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## Class 11 Applied Maths Model Papers 2023-24

Applied Maths is a newly introduced subject in senior secondary classes. The purpose of this subject is to give the most relevant option to commerce students. Earlier they had to study core maths that was not very relevant to their field.

The course curriculum of Applied Maths is designed in such as way that commerce students will get the topics related to their areas. Here, we are providing model question papers for class 11 Applied Maths. Our team of expert teachers has prepared these sample question papers as per the latest blueprint issued by CBSE, New Delhi.

## New Sample Papers for Applied Maths

As CBSE doesn’t issue sample papers for class 11th officially, you will get CBSE Class 11 Applied Maths Sample Papers 2024 from other sources only. So, the authenticity of the source is very important. You must download them from reputed websites only. That’s why we recommend myCBSEguide app for class XI Applied Maths model papers.

**Class 11 – Applied Mathematics Sample Paper (2023-24)**

**Maximum Marks: 80 Time Allowed: : 3 hours**

**General Instructions:**

- This Question paper contains –
**five sections**A, B, C, D and E. Each section is compulsory. However, there is some internal choice in some questions. - Section A has 18 MCQ’s and 02 Assertion Reason based questions of 1 mark each.
- Section B has 5 Very Short Answer(VSA) questions of 2 marks each.
- Section C has 6 Short Answer(SA) questions of 3 marks each.
- Section D has 4 Long Answer(LA) questions of 5 marks each.
- Section E has 3 source based/case based/passage based/integrated units of assessment (04 marks each) with sub parts.
- 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 alternatives in all such questions.

**CBSE Class 11 Applied Maths Sample Papers 2024**

**Section A**- If three events A, B and C are mutually exclusive, then which one of the following is correct?a) P(A {tex}\cap{/tex} B {tex}\cap{/tex} C) = 0b) P(A {tex}\cup{/tex} B {tex}\cup{/tex} C) = 0c) P(A {tex}\cap{/tex} B {tex}\cap{/tex} C) = 1d) P(A {tex}\cup{/tex} B {tex}\cup{/tex} C) = 1
- Which one of the following measures of central tendency is used in construction of index numbers?a) Medianb) Harmonic meanc) Moded) Geometric mean
- Section 80E refers to the deduction of:a) interest on fixed depositb) interest on education loanc) interest on savings accountd) interest on a home loan
- If log
_{10}2 = 0.3010, then log_{5}64 =a) {tex}\frac {202}{633}{/tex}b) {tex}\frac {633}{202}{/tex}c) {tex}\frac {233}{602}{/tex}d) {tex}\frac {602}{233}{/tex} - If f(x) = {tex}\frac{1}{\sqrt{x-|x|}}{/tex}, then D
_{f}is equal toa) {tex}\phi {/tex}b) Rc) {tex}(0,\infty ){/tex}d) {tex}( – \infty ,{\rm{ }}0){/tex} - The value of log
_{5}(1 + {tex}\frac 15{/tex}) + log_{5}(1 + {tex}\frac 16{/tex}) + log_{5}(1 + {tex}\frac 17{/tex}) + … + log_{5}(1 + {tex}\frac 1{624}{/tex}) is:a) 2b) 4c) 5d) 3 - A bag contains 5 black and 3 white balls. Two balls are drawn at random one after the other without replacement. What is the probability that both are white?a) {tex}\frac{3}{28}{/tex}b) {tex}\frac{1}{14}{/tex}c) {tex}\frac{1}{21}{/tex}d) {tex}\frac{1}{28}{/tex}
- If 3x + y = 0 is a tangent to the circle which has its center at the point (2, – 1), then the equation of the other tangent to the circle from the origin isa) x + 2y = 0b) x + 3y = 0c) 3x – y = 0d) x – 3y = 0
- A woman introduces a man as the son of the brother of her mother. How is that man related to the woman?a) Nephewb) Sonc) Uncled) Cousin
- The mean of 100 observations is 50 and their standard deviation is 10. If 5 is added to each observation, then new mean and new standard deviation respectively will bea) 50, 15b) 50, 10c) 60, 10d) 55, 10
- If log 0.0007392 = -3.1313, then log 73.92 isa) 1.8687b) 2.8687c) 1.1313d) 2.1313
- A sum of money at compound interest amounts to thrice of itself in 5 years. In how many years will it be 9 times of itself?a) 15 yearsb) 12 yearsc) 9 yearsd) 10 years
- A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of ₹ 32000. The shopkeeper sells the TV to a consumer at the listed price. If the sales are intra-state and the rate of GST is 18%, the tax (under GST) received by the State Government is:a) ₹ 1440b) ₹ 5760c) ₹ 2880d) ₹ 4320
- Five boys and four girls sit in a row randomly. The probability that no two girls sit together is:a) {tex}\frac{3}{21}{/tex}b) {tex}\frac{5}{42}{/tex}c) {tex}\frac{41}{42}{/tex}d) {tex}\frac{11}{42}{/tex}
- Two dice each numbered from 1 to 6 are thrown together. Let A and B be two events given by

A: even number on the first die

B: number on the second die is greater than 4

What is the value of P(A {tex}\cup{/tex} B)?a) {tex}\frac{1}{4}{/tex}b) {tex}\frac{1}{2}{/tex}c) {tex}\frac{1}{6}{/tex}d) {tex}\frac{2}{3}{/tex} - Sanjay buys land for 200000 and agrees to pay an equal amount at the end of each year for 3 years. If the money is worth 8%, then the amount of each instalment is: (Given (1.08)
^{-3}= 0.7938).a) ₹ 78895.12b) ₹ 75323.48c) ₹ 75428.56d) ₹ 77594.56 - How many different committees of 5 can be formed from 6 men and 4 women on which exact 3 men and 2 women serve?a) 60b) 6c) 20d) 120
- Let A be the set of all students of a boys school. The relation R on A is given by R = {(a, b) : a is sister of b} isa) Emptyb) Universalc) Triviald) Reflexive
**Assertion (A):**The mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 is 3.

**Reason (R):**The mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 is 8.5.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.**Assertion (A):**The sum of infinite terms of a geometric progression is given by {tex}S_{\infty}=\frac{a}{1-r}{/tex}, provided |r| < 1.

**Reason (R):**The sum of n terms of Geometric progression is S_{n }= {tex}\frac{a\left(r^n-1\right)}{r-1}{/tex}.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.**Section B**- At what time do the hands of the clock meet between 6: 00 to 7: 00?
- A and B are brothers. C and D are sisters. A’s son is D’s brother. How B is related to C?
**OR**If

**RAHUL**is coded as 22 – 5 -1 2 – 25 -16, then how will you code**VIRAT**? - At what time between 4:00 and 5:00 will the hands of the clock be at right angles?
- If x
^{y}= e^{x-y}, prove that {tex}\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}{/tex}**OR**Differentiate the functions with respect to x: log

_{x}3 - Convert the decimal number 15.6875 to binary.
**Section C**- If a, b, c are in A.P. show that following are also in A.P.
- {tex}\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}{/tex}
- b + c, c + a, a + b

**OR**Find three numbers in G.P. whose product is 216 and the sum of their products in pairs is 156.

- Arrange each of the following words in meaningful logical order.
- 1. Probation, 2. Interview, 3. Selection, 4. Appointment, 5. Advertisement, 6. Application
- 1. Gold, 2. Iron, 3. Sand, 4. Platinum, 5. Diamond

- Draw the graph of constant function f : R {tex}\rightarrow{/tex} R; f(x) = 2 {tex}\forall{/tex} x {tex}\in{/tex} R. Also, find its domain and range.
- The compound interest on a sum of money for 2 years is ₹ 410 and the simple interest on the same sum for the same period and at the same rate is ₹ 400. Find the sum and the rate of interest.
- For an industrial connection monthly consumption of water is 40 Kl, calculate the Water bill. Tariff rates can be considered as the table given below:
**Monthly Consumption (in Kilolitre)****Service Charge (in ₹)****Volumetric Charge (Per Kl in ₹)**Upto 20 146.41 5.27 20-30 219.62 *26.36 > 30 292.82 43.93 Plus Sewer Maintenance Charges: 60% of water volumetric charge - Using properties of sets and their complements prove that:
- (A {tex}\cup{/tex} B) {tex}\cap{/tex} {tex}(A \cap B^\prime){/tex} = A
- A – (A {tex}\cap{/tex} B) = A – B

**Section D**- Determine n if

(i) {tex}^{2n}{C_3}{:^n}{C_2} = 12:1{/tex}

(ii) {tex}^{2n}{C_3}{:^n}{C_3} = 11:1{/tex}**OR**If

^{n}P_{4}= 360, find the value of n. - Evaluate: {tex}\lim \limits_{x \rightarrow-\infty}(\sqrt{4 x^{2}-7 x+2 x}){/tex}.
- Find the variance and standard deviation for the following data:

65, 68, 58, 44, 48, 45, 60, 62, 60, 50**OR**Calculate the mean and standard deviation of the following data:

Age 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 No. of persons 3 51 122 141 130 51 2 - Mr. Saxena from Bhopal, M.R has an electricity connection of 5 kW. He consumed 1264 units in one month. Calculate his electricity bill for that month. Tariff plan is given below:
**No. of Units (in kWh)**0 – 50 51 -100 101- 300 > 300 **Price per unit (in ₹)**4.05 4.95 6.30 6.50 Fixed charge = ₹ 250 per kW per month

Surcharge = Nil, Energy duty = ₹ 0.63 per unit **Section E****Read the text carefully and answer the questions:**

Population vs Year graph given below.

- In which year the population becomes 110 crores?
- Find the equation of line perpendicular to line AB and passing through (1995, 97).
- Write the equation of line AB?
**OR**Find the slope of line AB.

**Read the text carefully and answer the questions:**

Different organisations collect the data and analyse it quantitatively. During one such analysis some mistake crept in. The result given was that mean and variance of 100 observations as 40 and 5.1 but later on rechecking it was found that one observation was mistakenly taken as 50 instead of 40.- What is incorrect sum of variates?
- What is correct sum of observations?
- What is incorrect {tex}\Sigma x^2{/tex}?
**OR**What is corrected variance?

**Read the text carefully and answer the questions:**

Shama is studying in class XII. She wants do graduate in chemical engineering. Her main subjects are mathematics, physics, and chemistry. In the examination, her probabilities of getting grade A in these subjects are 0.2, 0.3, and 0.5 respectively.

- Find the probability that she gets grade A in all subjects.
- Find the probability that she gets grade A in no subjects.
- Find the probability that she gets grade A in two subjects.

**OR****Read the text carefully and answer the questions:**

A shopkeeper sells three types of flower seeds A_{1}, A_{2}, A_{3}. They are sold in the form of a mixture, where the proportions of these seeds are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.

Based on the above information:- Calculate the probability that a randomly chosen seed will germinate.
- Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates.
- A die is throw and a card is selected at random from a deck of 52 playing cards. Then find the probability of getting an even number on the die and a spade card.

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**Class 11 Applied Mathematics Sample Paper ****Solution**

Not only question paper but you can also download complete step-by-step solutions to each question here. As this page shows only one model paper of class 11 Applied Maths, you should visit myCBSEguide app for more solved papers. CBSE Class 11 Applied Maths Sample Papers 2024 will definitely help you score better grades in your exams.

**Section A (solution)**- (a) P(A {tex}\cap{/tex} B {tex}\cap{/tex} C) = 0

**Explanation:**P(A {tex}\cap{/tex} B {tex}\cap{/tex} C) = 0 - (d) Geometric mean

**Explanation:**Geometric mean is used in construction of index numbers. - (c) interest on savings account

**Explanation:**Interest on savings account - (d) {tex}\frac {602}{233}{/tex}

**Explanation:**{tex}\frac {602}{233}{/tex} - (a) {tex}\phi {/tex}

**Explanation:**This function is defined only if

x – |x| > 0

{tex}\Rightarrow{/tex} x > |x|

Which is not possible because

|x| {tex}\ge{/tex} x

so, {tex}x \in \phi {/tex} - (d) 3

**Explanation:**3 - (a) {tex}\frac{3}{28}{/tex}

**Explanation:**Required probability = {tex}\frac{3}{8} \times \frac{2}{7}=\frac{3}{28}{/tex} - (d) x – 3y = 0

**Explanation:**The other tangent will be perpendicular to the given tangent 3x + y = 0

Therefore its equation is x – 3y + k = 0

Since it passes through the origin, k = 0

Hence the equation of the other tangent is x – 3y = 0 - (a) Nephew

**Explanation:**Nephew - (d) 55, 10

**Explanation:**5 is added to 100 observation each.

So 5 {tex}\times{/tex} 100 = 500 is added.

{tex}\bar{x}=\frac{500}{100}{/tex} = 5

So mean increased by 5 hence, {tex}\bar{x}{/tex} becomes 55.

Standard deviation will remain same because every observation increased by 5 and mean also increased by 5.

So S.D = 10

i.e. Mean = 55

S.D = 10 - (a) 1.8687

**Explanation:**1.8687 - (d) 10 years

**Explanation:**10 years - (c) ₹ 2880

**Explanation:**₹ 2880 - (b) {tex}\frac{5}{42}{/tex}

**Explanation:**{tex}\frac{5}{42}{/tex} - (d) {tex}\frac{2}{3}{/tex}

**Explanation:**Let S be the sample space.

{tex}\therefore{/tex} n(S) = 36- even number on the first die
- number on the second die is greater that 4

{tex}\therefore{/tex} n(A) = 18, n(B) = 12,

P(A) = {tex}\frac{18}{36}=\frac{1}{2}{/tex} and P(B) = {tex}\frac{12}{36}=\frac{1}{3}{/tex}

Also, A {tex}\cap{/tex} B = {(2, 5), (2, 6), (4, 5), (4, 6), (6, 5), (6, 6)}

P(A {tex}\cap{/tex} B) = {tex}\frac{6}{36}=\frac{1}{6}{/tex}

P(A {tex}\cup{/tex} B) = P(A) + P(B) – P(A {tex}\cap{/tex} B)

{tex}=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}{/tex}

{tex}=\frac{3+2-1}{6}=\frac{4}{6}=\frac{2}{3}{/tex} - (d) ₹ 77594.56

**Explanation:**₹ 77594.56 - (d) 120

**Explanation:**Number of commottes that can be formed = {tex}^{6} C_{3} \times^{4} C_{2}{/tex}

= {tex}\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}{/tex}

= {tex}\frac{6 \times 5 \times 4}{3 \times 2} \times \frac{4 \times 3}{2}{/tex}

= 120. - (a) Empty

**Explanation:**Because, the relation is defined over the set A be the set of all students of a boys school. - (c) A is true but R is false.

**Explanation:****Assertion**Mean of the given series

{tex}\bar{x} =\frac{\text { Sum of terms }}{\text { Number of terms }}=\frac{\sum x_i}{n}{/tex}

{tex}=\frac{4+7+8+9+10+12+13+17}{8}{/tex} = 10xi |xi – {tex}\bar x{/tex}| 4 |4 – 10| = 6 7 |7 – 10| = 3 8 |8 – 10| = 2 9 |9 – 10| = 1 10 |10 – 10| = 0 12 |12 – 10| = 2 13 |13 – 10| = 3 17 |17 – 10| = 7 {tex}\sum x_i{/tex} = 80 {tex}\sum \left|x_i-\bar{x}\right|{/tex} = 24 {tex}\therefore{/tex} Mean deviation about mean

{tex}=\frac{\Sigma\left|x_i-\bar{x}\right|}{n}=\frac{24}{8}{/tex} = 3

**Reason**Mean of the given series

{tex}\bar{x}= \frac{\text { Sum of terms }}{\text { Number of terms }}=\frac{\sum x_i}{n}{/tex}

{tex}= \frac{38+70+48+40+42+55}{+63+46+54+44}{/tex} = 50

{tex}\therefore{/tex} Mean deviation about mean

{tex}= \frac{\Sigma\left|x_i-\bar{x}\right|}{n}{/tex}

{tex}=\frac{84}{10}{/tex} = 8.4

Hence, Assertion is true and Reason is false. - (b) Both A and R are true but R is not the correct explanation of A.

**Explanation:**Both A and R are true but R is not the correct explanation of A. **Section B (solution)**- At 6 o’clock, the hour hand is at 6 and the minute hand is at 12. It means the angle between the two hands of the clock is 180°.

To catch the hour hand, the minute hand has to cover a relative distance of 180° at a relative speed of 5.5° per minute.

The time required to coincide the hands = {tex}\frac{180}{5.5}{/tex} minutes = {tex}\frac{180 \ \times \ 2}{11}{/tex} minutes

= {tex}32 \frac{8}{11}{/tex} minutes

= 32 min 44 sec

Hence, the hands of the clock coincide at 6:32:44 - A and B are brothers

C and D are sisters

A’s son is D’s brother

{tex}\therefore{/tex} B is C’s uncle**OR**Here R is coded as 22 which is its actual positions 18 + 4.

Similarly, A is coded as 1 + 4 i.e. 5.

H is coded as 8 + 4 = 12, U is coded as 21 + 4 = 25 and L is coded as 12 + 4 = 16.

V is equivalent to 22 + 4 = 26

I is equivalent to 9 + 4 = 13

R is equivalent to 18 + 4 = 22

A is equivalent to 1 + 4 = 5

and T is equivalent to 20 + 4 = 24

{tex}\therefore{/tex} ‘VIRAT’ will be coded as 26 – 13 – 22 – 5 – 24 - At 4 o’clock, the hour hand is at 4 and the minute hand is at 12. It means the angle between the two hands of the clock is 120°.

The hands will be at the right angles twice between 4:00 and 5:00. First time while minute hand approaching towards hour hand i.e. near 1 and the second time after crossing the hour hand i.e. near 7

**For the first time:**The minute hand has to cover a relative distance of 30°

So, the time required = {tex}\frac{30}{5.5}{/tex} minutes = {tex}\frac{30 \ \times \ 2}{11}{/tex} minutes = {tex}5\frac{5}{11}{/tex} minutes = 5 minutes 27 seconds.

**For the second time:**The minute hand has to cover a relative distance of 210°

So, the time required = {tex}\frac{210}{5.5}=\frac{210 \ \times \ 2}{11}{/tex} minutes = {tex}38 \frac{2}{11}{/tex} minutes = 38 minutes 11 seconds.

Hence, the hands of the clock are at the right angles at 4:05:27 and 4:38:11 - Given x
^{y}= e^{x-y}, taking logarithm of both sides, we get

y log x = (x – y) log e = (x – y) {tex}\cdot{/tex} 1 = x – y

{tex}\Rightarrow{/tex} y + y log x = x

{tex}\Rightarrow{/tex} (1 + log x) y = x

{tex}\Rightarrow{/tex} y = {tex}\frac{x}{1+\log x}{/tex}, differentiating w.r.t. x, we get{tex}\frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x \cdot\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}}=\frac{1+\log x-1}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}{/tex}**OR**Let y = log

_{x}3

{tex}\Rightarrow{/tex} y = {tex}\frac{\log 3}{\log x}{/tex} [{tex}\because{/tex} log_{a}b = {tex}\frac{\log b}{\log a}{/tex}] Differentiate it with respect to x, we get,

{tex}\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log 3}{\log x}\right){/tex}

= log 3{tex}\frac{d}{d x}{/tex}(log x)^{-1}

= log 3 {tex} \times {/tex} [-1(log x)^{-2}] {tex}\frac{d}{d x}{/tex}(log x) [using chain rule] = {tex}-\frac{\log 3}{(\log x)^{2}} \times \frac{1}{x}{/tex}

= {tex}-\left(\frac{\log 3}{\log x}\right)^{2} \times \frac{1}{x} \times \frac{1}{\log 3}{/tex}

= {tex}-\frac{1}{x \log 3\left(\log _{3} x\right)^{2}}\left[\because \frac{\log b}{\log a}=\log _{a} b\right]{/tex}

So, {tex}\frac{d}{d x}{/tex}(log_{x}3) = –{tex}\frac{1}{x \log 3\left(\log _{3} x\right)^{2}} {/tex} - Let us convert 15 and 0.6875 into their binary equivalents separately as given below.

0.6875 {tex}\times{/tex} 2 = 0.375 with a carry of 1

0.3750 {tex}\times{/tex} 2 = 0.750 with a carry of 0

0.750 {tex}\times{/tex} 2 = 0.50 with a carry of 1

0.50 {tex}\times{/tex} 2 = 0.00 with a carry of 1

{tex}\therefore{/tex} (15)_{10}= (1111)_{2}and (0.6875)_{10}= (0.1011)_{2}

Hence, (15.6875)_{10}= (1111.1011)_{2} **Section C (solution)**- Given a, b, c are in A.P

{tex}\Rightarrow{/tex} {tex}\frac{a}{a b c}, \frac{b}{a b c}, \frac{c}{a b c}{/tex} are also in A.P

[On dividing each term by abc] {tex}\Rightarrow{/tex} {tex}\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}{/tex} are also in A.P. - b + c, c + a, a + b will be in A.P.

if (c + a ) – (b + c) = (a + b) – (c + a)

i.e. if a – b = b – c

i.e. if 2b = a + c

i.e. if a, b, c are in A.P.

Thus, a, b, c are in A.P. {tex}\Rightarrow{/tex} b + c, c + a, a + b are in A.P.

**OR**Let three numbers in G.P. be {tex}\frac ar{/tex}, a, ar

{tex}\therefore{/tex} Their product = {tex}\frac ar \cdot{/tex}a{tex}\cdot{/tex}ar = 216 (given)

{tex}\Rightarrow{/tex} a^{3}= 216 = (6)^{3}{tex}\Rightarrow{/tex} a = 6.

Also sum of their products in pairs = 156 (given)

{tex}\Rightarrow \frac{a}{r} \cdot {/tex} a + a {tex}\cdot{/tex} ar + ar {tex}\cdot \frac{a}{r}{/tex} = 156

{tex}\Rightarrow a^{2}\left(\frac{1}{r}+r+1\right){/tex} = 156

{tex}\Rightarrow6^{2} \cdot \frac{1 + r^{2} + r}{r}{/tex} = 156

{tex}\Rightarrow 3 \cdot \frac{r^{2} + r +1}{r}{/tex} = 13

{tex}\Rightarrow{/tex} 3r^{2}+ 3r + 3 = 13r

{tex}\Rightarrow{/tex} 3r^{2}– 10r + 3 = 0

{tex}\Rightarrow{/tex} (r – 3){tex}\left(r-\frac{1}{3}\right){/tex} = 0 {tex}\Rightarrow{/tex} r = 3, {tex}\frac{1}{3}{/tex}

When r = 3, numbers are 2, 6,18 and when r = {tex}\frac 13{/tex}, numbers are 18, 6, 2- Given a, b, c are in A.P
- As for a job, a person first see an advertisement, then fill application form and go for interview. If he/she is selected, got appointment letter and final stage is of probation.

Hence, the Correct arrangement of given words is 5, 6, 2, 3, 4, 1. - All the given words represent substances which can be arranged in the increasing order of their cost. The least costly is sand after which comes the cost of iron, followed by gold, diamond and costliest among all is platinum. Hence, the Correct arrangement of given words is 3, 2, 1, 5, 4.

- As for a job, a person first see an advertisement, then fill application form and go for interview. If he/she is selected, got appointment letter and final stage is of probation.
- Given, f : R {tex}\rightarrow{/tex} R; f(x) = 2 {tex}\forall{/tex} x {tex}\in{/tex} R

Domain = R and Range = {2} - S.I. for 2 years = ₹ 400

{tex}\therefore{/tex} S.I. for 1 year = ₹ {tex}\frac{400}{2}{/tex} = ₹ 200

{tex}\therefore{/tex} C.I. for first-year = ₹ 200 ({tex}\because{/tex} for first year, C.I. = S.I.)

Given, C.I. for 2 years = ₹ 410

C.I. for the second year = ₹ 410 – ₹ 200 = ₹ 210

Difference of interests = ₹ 210 – ₹ 200 = ₹ 10

{tex}\Rightarrow{/tex} ₹ 10 is the interest on ₹ 200 for one year.

Rate of interest = {tex}\frac{10 \ \times \ 100}{200 \ \times \ 1}{/tex} = 5% {tex}(R=\frac{\text { S.I. } \ \times \ 100}{P \ \times \ T}){/tex}

Also P = {tex}\frac{\text { S.I. } \ \times \ 100}{R \ \times \ T} \Rightarrow P=₹ \frac{400 \ \times \ 100}{5 \ \times \ 2} \Rightarrow{/tex} ₹ 4000

The sum is ₹ 4000 - Volumetric Charge for consumption upto 20kl = ₹ 20 {tex}\times{/tex} 5.27 = ₹ 105.4

Volumetric Charge for consumption between 20-30 kl = ₹ 10 {tex}\times{/tex} 26.36 = ₹ 263.6

Volumetric Charge for consumption between 30-40 kl = ₹ 10 {tex}\times{/tex} 43.93 = ₹ 439.3

Total volumetric Charge for consumption of 40 kl = ₹(105.4 + 263.6 + 439.3) = ₹ 808.3

Service Charge = ₹ 292.82

Sewage Charges = 60% of Volumetric Charges

= 808.3 {tex}\times{/tex} 60% = ₹ 484.98

Amount of water bill for the given month = ₹(808.3 + 292.82 + 484.98) = ₹ 1586.1

Thus, amount of domestic water bill is ₹ 1586. - (A {tex}\cup{/tex} B) {tex}\cap{/tex} {tex}(A \cap B^\prime){/tex} = A

L.H.S. = (A {tex}\cup{/tex} B) {tex}\cap{/tex} {tex}(A \cap B^\prime){/tex}

= A {tex}\cup{/tex} {tex}(B \cap B^\prime){/tex} (By distributive law)

= A {tex}\cup{/tex} {tex}\phi{/tex} ({tex}\therefore{/tex} B {tex}\cap{/tex} {tex}B^\prime{/tex} = {tex}\phi{/tex})

= A

= R.H.S. Hence proved - A – (A {tex}\cap{/tex} B) = A – B

L.H.S. = A – (A {tex}\cap{/tex} B)

= A {tex}\cap{/tex} {tex}(A \cap B)^\prime{/tex} [{tex}\therefore{/tex} A – B = A {tex}\cap{/tex} {tex}B^\prime{/tex}] = A {tex}\cap{/tex} {tex}(A^\prime\cup B^\prime){/tex} (By Demorgan’s law)

= {tex}(A\cap A^\prime){/tex} {tex}\cup{/tex} (A {tex}\cap{/tex} B) (By distributive law)

= {tex}\phi{/tex} {tex}\cup{/tex} A {tex}\cap{/tex} {tex}B^\prime{/tex} ({tex}\therefore{/tex} A {tex}\cap{/tex} {tex}A^\prime{/tex} = {tex}\phi{/tex})

= A {tex}\cap{/tex} {tex}B^\prime{/tex}

= A – B

= R.H.S. Hence proved

- (A {tex}\cup{/tex} B) {tex}\cap{/tex} {tex}(A \cap B^\prime){/tex} = A
**Section D (solution)**- Here {tex}^{2n}{C_3}{:^n}{C_2} = 12:1{/tex}

{tex} \Rightarrow \frac{{(2n)!}}{{3!(2n – 3)!}} \times \frac{{2!(n – 2)!}}{{n!}} = \frac{{12}}{1}{/tex}

{tex} \Rightarrow \frac{{(2n)(2n – 1)(2n – 2)(2n – 3)!}}{{3 \times 2!(2n – 3)!}}{/tex}{tex} \times \frac{{2!(n – 2)!}}{{n(n – 1)(n – 2)!}} = \frac{12}{1}{/tex}

{tex} \Rightarrow \frac{{(2n)(2n – 1)(2n – 2)}}{3} \times \frac{1}{{n(n – 1)}} = \frac{{12}}{1}{/tex}.

{tex} \Rightarrow \frac{{4(2n – 1)}}{3} = \frac{{12}}{1}{/tex}{tex} \Rightarrow 8n – 4 = 36 \Rightarrow n = 5{/tex}

(ii) Here {tex}^{2n}{C_3}{:^n}{C_3} = 11:1{/tex}

{tex} \Rightarrow \frac{{(2n)!}}{{3!(2n – 3)!}} \times \frac{{3!(n – 3)!}}{{n!}} = \frac{{11}}{1}{/tex}

{tex} \Rightarrow \frac{{(2n)(2n – 1)(2n – 2)(2n – 3)!}}{{3!(2n – 3)!}}{/tex}{tex} \times \frac{{3!(n – 3)!}}{{n(n – 1)(n – 2)(n – 3)!}} = \frac{{11}}{1}{/tex}

{tex} \Rightarrow \frac{{(2n)(2n – 1)(2n – 2)}}{{n(n – 1)(n – 2)}} = \frac{{11}}{1}{/tex}

{tex} \Rightarrow \frac{{4(2n – 1)}}{{n – 2}} = \frac{{11}}{1}{/tex}{tex} \Rightarrow 8n – 4 = 11n – 22{/tex}

{tex} \Rightarrow {/tex} 3n = 18 {tex} \Rightarrow {/tex} n = 6**OR**Given:

^{n}P_{4}= 360

To find: value of n

^{n}P_{4}can be written as P(n, 4)

We know,

P(n, r) = {tex}\frac{n !}{(n-r) !}{/tex}

P(n, 4) = {tex}\frac{n !}{(n-4) !}{/tex}

So, according to question:

^{n}P_{4}= P(n, 4) = 360

{tex}\Rightarrow \frac{n !}{(n-4) !}{/tex} = 360

{tex}\Rightarrow \frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}{/tex} = 360

{tex}\Rightarrow{/tex} n (n – 1) (n – 2) (n – 3) = 360

{tex}\Rightarrow{/tex} n (n – 1) (n – 2) (n – 3) = 6 {tex}\times{/tex} 5 {tex}\times{/tex} 4 {tex}\times{/tex} 3

On comparing:

The value of n = 6 - {tex}\lim \limits_{x \rightarrow \infty}(\sqrt{4 x^{2}-7 x}+2 x){/tex}

Let x = -m, When {tex}n \rightarrow-\infty{/tex}, then {tex}m \rightarrow \infty{/tex}

{tex}\Rightarrow \lim \limits_{m \rightarrow \infty}[\sqrt{4 m^{2}+7 m}-2 m]{/tex}

{tex}=\lim \limits_{m \rightarrow \infty}[(\sqrt{4 m^{2}=7 m}-2 m){/tex} {tex}\left.\times \frac{(\sqrt{4 m^{2}+7 m}+2 m)}{(\sqrt{4 m^{2}+7 m}+2 m)}\right]{/tex}

{tex}=\lim \limits_{m \rightarrow \infty}\left[\frac{\left(4 m^{2}+7 m\right)-(2 m)^{2}}{\sqrt{4 m^{2}+7 m}+2 m}\right]{/tex}

{tex}=\lim \limits_{m \rightarrow \infty}\left[\frac{4 m^{2}+7 m-4 m^{2}}{\sqrt{4 m^{2}+7 m}+2 m}\right]{/tex}

Dividing the numerator and the denominator by m, we get,

{tex}\lim \limits_{m \rightarrow \infty}\left[\frac{7}{\sqrt{\frac{4 m^{2}+7 m}{m^{2}}}+\frac{2 m}{m}}\right]{/tex}

{tex}=\lim\limits _{m \rightarrow \infty}\left[\frac{7}{\sqrt{\frac{4 m^{2}}{m^{2}}+\frac{7 m}{m^{2}}}+2}\right]{/tex}

{tex}=\lim\limits_{m \rightarrow \infty}\left[\frac{7}{\sqrt{4+\frac{7}{m}}+2}\right]{/tex}

As {tex}n \rightarrow \infty, \frac{1}{m} \rightarrow 0{/tex}

{tex}=\frac{7}{\sqrt{4}+2}{/tex}

{tex}=\frac{7}{4}{/tex} - {tex}\bar x = \frac{{65 + 68 + 58 + 44 + 48 + 45 + 60 + 62 + 60 + 50}}{{10}}{/tex}

= {tex}\frac{{560}}{{10}}{/tex} = 56x x – 56 (x – 56) ^{2}65 9 81 58 2 4 68 12 144 44 – 12 144 48 – 8 64 45 – 11 121 60 4 16 62 6 36 60 4 16 50 – 6 36 662 Now variance = {tex}\frac{1}{n}{\left( {\Sigma x – \bar x} \right)^2} = \frac{{662}}{{10}}{/tex} = 66.2

S.D. = {tex}\sqrt {{\text{variance}}} = \sqrt {66.2}{/tex} = 8.13**OR**Here h = 10 and A = 55

Age Mid value f {tex}u = \frac{{x – A}}{h}{/tex} fu fu ^{2}20 – 30 25 3 – 3 – 9 27 30 – 40 35 51 – 2 – 102 204 40 – 50 45 122 – 1 – 122 122 50 – 60 55 141 0 0 0 60 – 70 65 130 1 130 130 70 – 80 75 51 2 102 204 80 – 90 85 2 3 6 18 **500****5****705**Here {tex}\bar x = A + \frac{{\Sigma fu}}{N} \times h = 55 + \frac{5}{{500}} \times 10 = 55.1{/tex}

{tex}{\sigma ^2}=\left[ {\frac{1}{N}\Sigma f{u^2} – {{\left( {\frac{1}{N}\Sigma fu} \right)}^2}} \right] \times {h^2}{/tex}

{tex} = \left[ {\frac{{705}}{{500}} – {{\left( {\frac{5}{{500}}} \right)}^2}} \right] \times 100 = 100\left[ {\frac{{705}}{{500}} – \frac{1}{{10000}}} \right]{/tex}

{tex}= 100\left[ {\frac{{14100 – 1}}{{10000}}} \right] = \frac{{100 \times 14099}}{{10000}}{/tex} {tex}= \frac{{14099}}{{100}} = 140.99{/tex}, which gives,

{tex} \sigma = \sqrt {140.99} = 11.87{/tex}. - Electricity consumed = 1264 unit (Kwh)
Unit Range Unit (KWh) Rate (₹/Kwh) Charge (₹) 0-50 50 4.05 202.5 51-100 50 4.95 247.5 101-300 200 6.13 1226 >300 964 6.50 6266 Total Electricity Charges (₹) 1264 7942 Fixed Charges (₹ 250/KW) = 250 {tex}\times{/tex} 5 = 1250

Energy duty (₹ 63/ Unit) = 0.63 {tex}\times{/tex} 1264 = 769.32

Final Bill = 1250 + 769.32 = 9961.32 **Section E (solution)**- Equation of line AB is,

x – 2y = 1801

Putting y = 110,

{tex}\therefore{/tex} x = 1801 + 220

{tex}\Rightarrow{/tex} x = 2021 - {tex}\because{/tex} Slope of AB = {tex}\frac{1}{2}{/tex}

Slope of the perpendicular of AB = {tex}\frac{-1}{\frac{1}{2}}{/tex} = -2

{tex}\therefore{/tex} Equation of line perpendicular to AB passing through (1995, 97) is

{tex}\Rightarrow{/tex} y – 97 = -2(x – 1995)

{tex}\Rightarrow{/tex} y – 97 = -2x + 3990

{tex}\Rightarrow{/tex} 2x + y = 4087 - Equation of line AB is,

{tex}\Rightarrow{/tex} y – y_{1}= m(x – x_{1})

{tex}\therefore{/tex} y – 92 = {tex}\frac{1}{2}{/tex}(x – 1985)

2y – 184 = x – 1985

{tex}\Rightarrow{/tex} x – 2y = 1801 **OR**Slope of line AB joining points A(1985, 92) and B(1995, 97)

m = {tex}\frac{97-92}{1995-1985}{/tex} = {tex}\frac{5}{10} = \frac{1}{2}{/tex}

- Equation of line AB is,
- Mean = 40, n = 100, sum = 100 {tex}\times{/tex} 40 = 4000
- Corrected sum = 4000 – 50 + 40 = 3990
- {tex}\sigma^2{/tex} = {tex}\frac{\Sigma x^2}{n}{/tex} – {tex}\left(\frac{\Sigma x}{n}\right)^2{/tex}

{tex}\Rightarrow{/tex} (5.1)^{2}= {tex}\frac{\Sigma x^2}{100}-(40)^2{/tex}

{tex}\Rightarrow{/tex} (26.01 + 1600)100 = {tex}\Sigma x^2{/tex} **OR**Corrected {tex}\Sigma x^2{/tex} = 162601 – (50)

^{2}+ (40)^{2}

= 162601 – 2500 + 1600 = 161701

Corrected {tex}\sigma^2{/tex} = {tex}\frac{161701}{100}-(39.9)^2{/tex}

= 1617.01 – 1592.01 = 25

- P (Grade A in Maths) = P(M) = 0.2

P (Grade A in Physics) = P(P) = 0.3

P (Grade A in Chemistry) = P(C) = 0.5

P(not A garde in Maths) = P({tex}\overline M{/tex}) = 1 – 0.2 = 0.8

P(not A garde in Physics) = P({tex}\overline P{/tex}) = 1 – 0.3 = 0.7

P(not A garde in Chemistry) = P({tex}\overline C{/tex}) = 1 – 0.5 = 0.5

P(getting grade A in all subjects) = P(M {tex}\cap P \cap{/tex} C)

= P(M) {tex}\times{/tex} P(P) {tex}\times{/tex} P(C)

= 0.2 {tex}\times{/tex} 0.3 {tex}\times{/tex} 0.5 = 0.03 - P (Grade A in Maths) = P(M) = 0.2

P (Grade A in Physics) = P(P) = 0.3

P (Grade A in Chemistry) = P(C) = 0.5

P(not A garde in Maths) = P({tex}\overline M{/tex}) = 1 – 0.2 = 0.8

P(not A garde in Physics) = P({tex}\overline P{/tex}) = 1 – 0.3 = 0.7

P(not A garde in Chemistry) = P({tex}\overline C{/tex}) = 1 – 0.5 = 0.5

P(getting grade A in on subjects) = {tex}P(\overline M \cap \overline P \cap \overline C){/tex}

= {tex}P(\overline M) \times P(\overline P) \times P(\overline C){/tex}

= 0.8 {tex}\times{/tex} 0.7 {tex}\times{/tex} 0.5 = 0.280 - P (Grade A in Maths) = P(M) = 0.2

P (Grade A in Physics) = P(P) = 0.3

P (Grade A in Chemistry) = P(C) = 0.5

P(not A garde in Maths) = P({tex}\overline M{/tex}) = 1 – 0.2 = 0.8

P(not A garde in Physics) = P({tex}\overline P{/tex}) = 1 – 0.3 = 0.7

P(not A garde in Chemistry) = P({tex}\overline C{/tex}) = 1 – 0.5 = 0.5

P(getting grade A in 2 subjects)

{tex}\Rightarrow{/tex} P(grade A in M and P not in C) + P(grade A in P & C not in M) + P(grade A in M & C not in P)

{tex}\Rightarrow{/tex} P{tex}\left( {M \cap P \cap \overline C } \right){/tex} + P{tex}\left( {P \cap C \cap \overline M } \right){/tex} + P{tex}(M \cap C \cap \overline P ){/tex}

{tex}\Rightarrow{/tex} 0.2 {tex}\times 0.3 \times {/tex}0.5 + 0.3 {tex}\times 0.5 \times{/tex} 0.8 + 0.2 {tex}\times 0.5 \times{/tex} 0.7 = 0.03 + 0.12 + 0.07

P(getting grade A in 2 subjects ) = 0.22

**OR**

Here, P(E_{1}) = {tex}\frac{4}{10}{/tex}, P(E_{2}) = {tex}\frac{4}{10}{/tex}, P(E_{3}) = {tex}\frac{2}{10}{/tex}

{tex}P\left(\frac{A}{E_1}\right){/tex} = {tex}\frac{45}{100}{/tex}, {tex}P\left(\frac{A}{E_2}\right){/tex} = {tex}\frac{60}{100}{/tex}, {tex}P\left(\frac{A}{E_3}\right){/tex} = {tex}\frac{35}{100}{/tex}

{tex}\therefore {/tex} P(A) = P(E_{1}) {tex}\cdot {/tex} P{tex}\left(\frac{A}{E_1}\right){/tex} + P(E_{2}) {tex}\cdot {/tex} {tex}P\left(\frac{A}{E_2}\right){/tex} + P(E_{3}) {tex}\cdot {/tex} {tex}P\left(\frac{A}{E_3}\right){/tex}

= {tex}\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}{/tex}

= {tex}\frac{180}{1000}+\frac{240}{1000}+\frac{70}{100}{/tex}

= {tex}\frac{490}{1000}{/tex} = 4.9- Required probability = {tex}P\left(\frac{E_2}{A}\right){/tex}

= {tex}\frac{P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)}{P(A)}{/tex}

= {tex}\frac{\frac{4}{10} \times \frac{60}{100}}{\frac{490}{1000}}{/tex}

= {tex}\frac{240}{490}{/tex} = {tex}\frac{24}{49}{/tex} - Let,

E_{1}= Event for getting an even number on die and

E_{2}= Event that a spade card is selected

{tex}\therefore {/tex} P(E_{1}) = {tex}\frac{3}{6}{/tex}

= {tex}\frac{1}{2}{/tex}

and P(E_{2}) = {tex}\frac{13}{52} = \frac 14{/tex}

Then, P(E_{1}{tex}\cap{/tex} E_{2}) = P(E_{1}) {tex}\cdot{/tex} P(E_{2})

= {tex}\frac 12{/tex}, {tex}\frac 14{/tex} = {tex}\frac 18{/tex}

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- P (Grade A in Maths) = P(M) = 0.2

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