CBSE Sample Papers for Class 12 Applied Maths Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Maths Set 7 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 present value of a perpetuity of ₹ 5000 payable at the end of each year, if money is worth 5% compounded annually is
(a) ₹20000
(b) ₹100000
(c) ₹10000
(d) ₹25000
Solution:
(b) We know that

Question 2.

(a) 1
(b) 0
(c) x
(d) x²
Solution:

Question 3.
Shantanu can row upstream at 10 km/h and downstream at 18 km/h. Shantanu’s rate in still water will be
(a) 7 km/h
(b) 14 km/h
(c) 8 km/h
(d) 4km/h
Solution:
(b) Given, speed upstream, v = 10 km/h
and speed downstream, u = 18 km/h
According to the formula,

Question 4.
In a 100 m race, Ajay runs at the speed of 4 km/h. Ajay gives Brijesh a start of 4 m and still beats him by 15 s. The speed of Brijesh is
(a) 3.89 km/h
(b) 3.54 km/h
(c) 3 km/h
(d) 3.29 km/h
Solution:
(d) Time taken by Ajay to cover 4000 m = 1 h
= 3600 s
∴ Time taken by Ajay to cover 100 m
= (\(\frac{60\times60}{4000}\) × 100)s = 90s
Now, Brijesh covers (100 – 4) m
i.e. 96 m = (90 + 15)s
= 105 s

Question 5.

(a) x = 1, y = 2
(b) x = -1, y = -2
(c) x = 1, y = – 2
(d) x = 1, y = 4
Solution:
(d) By equality of matrices, we have
1 + x = 2
⇒ x = 2 – 1
⇒ x = 1

Again, on comparing both the matrices, we get
y – 1 = 3
⇒ y = 3 + 1
⇒ y = 4

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

Question 7.
The feasible solution for a LPP is shown in following figure. Let Z = 3x-4y be the objective function. Minimum of Z occurs at

(a) (0, 0)
(b) (0, 8)
(c) (5, 0)
(d) (4, 10)
Solution:
(b)

Hence, the minimum of Z occurs at (0, 8) and its minimum value is (-32).

Question 8.
Seasonal variations are
(a) long term variation
(b) sudden variation
(c) short term variation
(d) None of these
Solution:
(c) Seasonal variations represents the variability in the data due to seasonal influence.

Question 9.
If \(\frac{x+1}{x+2}\) ≥ 1, then
(a) x ∈ [-∞, 2]
(b) x ∈ (-∞, -2)
(c) x ∈ (-∞, 2]
(d) x ∈ (-∞, 2)
Solution:

Question 10.

Solution:

Question 11.
The order and degree of the differential \(\frac{d^2y}{dx^2}+(\frac{dy}{dx})^{1/4}\) + x^1/5 = 0 respectively, are
(a) 2 and 4
(b) 2 and 2
(c) 2 and 3
(d) 3 and 3
Solution:

Hence, order is 2 and degree is 4.

Question 12.
In a Binomial distribution, the mean is 15 and variance is 10. Then, parameter n is
(a) 28
(b) 16
(c) 45
(d) 25
Solution:
(c) Given, mean (np) =15
and variance np(1 – p) = 10
∴ 1 – p = \(\frac{10}{15}=\frac{2}{3}\)
⇒ p = \(\frac{1}{3}\)
∴ n = 15 × 3 = 45

Question 13.
The assumed hypothesis which is tested for rejection considering it to be true is called
(a) true hypothesis
(b) alternative hypothesis
(c) null hypothesis
(d) simple hypothesis
Solution:
(c) Null hypothesis asserts that there is no true difference in sample statistics and population parameter under consideration.

Question 14.
The equation of the tangent to the curve y = x² – 4x – 5 at the point x = -2 is
(a) 8x + y + 9 = 0
(b) 8x – y + 9 = 0
(c) 8x + y – 9 = 0
(d) 8x – y – 9 = 0
Solution:
(a) Given, equation of curve is
y = x² – 4x – 5
When x = -2, then y = (-2)² – 4(-2) – 5
= 4 + 8 – 5 = 7
So, the required point is (-2, 7).
On differentiating Eq. (i), w.r.t. x, we get
\(\frac{dy}{dx}\) = 2x – 4
∴ Slope of tangent line at point (-2, 7) is

Question 15.
If f : R → R be defined by f(x) = 2x + cos x, then f is
(a) both increasing and decreasing
(b) neither increasing nor decreasing
(c) a decreasing function
(d) an increasing function
Solution:
(d) We have, f(x) = 2x + cosx
∴ f'(x) = 2 + (-sinx) = 2 -sinx
Since, f'(x) > 0, ∀ x [∵ sinx ∈ (-1, 1]]
Hence, f(x) is an increasing function.

Question 16.
If two coins are tossed 5 times, then the probability of getting 5 heads and 5 tails is

Solution:
(a) Let p be the probability of getting head = \(\frac{1}{2}\)
q be the probability of getting tail = \(\frac{1}{2}\)

Let getting a head is success and getting a tail is failure.
Let n = Number of trials = 10,
x = Number of successes

Question 17.
A person buys a house for which he agrees to pay ₹ 5000 at the end of each month for 8 yr. If money is worth 12% converted monthly. Then, the cash price of house is (given (1.01)^-96 = 0.3847)
(a) ₹ 307650
(b) ₹ 407650
(c) ₹ 507650
(d) None of these
Solution:
(a) Here, EMI = ₹ 5000, n = 12 × 8 = 96

Question 18.
The relation between ‘Marginal Cost’ and ‘Average Cost’ of producing ‘x unit of a product is

Solution:

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) A shopkeeper mix two types of oranges worth ₹30 per kg and ₹45 per kg respectively so as to get a mixture at ₹40 per kg in the ratio 2 : 1.
Reason (R) Let the price of cheaper ingredient is denoted by c and price of dearer ingredient is denoted by d and we mix these two types of ingredients in some ratio to produce a mixture at desired price, there

Solution:
(d) Assertion Ratio of quantity of 2 types of oranges

Hence, Assertion is false but Reason is true.

Question 20.
Assertion (A) In a binomial variate mean (µ) = 5 and variance (σ²) = 4. Then, number of trial is 25.
Reason (R) The variance of the binomial distribution is σ² = √npq
Solution:
(c) Assertion Given, µ = np = 5 and σ² = npq = 4
5q = 4 ⇒ q = \(\frac{4}{5}\)
⇒ p = \(\frac{1}{5}\) and n = 5 × 5 = 25
Reason The variance σ² = npq
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.
Evaluate

Or
Evaluate ∫log x dx.
Solution:

Question 22.
The cost and revenue function of a product are given by C(x) = 20x + 4000 and R(x) = 60 x + 2000 respectively, where x is the number of items produced and sold. Then, find the number of items must be sold to realise some profit.
Solution:
To earn some profit, we must have
R(x) > C(x)
⇒ 60x + 2000 > 20x + 4000
Subtracting 20x from both sides,
40x + 2000 > 4000
Subtracting 2000 from both sides,
40x > 2000
Dividing both sides by 40,
x > \(\frac{2000}{40}\) ⇒ x > 50
Hence, atleast 50 items must be sold to realise some profit.

Question 23.
Find the maximum value of Z for the problem maximise Z = 100x + 170y, subject to constraints 3x + 2 y ≤ 3600, x + 4y ≤ 1800, x, y ≥ 0.
Or
Find the maximum value of Z for the problem maximise Z = x + y, subject to constraints
x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0 and y ≥ 0.
Solution:
We have, Maximise Z = 100 x + 170y
Subject to the constraints are
3x + 2y ≤ 3600, x + 4y ≤ 1800
and x ≥ 0, y ≥ 0
From the shaded feasible region it is clear that the coordinates of corner points are
(0, 0), (1200, 0), (1080, 180) and (0, 450).
On solving x + 4y = 1800 and 3x + 2y = 3600, we get
x = 1080 and y = 180

Maximise Z = x + y,
Subject to the constraints are
x + 4y ≤ 8,2x + 3y ≤ 12,3x + y ≤ 9andx, y ≥ 0
On solving x + 4y = 8 and 3x + y = 9, we get
x = \(\frac{28}{11}\) and y = \(\frac{15}{11}\)
From the feasible region, it is clear that coordinates of corner points are (0, 0), (3, 0), (\(\frac{28}{11},\frac{15}{11}\)) and (0, 2).

Question 24.
A pipe can fill a tank in 5h, while another pipe can empty it in 6h. If both the pipes are opened simultaneously, then find the time will be taken to fill the tank.
Solution:

Hence, the tank will be filled completely in 30 h.

Question 25.
What sum of money is needed to invest now, so as to get ₹ 6000 at the beginning of every month forever, if the money is worth 9% per annum compounded monthly?
Solution:

Thus, ₹ 806000 are needed to invest to get ₹ 6000 at the beginning of every month forever.

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

Question 26.
A company anticipates a capital expenditure of X 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 27.

find a matrix D such that CD – AB = 0
Or
If A is skew-symmetric matrix of order 3, then prove that det A = 0.
Solution:

By equality of matrices,
2x + 5z = 3, 3x +8z = 43
and 2y + 5w = 0, 3y + 8w = 22
After solving, we get
x = -191, y = -110, z = 77 and w = 44

Question 28.
The mean and variance of a random sample of 81 observations were computed as 200 and 144, respectively. Compute the 95% confidence limits for population mean.
Or
The mean weekly sales of mango candy in candy stores was 225.4 mango candy per store. After an advertising campaign the mean weekly sales in 25 stores for a typical week increased to 237.6 and showed a standard deviation of 21.3 was the advertising campaign successful? Level of confidence 5%.
[given, t₂₄(0.05) = 1.711]
Solution:

Since, calculated value off > tabulated value of t.
Hence, H₀ can rejected.
⇒ Advertising campaign was successful.

Question 29.
Mr Narayan Sankar has invested ₹ 150000 in a financial plan whose compound annual growth rate (CAGR) is 8.5% and he received a final value of ₹ 300000. Find the period (completed) for which he has invested the amount, [given, log 2 = 0.3010 and log(1.09) = 0.0374]
S0lution:
Given, EV = ₹ 300000; BV = ₹ 150000
and CAGR = 8.5%
We know that

∴ The period for which he has invested the amount is 8 yr.

Question 30.
Find the number of pairs of consecutive even positive integers, both of which are larger than 8, such that their sum is less than 25, is?
Solution:
Let x be the smaller of two consecutive even positive integers.
Then, the other even integer is x + 2.
Given, x > 8 and x + x + 2 < 25
⇒ x > 8 and x + x + 2 < 25
⇒ x > 8 and 2x + 2 < 25
⇒ x > 8 and 2x < 23
⇒ x > 8 and x < \(\frac{23}{2}\)
Hence, there exists only one pair of even integer (10, 12).

Question 31.
From a lot of 15 bulbs which include 5 defectives, a sample of 2 bulbs is drawn at random (without replacement). Find the probability distribution of the number of defective bulbs.
Solution:
It is given that out of 15 bulbs, 5 are defective.
∴ Number of non-defective bulbs = 15 – 5 = 10
Let X be the random variable which denotes the defective bulbs. So, X may take values 0,1,2
P(X = 0) = P (No defective bulb)

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

Question 32.
Find the positive numbers x and y such that x + y = 60 and xy³ is maximum.
Or
An j Apache helicopter of enemy is flying along the curve given by y = x² + 7. A soldier placed at (3, 7) wants to shoot down the helicopter. When it is nearest to him. Find the nearest distance.
Solution:
Let x + y = 60 and P = xy³
Then, P =(60 – y)(y³) [∵ x = 60 – y]
On differentiating w.r.t. x, we get
\(\frac{dP}{dy}\) = 3y²(60 – y) – y³ = 180y² – 3y³ – y³
= 180y² – 4y³ = 4y²(45 – y)
Again, differentiating w.r.t. x, we get
\(\frac{d^2P}{dy^2}\) = 360y -12y² = 12y (30 – y) dy
For maxima and minima, put \(\frac{dP}{dy}\) = 0
⇒ 4y²(45-y) = 0
y = 0 or y = 45
Now, (\(\frac{d^2P}{dy^2}\))_y=45 = 12 × 45(30 – 45) = -8100 < 0
So, y = 45isapoint of maximum and another number is 60 – 45 = 15
Hence, number are 45 and 15.
Or
Let P(h, k) be any point on the curve y = x² + 7 and Q(3, 7) be the given point.
∴ Distance between PQ is

Let PQ² = f(h), then PQ is minimum or maximum.
According to f(h) is maximum or minimum
f(h) = h² – 6h + 9 + h⁴
f'(h) = 2h – 6 + 4h³
f”(h) = 2 + 12h²
For maxima or minima put f'(h) = 0
∴ 4h³ + 2h – 6 = 0
(h – 1)(4h² + 4h + 6) = 0
2(h -1) (2h² + 2h + 3) = 0
h = 1, 2h² + 2h + 3 ≠ 0
Now, f”(1) = 2 + 12 > 0 So, h = 1 is a point of local minima.
PQ = \(\sqrt{(1-3)^2+(1)^2}=\sqrt{4+1}\) = √5
∴ The nearest distance is √5 units.

Question 33.
A manufacturer produces two models of bikes-model Y and model Y. Model Y takes a 6 man-hours to make per unit, while model Y takes 10 man hours per unit. There is a total of 450 man-hour available per week. Handling and marketing costs are ₹ 2000 and ₹ 1000 per unit for models X and Y, respectively. The total funds available for these purposes are ₹ 80000 per week. Profits per unit for models X and Y are ₹ 1000 and ₹ 500, respectively. Find the maximum profit.
Or
Find the minimum value of Z for the problem minimise Z = 400x + 200 y, subject to constraints
5x + 2y ≥ 30, 2x + y ≤ 15, x ≤ y and x, y ≥ 0
Solution:
Let the manufacturer produces x number of models X and y number of model Y bikes.
Model X takes a 6 man-hours to make per unit and model Y takes a 10 man-hours to make per unit.
There is total of 450 man-hour available per week.
∴ 6x + 10y < 450
⇒ 3x + 5y < 225 … (i)
For models X and X handling and marketing costs are ₹ 2000 and ₹ 1000, respectively, total funds available for these purposes are ₹ 80000 per week.
∴ 2000x + 1000y < 80000
⇒ 2x + y < 80 …(ii)
Also, x > 0, y > 0
Since, the profits per unit for models X and Y are ₹ 1000 and ₹ 500, respectively.
∴ Required LPP is Maximise
Z = 1000x + 500y
Subject to constraints,
3x + 5y ≤ 225, 2x + y ≤ 80, x ≥ 0, y ≥ 0
On solving 3x + 5y = 225 and 2x + y = 80, we get
x = 25 and y = 30
From the shaded feasible region, it is clear that coordinates of corner points are (0, 0), (40, 0), (25, 30) and (0, 45).

The maximum profit is ₹ 40000.
Or
We have, minimise Z = 400x + 200y,
Subject to the constraints are
5x + 2y ≥ 30, 2x + y ≤ 15, x ≤ y, x ≥ 0, y ≥ 0.
On solving x – y = 0 and 5x + 2y = 30, we get

On solving x – y = 0 and 2x + y = 15, we get
x = 5, y = 5
So, from the shaded feasible region it is clear that coordinates of corner points are (0, 15), (5, 5) and

Hence, the minimum cost is ₹ 2571.43.

Question 34.
For a particular commodity the demand function and supply function are

respectively. Find the nearest rupee the consumer’s surplus and producer’s surplus, if market attains equilibrium.
Solution:
Given, the demand function

Hence, consumers surplus is 619 and producer surplus is 155.

Question 35.

Solution:

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

Question 36.
A urn contains 5 white, 7 red and 8 black balls. From which four balls are drawn successively with replacement.

Based on the above information, answer the following questions.
(i) Find the probability that all are white.
(ii) Find the probability that exactly one is white.
(iii) Find the mean of the distributions.
Or
Find the standard deviation of the distributions.
Solution:
(i) Total number of balls = 5 + 7 + 8 = 20
White balls = 5

Question 37.
EMI is a part of equally divided monthly outgoes to clear off an outstanding loan within a stipulated time frame. For a fixed interest rate loan, the EMI remain fixed for the entire tenure of the loan, provided there is no default or part payment in between. The EMI is used off both the principal and interest components of an outstanding loan. The first EMI has the highest interest component and the lowest principal component.

Rajesh purchased a house from a company for ₹2500000 and made a down payment of ₹500000. Fie repays the balance in 25 yr by monthly installments at the rate of 9% per annum compounded monthly, (given (1.0075 )^-300 = 0.1062)
(i) Find the number of payment.
(ii) Find the rate of interest per month.
(iii) What are the monthly payment.
Or
What is the total interest payment.
Solution:
(i) Here, time = 25 yr
∴ Total number of payment = 25 × 12 = 300
(ii) R = 9% per annum.
Rate of interest per month = \(\frac{9}{1200}\) = 0.0075

(iii) Cost of house = ₹ 2500000
Down payment = ₹ 500000
∴ Principal amount = ₹ (2500000 – 500000)
= ₹ 2000000,
n = 25 × 12 = 300
and i = \(\frac{9}{1200}\) = 0.0075

Hence, total interest is ₹ 3034681.

Question 38.
When observed over a long period of time, a time series data can predict trend can increase or decrease or stagnation of a variable under consideration. Such analytical studies can benefit a business for casting or prediction of future estimated sales or production.
The table below are given the figures of production (in thousand tonnes) of a sugar factory.

Fit a straight line trend by the method of least squares and find trend value for year 2005.
Or
Consider the following data

Calculate 3-days moving average and display these and the original figures on the same graph.
Solution:
Here, n = 7 (odd)
So, we shift the origin to the middle of the time period of the 2008.
Let the straight line trend of y on x be
y_t = a + bx …(i)
Now, construct the table as under.

Or
According to the question,
Let the 3-days moving average is m.