CBSE Class 11 Applied Mathematics — NCERT Solutions

Method: For a binary number $(x_n x_{n-1}\dots x_1 x_0)_2$, the decimal value is $N = x_0\cdot2^0 + x_1\cdot2^1 + \cdots + x_n\cdot2^n$.

(i) $(1010101100110)_2$

Label bits from right (position 0):
$$1\cdot2^{12}+0\cdot2^{11}+1\cdot2^{10}+0\cdot2^9+1\cdot2^8+0\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 4096+1024+256+64+32+4+2$$
$$= \boxed{5478}$$

(ii) $(101011000110)_2$

$$1\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+0\cdot2^8+1\cdot2^7+1\cdot2^6+0\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 2048+512+128+64+4+2$$
$$= \boxed{2758}$$

(iii) $(101111100110)_2$

$$1\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+1\cdot2^8+1\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 2048+512+256+128+64+32+4+2$$
$$= \boxed{3046}$$

(iv) $(1000000000110)_2$

$$1\cdot2^{12}+0+\cdots+0+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 4096+4+2$$
$$= \boxed{4102}$$

Method: Repeatedly divide by 2 and record remainders; the binary representation is the remainders read from bottom to top.

(i) $654321$

Successive divisions by 2:
$$654321 = 2\times327160+1$$
$$327160 = 2\times163580+0$$
$$163580 = 2\times81790+0$$
$$81790 = 2\times40895+0$$
$$40895 = 2\times20447+1$$
$$20447 = 2\times10223+1$$
$$10223 = 2\times5111+1$$
$$5111 = 2\times2555+1$$
$$2555 = 2\times1277+1$$
$$1277 = 2\times638+1$$
$$638 = 2\times319+0$$
$$319 = 2\times159+1$$
$$159 = 2\times79+1$$
$$79 = 2\times39+1$$
$$39 = 2\times19+1$$
$$19 = 2\times9+1$$
$$9 = 2\times4+1$$
$$4 = 2\times2+0$$
$$2 = 2\times1+0$$
$$1 = 2\times0+1$$

Reading remainders from bottom to top:
$$654321 = \boxed{(10011111101111110001)_2}$$

(ii) $1000001$

Note: $2^{19} = 524288$, $2^{20} = 1048576 > 1000001$.

Successive divisions:
$$1000001 \div 2: \text{remainder sequence (bottom to top gives binary)}$$

Performing repeated division:
$1000001 = 2(500000)+1$
$500000 = 2(250000)+0$
$250000 = 2(125000)+0$
$125000 = 2(62500)+0$
$62500 = 2(31250)+0$
$31250 = 2(15625)+0$
$15625 = 2(7812)+1$
$7812 = 2(3906)+0$
$3906 = 2(1953)+0$
$1953 = 2(976)+1$
$976 = 2(488)+0$
$488 = 2(244)+0$
$244 = 2(122)+0$
$122 = 2(61)+0$
$61 = 2(30)+1$
$30 = 2(15)+0$
$15 = 2(7)+1$
$7 = 2(3)+1$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$1000001 = \boxed{(11110100001001000001)_2}$$

(iii) $56237801$

Performing repeated division by 2:
$56237801 = 2(28118900)+1$
$28118900 = 2(14059450)+0$
$14059450 = 2(7029725)+0$
$7029725 = 2(3514862)+1$
$3514862 = 2(1757431)+0$
$1757431 = 2(878715)+1$
$878715 = 2(439357)+1$
$439357 = 2(219678)+1$
$219678 = 2(109839)+0$
$109839 = 2(54919)+1$
$54919 = 2(27459)+1$
$27459 = 2(13729)+1$
$13729 = 2(6864)+1$
$6864 = 2(3432)+0$
$3432 = 2(1716)+0$
$1716 = 2(858)+0$
$858 = 2(429)+0$
$429 = 2(214)+1$
$214 = 2(107)+0$
$107 = 2(53)+1$
$53 = 2(26)+1$
$26 = 2(13)+0$
$13 = 2(6)+1$
$6 = 2(3)+0$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$56237801 = \boxed{(11010110000111101101001)_2}$$

(iv) $2468097531$

Performing repeated division by 2:
$2468097531 = 2(1234048765)+1$
$1234048765 = 2(617024382)+1$
$617024382 = 2(308512191)+0$
$308512191 = 2(154256095)+1$
$154256095 = 2(77128047)+1$
$77128047 = 2(38564023)+1$
$38564023 = 2(19282011)+1$
$19282011 = 2(9641005)+1$
$9641005 = 2(4820502)+1$
$4820502 = 2(2410251)+0$
$2410251 = 2(1205125)+1$
$1205125 = 2(602562)+1$
$602562 = 2(301281)+0$
$301281 = 2(150640)+1$
$150640 = 2(75320)+0$
$75320 = 2(37660)+0$
$37660 = 2(18830)+0$
$18830 = 2(9415)+0$
$9415 = 2(4707)+1$
$4707 = 2(2353)+1$
$2353 = 2(1176)+1$
$1176 = 2(588)+0$
$588 = 2(294)+0$
$294 = 2(147)+0$
$147 = 2(73)+1$
$73 = 2(36)+1$
$36 = 2(18)+0$
$18 = 2(9)+0$
$9 = 2(4)+1$
$4 = 2(2)+0$
$2 = 2(1)+0$
$1 = 2(0)+1$

Reading from bottom to top:
$$2468097531 = \boxed{(10010011000010110111110111011)_2}$$

(v) $963258741$

Performing repeated division by 2:
$963258741 = 2(481629370)+1$
$481629370 = 2(240814685)+0$
$240814685 = 2(120407342)+1$
$120407342 = 2(60203671)+0$
$60203671 = 2(30101835)+1$
$30101835 = 2(15050917)+1$
$15050917 = 2(7525458)+1$
$7525458 = 2(3762729)+0$
$3762729 = 2(1881364)+1$
$1881364 = 2(940682)+0$
$940682 = 2(470341)+0$
$470341 = 2(235170)+1$
$235170 = 2(117585)+0$
$117585 = 2(58792)+1$
$58792 = 2(29396)+0$
$29396 = 2(14698)+0$
$14698 = 2(7349)+0$
$7349 = 2(3674)+1$
$3674 = 2(1837)+0$
$1837 = 2(918)+1$
$918 = 2(459)+0$
$459 = 2(229)+1$
$229 = 2(114)+1$
$114 = 2(57)+0$
$57 = 2(28)+1$
$28 = 2(14)+0$
$14 = 2(7)+0$
$7 = 2(3)+1$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$963258741 = \boxed{(111001101011010001001011101101)_2}$$

Given:

• Number of boys = 44, Average marks of boys = 40
• Number of girls = 36, Average marks of girls = 38

Formula used (Weighted Average):
$$\bar{x}_w = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

Working:
$$\text{Total marks of boys} = 44 \times 40 = 1760$$
$$\text{Total marks of girls} = 36 \times 38 = 1368$$
$$\text{Total students} = 44 + 36 = 80$$
$$\text{Average marks of class} = \frac{1760 + 1368}{80} = \frac{3128}{80} = 39.1$$

Answer: The average marks of the class = 39.1

Given:

• Average of 5 numbers = 87
• After excluding one number, average of remaining 4 numbers = 87 − 5 = 82

Working:
$$\text{Sum of 5 numbers} = 87 \times 5 = 435$$
$$\text{Sum of remaining 4 numbers} = 82 \times 4 = 328$$
$$\text{Excluded number} = 435 - 328 = 107$$

Answer: The excluded number = 107

A set is a well-defined collection of distinct objects. We check each collection for well-definedness.

(i) 'Most talented authors of India' is subjective and not well-defined. Not a set.

(ii) Months of a year beginning with 'M' are March and May — clearly defined. This is a set: {March, May}.

(iii) All integers from $-2$ to $20$ is clearly defined. This is a set: $\{-2, -1, 0, 1, 2, \ldots, 20\}$.

(iv) All even natural numbers is clearly defined. This is a set: $\{2, 4, 6, 8, \ldots\}$.

(v) 'Best tennis players of the world' is subjective and not well-defined. Not a set.

Conclusion: (ii), (iii), and (iv) are sets.

We identify the pattern in each set and express it using set-builder notation.

(i) Elements are multiples of 4: $4, 8, 12, 16, 20$, i.e., $4n$ for $n = 1, 2, 3, 4, 5$.
$$A = \{x : x = 4n,\ n \in \mathbb{N},\ n \leq 5\}$$

(ii) Elements are all prime numbers.
$$B = \{x : x \text{ is a prime number}\}$$

(iii) Elements are all consonants of the English alphabet.
$$C = \{x : x \text{ is a consonant of the English alphabet}\}$$

(iv) $D = \{-1, 1\}$. These are solutions of $x^2 - 1 = 0$.
$$D = \{x : x \text{ is a solution of } x^2 - 1 = 0\}$$

(v) $E = \{41, 43, 47\}$. These are prime numbers between 40 and 50.
$$E = \{x : x \text{ is a prime number between } 40 \text{ and } 50\}$$

(vi) Elements are $\frac{1}{n}$ for $n = 1, 2, 3, \ldots$
$$F = \left\{x : x = \frac{1}{n},\ n \in \mathbb{N}\right\}$$

(vii) Elements are $\frac{1}{n^2}$ for $n = 1, 2, 3, \ldots$
$$G = \left\{x : x = \frac{1}{n^2},\ n \in \mathbb{N}\right\}$$

Given: $S = \{1, 2, 3, 4, 5\}$, $R = \{(x, y) : x \mid y\}$.

First, list the ordered pairs in R:
$$R = \{(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,4),(3,3),(4,4),(5,5)\}$$

Reflexive: For every $x \in S$, $x$ divides $x$, so $(x, x) \in R$ for all $x \in S$. ✓ Hence R is reflexive.

Symmetric: Check whether $(x,y) \in R \Rightarrow (y,x) \in R$.
Counter-example: $(1, 2) \in R$ because 2 is divisible by 1, but $(2, 1) \notin R$ because 1 is not divisible by 2. ✗ Hence R is not symmetric.

Transitive: Suppose $(x, y) \in R$ and $(y, z) \in R$, i.e., $x \mid y$ and $y \mid z$.
Then $y = kx$ and $z = my$ for some integers $k, m$, so $z = mkx$, meaning $x \mid z$, i.e., $(x, z) \in R$. ✓ Hence R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

Given: $S_n = 2n - 3$

Concept: The $n^{\text{th}}$ term of a sequence is $a_n = S_n - S_{n-1}$.

Working:
$$a_n = S_n - S_{n-1} = (2n-3) - (2(n-1)-3) = (2n-3)-(2n-5) = 2$$

Since every term equals 2 (a constant), the common difference $d = a_n - a_{n-1} = 2$.

Answer: (b) 2

Given: $6!$

Formula: $n! = n \times (n-1) \times (n-2) \times \cdots \times 1$

Working:
$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$

Answer: $6! = 720$

Given sentence: "8 is less than 6."

Concept: A sentence is a mathematically acceptable statement if it is either true or false (but not both).

Working: The sentence "8 is less than 6" is false because $8 > 6$. Since it has a definite truth value (false), it qualifies as a statement.

Conclusion: Yes, it is a statement (a false statement).

Given: The table of values of $x$ and $y$.

Concept: A relation is a function if and only if every input (value of $x$) has exactly one output (value of $y$).

Working: Looking at the table, the input $x = 1$ appears twice with two different outputs: $y = 0$ and $y = 5$.

Since one input ($x = 1$) maps to two different outputs, this violates the definition of a function.

Conclusion: $y$ is NOT a function of $x$.

Given: We need an experiment whose sample space is the set of all non-negative integers $S = \{0, 1, 2, \ldots\}$.

Concept: The sample space lists all possible outcomes of a random experiment.

Answer: Observing the number of people who voted in a constituency is one such experiment. The number of voters can be 0, 1, 2, 3, … (any non-negative integer), so the sample space is $S = \{0, 1, 2, \ldots\}$.

Other valid examples: number of calls received at a call centre in a day, number of accidents on a highway in a week, etc.

Given: Two types of graphical representations — Bar Graph and Histogram.

Differences between Bar Graph and Histogram:

Conclusion: A bar graph is used for discrete/categorical data where gaps between bars indicate distinct categories, while a histogram is used for continuous frequency distributions where the area of each bar represents the frequency of that class interval.

Step 1: Calculate Gross Total Income

Step 2: Calculate Deductions under Section 80C

Maximum deduction allowed under Section 80C = ₹1,50,000

Since total = ₹1,50,000, deduction = ₹1,50,000

Step 3: Calculate Total Taxable Income

$$\text{Total Taxable Income} = \text{Gross Total Income} - \text{Deductions under 80C}$$

$$= ₹6,00,000 - ₹1,50,000 = ₹4,50,000$$

Step 4: Calculate Income Tax (as per Old Regime slabs)

Step 5: Add Health & Education Cess @ 4%

$$\text{Cess} = 4\% \times ₹10,000 = ₹400$$

Step 6: Total Income Tax Liability

$$\text{Total Tax Liability} = ₹10,000 + ₹400 = \boxed{₹10,400}$$

Note: A rebate under Section 87A of ₹12,500 is available if total taxable income does not exceed ₹5,00,000. Since taxable income here is ₹4,50,000 (≤ ₹5,00,000), the rebate applies and the tax payable becomes ₹NIL (tax of ₹10,000 < rebate limit of ₹12,500).

$$\therefore \text{Income Tax Liability of Mr X} = \mathbf{₹0}$$

Given: Two horizontal lines $y = 8$ and $y = -2$.

Concept: A line equidistant from two parallel lines lies exactly midway between them.

Working:
The required line is parallel to both given lines and passes through the midpoint of the perpendicular distance between them.

Midpoint of $y$-values:
$$y = \frac{8 + (-2)}{2} = \frac{6}{2} = 3$$

Answer: The required equation is $\boxed{y = 3}$.