Article

CBSE Sample Papers for Class 12 Applied Maths Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Maths Set 4 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 smallest non-negative integer congruent to 2796 (mod 7) is
(a) 1
(b) 2
(c) 5
(d) 3
Solution:

Question 2.
The rise in prices before Diwali is an example of
(a) seasonal trend
(b) cylic trend
(c) long term trend
(d) irregular trend
Solution:
(a) The rise in prices before Diwali is an example of seasonal trend.

Question 3.

is equal to
(a) 1
(b) 10
(c) 5
(d) 4
Solution:

On comparing both the matrices, we get
2x – 3 = x + 7
⇒ 2x – x = 7 + 3
⇒ x = 10

Question 4.
If the cost function of a firm is given by C(x) = 2x² – 4x + 5, then average cost, when x = 2 is
(a) 2.5
(b) 2
(c) 1.5
(d) 1
Solution:

Question 5.
A solution containing 15% spirit should be mixed with a solution containing 25% spirit, such that the resultant mixture contains 21% spirit in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 4
(d) 4 : 5
Solution:
(b) Concentration of spirit in 1st solution = 15%
Concentration of spirit in 2nd solution = 25%
and concentration of spirit in resultant mixture = 21 %
Using alligation rule,

Question 6.
The speed of motorboat in still water is 45 km/h. If the motorboat travels 80 km along the stream in 1 h 20 min, then the time taken by it to cover the same distance against the stream will be
(a) 1 h 20 min
(b) 3 h 40 min
(c) 2 h 40 min
(d) 2 h 55 min
Solution:
(c) Given, speed of motorboat in still water = 45 km/h
Let the speed of stream be a km/h.
∴ Downstream speed = Speed of boat in still water + Speed of stream

Question 7.
Let X is a normal distribution random variable with mean µ= 55 and standard deviation σ = 12. The value of probability P(X < 70) is [given, F (1.25) = 0.8944](a) 0.9332
(b) 0.9216
(c) 0.8944
(d) 0.1056
Solution:
(c) Given, µ = 55 and σ = 12

Question 8.
If λ = 1 in Poisson distribution, then the value of P(5) is
(a) 0.307
(b) 0.00307
(c) 0.0307
(d) 0.000307
Solution:
(b) For a Poisson distribution,

Question 9.
Mr. Ahuja borrowed ₹ 100000 from a bank to purchase a car and decided to repay the loan by equal monthly installments in 10 yr. If bank charges interest at 9% per annum compounded monthly, then the value of EMI is (given (1.0075)120 = 2.4514)
(a) ₹ 1250
(b) ₹ 1266.74
(c) ₹ 1300
(d) None of these
Solution:
(b) Given, P = ₹ 100000,

Question 10.
A vehicle costing ₹ 900000 has a scrap value of ₹ 270000, if annual depreciation charge is ₹ 70000, then its useful life in years is
(a) 8 yr
(b) 9 vr
(c) 10 yr
(d) 12 yr
Solution:
(b) Given original cost of vehicie = ₹ 900000
Scrap value = ₹ 270000
Annual depreciation = ₹ 70000

Question 11.
The interval in which f(x) = x4 – 2x² is increasing is
(a) (-1, 0)
(b) (1, ∞)
(c) (0, 1)
(d)(-1, 0) ∪ (1, ∞)
Solution:
(d) We have, f(x) = x4 – 2x²
⇒ f'(x) = 4x³ – 4x
= 4x(x² – 1)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 4x(x² – 1) > 0
⇒ x(x² – 1) > 0
⇒ x(x – 1) (x + 1) > 0
⇒ -1 < x < 0 or x > 1
⇒ x ∈ (-1, 0) ∪ (1, ∞)
So, f(x)is increasing on (-1, 0) ∪ (1, ∞)

Question 12.
The general pattern of increases or decreases in economics or social phenomena is known by
(a) irregular trend
(b) secular trend
(c) seasonal trend
(d) cylic trend
Solution:
(b) Secuiar trend movements are considered as long term movement.

Question 13.
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 14.

(a) (1, 1)
(b) (1, -1)
(c) (-1, 1)
(d) (-1, -1)
Solution:

On comparing both the matrices, we get
x + y = 0
⇒ x = -y
and x – y = -2
⇒ – y – y = – 2
⇒ y = 1 and x = -1
∴ (x, y) ≡ (-1, 1)

Question 15.
\(\int_0^{40}\frac{dx}{2x+1}\) = log k, then the value of k is
(a) 3
(b) \(\frac{9}{2}\)
(c) 9
(d) 81
Solution:

Question 16.
If the null hypothesis is false, then which of the following is accepted?
(a) Alternate hypothesis
(b) Null hypothesis
(c) Negative hypothesis
(d) Positive hypothesis
Solution:
(a) If the null hypothesis is false, then alternative hypothesis is accepted, it is also called as research hypothesis.

Question 17.
In a 100 m race, A takes 36 sec and B takes 45 sec, then A defeat B with the distance of
(a) 20 m
(b) 25 m
(c) 50 m
(d) 15 m
Solution:
(a) A takes 36 sec to cover 100 m.
S takes 45 sec to cover 100 m.
According to the question,
Beat time = 45 – 36 = 9
Distance cover in 4 sec = 100 m
Distance cover in 1 sec = \(\frac{100}{45}\) m
Distance cover in 9 sec = \(\frac{100}{45}\)× 9 = 20 m

Question 18.
At what rate of interest will the present value of a perpetuity of ₹ 300 payable at the end of each quarter ₹ 24000.
(a) 4%
(b) 5%
(c) 6%
(d) 8%
Solution:
(b) Let the rate of interest be r% per annum, then

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 present value perfect of ₹ 900 payable of the end of each year of money worth 5% annum is ₹ 20000.
Reason (R) The present value of ₹ R payable at the end of each payment period when i per period is \(\frac{R}{i}\).
Solution:

∴ Assertion is false but Reason is true.

Question 20.
Assertion (A) If a is any positive real number, then a + \(\frac{1}{a}\) ≥ 2.
Reason (R) Let a, b be distinct positive real number, then \(\frac{a+b}{2}\) ≥ √ab.
Solution:
(a) Assertion Using the inequality AM ≥ GM, we get

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.
What effective rate is equivalent to a nominal rate of 8% per annum compounded quarterly?
[given (1.02)4 = 1.0824]Or
Find the present value of an annuity of ₹ 1000 payable at the end of each year for 5 yr, if money is worth 6% compounded annually, [given (1.06)-5 = 0.7473]Solution:

Hence, present value of ordinary annuity is ₹ 4211.67.

Question 22.
The marginal cost function of manufacturing x units of a commodity is 6 + 10x – 6x². The total cost of producing one unit of commodity is ₹ 12. Find the total cost function.
Solution:
We have, MC = 6 + 10x – 6x²
∴ \(\frac{dC}{dx}\) = MC
⇒ dC = MC dx
On integrating both sides, we get
∫dC = ∫(6 + 10x – 6x²)dx
C = 6x + 5x² -2x³ + K …(i)
It is given that the cost of producing one unit of the commodity is ₹ 12,
i.e. x = 1 and C = 12
On putting x = 1 and C = 12 in Eq. (i), we get
12 = 6 + 5 – 2 + K
K = 3
On substituting the value of K in Eq. (i), we get
C = 6x + 5x² – 2x³ + 3

Question 23.

Solution:

Question 24.
Mr. Ravi invested ₹ 35000 in a shares of a company and reinvested the earning every year in buying the shares of same company at the end of 5 yr. The value of shares increased to ₹ 56000, calculate the CAGR of his investment.
[use (1.6)1/5 = 1.098]Solution:
Given, beginning value of the investment ₹ 35000.
Final value of the investment = ₹ 56000
Number of year = 5

Question 25.
A man can row upstream at 8 km/h and downstream at 13 km/h. Find the speed of the stream.
Or
Find the quantity of water must be added to 30L of milk at ₹ 60 per litre, so as to have mixture worth ₹ 50 per litre.
Solution:
Given, downstream speed = 13 km/h
and upstream speed = 8 km/h
∴ Speed of stream
= \(\frac{1}{2}\) (Downstream speed – Upstream speed)
= \(\frac{1}{2}\)(13 – 8) = \(\frac{5}{2}\)
= 2.5 km/h
Or
Let the quantity of water added = xL
Pure milk = 30 L
Cost of 30 L of milk = ₹ (30 × 60)
= ₹ 1800

Cost of (30 + x) L of mixtures = (30 + x) × 50
∴ 1800 = (30 + x) 50
⇒ 1800 = 1500 + 50x
⇒ 50x = 300 ⇒ x = 6 L

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

Question 26.
Find the maximum profit that a company can make, if the profit function is given by
P(x) = 41 + 24x – 18x², is
Solution:
We have, P (x) = 41 + 24x – 18x²

Question 27.

Solution:

On comparing both the matrices, we get
y + 1 = 0
⇒ y = -1
3(x – 1) = 0 .
⇒ x = 1
Now, x + y = 1 + (-1) = 0

Question 28.
Determine the maximum value of Z = 11x + 7y, subject to the constraints are
2x + y ≤ 6, x ≤ 2 and x, y ≥ 0.
Solution:
We have,
Maximise Z = 11x + 7y …(i)
Subject to the constraints,
2x + y ≤ 6 …(ii)
x ≤ 2 …(iii)
and x, y ≥ 0 …(iv)
We see that, the feasible region as shaded determined by the system of constraints (ii) to (iv) is OABC and is bounded.

So, now we shall use corner point method to determine the maximum value of Z.

Comer points Value of Z = 11x + 7y
O(0, 0) Z = 11(0) + 7(0) = 0
A(2, 0) Z = 11(2) + 7(0) = 22
B(2, 2) Z = 11(2) + 7(4) = 36
C(0, 6) Z = 11(0) + 7(6) = 42 (Maximum)

Hence, the maximum value of Z is 42 at C(0, 6).

Question 29.
Two batches of the same product are tested for their mean life. Assuming that, the lives of the product follow a normal distribution with an unknown variance, test the hypothesis that the mean life is the same for both the branches, given the following information.

Or
The Educational Testing Service conducted a study to investigate difference between the score of male and female students on the Scholastic Aptitude Test. The study identified a random sample of 562 female and 852 male students, who had achieved the same high score on the mathematics portion of the test. That is, female and male students were viewed as having similarly high abilities in mathematics.
The verbal scores for the two samples are as given
Female students, \(\overline{\mathrm{x_1}}\) = 557 and s1 = 83
Male students, \(\overline{\mathrm{x_2}}\) = 545 and s2 = 78
Do the data support the conclusion that given a population of female students and a population of male students with similarly high mathematics abilities, the female students will have a significantly higher verbal ability? Test at a 5% level of significance. What is your conclusion?
Solution:
Given, n1 =10, n2 = 8, \(\overline{\mathrm{x_1}}\) = 750, \(\overline{\mathrm{x_2}}\) = 820, s1 = ₹12 and s2 = ₹14
Consider, Null hypothesis H0 : Mean life is same for both the batches.
Alternate hypothesis H1 : Two batches have different mean lives.
Test statistics,

Since, calculated value |t| = 10.763 > tabulated value t16(0.05) = 2.120
So, rejected the null hypothesis at 5% level of significance.
Hence, the mean life for both the batches is not the same.
Or
We have, \(\overline{\mathrm{x_1}}\) = 557, \(\overline{\mathrm{x_2}}\) = 545 s1 = 83, s2= 78
n1 = 562 and n2 = 852
Consider, the null hypothesis be that there is no significant difference between male and female verbal ability,
H0 : µ1 – µ2 > 0 and H1 : µ1– µ2 < 0

Here, the computed value of Z is greater than the table value of Z i.e. 1.96.
So, reject the null hypothesis. Hence, there is a significant difference between the male and female verbal ability or female students have higher verbal ability.

Question 30.
Find the probability distribution of a random variable X is given as under

where n is a constant.
Or
If a die is thrown 5 times, then find the probability that an odd number will come up exactly three times.
Solution:
The probability distribution table for given function is

Question 31.
A firm anticipates an expenditure of ₹ 500000 for plant modernisation at the end of 10 yr from now. How much should the company deposit at the end of each year into a sinking fund earning interest 5% per annum, [use (1.05)10 = 1.629]Solution:
Given, A = ₹ 500000, r = 5%
⇒ i = 0.05 and n = 10

Hence, the company should deposit ₹ 39745.63 every year into the sinking fund.

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

Question 32.
If the marginal revenue function for output is given by MR = \(\frac{7}{(x+3)^2}\) + 10, then find the demand function.
Or
The marginal cost function of producing x units of a product is given by
MC = \(\frac{x}{\sqrt(x^2+2500)}\). Find the total cost function, if the fixed cost is ₹ 1000.
Solution:

On putting x² + 2500 = t² ⇒ xdx = tdt

Question 33.

Hence, solve the system of equations
x + 2y – 3z = -4, 2x + 3y + 2z = 2
and 3x – 3y – 4z = 11.
Solution:

On comparing corresponding elements, we get
x = 3, y = -2 and z = 1

Question 34.
Two cards are drawn simultaneously from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
Or
How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%.
Solution:
Let random variable X denotes the number of red card drawn in a draw of 2 cards from a pack of 52 cards the Xi = 0, 1, 2

Probability distribution are

Question 35.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution:
Let x passengers travel by executive class and y passengers travel by economy class.
We construct the following table

So, our problem is to maximise
Z = 1000x + 600y …(i)
Subject to constraints,
x + y ≤ 200 …(ii)
x ≥ 20 …(iii)
y – 4x ≥ 0 or y ≥ 4x … (iv)
and x, y ≥ 0 …(v)
Table for line x + y = 200 is

So, line passes, through the points (0, 200) and (200,0).
On putting (0,0) in the inequality x + y ≤ 200, we get 0 + 0 ≤ 200
⇒ 0 ≤ 200, which is true.
So, the half plane is towards the origin. Table for line y = 4x is

So, line passes through the points (0, 0) and (20, 80).
On putting (10, 0) in the inequality y ≥ 4x, we get 0 ≥ 4 × 10
⇒ 0 ≥ 40, which is not true.
So, the half plane is towards Y-axis.
Now, draw the graph of the line x = 20.
On putting (0, 0) in the inequality x ≥ 20, we get
0 ≥ 20, which is not true.
So, the half plane is away from the origin.

Also, x, y ≥ 0, so the region lies in the I quadrant.
On solving Eqs. (ii), (iii) and (iv), we get
A(20, 80), B(40, 160) and C (20, 180).
∴ Feasible region is ABCA.
The corner points of the feasible region are
A(20, 80), B(40, 160) and C(20, 180).

The value of Z at the corner points are given below

Comer points Value of Z = 1000x + 600y
A(20, 80) Z = 1000 × 20 + 600 × 80 = 68000
B(40, 160) Z = 1000 × 40 + 600 × 160 = 136000 (Maximum)
C (20,180) Z = 1000 × 20 + 600 × 180 = 128000

Thus, the maximum value of Z is 136000 at 6(40, 160).

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

Question 36.
A cistern has three pipes A, B and C. Pipes A and B can fill it in 5 h and 10 h respectively, while pipe C can empty the cistern in 30 h.
Based on the above information, answer the following questions.
(i) If the pipes A and B are opened, then find the time taken to fill the cistem.
(ii) If all the pipes are opened, then find the time taken to fill the cistern.
(iii) If the pipes A and B are opened alternatively and pipe C is opened all the time, then find the time taken to fill the cistern.
Or
If the tap A is opened partially in such a way that its efficiency of filling the cistern becomes half, then find the time taken by all the three taps to fill the tank.
Solution:

Hence, time taken to fill the tank is 6 h.

Question 37.
The cost of AC depreciates by ₹ 722 during the third year and by ₹ 685.90 during the fourth year.

Based on above information, answer the following questions.
(i) Find the rate of depreciation per annum.
(ii) Find the original cost of AC.
(iii) Find the total depreciation in its value at the end of 4 yr.
Or
Find the scrap value of AC of the estimated useful life is 15 yr.
(given, (0.95)15 = 0.4633)
Solution:
Let the original cost of AC = ₹P
and rate of depreciation = r%
The value of machine after n yr = P(1 – i)n,
where i = –\(\frac{r}{100}\).

(i) The cost of AC depreciate by ₹ 722 during third years
∴ P(1 – i)² – P(1 – i)³ = 722 …(i)
The cost of AC depreciate by ₹ 685.90 during fourth years.
∴ P(1 – i)³ – P( 1 – i)4 = 685,90 ….(ii)
Or
P( 1 – i)²i = 722 …(iii)
P(1 – i)³i = 685.90 … (iv)
From Eqs, (iii) and (iv), we get
1 – i = \(\frac{685.90}{722}\) = 0.95
⇒ i = 1 – 0.95 = 0.05
r = i × 100 = (0.05 × 100)% = 5%

∴ Original cost of AC is ₹ 16000.

(iii) Original cost of AC = ₹ 16000
Scrap value of AC at the end of fourth year
= ₹ 13032
Total depreciation value in 4 yr
= ₹ (16000 – 13032) = ₹ 2968
Or
Original cost of AC, P = ₹ 16000
Rate of depreciation, r = 5%
Number of years, n = 15
∴ Scrap value in 15 yr = P(1 – r)15
= 16000(1-0.05)15 [i]= 16000(0.95)15
= 16000(0.4633)
= 7412.80
= ₹ 7413
Hence, scrap value of AC estimated useful life in 15 yr is ₹ 7413.

Question 38.
Consider the following data.

and by using method of least square, answer the following questions.
Find the sum of the trend value for year 2009 and 2006.
Or
The difference of trend value for the year 2010 and 2005.
Solution:
Here, n = 7 odd

The equation of the trend line
yt = 138.86 + 7.64x
Slope of trend line, b = 7.64
For year 2006, x = -1
∴ yt = 138.86 + 7.64 × (-1)
= 138.86 – 7.64 = 13122
For year 2009, x = 2
yt = 138.86 + 7.64 × 2
= 138.86 + 15.28 = 154.14
∴ Sum of the trend value for year 2009 and 2006
= 154.14+ 131.22 =285.36
For year 2010, x = 3
∴ yt = 138.86 + 7.64 × 3
= 138.86 + 22.92 = 161.78
For year 2005, x = -2
∴ yt = 138.86 + 7.64 × (-2)
= 138.86 – 15.28 = 123.58
∴ Difference of the trend value for the year 2010 and 2005 = 161.78 – 123.58 = 38.20


Show More
यौगिक किसे कहते हैं? परिभाषा, प्रकार और विशेषताएं | Yogik Kise Kahate Hain Circuit Breaker Kya Hai Ohm ka Niyam Power Factor Kya hai Basic Electrical in Hindi Interview Questions In Hindi