CBSE Sample Papers for Class 12 Applied Maths Set 10 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Maths Set 10 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

1. This question paper contains five sections A.B, C, D and E. Each section is compulsory.
2. 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.
3. Sections-A comprises of 20 MCQs of 1 marks each.
4. Section-B comprise of 5 VSA Type Questions of 2 marks each.
5. Section-C comprises of 6 SA Type Questions of 3 marks each.
6. Section-D comprises of 4 LA Type Questions of 5 marks each.
7. 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.
8. 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 remainder when 862 × 783 × 671 × 549 × 411 × 395 × 217 is divisible by 8 is
(a) 5
(b) 6
(c) 4
(d) 3
Solution:
(b) We have to find
(862 × 783 × 671 × 549 × 411 × 395 × 217) mod 8
= 862 mod 8 × 783 mod 8 × 671 mod 8 × 549 mod 8 × 411 × mod 8 × 395 mod 8 × 217 mod 8
= 6 (mod8) × 7 (mod 8) × 7 (mod 8) × 5 (mod 8) × 3 (mod 8) × 3 (mod 8) × 1 (mod) 8
= (6 × 7 × 7 × 5 × 3 × 3 × 1) mod 8
= (42 × 35 × 9) and 8
= 42 (mod 8) × 35 (mod 8) × 9 (mod 8)
= 2 (mod 8) × 3 (mod 8) × 1 (mod 8)
= (2 × 3 × 1) mod 8
= 6 mod 8
Hence, the remainder when
(862 × 783 × 671 × 549 × 411 × 395 × 217) is divided by 8 is 6.

Question 2.
If 3 ≤ 3t – 18 ≤ 18, then which one of the following is true?
(a) 15 ≤ 2t + 1 ≤ 20
(b) 8 ≤ t ≤ 12
(c) 8 ≤ t + 1 ≤ 13
(d) 21 ≤ 3t ≤ 24
Solution:
(c) Given, 3 ≤ 3t – 18 ≤ 18
Adding 18 to each term,
3 + 18 ≤ 3f – 18 + 18 ≤ 18 + 18
⇒ 21 ≤ 3f ≤ 36
Dividing by 3 to each term,
\(\frac{21}{3}≤\frac{3t}{3}≤\frac{36}{3}\)
⇒ 7 ≤ t ≤ 12
Adding 1 to each term,
7 + 1 ≤ t + 1 ≤ 12 + 1 ⇒ 8 ≤ t + 1 ≤ 13

Question 3.

a² + b² is equal to
(a) 20
(b) 22
(c) 12
(d) 10
Solution:

Question 4.
A bag contains 100 bulbs, out of which 10 are defective. A sample of 5 bulbs is drawn. The probability that none is defective, is

Solution:
(c) Let probability of defective bulb, p = \(\frac{10}{100}\) = \(\frac{1}{10}\) = 0.1
and probability of non-defective bulb, q = 1 – 0.1 = 0.9
Here, n = 5
∴ P(none is defective) = P(X = 0)
= ⁵C₀(0.1)⁰(0.9)⁵
= 1 × (0.9)⁵ = (\(\frac{10}{100}\))⁵

Question 5.

Solution:

Question 6.
A member of the population is called
(a) data
(b) element
(c) family
(d) group
Solution:
(b) A member of the population is called element.

Question 7.
In a 400 m race, A runs at a speed of 16 m/s. If A gives B a start of 16 m and still beats him by 40 s. Then, what will be the speed of B?
(a) 6 m/s
(b) 8 m/s
(c) 15 m/s
(d) 5.9 m/s
Solution:

Question 8.
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 the points (15, 15) and (0, 20), is
(a) p = q
(b) p = 2q
(c) q = 2p
(d) q = 3p
Solution:
(d) The comer points of the feasible region are (0, 10), (5, 5), (15, 15), (0, 20)
The given objective function is Z = px + qy, where p, q > 0
It is given that the maximum value of Z occurs at the points (15,15) and (0, 20).
∴ We have, p – 15 + g – 15 = p – 0 + q – 20
⇒ (p + g)15 = g – 20 ⇒ 3(p + g) = 4q
∴ 3p = q

Question 9.
Ashwani holds a perpetual bond that generates an annual of ₹ 50000 each year. He believes that the borrower is credit worthy and that an 8% interest rate will be suitable for this bond. Then, present value of this perpetuity is
(a) ₹ 625000
(b) ₹ 400000
(c) ₹ 600000
(d) ₹ 62500
Solution:
(a) Here, R = ₹ 50000 and i = 8%
∴ P = \(\frac{R}{i}=\frac{50000\times100}{8}\) = 625000
∴ Present value = ₹ 625000

Question 10.
A vessel contain liquid A and liquid B in the ratio 5 : 7. If 10 L of liquid A is added to the vessel, the new ratio becomes 6 : 7. The quantity of liquid A in the original mixture is
(a) 45 L
(b) 30 L
(c) 50 L
(d) 55 L
Solution:
(c) Let the quantity of liquid A and 6 in the vessel initially be 5x and 7 x, respectively.
Then, according to the question,
\(\frac{5x+10}{7x}=\frac{6}{7}\) ⇒ 5x + 10 = 6x ⇒ x = 10
∴ Quantity of liquid A in original mixture
= 5 × 10 = 50L

Question 11.
Which of the following cannot be component for a time series?
(a) Seasonality
(b) Cyclical
(c) Noise
(d) None of these
Solution:
(d) A seasonal pattern exists when a series is influenced by seasonal factors, e.g. the month, the day or day of the week.
A cyclic pattern exists when data exhibit rises and falls that are not of fixed period.
In discrete time, white noise is a discrete signal whose samples are regarded as a sequence of serially uncorrelated random variables with zero mean and finite variance.

Question 12.
If A is a non-singular 3 × 3 matrix and B is its adjoint such that |B| = 64, then find |A|.
(a) 64
(b) ±64
(c) ±8
(d) 18
Solution:
(c) Given, A is a square matrix of order 3 and 6 is its adjoint such that |B| = 64 = |adj A|
⇒ |adj A| = 64
∵ |adj A| = |A|^n-1 for any non-singular square matrix A of order n.
For given matrix A of order 3 × 3, we have
|adj A| = |A|^3-1 = |A|²
⇒ |A|² = 64 ⇒ |A| = ±8

Question 13.
The data point of a normal variate with mean 10 and standard deviation 4 and Z-score 2 is
(a) 24
(b) 18
(c) 80
(d) 38
Solution:
(b) We know,

Question 14.
If log_e(1 + \(\frac{d^2y}{dx^2}\)) = x, then the order and degree of the given differential equation are
(a) 1 and 1
(b) 1 and 2
(c) 2 and 2
(d) 2 and 1
Solution:
(d) Given, differential equation is loge (1 + \(\frac{d^2y}{dx^2}\)) = x,
which can be rewritten as 1 + \(\frac{d^2y}{dx^2}\) = e^x.
Clearly, the order of differential equation order is 2 and its degree is 1.

Question 15.
A statement made about a population for testing purpose is called
(a) testing statistics
(b) level of significance
(c) statistics
(d) hypothesis
Solution:
(d) Hypothesis is a statement made about a population, it is tested and corresponding accepted, if true and rejected, if false.

Question 16.
If the function f be given by f(x) = xe^1-x, then choose the correct statement from the following.
(a) f is strictly increasing in (\(\frac{1}{2}\), 2)
(b) f is decreasing in (0, 2)
(c) f is increasing in (0, ∞)
(d) f is strictly decreasing in (1, ∞)
Solution:
(d) Since, f(x) = xe^1-x
f'(x) = -xe^1-x + e^1-x = e^1-x (1 – x)
⇒ f'(x) < 0, ∀ (1, ∞)
Hence, f(x) is strictly decreasing in (1, ∞).

Question 17.
The total cost C(x) of producing x items in a firm is given by
C(x) = 0.005x³ – 0.02 x² + 30x + 6000.
The marginal cost when 4 units are produced is
(a) ₹ 29.08
(b) ₹ 28.08
(c) ₹ 31.08
(d) ₹ 30.08
Solution:
(d) We have,
C(x) = 0.005x³ – 0.02x² + 30x + 6000
∴ Marginal cost,
MC = \(\frac{dC}{dx}\) = 0.005 × 3x² – 0.02 × 2x + 30
⇒ [MC]_x = 4= (0.005 × 3 × 16) – (0.02 × 2 × 4) + 30
= 0.24 – 0.16 + 30 = 30.08
Hence, the required marginal cost is ₹30.08.

Question 18.
A money-lender charges interest at the rate of ₹10 per ₹100 per half-year, payable in advance. The effective rate of interest per annum is
(a) 22%
(b) 23%
(c) 23.30%
(d) 23.45%
Solution:
(d) It is given that the money-lender charges interest at the rate of ₹ 10 per ₹ 100 half-year, payable in advance. This means that ₹ 10 is the interest of ₹ 90 for half year.
So, the interest rate per half-year is i = \(\frac{10}{90}=\frac{1}{9}\) = and the annual interest rate is \(\frac{2}{9}\) compounded half-yearly.
Thus, we have r = \(\frac{2}{9}\) and m = 2
∴ r_e = (1 + \(\frac{r}{3}\))^m – 1

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 experiment of tossing a coin three times is said to have binomial distribution.
Reason (R) A die is thrown 5 times, then the probability that an odd number will come up exactly three times is \(\frac{5}{16}\).
Solution:
(b) Assertion Since, in the experiment of tossing a coin three times a random variable X can take values,
0, 1, 2, or 3 with parameter n = 3, p = \(\frac{1}{2}\) and its probability distribution is given by
P(X = r) = ⁿC_r p^rq^n-1
Where, q = 1 – p and r = 0, 1, 2, 3.
∴ The experiment of tossing a coin three times is said to have binomial distribution.
Reason Here, n = 5

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

Question 20.

Reason (R) If A and B are matrices of order 3 and |A| = 4, |B| = 6, then |2 AB| = 192.
Solution:

Hence, both Assertion and Reason 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.
At 6% converted quarterly, find the present value of a perpetuity of ₹ 45000 payable at the end of each quarter.
Solution:
Let P be the present value of perpetuity.

Hence, the present value is ₹ 3000000.

Question 22.
The time taken by a boat to travel 117 km, downstream is 9 h and the same distance upstream is 13 h. The speed of the stream is \(\frac{1}{4}\) of the speed of the boat. Then, find the 4 distance travelled by the boat going upstream in 2 h.
Or
The ratio of milk to water in three containers of equal capacity is 3 : 2, 7 : 3 and 11 : 4, respectively. The three containers are mixed together. Find the ratio of milk to water after mixing.
Solution:
∵ Distance travelled = 117 km
and time = 13 h

Now, distance travelled by boat going upstream in 2h
= speed upstream × 2
= 9 × 2 = 18km
Or
Let the capacity of each container be x.
Quantity of milk in container after mixing is

Question 23.
Find the maximum value of Z = 4x + 2 y, subject to the constraints 2x + 3y < 18, x + y > 10, x, y > 0.
Solution:
The given LPP is, Maximise Z = 4x + 2y
Subject to the constraints are
2x + 3y ≤ 18
x + y ≥ 10
x, y ≥ 0
The graph of the above LPP is

From the above graph, we see that for the feasible solution, there is no common area in the first quadrant.
Hence, the objective function Z maximised.

Question 24.

Solution:

Question 25.

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

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

Question 26.
Mr Taneja purchase a house of ₹ 4500000 with a down payment of ₹ 500000 and balance in EMI for 25 yr. If bank charges 6% per annum compounded monthly. Calculate the EMI.
[given (1.005)³⁰⁰ = 4.4650],
Solution:
Cost of house = ₹ 4500000
Down payment = ₹ 500000
∴ Balance = ₹ (4500000 – 500000) = ₹ 4000000

Hence, EMI of Mr Taneja is ₹ 25772.

Question 27.
Consider the following hypothesis test
H₀ : p ≥ 0.55 ; H_a : p < 0.55
A sample of 250 provided a sample proportion of 0.48. Compute the p-value.
Solution:

∴ p-value of- 221 = Area under the standard normal curve to left of Z
= 0.136
∴ p-value = 0.136

Question 28.
If x + 4y = 14 is normal to the curve y² = αx³ – β at (2, 3), then find the value of α + β.
Solution:
Since, slope of normal line

Also, (2, 3) lies on the curve.
Therefore, 9 = 8α – β or β = 16 – 9 = 7
Now, α + β = 2 + 7 = 9

Question 29.
Amit purchased an old motorcycle for ₹ 16000. If the cost of the scooter after 2 yr depreciates to ₹ 14440. Find the rate of depreciation.
Solution:
Present value of motorcycle (P) = ₹ 16000
Scrap value of motorcycle (S) = ₹ 14440
n = Number of years = 2
Let rate of depreciation is r%.

∴ Rate of depreciation = (0.05 × 100)% = 5%

Question 30.
A population grows at the rate of 10% per year. How long does it take for the population to double? Use differential equation for it.
Or
Find the particular solution of the differential equation \(\frac{dy}{dx}\) = 1 + x + y + xy, given that y = 0, when x = 1.
Solution:
Let P0 be the initial population and P be the population after f yr.


which is the required particular solution of differential equation.

Question 31.
How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?
Or
If the mean and variance of a Binomial variate X are 8 and 4 respectively, then find the value of P( X < 3).
Solution:
Let man tosses the coin n times.
Now, given that,
P(having at least one head) > 80%

Inequality (i) is satisfied forn ≥ 3.
Hence, the coin must be tossed 3 or more times.
Or
Given, mean of Binomial variable, np = 8
and variance of Binomial variable, npq = 4

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

Question 32.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 h and that of second machine is 9 h per day.
Each unit of product A requires 3 h on both machines and each unit of product B requires 2 h on first machine and 1 h on second machine. Each unit of product A is sold at a profit of ₹ 7 and B at a profit of ₹ 4. Find the production level per day for maximum profit graphically.
Solution:
Let the manufacturer produces the products A and B x and y units, respectively.
We construct the following table

Here, total profit Z = 7x + 4y
i.e. maximise Z = 7x + 4y ,..(i)
Subject to the constraints are
3x + 2y ≤ 12,
3x + y ≤ 9 and x ≥ 0, y ≥ 0
Now, consider the given inequations as equations
3x + 2y = 12, …(ii)
3x + y = 9 …(iii)

Table for line 3x + 2y = 12 or y = \(\frac{12-3x}{2}\) is

It passes through the points (0, 6) and (4, 0).
On putting (0,0) in the inequality 3x +2y ≤ 12, we get
0 + 0 ≤ 12 ⇒ 0 ≤ 12 [true]
So, the half plane is towards the origin.
Table for line 3x + y = 9 or y = 9 – 3x is

It passes through the points (0, 9) and (3, 0).
On putting (0, 0) in the inequality 3x + y ≤ 9, we get
0 + 0 ≤ 9 ⇒ 0 ≤ 9 [true]
So, the half plane is towards the origin.
Also, x ≥ 0 and y ≥ 0, so the region lies in 1st quadrant.
Now, the intersection point of lines (ii) and (iii) is
(3x + 2y) – (3x + y) = 12 – 9
⇒ y = 3
and 3x = 12 – 2 × 3
⇒ 3x = 12 – 6
⇒ x = 2
Thus, the point of intersection is B (2, 3).
The graph of inequations is shown below

Here, we see that OABC is a required feasible region, whose corner points are O(0, 0), A(3, 0), B(2, 3) and C(0, 6).
The values of Z at these corner points are as follows

Hence, for maximum profit, manufacturer must produce 2 units of product A and 3 units of product B.

Question 33.
Fit a straight line trend for the following data and find the trend values. Estimate the sales for 2018.

Solution:
Table for fitting straight line trend

The estimated sales for the year 2018 is ₹ 149.893 lakh.

Question 34.
Two containers of equal capacity are full of mixture of oil and water. In the first, the ratio of oil to water is 4 : 7 and in the second, it is 7 : 11. Now, both the mixtures are mixed in a bigger container. Find the resulting ratio of oil to water.
Or
Two pipes A and B can fill a tank in 24 min and 32 min, respectively. If both the pipes are opened together, then after how much time pipe B should be closed so that the tank is full in 9 min?
Solution:
Let both have x L quantity of mixture.
Given, ratio of oil and water in 1st container = 4 : 7
∴ Quantity of oil in 1st container = \(\frac{4}{11}\) x
and quantity of water in 1st container = \(\frac{7}{11}\)x
Similarly, ratio of oil and water in llnd container = 7 : 11

Hence, B must be closed after 20 min.

Question 35.
Rohit has taken a loan of ₹ 600000 with interest rate 10% for the period of 12 yr.
Find (reducing) EMI for the above data, [given (1.0083)¹² = 1.1043]
Or
What nominal rate compounded quarterly will be equivalent to 12% compounded continuously? [given e^0.12 = 1.1274]
Solution:
Given, P = ₹ 600000, n = 12 yr
Interest rate = 10% (per annum)

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

Question 36.
Two schools P and Q decided to award their selected students for the value of discipline and honesty in the form of prizes at the rate of ₹ x and ₹ y, respectively school P decided to award, respectively 3, 2 students a total prize of ₹ 2300 and school Q decided to award respectively 5, 3 students a total prize money of ₹ 3700.

Based on the above information, answer the following questions.
(i) Find the linear equation representing above information.

(iii) If A is the matrix representing the coefficient of x and y, then find A(adjA).
Or
Find \(\frac{1}{2}\) (A + A^-1).
Solution:
(i) The linear equation representing the given situation is
3x + 2y = 2300
and 5x + 3y = 3700

Question 37.
A particle is moving along the curve represented by the polynomial f(x) = (x – 1) (x – 2)² as shown in figure given below.

Based on the above information, answer the following question.
(i) Find the critical point of polynomial f(x).
(ii) What is the point of local maxima for f(x).
(iii) Find the interval, where f(x) is strictly increasing.
Or
Find the interval, where fix) is strictly decreasing.
Solution:
(i) f(x) = (x – 1) (x – 2)²
f'(x) = (x – 2)² + 2(x – 1) (x – 2) = (x – 2)(3x – 4)
f'(x) = 0
⇒ (x – 2) (3x – 4) = 0

Question 38.
Let X denotes the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges it is given that

Where, k is a positive constant.
Based on the above information, answer the following questions.
Find the mean of the distribution
Or
Find the variance of the distribution.
Solution: