give the maximum value and these points are infinite.

Question 16.

In a statistical hypothesis test, then hypothesis is true but our test reject it, it is

(a) Type II error

(b) Type I error

(c) Both Type I and II error

(d) None of the above

Solution:

(b) In a statistical hypothesis test, when the hypothesis true but our test reject it is called type I error.

Question 17.

A statement made about a population parameter for testing purpose is called

(a) statistic

(b) parameter

(c) hypothesis

(d) level of significance

Solution:

(c) A-Statement made about a population parametes for testing purpose is called hypothesis.

Question 18.

A motorboat travelling with some speed, can cover 25 km upstream and 39 km downstream in 8 h. With the same speed, it can travel 35 km upstream and 52 km downstream in 11 h. The speed of the stream is

(a) 2 km/h

(b) 3 km/h

(c) 4 km/h

(d) 5 km/h

Solution:

(c) Let the downstream speed be a km/h

and upstream speed be b km/h.

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) The marginal cost of a total cost function, (x) = \(\frac{x^3}{3}\) + 3x² – 7x + 16 at 3 x = 2 is ₹ 10.

Reason (R) Marginal cost(MC) = \(\frac{d}{ dx}\) (C(x))

Solution:

(d) Reason is true.

Assertion is false.

Hence, Assertion is false but Reason is true.

Question 20.

Assertion (A) The effective rate which is equivalent to a stated rate of 6% compounded semi-annually is 6.09%

Reason (R) Effective rate of ration (per rupee ) = (1 + \(\frac{1}{2}\))^{m} – 1

where, r = annual rate of interest

m – number of compounding in a year.

Solution:

(a) Reason is true.

Hence, the effective rate = 0.0609 × 100% = 6.09 %

Hence, Assertion and Reason both 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.

A firm anticipates an expenditure of ₹ 500000 for plant modernization at the end of 10 yr from now. How much should the company deposit at the end of each year into a sinking find earning interest 5% per annum using ? [(1.05)^{10} = 1.629]

Solution:

Given, A = ₹ 500000,

r = 5%

i = 0.05

and n = 10, R = ?

Question 22.

Suppose a book of 614 pages contains 43 typographical errors. If these errors are randomly distributed throughout the book, find the probability that 10 pages, selected at random will be free from errors, (use e^{-07} = 0.497)

Or

A coin is biased so that the head is 2 times as likely to occur as tail. If the coin is tossed twice, get the probability distribution of number of tails. Then, find the mean of the distribution.

Solution:

Let p be the probability that a misprint appears on a page.

Question 23.

The cost price of Type I rice is ₹ 60/kg and that of Type II is ₹ 80/kg if both type of rice are mixed in the ratio 2:3 respectively. Then, find the price per kg of mixed rice.

Or

A chemist has prepared a solution in which the volume of water is 30% of the total volume. It is observed that on adding 5 L of water in the solution, the volume of water increases to 40%. Find the quantity of water (in L) in original solution.

Solution:

Given, cost of cheaper rice = ₹ 60 per kg.

Cost of deorer rice = ₹ 80 per kg.

Quantity of cheaper = Quality of deaser 2 : 3

Let the price of mixed rice be ₹x per kg.

Question 24.

An asset costing ₹ 95000 is expected to have a useful life of 8 yr, has final annual depreciation of ₹ 8000. Find the final scrap value of asset.

Solution:

Given, D = ₹ 8000, C = ₹ 95000 and n = 8yr

∴ Final scrap value of the asset is ₹ 31000.

Question 25.

then find matrix B.

Solution:

Section C

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

Question 26.

Consider the following hypothesis

H_{0} : µ = 315 and H_{a} : µ ≠ 315

A sample of 60 provided a sample mean of 324.6. The population standard is 14 and level of significance a =0.05. Check the hypothesis test given above, using interval estimation.

Solution:

Since, the hypothesis value of population mean µ_{0} = 315 does not lies in confidence interval, so H_{0} rejected.

Question 27.

Three pipes A, B and C together can fill a tank in 6 h. After working together for 2 h, C is closed and A and B fill the tank in 8 h. Find the time (in h) in which the tank

Solution:

Let the three pipes A, 6 and C can fill the tank in x, y and z hour, respectively.

Then, parts of tank filled by pipes A, B and C in 1 h are

Hence, pipe C alone can fill the tank in 12 h.

Question 28.

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

Find the maximum value of Z = 3x + 4y

Solution:

As clear from the graph corner points are O, A, E, D with coordinates (0, 0), (52, 0), (44, 16) and (0, 38).

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

O(0, 0) | 0 |

A(52, 0) | 156 |

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

D( 0, 38) | 152 |

Hence, the maximum value of Z is 196 at E (44, 16).

Question 29.

A manufacturer’s marginal cost is \(\frac{600}{\sqrt{5x+36}}\). Find the cost involved to increase production from 150 units to 450 units.

Or

The marginal revenue function of a commodity is given MR = 12 – 3x² +4x. Find the total revenue from the sale of 4 units.

Solution:

Or

We have, MR = 12 – 3x² + 4x

∵ ∫dR = ∫MRdx [∵ MR = \(\frac{dR}dx}\)]

⇒ R = ∫(12 – 3x² + 4x)dx

= 12x – x³ + 2x² + K

We know that when x = 0, then R = 0

∵ K = 0

R = 12x – x³ + 2x² [from Eq. (i)]

Now, the total revenue from the sale of 4 units is given by put x = 4

R = 48 – 64 + 32 = 16

Question 30.

There are three bottles of mixture of syrup and water in ratios 2 : 3, 3 : 4 and 7 : 5. 10L and 21 L mixture are taken out from first and second bottles, respectively. How much quantity from third bottle is to be taken so that final mixture from three bottles may contain syrup and water in the ratio of 1 : 1 ?

Solution:

Hence, the mixture taken out from third bottle is 30 L.

Question 31.

For the following series of observations

Calculate 5- yearly moving averages.

Or

The data below shows the figures of income of a municipal corporation over the years.

Calculate 4-yearly centred moving averages.

Solution:

∴ 5-yearly moving averages are 2.4, 2.8, 3.6 and 3.8

∴ 4-yearly centred moving averages are 5.375, 5.125 and 4.875.

Section D

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

Question 32.

A machine costs a company ₹ 52000 and its effective life is estimated to be 25 yr. A sinking fund is created for replacing the machine by a new model at the end of its life time, when its scrap realizes a sum of ₹ 2500 only. The price of the new model is estimated to be 25% more than the price of present one. Find what amount should be set aside at the end of each year out of the profit for the sinking fund, if it accumulates at 3.5% compound annually? [given (1.035)^{25} = 2.3632]

Solution:

Let ₹ R be the amount set aside each year.

Since, the cost of new machine is 25% more than the cost of present.

Cost of machine = ₹ 52000

Cost of new machine after increasing the 25% of the

cost of machine,

i.e. 52000 + 25% of 52000

= 50000(1 + \(\frac{25}{100}\)) = 52000 × \(\frac{5}{4}\) = ₹ 65000

Scrap value of the present machine = ₹ 2500

So, net amount required at the end of 25 yr to purchase the new model = ₹ (65000 – 2500)

= ₹ 62500

Thus, ₹ 1604.68 are set aside each year out of the pipfit to purchase the new model of the machine.

Question 33.

Solve the system of equations 2x – y + 3z -9, x + y + z = 6 and x-y + z -2.

Solution:

Given, system of equations is

2x – y + 3z = 9,

x + y + z = 6

and x – y + z = 2

Question 34.

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.

Or

The random variable X can take only the values 0, 1, 2. Given that,

P(X = 0) = P(X = 1) = p and that

E(X²) = E[X] then find value of p.

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.

Therefore, the required probability distribution is as follows.

Or

Given that, P(X = 0) = p

P(X = 1) = p

Let P(X = 2) = x

SP(X) = 1

⇒ P + P + x = 1 ⇒ x = 1 – 2P

Also, E(X) = ∑XP(X)

= 0 × P + 1 × P + 2 × x

= P + 2x

= P + 2(1 – 2P) [using Eq. (i)]

= P + 2 – 4P = 2 – 3P

E(X²) = ∑X²P(X)

= 0 + 1 × P + 4 × x

= P + 4x

= P + 4(1 – 2P) [using Eq. (i)]

= 4 – 7P

∵ E(X²) = E(X)

⇒ 4 – 7P = 2 – 3P

⇒ 2 = 4P

⇒ P = \(\frac{1}{2}\)

Question 35.

Solution:

Section E

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

Question 36.

Let subject to the constraints of LPP are x + 2y ≤ 40; 2x + y ≤ 50 and x ≥ 0, y ≥ 0, whose graph is shown below.

Based on the above information, answer the following questions.

(i) What is the point of intersection of line l_{1} and l_{2}.

(ii) Find the corner points of the feasible region shown in the above graph.

(iii) If Z – 2x + y be the objective function, then what is the minimum value of Z.

Or

If Z = 3x – y be the objective function, then what is the maximum value of Z?

Solution:

(i) Only solving equations 2x + y = 50 and

x + 2y = 40, we get 6(20, 10).

(ii) The corner points of the feasible region are

O(0, 0), A(25, 0), 6(20, 10) and C(0, 20)

(iii) Consider the table as,

Comer points | Value of Z = 2x +y |

(0, 0) | 0 (Minimum) |

(25, 0) | 50 |

(20, 10) | 50 |

(0, 20) | 20 |

Or

Consider the table as

Corner points | Value of Z = 3x – y |

(0, 0) | 0 |

(25 0) | 75 (Maximum) |

(20, 10) | 50 |

(0, 20) | -20 |

Question 37.

P(x) = -5x² + 125x + 37500, x > 0 is the total profit function of a company where x is the production of the company.

Based on the above information, answer the following questions.

(i) What will be the production when the profit is maximum.

(ii) What will be the maximum profit.

(iii) When the production is 2 units what will be the profit of the company.

Or

What will be production of the company when the profit is 38250?

Solution:

Given, total profit function

P(x) = – 5x² + 125x+ 37500

(i) We have, P(x) = – 5x² + 125x + 37500

⇒ P'(at) = -10x + 125

⇒ P”(x) = -10 < 0

∴ Profit is maximum when P'(x) = 0

⇒ -10x + 125 = 0 ⇒ x = \(\frac{125}{10}\) = 12.5

⇒ x = = 12.5

The production of company at 12.5 units, gives maximum profit.

(ii) Since, profit is maximum at 12.5.

∴ P(12.5) = -5(12.5)² + 125(12.5) + 37500

= -781.25 + 1562.5 + 37500

= ₹ 38281.25

(iii) We have, P(x) = -5x² + 125x + 37500

P(2) = – 5(2)² + 125 × 2 + 37500

= -20 + 250 + 37500

= ₹37730

Or

We have, P(x) = – 5x² + 125x + 37500

and P(x) = 38250

⇒ 38250 = -5x² + 125x + 37500

⇒ 5x² – 125 + 750 = 0

⇒ x² – 25x + 150 = 0

⇒ (x – 15) (x – 10) = 0

⇒ x = 10, 15

The production of company be 15 units.

Question 38.

A loan is the leading of money by one or more individual, organisations or other entities to other individuals, organisations etc. The recepient incurs a debt and is usually liable to pay interest on that debt until it is repaid as well as to repay the principal amount borrowed.

A loan of ₹ 400000 at the interest rate of 6.75% per annum compounded monthly is to be amortized by equal payment at the end of each month for 10 yr,

Based on the above information, answer the following questions.

Find the size of each monthly payment.

Or

Find the principal outstanding at the beginning of 61st month.

[given (1.005625)^{120} = 1.9603 and (1.005625)^{60} = 1.4001]

Solution:

Given, P = ₹ 400000, n = 12 × 10 = 120

Hence, principal outstanding at beginning of 61st month is ₹ 233336.89.