Electrical

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Maths with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Maths Set 11 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) 4
(b) 2
(c) 5
(d) 3
Solution:
(d) Let x be a non-negative integer congruent to 2796 (mod 7)
x ≡ 2796 (mod 7)
x = 3 (mod 7)

Question 2.
If -3 ≤ 4 – \(\frac{7x}{2}\) ≤ 18, then
(a) x ∈ (-∞, -4](b) x ∈ [-4, ∞)
(c) x ∈ [-4, 2](d) x ∈ [2, ∞)
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 1
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 2
∴ Solution set is [- 4,2].

Question 3.
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:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 3
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 4
So, the declared rate of interest
= (0.10 × 100)% = 10%

Question 4.
A mixture of 150 L of wine and water contains 20% water. The amount of water which should be added so that the water becomes 25% of the new mixture is
(a) 8 L
(b) 10 L
(c) 12 L
(d) 14 L
Solution:
(b) In 150 L of mixture, quantity of water
= 20% of 150
= 30
∴ Quantity of wine = (150 – 30) L
= 120 L
Let the water added in the mixture be x.
According to the question,
30 + x = 25% of (150 + x)
⇒ 30 + x = \(\frac{150+x}{4}\)
⇒ 120 + 4x = 150 + x
⇒ 3x = 30
⇒ x = 10
Hence, 10 L of water is added to the mixture.

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 5.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 5
what is the value of x?
(a) 2
(b) 3
(c) -1
(d) 5
Solution:
(d) ∵ A = A’
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 6
⇒ 2x – 3 = x + 2
⇒ x = 5

Question 6.
The area of the triangle, whose vertices are (3, 8), (-4, 2) and (5, 1), is
(a) 60 sq units
(b) 61 sq units
(c) \(\frac{61}{2}\) sq units
(d) 30 sq units
Solution:
(c) The area of triangle is given by
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 7

Question 7.
A coin is tossed 4 times. The probability of getting atleast one head is
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{7}{8}\)
(d) \(\frac{15}{16}\)
Solution:
(d) ∴ Required probability
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 8

Question 8.
The value of ∫\(\frac{1}{x(5+logx)}\) dx is
(a) \(\frac{1}{5}\) log |5 + logx| + C
(b) 5log|5 + logx| + C
(c) \(\sqrt{5+logx}\) + C
(d) log|5 + logx| + C
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 9

Question 9.
A random variable ‘X’ has the following distribution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 10
The value of k is
(a) -1
(b) –\(\frac{1}{10}\)
(c) 1
(d) \(\frac{1}{10}\)
Solution:
(d) ∑Pi = 1
∴ k + k + 2k + 3k + k² + 2k² + 7k² + 2k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ (10k – 1)(k + 1)= 0
⇒ k = \(\frac{1}{10}\), k ≠ 1

Question 10.
A machine costing ₹ 50000 has a useful life of 4 yr. The estimated scrap value is ₹ 10000. The rate of depreciation per annum is
(a) 20%
(b) 30%
(c) 25%
(d) 15%
Solution:
(c) Annual depreciation amount = ₹ 10000
Original cost of machine – Salvage value of machine
= ₹ (50000 – 10000)
= ₹40000
∴ Depreciation rate percent
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 11

Question 11.
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) Variance = E(X)² – (E(X))² = 11 – (3)² = 11 – 9 = 2

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 12.
The function f(x) = x4 – 4x is strictly
(a) decreasing in [1, ∞)
(b) increasing in (1, ∞)
(c) increasing in (-∞, 1)
(d) increasing in (-1, 1)
Solution:
(b) f(x) = x4 – 4x
⇒ f'(x) = 4x³ – 4
⇒ f'(x) = 0
⇒ 4x³ – 4 = 0
⇒ 4(x³ – 1) = 0 , x = 1
∴ f(x) is strictly increasing in (1, ∞) and decreasing in (-∞, 1).

Question 13.
The time series analysis helps
(a) to make predictions
(b) to compare the two or more series
(c) to know the behaviour of business
(d) All of the above
Solution:
(d) The time series analysis helps to make predictions, to compare the two or more series and to know the behaviour of business.

Question 14.
Random sampling is useful as it is
(a) reasonably more accurate as compared to other methods
(b) economical in nature
(c) free from personal biases of the investigator
(d) All of the above
Solution:
(d) All the given option are correct.

Question 15.
The degree of the differential equation
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 12
(a) 1
(b) 2
(c) 3
(d) not defined
Solution:
(d) The degree of given differential equation is not defined because when we expand sin (\(\frac{dy}{dx}\)), we get
an infinite series in the increasing powers of \(\frac{dy}{dx}\). Therefore, its degree is not defined.

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.
The variables x and y in a linear programming problem are called
(a) decision variables
(b) linear variables
(c) optimal variables
(d) None of these
Solution:
(a) The variables x and y in a linear programming problem are called decision variables.

Question 18.
In a 100 m race, A runs at a speed of \(\frac{5}{6}\) m/s. If A gives a start of 4 m to B and still beats him by 12 s. Then, the speed of B is
(a) \(\frac{5}{4}\) m/s
(b) \(\frac{7}{5}\) m/s
(c) \(\frac{4}{3}\) m/s
(d) \(\frac{6}{5}\) m/s
Solution:
(c) Let the speed of 6 be x m/s.
According to the question,
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 13

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) In 30 L mixture of acid, the ratio of acid and water is 2 : 3. The amount of water should be added to the mixture so that the ratio of acid and water becomes 2 : 5 is 12 L.
Reason (R) Mean price is always lesser than cost price of cheaper quantity and higher than the cost price of dearer quantity.
Solution:
(c) Assertion Quantity of acid in the mixture
\(\frac{2}{5}\) × 30 = 12 L
Quantity of water in the mixture = 30 -12 = 18 L Let the amount of water added to the mixture be x. According to the question,
\(\frac{12}{18x+x}=\frac{2}{5}\)
⇒ 36 + 2x = 60
⇒ 2x =24
⇒ x = 12 L
Reason The mean price of the quantity will be always higher than the cost price of cheaper quantity and lower than the cost price of dearer quantity.
Hence, Assertion is true but the Reason is false.

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 20.
Assertion (A) If x is real, then the minimum value of x² – 8x + 17 is 1.
Reason (R) If f”(x) > 0 at critical point, then the value of the function at critical point will be the minimum value of the function.
Solution:
(a) Assertion Let f(x) = x² – 8x + 17
∴ f'(x) = 2x – 8
So, f'(x) = 0, gives x = 4
Here, x = 4 is the critical number.
Now, f”(x) = 2 > 0, ∀ x
So, x = 4 is the point of local minima.
∴ Minimum value of f(x) at x = 4,
f (4) = 4 × 4 – 8 × 4 + 17 = 1
Hence, we can say that both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

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

Question 21.
A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then find the time taken to fill the tank.
Or
Find the last digit of 1212.
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 14
Net part filled in 1 h when both (tap and leakage) work
∴ Required time to fill the tank is 30 h.
Or
To find the last digit of 1212, we find 1212 (mod 10)
Since, 12 ≡ 2 (mod 10)
⇒ 124 = 24 (mod 10) = 16 (mod 10)
⇒ 124 = 6(mod 10)
⇒ (124)³ = (6)³ mod 10
⇒ 1212 = 216 mod 10
⇒ 1212 = 6mod 10
Hence, the last digit of 1212 is 6.

Question 22.
Find the maximum value of Z = 4x + 3y, if the feasible region for an LPP is shown in following figure.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 15
Solution:
The feasible region is bounded. Therefore, maximum of Z must occurs at the corner points of the feasible region.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 16
The corner points of the feasible region are given below

Corner points Value of Z = 4 x + 3y
O(0, 0) 4(0) + 3(0) = 0
A(25, 0) 4(25) + 3(0) = 100
B(16, 16) 4(16) + 3(16) = 112 ← Maximum
C(0, 24) 4(0) + 3 (24) = 72

Hence, the maximum value of Z is 112 at 6 (16, 16).

Question 23.
Consider the following hypothesis tests H0 : µ = 22; Ha : µ ≠ 22
A sample of 65 provided a sample mean, \(\overline{\mathrm{x}}\) = 20 and sample standard deviation, S = 6.4. Compute the value of the test statistic.
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 17

Question 24.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 18
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 19
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 20

Question 25.
At what rate of interest will the present value of ₹ 300 payable at the end of each quarter be ₹ 24000?
Solution:
Let the rate of interest be r% per annum.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 21
Hence, rate of interest is 5% per annum.

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

Question 26.
Find the matrix X for which
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 22
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 23
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 24

Question 27.
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
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 25
Hence, the cash price of house is ₹ 1538250.

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 28.
In a binomial distribution, the sum of its mean and variance is 1.8 find the probability of two success if the events was conducted 5 times.
Or
Two biased dice are thrown together. For the first die P(6) = \(\frac{1}{2}\), the other scores being equally likely while for the second die P(1) = \(\frac{2}{5}\) and the other scores are equally likely. Find the probability distribution of ‘the number of one’s seen’.
Solution:
Here, np + npq = 1.8 and n = 5
5p + 5pq = 1.8
⇒ 5p(1 + qr) = 1.8
⇒ 5(1 – q)(1 + q) = 1.8 [∵ p + q = 1]⇒ 1 – q² = \(\frac{1.8}{5}\)
⇒ q² = 1 – 0.36 = 0.64
⇒ q = 0.8 [∵ 0 ≤ q ≤ 1]∴ p = 1 – 0.8 = 02
n = 5, p = 0.2, q = 0.8
∴ p(r = 2) = 5C2p²q³
= 10(0.2)² (0.8)³
= 0.2048
Or
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 26
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 27

Question 29.
A monopolistic demand function for one of its product is p(x) = ax + b. He knows that he can sell 1400 units when the price is ₹ 4 per unit and he can sell 1800 units at a price of ₹ 2 per unit. Find the marginal revenue function.
Or
A manufacturer’s marginal revenue function is given by MR = 275 – x – 0.3x². Find the increase in the manufacturer’s total increase in revenue, if the production is increased from 10 to 20 units.
Solution:
We have, p(x) = ax + b
It is given that p = ₹ 4 when x = 1400 and p = ₹ 2 when x = 1800
Substituting these values in Eq. (i), we get
∴ 4 = 1400 a+b
2 = 1800a + b
Solving these two equations, we get
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 28

We have to find the total increase in revenue, if the production is increased from 10 to 20 units
i.e. we have to find R(20) – R (10)
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 29
Hence, increase in revenue is ₹ 1900

Question 30.
Find the general solution of
log(\(\frac{dy}{dx}\)) = 3x +4y
Solution:
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 30
which is the required general solution of given differential equation.

Question 31.
A person invested of ₹ 15000 in a mutual fund and the value of investment at the time of redemption was ₹ 25000. If CAGR for this investment is 8.88% calculate the time period for which the amount was invested? [log(L667) = 0.2219 and log(1.089) = 0.037]Solution:
Given, beginning value = ₹ 15000
ending value = ₹ 25000
CAGR = 8.88%
Let number of years = x
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 31
On taking log both sides., we get
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 32
Hence, time period = 6 yrs

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

Question 32.
The table given below shows the daily attendance in thousand at a certain exhibition over a period of two weeks
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 33
Calculate seven day moving averages and illustrate these and original information on the same graph using the same scales.
Solution:
According to the question,
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 34
On the basis of above data we can draw the following moving average graph
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 35

Question 33.
The sum of three numbers is 20. If we multiply the first number by 2 and add the second number to result and subtract the third number, we get 23. By adding second and third numbers to three times the first number, we get 46. Represent the above problem algebraically and use Cramer’s rule to find the number of those equations.
Or
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 36
Find AB, use this to solve the system of equations x – y = 3, 2x + 3y + 4z – 17 and y + 2z = 7.
Solution:
Let three numbers be x, y and z.
Given, x + y + z = 20
2x + y – z = 23
3x + y + z = 46
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 37
Hence, three numbers are 13, 2, 5.
Or
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 38
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 46
On comparing corresponding elements, we get
x = 2, y = -1 and z = 4

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 34.
A person borrows ₹ 68962 on the condition that he will repay money with compound interest at 5% per annum in 4 equal annual installment the first one being payable at the end of the first year. Find the value of each installment.
[given (1.05)-4 = 0.8227]Solution:
According to the question, the borrow amount (principal amount = P) is ₹ 68962.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 47
So, the value of each installment is ₹ 19447.82.

Question 35.
Solve the following LPP graphically. Maximise Z =3x + 2y,
Subject to constraints are x + 2y ≤ 10, 3x + y ≤ 15 and x ≥ 0, y ≥ 0.
Also, determine the area of the feasible region.
Or
Find the maximum value of Z for the problem maximise Z =2x + 3y,
Subject to constraints
x + 2y ≤ 10, 2x + y ≤ 14, x,y ≥ 0.
Solution:
Our problem is to maximise Z = 3x + 2y … (i)
Subject to constraints are
x + 2y ≤ 10 …(ii)
3x + y ≤ 15 …(iii)
and x ≥ 0, y ≥ 0 …(iv)
Table for line x + 2y = 10 is

So, the line passes through the points (0, 5) and (4,3).
On putting (0, 0) in the inequality x + 2y ≤ 10, we get
0 + 2 × 0 ≤ 10 ⇒ 0 ≤ 10, which is true.
So, the half plane is towards the origin.
Table for line 3x + y = 15 is

So, the line passes through the points (4, 3) and (5, 0).
On putting (0, 0) in the inequality 3x + y ≤ 15, we get
3 × 0 + 0 ≤ 15 ⇒ 0 ≤ 15, which is true.
So, the half plane is towards the origin.
Also, x, y ≥ 0, so the region lies in the I quadrant.
On solving equations x + 2y = 10 and 3x + y = 15, we get
x = 4 and y = 3
So, the intersection point is B (4, 3).
∴ Feasible region is OABCO.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 48

The corner points of the feasible region are 0(0, 0), A(5, 0), 6(4, 3) and C(0, 5).
The values of Z at the corner points are given below

Corner points Value of Z = 3x + 2y
O(0, 0) Z = 3 × 0 + 2 × 0 = 0
A(5, 0) Z = 3 × 5 + 2 × 0 = 15
B(4, 3) Z = 3 × 4 + 2 × 3 = 18
C(0, 5) Z = 3 × 0 + 2 × 5 = 10

Therefore, the maximum value of Z is 18 at the point B(4, 3).
∴ Area of feasible region
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 39
Or
Given, maximise Z = 2x + 3y
Subject to constraints are
x + 2y ≤ 10
2x + y ≤ 14
and x ≥ 0, y ≥ 0
Shade the region to the right of Y-axis to show x ≥ 0 and above X-axis to show y ≥ 0.
Table for line x + 2y = 10 is
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 40
So, the line is passing through the points (0, 5), (6,2) and (10,0).
On putting (0, 0) in the inequality x + 2y ≤ 10, we get 0 + 0 ≤ 10, which is true.
So, the half plane is towards the origin.
Table for line 2x + y = 14 is
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 41
So, the line is passing through the points (4, 6), (6,2) and (7, 0).
On putting (0, 0) in the inequality 2x + y ≤ 14, we get
0 + 0 ≤ 14, which is true.
So, the half plane is towards the origin.
The intersection point of lines corresponding to
Eqs. (i) and (ii) is 6(6,2).
On shading the common region, we get the feasible region OABD.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 42
The values of Z at corner points are given below

Corner points Value of Z = 2x + 3y
0(0, 0) Z = 2 × 0 + 3 × 0 = 0
A(7, 0) Z = 2 × 7 + 3 × 0 = 14
8(6, 2) Z = 2 × 6 + 3 × 2 = 18
D(0, 5) Z = 2 × 0 + 3 × 5 = 15

Hence, the maximum value of Z is 18 at the point 6(6, 2).

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

Question 36.
An urn contains 25 balls of which 10 balls bear a mark X and remaining 15 bear a mark Y. A balls is drawn at random from the urn, its mark noted down and it is replace. In this way 6 balls are drawn.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 43
Based on the above information, answer the following questions:
(i) Find the probability that all balls will bear X marks.
(ii) Find the probability that all balls will not bear X mark.
(iii) Find the probability that at most 2 balls will bear mark Y.
Or
Find the probability that at least 2 balls will bear mark Y.
Solution:
Given total balls = 25
n(X) = 10, n(Y) = 15
Let p = probability of getting mark X ball
q = probability of getting mark Y ball
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 44

Question 37.
There are 24 h in a day. There are mainly two types of clocks used 12 h clock and 24 h clocks 12 h clock repeat itself twice in a day i.e. 24 h of a day are divided into two periods called am and pm. Each period consists of 12 h numbered 12 (acting as 0), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Then, 24 h per day cycle start at 12 mid-night (usually indicated as 12 am) runs through 12 noon (usually indicated as 12 pm) and continues before mid-night at the end of the day.
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 45
Based on above information, answer the following questions
(i) It is currently 6:00 pm in 12 h clock. What is the time after 375 h.
(ii) What time is 24 h clock in equivalent to 5:00 pm in 12 h clocks.
(iii) If the time after 490 h from now will be 7:00 pm. Then, what is current time?
Or
What time in 12h clock is equivalent to 22:00 in 24 h clock?
Solution:
(i) 375 = 24 × 15 + 15
∴ 375 h = (24 × 15 + 15) h
Time = 6 pm + 15 h = 6 am + 3h = 9 am

(ii) In 24 h clock 5:00 in 12 h equivalent = 17:00

(iii) 490 = 24 × 20 + 10
The current time = 7:00 pm – 10 h
= 9:00 am
Or
22:00 = 10:00 pm in 12 h clock.

CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions

Question 38.
An architect designs a building for a multi-national company The floor consists of a rectangular region with semi-circular ends having a perimeter of 200 m as shown below
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 49
Based on the above information answer the following
Find the maximum value of area A.
Or
The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen find the value of x.
Solution:
From the figure, it is clear that perimeter of floor = 200 m
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 50
CBSE Sample Papers for Class 12 Applied Maths Set 11 with Solutions 51


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