# CBSE Sample Papers for Class 12 Applied Maths Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 5 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 12 Applied Maths Set 5 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 value of 14 mod 5 is

(a) 4

(b) 2

(c) 3

(d) 1

Solution:

(a) To find 14 mod 5, let us divide 14 by 5

So, 14 mod 5 = 4

Question 2.

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 3c and margin of error be E.

Question 3.

An asset costing ₹ 80000 has a useful life of 8 yr, if annual depreciation is ₹ 9000, then scrap value of asset is

(a) ₹ 9000

(b) ₹ 8000

(c) ₹ 10000

(d) ₹ 7000

Solution:

(b) Here, original cost = ₹ 80000

Annual depreciation = ₹ 9000

Useful life = 8 yr

Question 4.

Two pipes A and B can fill a tank in 18 h and 12 h, respectively. If both the pipes are opened simultaneously, then the time will be taken to fill the tank is

(a) 6h

(b) 7\(\frac{1}{5}\)h

(c) 7h

(d) 6\(\frac{1}{5}\)h

Solution:

Question 5.

and y are

(a) -4, 3

(b) 3, -4

(c) 3, -3

(d) 4,-4

Solution:

By definition of equality of matrices as the given matrices are equal, their corresponding elements are equal.

Now, comparing the corresponding elements, we get

2x – y = 10 ,..(i)

and 3x + y = 5 …(ii)

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

5x = 15 ⇒ x = 3

On substituting x = 3 in Eq. (i), we get

2 × 3 – y = 10 ⇒ y = 6 – 10 = -4

Question 6.

The total revenue (in₹) received from the sale of x units of a product is given by R(x) – 3x² + 6x + 5. The marginal revenue, when x = 5 is

(a) ₹ 36

(b) ₹ 35

(c) ₹ 34

(d) ₹ 33

Solution:

(a) Given, R(x) = 3x² + 6x + 5.

We know that marginal revenue is the rate of change ‘of total revenue with respect to the number of units sold.

∴ Marginal Revenue, MR = \(\frac{dR}{dx}\) = 6x + 6

When x = 5, then MR = 6(5) + 6 = 36

Hence, the required marginal revenue is ₹ 36.

Question 7.

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

(a) 21

(b) 47

(c) 20

(d) 31

Solution:

(a) The given feasible region is

Let A B and C be the corner points of the feasible region which is bounded.

The coordinates of A are (3, 2).

The coordinates of B are (0, 3).

and the coordinates of C are (0, 5).

The given objective function is Z = 11x + 7y.

The values of Z at the corner points are given by

Comer point | Value of Z = 11x + 7y |

(3, 2) | 11 × 3 + 7 × 2 = 47 |

(0, 3) | 11 × 0 + 7 × 3 = 21 (Minimum) |

(0, 5) | 11 × 0 + 7 × 5 = 35 |

From the above table, we see that the minimum value of Z is 21 at (0, 3).

Question 8.

If X has a poisson variable such that P(X = 1) = 2P(X = 2), then P(X = 0) is

(a) e

(b) e^{-1}

(c) 1

(d) e²

Solution:

(b) We have, in poisson distribution

Question 9.

A person has set up a sinking fund in order to have ₹ 100000 after 10 yr for his children’s college education. How much amount should be set aside bi-annually into an account paying 5% per annum compounded half-yearly?

(given (1.025)^{20} = 1.6386)

(a) ₹ 3914.81

(b) ₹ 2914.81

(c) ₹ 3614.81

(d) ₹ 3814.81

Solution:

(a) Let ₹ R be set a side bi-annually for 10 yr in order to have ₹ 100000 after 10 yr.

Then, S = ₹ 100000, n = 10 × 2 = 20 yr

Question 10.

A random variable X has the following probability distribution

Then, the mean of X is

(a) 3

(b) 1

(c) 4

(d) 2

Solution:

(a) We know that sum of probabilities distribution is 1.

∴ k + 2k + 3k + 4k = 1 ⇒ k = \(\frac{1}{10}\)

Now, mean \(\overline{\mathrm{X}}\) = k × 1 + 2k × 2 + 3k × 3 + 4k × 4

= k + 4k + 9k + 16k = 30k

⇒ \(\overline{\mathrm{X}}\) = 30 × \(\frac{1}{10}\) = 3

Question 11.

Ravi can row upstream 8 km/h and downstream 16 km/h. The rate of current is

(a) 2 km/h

(b) 3 km/h

(c) 4 km/h

(d) 5 km/h

Solution:

(c) Given, speed upstream, v = 8 km/h

and speed downstream, u = 16 km/h

According to the formula,

The rate of current, y = \(\frac{1}{2}\)(u – v)

⇒ y = \(\frac{1}{2}\)(16 – 8) = \(\frac{8}{2}\) = 4 km/h

Question 12.

The effective rate of interest which is equivalent to a nominal rate of 8% compounded semi-annually is

(a) 8.24%

(b) 8.33%

(c) 8.16%

(d) 8.06%

Solution:

∴ Effective rate of interest = (0.0816 × 100)% = 8.16%

Question 13.

The order and degree of a differential

(a) 2, 3

(b) 3, 2

(c) 2, 2

(d) 3, 3

Solution:

(b) Given, differential equation is

Question 14.

In testing the statistical hypothesis, which of the following statement is false?

(a) The critical region is the values of the test statistics for which we reject the null hypothesis.

(b) The level of significance is the probability of Type-I error.

(c) In testing H_{0}: µ, H_{a} : µ ≠ µ_{0} the critical region is two sided.

(d) The p-value measure the probability that the null hypothesis is true.

Solution:

(d) A p-value is a probability that provides a measure of the evidence against the null hypothesis provided by the sample.

Question 15.

If the critical region is evenly distributed, then the test is referred as

(a) zero tailed

(b) one-tailed

(c) two-tailed

(d) three-tailed

Solution:

(c) In two-tailed test the critical region is evenly distributed. One region contains the area, where null hypothesis is accepted and another contains the area, where it is rejected.

Question 16.

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) We know that,

σ² = E(X²) – (E(X))²

= 11 – (3)² = 11 – 9 = 2

Question 17.

Solution:

Question 18.

f(x) = 2x³ – 9x² + 12x – 3 is increasing

(a) inside the interval (1, 2)

(b) inside the interval (2, 3)

(c) outside the interval (1, 2)

(d) outside the interval (2, 3)

Solution:

(c) We have, f(x) = 2x³ – 9x² + 12x – 3

=> P(x) = 6x² – 18x + 12

For increasing function, f (x) ≥ 0

∴ 6(x² – 3x + 2) ≥ 0

⇒ 6(x – 2)(x -1) ≥ 0

⇒ x ≤ 1 and x ≥ 2

∴ f(x) is increasing outside the interval (1, 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) If 2x + 1 < |2x + 1|, x ∈ R, then x ∈ (-∞, –\(\frac{1}{2}\)).

Reason (R) |x| ≥ |y| iff x² ≥ y².

Solution:

(b) Assertion Given,

2x + 1 < |2x + 1|

⇒ 2x + 1 < 0 [∵ x < |x|, iff x < 0]⇒ 2x < -1

⇒ x < –\(\frac{1}{2}\)

Hence, the solution set is (-∞, –\(\frac{1}{2}\))

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

Question 20.

Assertion (A) Scalar matrix

an identity matrix when k = 1.

Reason (R) Every identity matrix is not a scalar matrix.

Solution:

(c) Assertion A scalar matrix

is an identity matrix when k = 1.

Reason But every identity matrix is clearly a scalar matrix.

Hence, Assertion is true but Reason is false.

Section B

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

Question 21.

To what sum will ₹ 5000 accumulate in 6 yr, if invested at an effective rate of 6%? (given (1.06)^{6} = 1.41851)

Solution:

We know that an effective rate is the actual rate compounded annually, So, the sum S accumulated is the compound amount of the sum f 5000 invested at 6% compounded annually.

We have, P = ₹5000, i = \(\frac{6}{100}\) = 0.06 and n = 6yr

∴ S = P(1 + i)^{n}

⇒ S = 5000(1 + 0.06)^{6}

= 5000(1.06)^{6}

= 5000 × 1.4185 = 7092.55

Hence, required sum = ₹ 7092.55

Question 22.

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to 1.

Solution:

Let X follows binomial distribution with parameters n and p i.e. X ~ B(n, p).

We have, E(X) = np = 2 ,..(i)

and Var(X) = npq = 1 …(ii)

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

Question 23.

Evaluate ∫\(\frac{x^3+3x+4}{\sqrt{x}}\) dx.

Or

Evaluate ∫\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\) dx.

Solution:

Question 24.

Find the general solution of xdy – ydx = 0.

Or

The supply function for a commodity is p = 8 + x. If 10 units of goods are sold, then find the producers surplus.

Solution:

Given that, xdy – ydx = 0

Question 25.

Find the maximum value of Z for the problem maximise Z = 11x + 7y, subject to constraints x < 3, y < 2, x, y > 0.

Solution:

We have, maximise Z = 11x + 7y

Subject to the contraints are

The shaded region as shown in the figure as OABC is bounded and the coordinates of corner points are (0, 0), (3, 0), (3, 2) and (0, 2), respectively.

Corner points | Corresponding value of Z = 11x + 7y |

(0, 0) | 0 |

(3, 0) | 33 |

(3, 2) | 47 (Maximum) |

(0, 2) | 14 |

Hence, Z is maximum at (3, 2) and its maximum value is 47.

Section C

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

Question 26.

Consider the following hypothesis test

H_{0} : µ = 15

H_{a} : µ ≠ 15

A sample of 50 provided a sample mean of 14.15, the population standard deviation is 3, if α = 0.05, then what is the p-value?

Solution:

∴ p-value = 2 (Area under the standard normal curve to the left of Z)

= 2 × (0.0228) = 0,0456

∴ p-value = 0.0456

Question 27.

A company anticipates a capital expenditure of ₹ 40000 for a new equipment in 7 yr. How much should be deposited quarterly in a sinking fund carrying 15% per annum compounded quarterly to provide for the purpose?

(given (1.0375)^{28} = 2.80)

Solution:

Let R be deposited quarterly to accumulate ₹ 40000 in 7 yr.

Given, S = ₹ 40000, n = 7 × 4 = 28yr

Thus, the required amount is ₹ 833.33

Question 28.

A mixture of a certain quantity of milk with 16 L of water is worth ₹ 0.75 per litre. If pure milk is worth ₹ 2.25 per litre, then find the amount of milk in the mixture.

Solution:

Water is available free of cost, so its cost price = ₹0

Now, according to the rule of alligation,

Question 29.

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 30.

Find the maximum value of

Or

Find the value of k, if the points (k + 1, 1), (2k +1, 3) and (2k+ 2, 2k) are collinear.

Solution:

⇒ 2k(k – 2) + 1(k – 2) = 0

⇒ (k – 2) (2k + 1) = 0

⇒ k – 2 = 0 or 2k + 1 = 0

∴ k = 2 or k = –\(\frac{1}{2}\)

Question 31.

The radius r of a right circular cone is decreasing at the rate of 3 cm/min and the height h is increasing at the rate of 2 cm/min. When, r = 9 cm and h = 6 cm, find the rate of change of its volume.

Or

Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24x – 18x².

Solution:

Thus, the volume is decreasing at the rate 54π cm³/min.

Or

We have, P(x) = 41 + 24x – 18x²

⇒ P'(x) = 24 – 36x

and P”(x) = – 36 < 0

For maximum or minimum, we must have

P'(x) = 0

⇒ 24 – 36x = 0

⇒ 36x = 24

⇒ x = \(\frac{24}{36}\) = \(\frac{2}{3}\)

Also, P”(x) < 0

So, profit is maximum when x = \(\frac{2}{3}\)

Maximum profit 1 value P(x) at x = \(\frac{2}{3}\)

Section D

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

Question 32.

of order 2, then show that A² = 4A – 3I. Hence, find A^{-1}.

Solution:

Question 33.

One kind of cake requires 200 g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat. Then, find the maximum number of cakes which can be made from 5 kg of flour and 1kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Solution:

Let x be the number of cakes of one kind and y be the number of cakes of other kind. Construct the following table.

Our problem is to maximise Z = x + y

Subject to constraints are

200x+ 100y ≤ 5000

⇔ 2x + y ≤ 50

25x + 50y ≤ 1000 ⇔ x + 2y ≤ 40

and x, y ≥ 0

Firstly, draw the graph of the line 2x + y = 50

On putting (0, 0) in the inequality 2x + y ≤ 50, we get

2 × 0 + 0 ≤ 50 ⇒ 0 ≤ 50 (which is true)

So, the half plane is towards the origin.

Secondly, draw the graph of the line x + 2y = 40

On putting (0, 0) in the inequality x + 2y ≤ 40, we get

0 + 2 × 0 ≤ 40

⇒ 0 ≤ 40 (which is true)

So, the half plane is towards the origin.

Since, x, y ≥ 0

So, the feasible region lies in the first quadrant.

On solving equations 2x + y = 50 and x + 2y = 40, we get

B(20, 10)

∴ Feasible region is OABCO.

The corner points of the feasible region are

O(0, 0), A(25, 0), B(20, 10) and C(0, 20).

The values of Z at these points are as follows

Comer points | Value of Z = x + y |

0(0, 0) | 0 |

A(25, 0) | 25 |

B(20, 10) | 30 (Maximum) |

C(0, 20) | 20 |

Thus, the maximum number of cakes that can be made is 30 i.e. 20 of one kind and 10 of the other kind.

Question 34.

A piece of machinery costing ₹ 100000 is expected to have a useful life of 5 yr and, scrap value ₹ 20000. Using the straight line method, find the annual depreciation and construct a schedule for depreciation. Also, find the depreciation rate percent.

Solution:

We have, C = Original value = ₹ 100000

S = Scrap value = ₹ 20000

n = Useful life in years = 5yr

The annual depreciation D is given by

At the begining of the first year, the book value of the machine is ₹ 100000. At the end of the first year, the accumulated depreciation is ₹ 16000. Hence, the depreciation charge for the first year is ₹ 16000.

The book value at the end of the first year or in the begining of the second year is ₹ (100000 – 16000)

= ₹84000

At the end of second year, we have

Accumulated depreciation = ₹32000

∴ Depreciation charge = ₹(32000 – 16000) = ₹16000

⇒ Book value at the end of second year

= ₹(84000 – 16000) = ₹ 68000

At the end of third year, we have

Accumulated depreciation = ₹ 48000

∴ Depreciation charge = ₹ 16000

⇒ Book value at the end of third year

= ₹(68000 – 16000)

= ₹52000

Similarly, at the end of fourth and fifth year.

Book values are ₹ 36000 and ₹ 20000 respectively.

Clearly, ₹ 20000 is the scrap value of the machine. These values can be represented in the following tabular form which is known as the depreciation schedule of the machine.

We find that,

Annual depreciation amount = ₹ 16000

Original cost of machine – Scrap value of machine = ₹(100000 – 20000) = ₹ 80000

∴ Depreciation rate per cent

= (\(\frac{16000}{80000}\) × 100)% = 20%

Question 35.

The random variable X can be take only the values 0, 1, 2, 3. Given that

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

and P(X = 2) = P(X = 3)

such that ∑p_{i}x²_{i} = 2∑p_{i}x_{i}, then find the value of p.

Or

For 6 trials of an experiment, let X be a binomial variate which satisfies the relation 9P(X = 4) = P(X = 2). Find the probability of success.

Solution:

Given, X = 0, 1, 2, 3 and

P(X = 0) = P(X = 1) = p, P(X – 2) = P(X = 3)

The probability distribution of X is given by

Or

Let p = Probability of success, q = Probability of failure and X be a random variable that denote the number of success in 6 trials.

As, X follows binomial distribution with parameters, n = 6, p and q.

∴ p = \(\frac{-1}{2}\) or p = \(\frac{1}{4}\)

Thus, p = \(\frac{1}{4}\) [∵ probability cannot be negative]

Section E

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

Question 36.

The average number (in lakh) of working days lost in strikes during each year of the period 1981-90 was

Using 3-yearly moving averages answer the following question

(i) Determine the trend value for the year 1982.

(ii) Determine the moving averages for the year 1985.

(iii) The sum of the moving averages for the year 1984 and 1986.

Or

Draw the moving average graph.

Solution:

According to the question,

(i) Clearly m_{1} = 1.73

(ii) Clearly m_{4} = 2.83

(iii) Clearly m_{3} = 2.23 + 2.83 = 5.06

Or

Moving average graph is as follows

Question 37.

Consumer surplus and producer surplus.

The above graph showing the demand and supply curves of a tyre manufacturer company are linear.

‘ABC’ tyre manufacturer sold 25 units every month when the price of a tyre was ₹ 20000 per units and ‘ABC’ tyre fnanufacturer sold 125 units every month when, the prize dropped to ₹ 15000 per unit. When, the price was ₹ 25000 per unit, 180 tyre were, available per month for sale and, when the price was only ₹ 15000 per unit, 80 tyres remained.

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

(i) Find the demand function D(x)

(ii) Find the supply function S(v).

(iii) ‘Find the consumer surplus C(s)

Or

Find the producer surplus P(s).

Solution:

(i) Let us consider demand function be

D(x) = ax + b …(i)

When x = 25, then p = 20000

From Eq. (i), we have

20000= 25a + b …(ii)

and when x = 125, then p = 15000

From Eq. (i), we have

15000 = 125a + b …(iii)

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

a = -50 and b = 21250

∴ Demand function D(x) = – 50x + 21250

(ii) Let us consider supply function be

S(x) = cx + b …(i)

When x = 180, then p = 25000

From Eq. (i), we have

25000 = 180c +d …(ii)

and when x = 80, then p = 15000

From Eq. (i), we have

15000 = 80c +d …(iii)

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

c = 100 and d = 7000

∴ Supply function S (x) = 100x + 7000

(iii) For equilibrium point D(x) = S(x)

⇒ 50x_{0} + 21250 = 100x_{0} + 7000

⇒ 150x_{0} = -14250

⇒ x_{0} = 95

On putting value of x_{0} in demand function dr supply function, we get

Question 38.

Suppose A and B are two participants in a race. “A gives B a start of x m” means A starts the race from the starting points whereas B starts the race x m ahead of A i.e. to cover a distance of 100 m. A will cover 100 m while B will cover (100 – x) m.

“A beats B by x m’ means when A reached the finish point. B behind by A by x m i.e. in a 100 m race. A has covered 100 m while B has covered only by (100 – x) m.

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

In a 100 m race. A car give a start of 10 m to B and C starts of 28 m to A. In the same race how much start can B give to C.

Or

In a 10 km race, A, B and C each running at uniform speed get the gold, silver and bronze metals respectively. If A beats B by 1 km and B beats C by 1 km. Then, find how many meters does A beat C?

Solution:

In the same time. A covers 100 m,

B covers (100 -10) m = 90 m

and C covers (100 – 28) = 72 m

If the race is between B and C only

When B covers 90 m, then C covers = 72 m

When B cover 1 m, then C covers = \(\frac{72}{80}\) m

When B covers 100 m, then C covers = \(\frac{72}{80}\) × 100 = 80 m

Hence, B can give a starts of 20 m to C.

Or

A beats B by 1 km means in the same time A travels 10 km and B travels 9 km.

∴ Speed of A : speed of B = 10 : 9= 100 :90

Similarly, B beats C by 1 km

Means, speed of B : speed of C = 10 : 9 = 90 : 81

∴ Speed of A: speed of B: speed of C = 100 : 90 : 81

It means in sometime, A travels 100 m, C travels 81 m = A travel 10000 m then C travels 8100 m

⇒ A beats C by 1900 m.