Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 2 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Applied Maths Set 2 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.
(62 +53)mod 7 is equal to
(a) 3
(b) 5
(c) 7
(d) 9
Solution:
(a) To find (62 + 53)mod7, let us divide 62 + 53 i.e. 115 by 7.
So, (62 + 53)mod7 = 3
Question 2.
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. The distance travelled by the boat going upstream in 2 h is
(a) 32 km
(b) 14 km
(c) 18 km
(d) 20 km
Solution:
(c) Given, time taken to cover distance downstream
∴ Distance travelled by boat going upstream in 2 h
= 2 × 9 [∵ distance = time x speed]
= 18km
Question 3.
In a game of 600 points, A scores points 407 points while B scores 307 points. In this game, the point given by A to B is
(a) 105
(b) 200
(c) 100
(d) 150
Solution:
(p) Total score points = 600
A score’s = 407
B score’s = 307
∴ A give (407 – 307) = 100 points to B.
Question 4.
The slope of normal of the curve xy = 6 at (1, 6) is equal to
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{4}\)
Solution:
(a) We have,
Question 5.
The standard deviation of a Poisson variate X is √3. The probability that X = 2 is (given, e-3 = 0.0498)
(a) 0.3421
(b) 0.2341
(c) 0.2241
(d) 0.3241
Solution:
(c) Since, standard deviation of Poisson distribution is √λ.
Question 6.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Question 7.
A factory produces bulbs. The probability that any one bulb is defective is \(\frac{1}{50}\) and they are packed in 10 boxes. From a single box, then the probability that none of the bulbs is defective is
Solution:
(b) Let X is the random variable which denotes that a bulb is defective.
Also, n = 10,
Question 8.
Additive model for time series O is equal to
(a) r + S + C + I
(b) T – S + C – I
(c) T + S – C – I
(d) -T + S – C + I
Solution:
(a) In this model, we represents a particular observation in a time series as the addition of these four components.
Question 9.
In a time series values of the variables y(say) depend on time t represented as y = F(t), has
(a) four components
(b) two components
(c) three components
(d) five components
Solution:
(a) There can be four types of components in the values of the variables
(i) trend
(ii) cyclical
(iii) seasonal
(iv) irregular
Question 10.
A person invested ₹20000 in a mutual fund in year 2016. The value of mutual fund in increased to ₹32000 in year 2021. The compound annual growth rate of his investment is (given, (16)1/5 = 1098)
(a) 9.5%
(b) 9.8%
(c) 19.8%
(d) None of these
Solution:
(b) Given, beginning value of investment (BV)= ₹20000
Ending value of an investment (EV) = ₹32000
Number of years, n = 5
Hence, CAGR is (0.098 × 100)% = 9.8%
Question 11.
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 12.
If a furniture dealer estimates that from the sale of one table he can make a profit of ₹ 250 and from the sale of one chair a profit of ₹ 75 and if x is the number of chairs and y is the number of tables, then its linear objective function is
(a) Z = 75x + 250y
(b) Z = 75x + 25y
(c) Z = 250x + 75y
(d) Z = 25x + 75y
Solution:
(a) Z = 75x + 250y is the linear objective function for the given problem.
Question 13.
The process of making estimates about the population parameter from a sample is called statistical
(a) decision
(b) inference
(c) hypothesis
(d) independence
Solution:
(b) The process through which inference about the population are drawn which is based on population parameter is called statistical inference.
Question 14.
symmetric matrix, then x + 3y is equal to
(a) 1
(b) 2
(c) -1
(d) 0
Solution:
On comparing the corresponding elements, we get
y = 1 and x + 1 = -2 ⇒ x = -3
∴ x + 3y = – 3 + 3(1) = -3 + 3 = 0
Question 15.
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 16.
If A² – 5A + 7I = O, then A-1 is equal to
(a) \(\frac{1}{7}\)(A – 5I)
(b) \(\frac{1}{7}\)(5I – A)
(Question c)
latex]\frac{-1}{7}[/latex](5I – A)
(d) None of these
Solution:
(b) Given, A² – 5A + 7I = 0
On post-multiplying by A-1 both sides, we get
A²A-1 – 5AA-1 + 7IA-1 = OA-1
⇒ A(AA-1) – 5I + 7A-1 = 0
⇒ AI – 5I + 7A-1 = O
⇒ A – 5I + 7A-1 = O ⇒ 7A-1 = 5I – A
∴ A-1 = \(\frac{1}{7}\)(5I – A)
Question 17.
The value of \(\int_2^3\)3x dx is
(a) 3 log 3
(b) \(\frac{3}{log 3}\)
(c) \(\frac{18}{log 3}\)
(d) 18 log 3
Solution:
Question 18.
The unit digit in 1124 is
(a) 7
(b) 8
(c) 9
(d) 1
Solution:
(d) We find that, 11 =1 (mod 10)
⇒ 1124 =(1)24 (mod 10)
⇒ 1124 = 1(mod 10)
Hence, the digit at units places of 1124 is 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 js false but R is true
Question 19.
Assertion (A) If |x – 5| < 2 x ∈ R, then x ∈ [3, 7].
Reason (R) |x| < a, iff – a < x < a
Solution:
(d) We know, |x| < a, then – a < x < a
Now, |x – 5| < 2
⇒ -2 < x – 5 < 2 ⇒ 3 < x < 7
∴ x ∈ (3, 7)
Assertion is false and Reason is true.
Question 20.
Assertion (A) For a Poisson distributions, 3P(X = 2) = P(X = 4), then P(X = 3) = 36e-6
Reason (R) In Poisson distribution
Solution:
(a) Assertion Given,
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.
Section B
All questions are compulsory. In case of Internal choice, attempty any one question only
Question 21.
Solution:
Question 22.
1000 light bulbs with a mean life of 120 days are installed in a new factory; their length of life is normally distributed with standard deviation 20 days. Find the number of bulbs will expire in less than 90 days are [given, F(1.5) = 0.9332]
Solution:
Given, µ = 120 days and σ = 20 days
P(X < 90) = P(Z < \(\frac{90-120}{20}\))
= P(Z< – 1.5) = F(- 1.5)
= 1 – F(1.5) = 1 – 0.9332
= 0.0668
Therefore, the number of bulbs that will expire in less than 90 days is
0.0668 × 1000 = 66.8 – 67
Question 23.
Find the effective rate of interest . corresponding to 10% nominal rate compounded quarterly.
[given (1.025)4 = 1.1038]
Or
Find the present value of a sequence of payments of ₹ 4500 made at the end of every 6 months continuing, if money is worth 8% converted half-yearly.
Solution:
Given, nominal rate of interest,
r = 10% = 0.10
Since, interest is compounded quarterly.
∴ Number of conversion per year (m) = 4
We know that effective rate of interest
Given, R = ₹ 4500
and rate of interest
(r) = 8% = 0.08
Since, interest is compounded semi-annually.
Question 24.
Two tailors A and B earn ₹ 300 and ₹ 400 per day, respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produced at least 60 shirts and 32 pairs of trousers at a minimum labour cost. Formulate this problem as LPP.
Solution:
Suppose, tailor A works for x days and tailor S works for y days.
The given data can be written in the tabulor form as follows
Required linear programming problem is
Minimum Z = 300x + 400y
Subject to constraints are
6x + 10y ≥ 60,
4x + 4y ≥ 32
⇒ x + y ≥ 8
and x, y ≥ 0
Question 25.
The value of a machine purchased 2 yr ago, depreciates at the annual rate of 8%. If its present value is ₹ 85100, find its value, when it was purchased.
Solution:
∴ Present value = Value of 2 yr ago × (1 – i)²
⇒ 85100 = Value of 2 yr ago (1 – 0.08)²
⇒ 85100 = Value of 2 yr ago × (0.92)²
⇒ Value of 2 yr ago = \(\frac{85100}{(0.92)^2}\)
= \(\frac{85100}{0.8464}\)
= ₹ 100543.4783
Section C
All questions are compulsory. In case of internal choice, attempt any one questions only
Question 26.
If C(x) = ax² +bx + k is the total cost function, then find the slope of average cost curve.
Solution:
We have, C(x) = ax² + bx + k
Question 27.
Or
If A is a square matrix such that A² – A, then write the value of 7A – (I + A)³, where I is an identity matrix.
Solution:
Or
Given, A² = A
Now, 7A – (I + A)³ = 7A – [I³ + A³ + 3IA(I + A)]
[∵ (x + y)³ = x³ + y³ + 3xy (x + y)]
= 7A – [I + A².A + 3A(I + A)]
[∵ I³ = I and IA = A]
= 7A – (I + A . A + 3 . AI +3A²) [∵ A² = A]
= 7A – (I + A + 3A + 3A) [∵ AI = Aand A² = A]
= 7A – (I + 7 A) = – I
Question 28.
A random sample of 12 families in one city showed an average weekly food expenditure of ₹1380 with a standard deviation of ₹100 and a random sample of 15 families in another city showed an average weekly food expenditure of ₹1320 with a standard deviation of ₹120. Test, whether the difference between the two means is significant at a level of significance of 0.01.
Or
A population consisting of 6 units 1,2, 3,4, 5 and 9. Find 95% confidence interval for estimating the population mean.
Solution:
Given, n1 = 12, n2 = 15, \(\overline{\mathrm{x_1}}\) = 1380, \(\overline{\mathrm{x_2}}\) = 1320,
s1 = ₹100 and s2 = ₹120
Consider, the null hypothesis that there is no significant difference in the mean expenditure of the families in the two cities.
We know that test statistics
Here, degree of freedom v = 12 + 15 – 2 = 25
From v = 25, test statistics, f0.01 = 2.485.
Here, the calculated value t is less than the table value. Therefore, the null hypothesis is accepted. Hence, the difference between the two means is not significant.
Question 29.
A man can row at 16 km/h in still water and finds that it takes him thrice as much time to row up than as the row down the same distance in the river. The speed of the current is
Solution:
Since, speed ∝ \(\frac{1}{time}\)
∴ Ratio of downstream speed and upstream speed = 3 : 1
Let downstream speed = 3a km/h and upstream speed = a km/h
∵ Speed in still water
= \(\frac{1}{2}\) (Downstream speed + Upstream speed)
Question 30.
From a lot of 30 bulbs which includes 6 defectives, a sample of 4 bulbs are drawn at random one-by-one with replacement. Find the probability distribution of the number of defective bulbs. Find the mean of the distributions.
Solution:
Let X = Number of defective bulbs out of 4 bulbs drawn with replacement
∴ X = 0, 1, 2, 3, 4
p = Probability of defective bulb = \(\frac{6}{30}=\frac{1}{5}\)
q = Probability of good bulb = \(\frac{4}{5}\)
Probability distribution of the defective bulb when r bulb are defective will be
p(x) = 4Crprqn-r (where r = 0, 1, 2, 3, 4))
Question 31.
A man borrows ₹30000 at 12% per annum compound interest from a bank and promises to pay off the loans in 20 annual installments beginning at the end of first year. What is the annual payment necessary? [given (1.12)20 = 9.638]
Solution:
Let ₹ R be the annual payment and ₹P be the present worth.
Hence, the annual payment should be ₹ 4016.76.
Section D
All questions are compulsory. In case of internal choice, attempt any one questions only
Question 32.
Fit a straight line trend by the method of least squares to the following data on sales (₹ in lakh) for the period 2005 – 2012.
Also,
(i) calculate the trend values from 2005 to 2012.
(ii) what will be predicted sales for 2015, assuming that the same rate of change continues?
Solution:
Here, n = 8 (even)
So, origin is mean of two middle years
Question 33.
Amit has taken a loan of ₹ 1000000 with interest rate 11% per annum for the period of 15 yr. Calculate (reducing) EMI for the above data, [given (1.0091)180 = 5.1068]
Solution:
We have, loan amount (P) = ₹ 1000000
Interest rate = 11% (per annum)
∴ Monthly interest rate
Hence, EMI is ₹11314.94 (approximately).
Question 34.
Find the minimum value of Z for the problem, minimise Z = x + 2y, subject to constraints 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Or
Solve the following linear programming problem graphically.
Maximise Z = 34x + 45y,
Subject to constraints
x + y ≤ 300, 2x + 3y ≤ 70; x, y ≥ 0
Solution:
Our problem is to minimise,
Z = x + 2y ,..(i)
Subject to constraints are,
2x + y ≥ 3 …(ii)
x + 2y ≥ 6 …(iii)
and x ≥ 0, y ≥ 0 …(iv)
Table for line 2x + y = 3 is
x | 0 | 3/2 |
y | 3 | 0 |
So, line passes through the points (0, 3) and (\(\frac{3}{2}\), 0).
On putting (0, 0) in the inequality 2x+ y ≥ 3, we get 2 × 0 + 0 ≥ 3 ⇒ 0 ≥ 3, which is not true. So, the half plane is away from the origin.
Table for line x + 2y = 6 is
x | 0 | 6 |
y | 3 | 0 |
So, line passes through the points (0, 3) and (3 0).
On putting (0, 0) in the inequality x + 2y ≥ 6, we get 0 + 2 × 0 ≥ 6
⇒ 0 ≥ 6, which is not true.
So, the half plane is away from the origin.
Also, x ≥ 0 and y ≥ 0, so the region lies in the I quadrant.
The intersection point of the lines x + 2y = 6 and 2x + y = 3 is B(0, 3).
The corner points of the feasible region are A(6, 0) and B(0, 3).
The values of Z at the corner points are given below
Corner points | Value of Z = x + 2y |
A(6, 0) | Z = 6 + 2 × 0 = 6 |
B(0, 3) | Z = 0 + 2 × 3 = 6 |
As the feasible region is unbounded, therefore 6 may or may not be the minimum value of Z. For this we draw a dotted graph of the inequality x + 2y < 6 and check whether the resulting half plane has points in common with a feasible region or not. It can be seen that the feasible region has no common point with x + 2y < 6.
Therefore, the minimum value of Z is 6.
Or
We have the following LPP
Maximise Z = 34x + 45y,
Subject to the constraints are
x + y ≤ 300, 2x + 3y ≤ 70; x, y ≥ 0
Now, considering the inequations as equations, we get
x + y = 300 …(i)
2x + 3y = 70 …(ii)
Table for line x + y = 300
x | 0 | 300 |
y | 300 | 0 |
So, the line passes through the points (0,300) and (300, 0).
On putting (0, 0) in the inequality x + y ≤ 300, we get
0 + 0 ≤ 300, which is true.
So, the half plane is towards the origin.
Table for line 2x + 3y = 70 is
x | 35 | 0 |
y | 0 | 70/3 |
So, the line passes through the points (35, 0) and (0, 70/3).
On putting (0,0) in the inequality 2x + 3y ≤ 70, we get
0 + 0 ≤ 70, which is true.
So, the half plane is towards the origin.
Also, x, y ≥ 0, so the region lies in the 1st quadrant.
The graphical representation of the system of inequations is as given below
Clearly, the feasible region is 0A80, where the corner points areO(0, 0), A(35, 0)and 8(0, 70/3).
Now, the values of Z at corner points are as follow
Corner points | Value of Z = 34 x + 45y |
O(0, 0) | z = 0 + 0 = 0 |
A(35, 0) | Z = 34 × 35 + 45 × 0 = 1190 |
B(0, 70/3) | Z = 34 × 0 + 45 × \(\frac{70}{3}\) = 1050 |
Hence, the maximum value of Z is 1190 at A(35, 0).
Question 35.
The demand and supply function for a commodity are Pd = 56 – x² and Ps = 8 + \(\frac{x^2}{3}\). Find the consumer’s surplus and producer’s surplus at equilibrium price.
Or
Find the particular solution of the differential equation \(\frac{dy}{dx}\) = 1 + x + y + xy, given that y = 0, when x = 1.
Solution:
Given, the demand function Pd = 56 – x²
and supply function Ps = 8 + \(\frac{x^2}{3}\)
At equilibrium, demand = Supply
Hence, consumer’s surplus is 144 and producer surplus is 48.
which is the required particular solution of differential equation.
Section E
All questions are compulsory. In case of internal choice, attempt any one questions only
Question 36.
A farmer has a field in the shape of quadrilateral, where he grews different crops in the shape of triangle. For this, he needs to measure the area of field, which is given below.
Area of a triangle whose vertices are(x1, y1) , (x2, y2) and (x3, y3) is given by the determinant
Since, area is a positive quantity, so we always take the absolute value of the determinant ∆. Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions.
(i) Find the area of the triangle whose vertices are (-1, 2), (2, 3) and (6, 1).
(ii) Using determinants, find the equation of the line joining the points A(2, 1) and B(3, 5).
(iii) If the points (2, -3), (k, -1) and (0, 4) are collinear, then find the value of k.
Or
If the area of a ∆ABC, with vertices A( 1, 3), B(0, 0) and C(k, 0) is 3 sq units, then find the value of k.
Solution:
(i) Area of triangle, is given by the determinant
Question 37.
A man can row against the current, three fifth of a kilometer in 15 min and returns same distance in 10 min.
Based on the above information answer the following questions.
(i) Find the speed of the current.
(ii) Find the speed of men in still water.
(iii) If a man can row a distance of 3 km against the stream, then find the time he would cover the distance.
Or
If the time taken by a man to cover 4.5 km downstream and x km upstream is same, then find the value of x.
Solution:
(i) Let the speed of man in still water be x and speed of current be y.
Question 38.
Western Music concert organised every year in the stadium that can hold 36000 spectators with ticket price of ₹10 the average attendance has been 24000. Some principal expert estimated that price of a ticket should be determined by the
function P(x) = 15 – \(\frac{x}{3000}\), where x is the number of ticket sold.
Based on above information answer the following question.
Find the value of x when revenue is maximum.
Or
Find the price of ticket when revenue is maximum and how many spectators should be present for revenue to be maximum.
Solution:
Let P be the prize per ticket and x be the number of ticket sold.
Then, revenue function R(x) = Px
Number of spectator will be equal to number of ticket sold.
Required number of spectators = 22500