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Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 7 Linear Inequalities will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 7 Linear Inequalities

Inequality:
A statement with symbols like >, ≥, <, ≤ is an inequality.

Different types of inequality:

(a) Numerical inequality: It is an inequality involving numbers not variables.
(b) Literal inequality: It is the inequality involving literal numbers(variable).
(c) Strict inequality: An inequality with only > or < symbols is a strict inequality.
(d) Slack inequality: An inequality with only ≥ or ≤ symbols is a slack inequality.

Linear inequality:
An inequality involving variables in the first degree is called linear inequalities.
(a) General form of inequalities:
(i) In one variable: ax + b > or ≥ or < or ≤ 0
(ii) In two variables: ax + by + c > or ≥ or < or ≤ 0.

Intervals:

• Closed Interval: [a, b] = {x ∈ R: a ≤ x ≤ b}
• Open Interval: (a, b) = {x ∈ R: a < x < b}
• Semi-open or semi-closed interval:
  ⇒ [a, b) = {x ∈ R: a ≤ x < b}
  ⇒ (a, b] = {x ∈ R: a < x ≤ b}

Basic properties of inequalities:
(1) a > b, b > c ⇒ a > c
(2) a > b ⇒ a ± c > b ± c
(3) a > b

• m > 0 ⇒ am > bm, \(\frac{a}{m}>\frac{b}{m}\)
• m < 0 ⇒ am < bm, \(\frac{a}{m}<\frac{b}{m}\)

(4) If a > b > 0, then
a² > b², |a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)
If a < b < 0, then
|a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)

Graphical solution of linear inequalities in two variables:
Working rule:

Let the inequality is ax + by + c < or ≤ or > or ≥ 0

Step – 1: Consider the equation ax + by + c = 0 in place of the inequality and draw its graph (Draw a dotted line for > or < and a bold line for ≥ or ≤).
Step – 2: Take any point that does not lie on the graph, and put the coordinate in the inequality.
If you get true then the inequality is satisfied. Shade the half-plane containing that point otherwise the inequality is not satisfied. In this case shade the half plane region that does not contain the point.
Step – 3: The shaded region is the required solution.

Solution of a system of linear inequalities in two variables:

Step – 1: Draw the graph of all lines.
Step – 2: Shade the appropriate region for each inequality.
Step – 3: The common region is the required solution.

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 5 ପରିସଂଖ୍ୟାନ Ex 5(a)

(କ – ବିଭାଗ )

Question 1.
ନିମ୍ନ ଉକ୍ତିମାନଙ୍କ ମଧ୍ୟରୁ ଯେଉଁଟି ଠିକ୍ ତା’ ପାଖରେ T ଓ ଯେଉଁଟି ଭୁଲ ତା’ ପାଖରେ F ଲେଖ ।
(i) ଦୁଇଟି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ ସେ ଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଯୁଗ୍ମସଂଖ୍ୟା ସଙ୍ଗେ ସମାନ ।
(ii) ଏକ ସମାନ୍ତର ପ୍ରଗତିରେ ଥିବା ତିନୋଟି କ୍ରମିକ ପଦର ମାଧ୍ଯମାନ ସେମାନଙ୍କର ମଧ୍ଯମପଦ ସଙ୍ଗେ ସମାନ ।
(iv) ଭିନ୍ନ ଭିନ୍ନ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ ଦତ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କଲେ ଭିନ୍ନ ଭିନ୍ନ ଉତ୍ତର ମିଳିବ ।
(v) କୌଣସି ତଥ୍ୟାବଳୀର ଆରମ୍ଭ ବିନ୍ଦୁ 20 ହେଲେ ଅନ୍ତର୍ଭୁକ୍ତ ଲବ୍‌ଧାଙ୍କ 15ର ବିଚ୍ୟୁତି 5 ।
(vi) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ \(\frac{n+2}{2}\)।
(vii) ପ୍ରଥମ n ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 2n + 2 ।
(viii) ପ୍ରଥମ ଦଶଗୋଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 10 ।
(ix) 15 ଗୋଟି ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 17 । ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ 2 ଦ୍ୱାରା ଗୁଣି ସେମାନଙ୍କର ମାଧ୍ଯମାନ ସ୍ଥିର କଲେ ମାଧ୍ୟମାନ 8.5 ହେବ ।
(x) ପ୍ରଥମ 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ, ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧମାନର ଦୁଇ ଗୁଣ ।
ଉ :
(i) ଦୁଇଟି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ ସେ ଦ୍ଵୟ ମଧ୍ୟବର୍ତ୍ତୀ ଯୁଗ୍ମସଂଖ୍ୟା ସଙ୍ଗେ ସମାନ । (T)
(ii) ଏକ ସମାନ୍ତର ପ୍ରଗତିରେ ଥିବା ତିନୋଟି କ୍ରମିକ ପଦର ମାଧ୍ଯମାନ ସେମାନଙ୍କର ମଧ୍ଯମପଦ ସଙ୍ଗେ ସମାନ । (T)
(iv) ଭିନ୍ନ ଭିନ୍ନ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ ଦତ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କଲେ ଭିନ୍ନ ଭିନ୍ନ ଉତ୍ତର ମିଳିବ । (T)
(v) କୌଣସି ତଥ୍ୟାବଳୀର ଆରମ୍ଭ ବିନ୍ଦୁ 20 ହେଲେ ଅନ୍ତର୍ଭୁକ୍ତ ଲବ୍‌ଧାଙ୍କ 15ର ବିଚ୍ୟୁତି 5 । (F)
(vi) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ \(\frac{n+1}{2}\)। (T)
(vii) ପ୍ରଥମ n ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 2n + 2 । (F)
(viii) ପ୍ରଥମ ଦଶଗୋଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 10 । (T)
(ix) 15 ଗୋଟି ସଂଖ୍ୟାର ମାଧ୍ୟମାନ 17 । ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ 2 ଦ୍ୱାରା ଗୁଣି ସେମାନଙ୍କର ମାଧ୍ଯମାନ ସ୍ଥିର କଲେ ମାଧ୍ୟମାନ 8.5 ହେବ । (F)
(x) ପ୍ରଥମ 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ, ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧମାନର ଦୁଇ ଗୁଣ । (F)

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) (T) (କାରଣ 3 ଓ 5ର ମାଧ୍ୟମାନ \(\frac{3+5}{2}=4\))
(ii) (T) (କାରଣ AM \(\frac{a+b}{2}\))
(iii) (T) (କାରଣ ମାଧ୍ଯମାନର ପ୍ରତିଶବ୍ଦ ହାରାହାରି ଅଟେ ।)
(iv) (F) (ସର୍ବଦା ବିଚ୍ୟୁତିର ମାଧ୍ଯମାନ ସହିତ ଆରମ୍ଭ ବିନ୍ଦୁ ଯୋଗ କରାଯାଏ, ତେଣୁ ଉତ୍ତର ସର୍ବଦା ସମାନ ହେବ ।)
(v) (F) (କାରଣ ବିଚ୍ୟୁତି = ଲବ୍‌ଧାଙ୍କ – ଆରମ୍ଭ ବିନ୍ଦୁ = 15 – 20 = – 5)
(vi) (T) (କାରଣ ପ୍ରଥମ n ସଂଖ୍ୟକ ସଂଖ୍ୟାର ସମଷ୍ଟି = \(\frac{n(n+1)}{2}\)
∴ ମାଧ୍ୟମାନ = \(\frac{n(n+1)}{2n}=\frac{n+1}{2}\))
(vii) (F) (ସୂତ୍ର ଅନୁସାରେ n + 1 ହେବ ।)
(viii) (T) (କାରଣ ପ୍ରଥମ ଦଶଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି = 10², ମାଧ୍ୟମାନ = \(\frac{10²}{10}\) = 10)
(ix) (F) (କାରଣ ମାଧମାନ 2 ଗୁଣ ହେବ ।)
(x) (F) (କାରଣ, ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାରେ 2 ଗୁଣିଲେ ତା 20ଟି ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟା ହେବ ।)

Question 2.
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନ ପାଇଁ ପ୍ରଦତ୍ତ ସମ୍ଭାବ୍ୟ ଉତ୍ତରମାନଙ୍କ ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛ ।
(i) 61, 62, 68, 56, 64, 72, 69, 51, 71, 67, 70, 55, 63 ଏହି ଲବ୍ଧାଙ୍କମାନଙ୍କର ମାଧ୍ୟମାନ ନିରୂପଣ ଲାଗି ନିମ୍ନସ୍ଥ ସଂଖ୍ୟାମାନଙ୍କ ମଧ୍ୟରୁ କେଉଁଟି ଉପଯୁକ୍ତ ଆରମ୍ଭ ବିନ୍ଦୁ ହେବ ?
(A) 55
(B) 60
(C) 70
(D) 72

(ii) ପ୍ରଥମ 20ଟି ଗଣନ ସଂଖ୍ୟାର ମାଧ୍ୟମାନ କେତେ ?
(A) 10
(B) 10½
(C) \(\frac{21}{20}\)
(D) 210

(iii) ପ୍ରଥମ ‘n’ ସଂଖ୍ୟକ ସଂପ୍ରସାରିତ ସ୍ଵାଭାବିକ ସଂଖ୍ୟା (Whole number)ର ମାଧ୍ଯମାନ କେତେ ?
(A) \(\frac{n-1)}{2}\)
(B) \(\frac{n}{2}\)
(C) \(\frac{n+1}{2}\)
(D) n

(iv) ପ୍ରଥମ ‘n’ ସଂଖ୍ୟକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ଯମାନ କେତେ ?
(A) (n – 1)
(B) n
(C) n + 1
(D) n + 2

(v) ପ୍ରଥମ n ସଂଖ୍ୟକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ମାଧ୍ଯମାନ କେତେ ?
(A) (n – 11)
(B) n
(C) n + 1
(D) n + 2

(vi) ‘m’ ମାଧମାନ ବିଶିଷ୍ଟ 10ଟି ଲବ୍‌ଧାଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 2 ବଢ଼ାଇଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କ 10ଟିର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) (n – 11)
(B) n
(C) n + 1
(D) n + 2

(vii) ‘M’ ମାଧ୍ୟମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 4 ଗୁଣ କରିଦେଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) \(\frac{M)}{4}\)
(B) M
(C) 4M
(D) \(\frac{4}{M}\)

(viii) ‘M’ ମାଧ୍ଯମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକରୁ x ବିୟୋଗ କଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) M
(B) (M + x)
(C) Mx
(D) (M – x)

(ix) ‘M’ ମାଧ୍ଯମାନ ବିଶିଷ୍ଟ n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକକୁ 5 ଦ୍ଵାରା ଭାଗକଲେ ନୂତନ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ କେତେ ହେବ ?
(A) M
(B) \(\frac{M}{5}\)
(C) 5M
(D) M – 5

(x) ଯଦି à ସଂଖ୍ୟକ ବାଳକମାନଙ୍କର ମାଧ୍ଯମାନ ବୟସ 12 ବର୍ଷ ଓ b ସଂଖ୍ୟକ ବାଳିକାଙ୍କର ମାଧ୍ଯମାନ ବୟସ 10 ବର୍ଷ ହୁଏ, ତେବେ ଉପରୋକ୍ତ ସମସ୍ତ ବାଳକ ବାଳିକାଙ୍କର ମାଧ୍ଯମାନ ବୟସ କେତେ ବର୍ଷ ହେବ ?
(A) \(\frac{10a+12b}{a+b}\)
(B) \(\frac{12a+10b}{a+b}\)
(C) \(\frac{10a+12b}{10+12}\)
(D) \(\frac{12a+10b}{10+12}\)

(xi) 998.9, 999.1, 1000-3, 1000-6, 1000.1 ର ମାଧ୍ୟମାନ କେତେ?
(A) 998
(B) 999
(C) 1000
(D) 1001

(xii) 6,8, 5, 7, x ଏବଂ 4 ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧ୍ଯମାନ 7 ହେଲେ xର ମାନ କେତେ ହେବ ?
(A) 10
(B) 11
(C) 12
(D) 13

(xiii) E₁, E₂, E₃, E₄, E₅, E₆ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧ୍ଯମାନ M ହେଲେ ⁶Σ_i=1(x₁ – M)ର ମାନ କେତେ ହେବ ?
(A) 0
(B) 6
(C) 36
(D) -6

(xiv) x, x + 2, x + 4, x + 6, x + 8ର ମାଧ୍ୟମାନ କେତେ ?
(A) x+2
(B) x + 4
(C) x+6
(D) x

(xv) 18ର ସମସ୍ତ୍ର ଗୁଣନୀୟକମାନଙ୍କର ମାଧ୍ୟମାନ କେତେ
(A) 5
(B) 6
(C) 6.5
(D) 7

ଉତ୍ତର:
(i) 69
(ii) 10½
(iii) \(\frac{n-1}{2}\)
(iv) n + 1
(v) n
(vi) m + 2
(vii) 4M
(viii) (M – x)
(ix) \(\frac{M}{5}\)
(x) \(\frac{12a+10b}{a+b}\)
(xi) 1000
(xi) 1000
(xii) 12
(xiii) 0
(xiv) x + 4
(xv) 6.5

(ଖ – ବିଭାଗ )

Question 3.
ଦଶଥର ଖେଳି ଜଣେ କ୍ରିକେଟ୍ ଖେଳାଳୀ ସଂଗ୍ରହ କରିଥିବା ରଗୁଡ଼ିକ ହେଲା – 47, 41, 50, 39, 45, 48,
42, 32, 60 ଏବଂ 20 । ତାଙ୍କଦ୍ୱାରା ସଂଗୃହୀତ ରନ୍‌ର ମାଧ୍ଯମାନ ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀରେ (ଉପଯୁକ୍ତ ଆରମ୍ଭ ବିନ୍ଦୁ ନେଇ) ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର ଆରମ୍ଭ ବିନ୍ଦୁ 45 1 ( ∵ ସର୍ବନିମ୍ନ ଏବଂ ସର୍ବାଧ‌ିକ ରନ୍ ଯଥାକ୍ରମେ 20 ଏବଂ 60) ।
∴ ଲବ୍ଧାଙ୍କମାନଙ୍କର ବିଚ୍ୟୁତିମାନ 2, − 4, 5, 6, 0, 3, – 3, −13, 15, – 25
ବିଚ୍ୟୁତିମାନଙ୍କର ସମଷ୍ଟି = – 26

∴ ଦଶଥର ଖେଳି ସଂଗୃହୀତ ରନ୍‌ର ମାଧ୍ୟମାନ = 42.4

Question 4.
କିଲୋଗ୍ରାମ୍ ଓଜନରେ 30 ଜଣ ପିଲାଙ୍କର ଓଜନ ହେଲା 21, 30, 40, 25, 26, 22, 26, 31, 22, 36, 30, 25, 25, 33, 30, 25, 27, 27, 25, 31, 33, 22, 21, 36, 40, 31, 33, 30, 37, 36 | ଏହି ତଥ୍ୟାବଳୀକୁ ବାରମ୍ବାରତା ବଣ୍ଟନରେ ସଜ୍ଜିତ କରି ମାଧ୍ଯମାନ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଓଜନ କିଲୋଗ୍ରାମ୍ ମାପରେ ଥ‌ିବା ଲବ୍‌ଧାଙ୍କମାନଙ୍କୁ ବାରମ୍ବାରତା ବଣ୍ଟନ ସାରଣୀରେ ରଖିଲେ –

ଉକ୍ତ ସାରଣୀରୁ 2f = 30 ଏବଂ Efx = 876.. ମାଧ୍ୟମାନ = \(\frac{Σf_x}{Σf}\)

Question 5.
କିଛି ରାସାୟନିକ ପଦାର୍ଥର ଓଜନ 30 ଥର ନିଆଯାଇ ଫଳାଫଳକୁ ନିମ୍ନ ସାରଣୀରେ ସଜାଯାଇଛି । ମାଧ୍ଯମାନ ଓଜନ ନିର୍ଣ୍ଣୟ କର ।

ଓଜନ (ଗ୍ରାମ୍‌ରେ)	3.8	3.9	4.0	4.1	4.2	4.3	4.4	4.5	4.6
ବାରମ୍ବାରତା	1	1	6	6	7	5	2	1	1

ସମାଧାନ :

ଓଜନ (ଗ୍ରାମ୍‌ରେ) (x)	ବାରମ୍ବାରତା (f)	ଓଜନ × ବାରମ୍ବାରତା (fx)
3.8	1	3.8
3.9	1	3.9
4.0	6	24.0
4.1	6	24.6
4.2	7	29.4
4.3	5	21.5
4.4	2	8.8
4.5	1	4.5
4.6	1	4.6
	Σf=30	Σfx=125.1

∴ ମାଧ୍ଯମାନ = \(\frac{Σf_x}{Σf}=\frac{125.1}{30}=4.17\)
∴ ମାଧ୍ୟମାନ ଓଜନ 4.17 ଗ୍ରାମ୍ ।

Question 6.
ଏକ ଶ୍ରେଣୀରେ 30 ଜଣ ଛାତ୍ରଙ୍କର ହାରାହାରି ବୟସ 12 ବର୍ଷ । ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ସହିତ ସେମାନଙ୍କର ହାରାହାରି ବୟସ 13 ବର୍ଷ ହେଲେ, ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ଶ୍ରେଣୀରେ 30 ଜଣ ଛାତ୍ରଙ୍କର ହାରାହାରି ବୟସ 12 ବର୍ଷ ।
30 ଜଣ ଛାତ୍ରଙ୍କର ମୋଟ ବୟସ = 30 × 12 = 360 ବର୍ଷ ।
ଛାତ୍ରମାନଙ୍କ ସହ ତାଙ୍କର ଶ୍ରେଣୀଶିକ୍ଷକ ମିଶିବାରୁ ହାରାହାରି ବୟସ 13 ବର୍ଷ ହେଲା ।
∴ 31 ଜଣ ଅର୍ଥାତ୍ 30 ଜଣ ଛାତ୍ର ଓ ଜଣେ ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ମୋଟ ବୟସ = 31 × 13 = 403 ବର୍ଷ ।
ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ = 403 – 360 = 43 ବର୍ଷ ।
∴ ଶ୍ରେଣୀ ଶିକ୍ଷକଙ୍କ ବୟସ 43 ବର୍ଷ ।

Question 7.
x₁, x₂, x₃ …… ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧ୍ଯମାନ m । ଯଦି ପ୍ରତ୍ୟେକ ଲବ୍‌ଧାଙ୍କରେ (a + b) ଯୋଗ କରାଯାଏ ଦର୍ଶାଅ ଯେ, ନୂତନ ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକର ମାଧମାନ (m + a + b) ହେବ ।
ସମାଧାନ :
ଲବ୍‌ଧାଙ୍କଗୁଡ଼ିକ ହେଲେ x₁, x₂, x₃ ……… x_n
ଉକ୍ତ n-ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କମାନଙ୍କର ମାଧ୍ଯମାନ (m) = \(\frac{x_1+x_2+x_3+…..x_n}{n}\)

(ଗ – ବିଭାଗ )

Question 8.

ସମାଧାନ :

ଉଚ୍ଚତା (x)	ବାରମ୍ବାରତା (f)	ଫଭାଗର ମଧ୍ୟବିନ୍ଦୁ	ମଧ୍ୟବିନ୍ଦୁ × ବାରମ୍ବାରତା (fy)
70-65	4	67.5	270.0
65-60	7	62.5	437.5
60-55	8	57.5	460.0
55-50	10	52.5	525.0
50-45	5	47.5	237.5
45-40	6	42.5	255.0
40-35	3	37.5	112.5
35-30	7	32.5	227.5
30-25	2	27.5	55.0
	Σf = 52		Σfy = 2580.00

∴ ମାଧ୍ଯମାନ = \(\frac{Σfy}{Σf}=\frac{2580}{52}=49.6\)
ବିକଳ୍ପ ପ୍ରଣାଳୀ : (ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀ)

ଆରମ୍ଭ ବିନ୍ଦୁ = 47.5, ସଂଭାଗ ବିସ୍ତାର (i) = 5
ମାଧ୍ୟମାନ = ଆରମ୍ଭ ବିନ୍ଦୁ + \(\frac{Σfy’}{Σf}\) × i = 47.5 + \(\frac{22×5}{52}\) (y’ = ବିଚ୍ୟୁତି) = 47.5 + \(\frac{110}{52}\) = 47.5+2.1 = 49.6
ମଧ୍ୟବିନ୍ଦୁ = \(\frac{\text { ସଂଭାଗର ନିମ୍ନସୀମା + ସଂଭାଗର ଉଚ୍ଚସୀମା }}{2}\)
ଅନ୍ତର୍ଭୁକ୍ତ ସଂଭାଗୀକରଣରେ ସଂଭାଗ ବିସ୍ତାର = ସଂଭାଗର ଉଚ୍ଚସୀମା – ସଂଭାଗର ନିମ୍ନସୀମା

Question 9.
ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀର ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ନିରୂପଣ କର ।

ସଂଭାଗ	84-90	90-96	96-102	102-108	108-114	114-120
ବାରମ୍ବାରତା	8	10	16	23	12	11

ସମାଧାନ :

∴ ମାଧ୍ଯମାନ = A + \(\frac{Σfy}{Σf}=100+\frac{244}{80}\) = 100 + 3.05 = 103.05

Question 10.
ନିମ୍ନ ଭାଗ-ବିଭକ୍ତ ବାରମ୍ବାରତା ବିତରଣ ସାରଣୀରେ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସୋପାନ-ବିଦ୍ୟୁତ ପ୍ରଣାଳୀରେ ସ୍ଥିର କର ।

ସଂଭାଗ	0-4	4-8	8-12	12-16	16-20	20-24
ବାରମ୍ବାରତା	5	7	10	15	9	4

ସମାଧାନ :

∴ ମାଧ୍ୟମାନ = ଆରମ୍ଭ ବିନ୍ଦୁ + \(\frac{Σfy’}{Σf}\) × c = 12 + \(\frac{6}{50}\) × 2 = 12 + 0.24 = 12.24
ବିକଳ୍ପ ପ୍ରଣାଳୀ : ମାଧମାନ (M) = A + \(\frac{Σfy}{Σf}\) × i
ସୂତ୍ରର ପ୍ରୟୋଗ କରି ସମାଧାନ କରାଯାଇପାରିବ । ଯେଉଁଠାରେ i = ସଂଭାଗବିସ୍ତାର ହେବ ।

Question 11.
ନିମ୍ନ ସାରଣୀରେ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ଉଭୟ ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀ ଓ ସୋପାନ-ବିଦ୍ୟୁତ ପ୍ରଣାଳୀ ଅବକମୂଳରେ ସ୍ଥିର କର ।

ସଂଭାଗ (C.I.)	0-50	50-100	100-150	150-200	200-250	250-300
ବାରମ୍ବାରତା (f)	4	10	12	10	8	8

ସମାଧାନ :

∴ ମାଧ୍ଯମାନ = A + \(\frac{Σfy}{Σf}=150+\frac{300}{52}\) = 150 + 5.77 = 155.77

∴ ମାଧ୍ୟମାନ (M)= A + \(\frac{Σfy’}{Σf}\) × c = 150 + \(\frac{12}{52}\) × 25 = 150 + 5.77 = 155.77

Question 12.
ସୋପାନ ବିଚ୍ୟୁତି ପ୍ରଣାଳୀ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ, ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସ୍ଥିର କର ।

ସଂଭାଗ	20-30	30-40	40-50	50-60	60-70	70-80
ବାରମ୍ବାରତା	10	6	8	12	5	9

ସମାଧାନ :

ଏଠାରେ A = 55, i = 60 – 50 = 10
∴ ମାଧ୍ୟମାନ (M)= A + \(\frac{Σfy’}{Σf}\) × i = 55 + \(\frac{-27}{50}\) × 10 = 55 + (-5.4) = 49.6

Question 13.
(i) ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ 7.5 ହେଲେ ‘f” ର ନିରୂପଣ କର ।

ସଂଭାଗ	5	6	7	8	9	10	11	12
ବାରମ୍ବାରତା	20	17	f	10	8	6	7	6

(ii) ମୂଲ୍ୟ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ 6 ହେଲେ ‘P’ ର ମୂଲ୍ୟ ନିରୂପଣ କର ।

ସଂଭାଗ	3	6	7	4	P+3	8
ବାରମ୍ବାରତା	5	2	3	2	4	6

ସମାଧାନ :
(i) ବଡ ତଥ୍ୟାବଳୀର ମାଧ୍ୟମାନ = 7.5

ମାଧ୍ୟମାନ (M) = \(\frac{Σfy}{Σf}\) ⇒ 7.5 = \(\frac{563+7f}{74+f}\)
⇒ 555 + 7.5f = 563 + 7f ⇒ 7.5f – 7f = 563 – 555
⇒ 0.5f = 8 ⇒ f = 8 ⇒ \(\frac{1}{2}\)f = 8 × 2 = 16

(ii) ବଡ ତଥ୍ୟାବଳୀର ମାଧ୍ୟମାନ = 6

ମାଧ୍ୟମାନ (M) = \(\frac{Σfy}{Σf}\) ⇒ 6 = \(\frac{116+4p}{22}\)
⇒ 4p + 116 = 132 ⇒ 4p = 16
⇒ p = \(\frac{16}{4}\) = 4

Question 14.
ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧମାନ 50 ଏବଂ ବାରମ୍ବାରତାଗୁଡ଼ିକର ସମଷ୍ଟି 120 ହେଲେ f₁ ଓ f₂ ନିର୍ଣ୍ଣୟ କର ।

ସଂଭାଗ	0-20	20-40	40-60	60-80	80-100
ବାରମ୍ବାରତା	17	f1	32	F2	19

ସମାଧାନ :
ବଡ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧମାନ = 50, ବାରମ୍ବାରତାଗୁଡ଼ିକର ସମଷ୍ଟି = 120

ପ୍ରଶ୍ନନୁସାରେ, 68 + f₁ + f₂ = 120
f₁ + f₂ = 52
Σfx = 3480 + 30f₁ + 70f₂ = 3480 + 30(f₁ + f₂) + 40f₂
=3480 + 30 × 52 ÷ 40f₂ = 3480+ 1560 + 40f₂ = 5040 + 40f₂
∴ ମାଧ୍ୟମାନ (m) = \(\frac{Σfx}{Σf}=\frac{5040+4f_2}{120}\)
⇒ 50= \(\frac{5040+4f_2}{120}\) ⇒ 40f₂ = 6000 – 5040 ⇒ f₂ = \(\frac{960}{40}\) = 24

ଆଗରୁ ପ୍ରମାଣିତ f₁ + f₂ =52 f₁ = 52 – 24 = 28
∴ f₁ = 28 ଏବଂ f₂ = 24

Question 15.
ସୋପାନ-ବିଚ୍ୟୁତି ପ୍ରଣାଳୀ ଅବଲମ୍ବନରେ ନିମ୍ନ ସାରଣୀ ଅନ୍ତର୍ଭୁକ୍ତ ତଥ୍ୟାବଳୀର ମାଧ୍ଯମାନ ସ୍ଥିର କର ।

ସଂଭାଗ	10-19	20-29	30-39	40-49	50-59	60-69	70-79
ବାରମ୍ବାରତା	5	65	222	112	53	40	3

ସମାଧାନ :

∴ ମାଧ୍ୟମାନ = A + \(\frac{Σfy’}{Σf}\) × i = 44.5 + \(\frac{-225}{500}\) × 10 = 44.5 + \(\frac{-450}{100}\) = 44.5 – 4.5 = 40

Question 16.
x₁, x₂, x₃ ……. ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧ୍ଯମାନ M । ଯଦି \(\sum_{i=1}^n\left(x_i-5\right)=60\) ଏବଂ \(\sum_{i=1}^n\left(x_i-8\right)\) = 24 ହୁଏ ତେବେ ‘n’ ଓ M ସ୍ଥିର କର ।
ସମାଧାନ :
x₁, x₂, x₃ ………. ପ୍ରଭୃତି n ସଂଖ୍ୟକ ଲବ୍‌ଧାଙ୍କର ମାଧମାନ M ।
⇒ \(\frac{x_1+x_2+x_3+…..x_n}{n}=M\)
⇒ x₁ + x₂ + x₃ ……. + x_n = nM
\(\sum_{i=1}^n\left(x_i-5\right)=60\)
⇒ (x₁ – 5) + (x₂ – 5) + (x₃ – 5) ……. + (x_n – 5) = 60
⇒ (x₁ + x₂ + x₃ ……. + x_n) – 5n = 60
⇒ nM – 5n = 60 ………(i)
⇒ \(\sum_{i=1}^n\left(x_i-8\right)\) = 24 ⇒ nM – 8n = 24 ………(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ

‘n’ ର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ nM – 5n = 60
⇒ 12M – 60 = 60 ⇒ 12M = 120
⇒ M = \(\frac{120}{12}\) = 10
∴ n = 12 ଓ M = 10

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text C: Psychobabble School Textbook Activity Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Unit 4 Text C: Psychobabble

Activity-10

a) Psychoanalysis is no longer used for curing mental diseases. (✓)
b) There is no end to an analysis. (✓)
c) Change in behavior ¡s only produced by self-knowledge. (✓)
d) Psychoanalysis is a waste of time. (✓)
e) Brief counseling is an honest form of talking cure. (✓)
f) Only doctors can become analysts. (✓)
g) Freud used psychoanalysis to cure a wide variety of psychological problems. (✓)

Activity-11

a)“ __________ it goes without saying that his research contributed enormously to our understanding of the subconscious.” (approval).
b) “But the analysis was then adopted for all sorts of psychological problems to which it was entirely insulted” (disapproval)
c) “ __________ if your problem is morbid introspection then the worst thing you can do is to spend hours talking about yourself. (disapproval)
d) “You create new problems for yourself as fast as you solve them, and the phony sense of progress is one of the things that makes it so addictive.” (disapproved)
e) “And all you get rid of ¡s the fee for another two years of treatment.” (approval)
f) In America it was finally the health insurance companies who called a halt to this madness. (approval)
g) “ __________ it involves a maximum of 25 sessions and sometimes just one.” (disapproval).

Activity-12

a) Which of the following in your opinion is the main purpose of the article?
(i) to describe a typical analysis session.
(ii) to amuse the reader.
(iii) to shock the reader.
(iv) to criticize psychoanalysis.
(v) to convince the reader that psychoanalysis is a waste of memory.
Answer:
(v) to convince the reader that psychoanalysis is a waste of memory.

b) How would you describe the writer’s attitude towards
(i) psychoanalysis,
(ii) Brief counseling.
Disapproving       Admiring
Approving            Indifferent
Contemptuous     Prejudiced
Uncompromising.
Answer:
Critical.

Activity-13
Cohesive Devices: Link Words

In Unit III you have looked at reference aa a device that binds the sentences of a text together. There are hints at the use of discourse markers as the author’s important device of text cohesion. Discourse markers (also called indicators in discourse) are easily recognized. “Signposts indicate how the writer has organized the text and what’s” he intends to say. They include link words such as ‘however, although furthermore but, newly’. They also include expressions such as “the second fact is ‘that’, which shows that the writer is introducing a second point in his discourse. In the following text, some link words are missing. Put in the link words from the

Instead of      When         But         Then         Yet        However      That’s how.

Television was invented by John Logie Baird. When he was young he built an airplane. He tried to fly in it, But it crashed down below. Baird was fortunate not to be killed. __________ he was older, he became a businessman. __________ his business failed, __________ he thought of working at television. His family advised him not to do it. He did not listen to them. __________ he rented an attic and brought the apparatus he needed. He started working. One day, he saw a picture on his screen. He rushed out to get someone he could ‘televise’. He found an office boy and took him back to the office. __________ no image of the boy appeared on the screen. The boy terrified, had put his head down. He put it up again. His picture appeared on the screen. __________ television had been invented.
Answer:
Television was invented by John Logic Baird. When he was young he built an airplane. He tried to fly in it. But it crashed down below. Baird was fortunate not to be killed. When he was older, he became a businessman. But his business failed, Then he thought of working at television. His family advised him not to do it instead. He did not listen to them. However, he rented an attic and bought the apparatus he needed. He started working. One day, he saw a picture on his screen. He rushed out to get someone he could ‘televise’. He found an office boy and took him back to the office.  Yet no image of the boy appeared on the screen. The boy terrified, had put his head down. He put it up again. His picture appeared on the screen. That’s how television had been invented.

Section – C
Introduction:
In this section, you will have the pleasure of reading an interesting article, ‘Psychobabble’. For the present, however, read-only its opening paragraph and guess what the article is about.

Psychobabble Summary in English

An unhappy man lies on a sofa. He is allowed to ramble on for an hour about thinking about something. He thinks about how an amount of $35 will be $30,000 after four or five years. It is actually, very odd. But people have been falling for it for a century. Freud invented Psychoanalysis in 1895. His research contributed enormously to the understanding of the subconscious. But it is obvious whether this analysis has any place in modern medical treatment.

Fraud and his co-worker’s ‘The Talking Cure’ was designed specifically to uncover the cause of hysterical symptoms and had a few successes. George Gershwin who was psychoanalyzed by doctors died, at the age of 39. Psychoanalysis was also administered as a cure for Schizophrenia and mental deficiency on which there was no effect at all. Woody Allen, a Western intellectual who is himself living proof that you can be analyzed until you are semicomatose and still end up with your personal life.

They believed that understanding will produce change which is highly doubtful. Any drunk driver who gets pulled over may well understand that he has behaved irresponsibly. But this understanding does not reduce the pleasure of drinking. It is considered a bad form to talk about what .you will achieve. The other thing that hooks people in the analysis is the phenomenon of transference. Psychoanalysts who expect and even encourage this will tell you it’s how the patient ultimately gets rid of those feelings.

In America, it was finally the Health Insurance Companies who called a halt to all this madness. The analysts were forced to admit that treatment was open-ended and the benefits uncertain. The dominant psychological problems are identified right from the start and a time limit is set for sorting them out. Learning from experiences is encouraged and strategies are worked out that will stop one from repeating self-destructive behavior. Most of our problems arise from making the same stupid mistake again and again.

Analytical outlines of the text:

• An unhappy man lies on a sofa.
• He is allowed to think for one hour.
• He thinks about how an amount of $35 will be $30,000 after four or five years.
• It is, actually, very odd.
• But people have been falling for it for a century.
• Freud invented psychoanalysis in 1895.
• His research helps to understand the subconscious.
• But it is obvious whether this analysis has any place in modern medical treatment.
• Freud and his coworkers produced “The talking cure”.
• It was specially designed to uncover the cause of hysterical symptoms.
• It had, however, a few successes.
• George Gershwin was psychoanalyzed by doctors.
• But he died at the age of 39.
• Psychoanalysis was administered as a cure for Schizophrenia.
• It is also applicable to mental deficiency.

• Actually, there was no effect at all.
• Woody Allen, a Western intellectual is himself living proof.
• One can be analyzed until one is semi-comatose.
• One can analyze it till ends up with his personal life.
• They believe that understanding will produce change.
• Actually, it is highly doubtful.
• Any drunk driver can understand his irresponsible behavior.
• But this understanding does not reduce the pleasure of drinking.
• It is considered a bad form.
• The other thing that hooks people in the analysis is the phenomenon of transference.
• Psychoanalysts encourage the patients to get rid of these feelings ultimately.
• The health insurance companies in America have stopped all this madness.
• To analysts, treatment is open-ended.
• But benefits are uncertain to them.
• The dominant psychological problems are identified.
• Learning from experiences is encouraged.
• Strategies are worked out not to repeat self-destructive behavior.
• Most of our problems arise from making the same stupid mistake again and again.

Meaning of difficult words:

ramble – to travel, to wander, to trail, here talk about ceaselessly.
confused – perplexed, disordered, disturbed.
enormously – immensely, atrociously, greatly.
hysterical – excitement, morbidity, terrible mental excitement.
symptoms – signs, characteristics, and traits of a desire.
adopted – taken up, received, used, employed.

Schizophrenia – violent mental problem.
depression – pressing down, saddening, mental frustration.
diminish – lessen, reduce, decrease.
compensate – make amounts for, replace the loss with something.
diagnosis – finding out the cause of an ailment, and identification of disease by symptoms.
therapists – persons treating diseases in a certain way.

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text D: The Case against Man Textbook Activity Questions and Answers.

CHSE Odisha 11th Class Alternative English Solutions Unit 4 Text D: The Case against Man

Activity-14

a) It means that mankind is although a living organism, it is also a thing or inanimate object.
b) The unrestrained population growth is compared with cancer. This is a good comparison because its growth will kill mankind as cancer.
c) If the present rate of population growth continues the ecology will be spoilt.
d) the thesis of the essay is increasing the birth rate and its control He waits to describe things and then concludes.
e) This is really a problem that has been shown by the author perfectly and which does not need any other way of description.
f) Interrelation and interdependence are in common among the living and nonliving things on earth.
g) The conclusion of the essay is— At the rate, we are going without birth control, then even if science serves us in an absolutely ideal way, we will reach the planetary high-rise with no animals but men, with no plants but algae, with no room for even one more person by AD. 2430.
h) The essay starts with the interrelation of the living and the nonliving and ends in control of the birth rate which will help the organism to double itself.

Activity-15

	Headings	Paragraph Numbers
a)	The thesis of the problem: “birth control”	21
b)	Reasons/causes Lowering death rate	19
c)	Examples: The number of Homo Sapiens increase	16
d)	Suggested solution: Birth control	19
e)	Special Features of the Development of the Argument (if any)	19, 20
f)	Conclusion Ready birth control without delay	20, 21, 22, 23

Activity – 16

Rather than exploiting the environment shouldn’t we be in a partnership? If we continue to waste the earth’s resources as if there were no tomorrow, there could well be no tomorrow. By the year 2010, one-third of the world’s cropland will have turned to dust, of people, will face starvation. All this is happening since our civilization has kept on expanding, on the assumption that the world’s resources are limitless. But merely stopping growth is not the answer. What we need is development that works in partnership with the environment that uses the earth’s resources more productively and after all is suitable at the same time. This is the reason why our organization Earth life exists.

Activity-17

Anita : Hello, Banita. You have a debate today, haven’t you?
Banita: Yes Anni. It’s at 2-30 p.m.
Anita: What’s it about?
Banita: It’s about population explosion.
Anita : Population explosion ! It’s a burning topic, isn’t it?
Banita: Yes, it is. But it’s a topic that needs many things to incorporate.
Anita: What’re you going to hint at?
Banita: Just the causes and consequences of population growth.
Anita: Won’t you suggest any solution?
Banita: Yes, I will.
Anita: Why’s the population on the rise now?
Banita: It’s owing to the lower death rate.
Anita: Lower death rate! Aren’t people dying now? You’ll see in the papers hundreds of people are dying every day.
Banita: No, no. People are dying but their number is eye-catching due to the high population and media network.
Anita: Do you think that the death rate has really come down?
Banita: Is there doubt about it? The death rate is very much lower than before. Thousands of people were dying of starvation, Cholera, and Smallpox in the past. But we don’t see these diseases active now. A number of villages were having mass funerals with the approach of such a disease.
Anita: You’re quite right Banita. This was a usual case that is not seen these days. Thank you.

Activity -18

Pranati: Hello. This is 250845
Minati: Can I speak to Pranati, please?
Pranati: Yes, speaking
Minati: Hi Pranati, it’s Minati here.
Pranati: Listen, Minu, We’d proposed to go to the cinema this afternoon, hadn’t we?
Minati: Yes, we had. You told me to book a pair of tickets and inform me earlier what’s about.
Pranati: I’m quite sorry. I failed to book tickets at the counter. I’d gone to do it, but I wasn’t able to.
Minati: What’s really happened?
Pranati: The counters were overcrowded. None of the counters was free to buy a ticket at.
Minati: The film has recently been released. People must be thronging to see it.
Pranati: Yes, Blakers are moving about. They are charging very high. I didn’t feel like purchasing a ticket from them.
Minati: OK. Don’t mind. We’ll see the cinema within a couple of days. The rush will be subsiding. Thank you.
Pranati: Welcome

Activity – 19

• In Kingston, Jamaica’s capital, RSLs own cruiser is waiting to introduce them to the unique world of the Caribbean.
• Every Tuesday a British Airways flight leaves Heathrow for Jamaica.
• Like all our ships, this cruiser has been specially designed to give you maximum comfort, luxury, and enjoyment.
• For this lucky one it’s the beginning of an unforgettable air-sea holiday with the world’s leading cruiser company. The Royal Seafaring Line.
• For many of the passengers, it’s just a normal scheduled flight, but for some, it’s the start of something very special.
• Whether you choose relaxation on board or stimulation on land,’ you will have the holiday of a lifetime.
• And it’s all included in the price — Just 1,995 for 21 days.
• While you can thus spend a perfect holiday without leaving the ship, there is also the added attraction of fascinating store visits at each of our parts of the cell.
• So you can relax on the vest sun deck, bide your time with a cocktail, or dance till dawn in the nightclub or in the discotheque.
• So don’t delay- See your travel agents today.

The Case Against Man Summary in English

Section – D
Part – One

Summary:
The first mistake is to think of man as a thing in itself. It is, however, a part of an intricate problem of life. Life gets its energy from the sun. Five billion years back, the earth had undergone a vast revolution. On its first appearance, it lacked an ocean and an atmosphere. Far within the solid crust, there are slow continual changes whose hot springs, volcanoes, and earthquakes are the more noticeable manifestations here on the surface.

Portions of the surface water with solar radiation developed complicated compounds called ‘life’. It has assumed a complex proportion. But, life forms are as much part of the structure of the Earth as any inanimate portion is. It is all an inseparable part of a whole if any animal is isolated totally from other forms of life, and death by starvation will surely follow. If isolated from water, death by dehydration will follow even faster.

If isolated from air, death by asphyxiation will take place. Isolation from the sun will bring death to the animal world. The inanimate portion also suffers. The entire planet and solar system are closely interrelated. A planet is a life form made up of nonliving portions. For instance, a man is composed of 50 trillion cells of a variety of types, all interrelated and interdependent.

Part – Two

Summary:
Sometimes, the neat economy of growth within an organism such as a human being is disrupted. The growing of a group of cells is stopped. If one type of organism began to multiply without limit killing its competitors, the same thing would happen in ecology. The earth’s human population is estimated to have been 150 million al the time of Julius Caesar. This population since then has been on the rise. It is really an alarming proportion.

The current increase of the human population qualifies Homo Sapiens as ecological cancer. However, this cancerous growth must be stopped. It can be done by raising the death rate or towering the birth rate. There is no other alternative. If we do nothing, the death rate will rise fabulously. Lowering the birth rate is surely the preferable way.

Analytical outlines of the Text

• The first mistake is to think of man as a thing in itself.
• It is, however, a part of an intricate problem of life.
• Life gets its energy from the sun.
• Five billion years back, the earth had undergone a vast revolution.
• On its first appearance. it had lacked an ocean and an atmosphere.
• Far within the solid crust, there are slow continual changes.
• The hot springs, volcanoes, and earthquakes are the more noticeable manifestations here on the surface.
• Portions of the surface water with solar radiation developed complicated compounds called life.
• It has assumed a complex proportion.
• But life forms are as much part of the structure of the Earth as any inanimate portion is.
• It is all an inseparable part of a whole.
• Any animal is isolated totally from other forms of life.
• It will surely follow death by i$arvation.
• Any animal is isolated from water.
• It will follow death by dehydration.
• Any animal is isolated from the air.
• It will take place death by asphyxiation.
• Isolation from the sun will bring death to the animal world.
• The inanimate portion also suffers.
• The entire planet and solar system are closely interrelated.
• A planet is a life form made up of nonliving, portions.

• For instance, a man is composed of 50 trillion cells of a variety of types all interrelated and interdependent.
• Sometimes, the net economy of growth within an organism such as a human being is disrupted.
• The growing of a group of cells is stopped.
• One type of organism began to multiply without limit killing its competitors.
• The disruption will happen in ecology.
• The earth’s human population is estimated to have been 150 million at the time of Julius Caesar.
• This population since then has been on the rise.
• It is really an alarming proportion.
• The current increase in the human population qualifies Homo Sapiens as ecological cancer.
• However, this cancerous growth must be stopped.
• It can be achieved in two ways.
• One is by raising the death rate.
• The other is by lowering the birth rate.
• There is no other alternative.
• We have to do something.
• Otherwise, the death rate will rise fabulously.
• Lowering the birth rate is certainly the preferable way.

Meaning of difficult words

crust – the thin hard surface of the earth.
versatile – clever to do a number of things, good at doing a lot of different things.

asphyxiation – death by choking.
reefs – a line of sharp rocks, often made of coral.
quiescent – becoming quiet or silent, not developing or doing anything.
cougar – a puma, a large brown wild cat of North West America.
decimated – killed large numbers of ruined a large part of something.
predators – animals that live by killing and eating other animals.
ecology – the study of living things in their surroundings.
Homo Sapiens – the type of human beings that inhabit the earth now.
catastrophically- in a terribly destructive manner.

 

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text B: What is Art? Textbook Activity Questions and Answers.

CHSE Odisha 11th Class Alternative English Solutions Unit 4 Text B: What is Art?

Activity-5
Getting The Main Idea Of The Paragraph

Find out a suitable title for each of the paragraphs in Text-B (Part one)

Paragraph     Title
1                 :
2                 :
3                 :
4                 :
5                 :
6                 :
7                 :

Answer:
Paragraph -1 : Title – Defining Art.
Paragraph -2: Title – Relationship of Art.
Paragraph -3 : Title – Art Transmitting Human Thought.
Paragraph -4 : Title – Activity of Art.
Paragraph -5: Title – Man’s capacity of Receiving other’s Emotional Dimensions.
Paragraph -6 : Title – Infecting feelings.
Paragraph -7 : Title – Object of Joining Another.

Descriptive Sequence

After going through part two Text-B, read the whole of Text-B (That is, both the parts) and arrange the following items in the sequence in which they are presented in the Text.
a) Discussing what art is not.
b) Talking about defining art.
c) Speaking about the characteristics of art.
d) Citing examples that does not amount to art.
e) Comparing art with speech.
f) Arriving at a definition of art.
g) Speaking of the variety of feelings on which art is based.

Answer:
a) Talking about defining art.
b) Comparing art with speech.
c) Speaking of the characteristics of art.
d) Speaking of the variety of feelings on which art is based.
e) Arriving at a definition of art.
f) Discussing what art is not.
g) Citing examples of what does not amount to art.

Activity-7
Reacting To The Ideas In The Text

• Art is superior to speech because it transmits feelings as well as thoughts because a man transmits his thoughts to another by words but by art, he transmits his feelings.
• Tolstoy speaks of the essential elements of art in paragraph 5 but seems to contradict himself in the next paragraph. The views presented in these two paragraphs can, however, be concealed.
• The analogy between the boy who encounters- a wolf and the artist who recreates his emotions in a work of art is now appropriate because the feelings and emotions are equally infected with one another.
• The writer begins his essay by saying that art should not be considered “as a means to pleasure” but should be considered “as one of the conditions of human life”. And he has proved this in his essay taking suggestive examples from various lores of life.

Activity – 8

a) A direct approach is chosen to define the term ‘renaissance’ in passage 1, but a descriptive technique is followed in passage 2 to define the term ‘elegance’.
b) An etymological analysis of the term ‘renaissance’ finds an outlet in passage 1 but the implied meaning of the term ‘elegance’ is given in passage 2.
c) A general meaning of the word ‘renaissance’ is reflected in passage 1 whereas the views and considerations of the word ‘elegance’ have been found in passage 2.

Activity-9
Remedial Grammar

My town is an excellent place to live in, I think it is wonderful. It is an important town, because, it is the center of the district administration. It is also great because of the two very famous museums. The weather here is nice. It is hot in summer with occasional rains and is cool in winter. I like my home town very much.

What is Art? Summary in English

Section – B
Part – One
Read below the first paragraph of Leo Tolstoy’s “What is Art ?” and try to guess the writer’s purpose.
In order to define art correctly it is necessary first of all to cease to consider it as a means to pleasure and to consider it as one of the conditions of human life viewing it in this way, we can’t fail to observe that art is one of the means of intercourse between man and man. Now read part one of Text B and note how Tolstoy develops his idea of art across the paragraph.

Summary:
Leo Tolstoy defines art to cease to consider it as a means to pleasure and to consider it as one of the conditions of human life. Art is an intercourse between man and man. The receiver of every work of art enters into a certain kind of relationship both with him who produced or is producing the art and with all those who simultaneously, previously or subsequently receive the same artistic impression, speech transmitting the thoughts and experiences of man serve as a means of union among them and art serves a similar purpose.

A man communicates himself with another by means of words and by it he transmits his feelings. A man shares his feelings by listening to another man. When one man laughs, another becomes merry to hear it. But when a man cries, another feels sorry. A man is excited or irritated and another man who sees him is brought to a similar state of mind by his movements or by the sounds of his voice. A man expresses courage and determination or sadness and calmness and this state of mind passes on to others.

A man suffers expressing his suffering through groans and spasms and this suffering transmits itself to other people. A man expresses his feelings of admiration, devotion, fear, respect, or love to certain objects, persons, or phenomena and others infected by the same feelings of administration, devotion, fear, respect or love to some objects, persons or phenomena. Art begins when one person expresses his feelings by certain external indications in order to join others or others.

For instance, a boy having experienced fear of encountering a wolf relates the encounters, and in order to evoke in others the feelings he has experienced describes his conditions before the encounter, the surrounding of the world, his own lightheartedness, and then, the wolf’s appearance, its movements, the distance between himself and the wolf and so forth. If only the boy when telling the story again experiences the feelings he has lived through and infects the heart and compels them to feel that he had experienced is art.

It is also art if a man having experienced either the fear of suffering or the attraction of enjoyment expresses these feelings on canvas or in marble so that others are infected by them. It is again art of a man who feels or imagines to himself feelings of delight gladness, sorrow, despair, courage or despondency and the transition by sounds from me to another of those feelings and expresses them by sounds so that the hearers are inflected by them and experience them as they were experienced by the composer.

Analytical outlines of the text:

• According to Leo Tolstoy, art is a means to provide pleasure.
• He also considers it as one of the conditions of human life.
• Art is an intercourse between man and man.
• The receiver of every work of art enters into a certain kind of relationship both with him.
• It is a relationship with him who produced or is producing the art.
• It relates to those who simultaneously, previously or subsequently receive the same artistic impression.
• Speeches transmitting the thoughts and experiences .of men serve as a means of union, among them and art serves a similar purpose.
• A man communicates himself with another by means of words and by it he transmits his feelings.

• A man shares his feelings by listening to another.
• One man laughs, and another becomes merry to hear it.
• One man cries and another becomes feel sorry.
• A man as excited or irritated.
• Another is brought to the same state by seeing it.
• He acquires it by his movements or the sounds of his voice.
• A man expresses his courage or determination.
• This state of mind passes to another.
• A man expresses his sadness or calmness.
• This state of mind passes to another.
• A man expresses his suffering through groans or spasms.
• It transmits itself to other people.
• A man expresses his admiration, devotion, fear, respect, and love to certain objects, persons or phenomena.
• Others are infected by the same feelings.
• Art begins when one joins others with the same feelings.
• For instance, a boy experiences of fear by encountering a wolf.
• He expresses this fear in order to evoke a feeling in others.

• He provides an elaborate description of it.
• He expresses the conditions before the encounter.
• He also expresses the surrounding of the words.
• He also expresses his distance from the wolf.
• On the other hand, he also expresses the wolf’s appearance, its movement, distance from him, etc.
• When he compels others to feel his experiences, it is called art.
• Hence, to be an art, the feelings of suffering and enjoyment should be infected by them.
• When a man feels or imagines those feelings of delight, gladness, sorrow, despair, etc. it is called art.
• Therefore, art refers to the transmission of the sounds of those feelings from one man to other so that one must be infected by them and also experiences them by themselves.

Meaning of difficult words:

simultaneously – happening at the same time.
previously – formerly.
subsequently – followingly.
intercourse – deal with, interact, and communicate.
transmit – sends, communicates.
groan – moan, lamentation.
spasms – muscular contraction, stiffness of muscles.
encountering – facing, confronting, meeting
despondency – misery, sorrow, unhappiness.
transition – change, transformation, movement.

Text-B
Part – Two

Summary:
The feelings which the artist transmits to others are varied and many. Some are very strong and some are very weak, some significant and others insignificant, some very bad, and others very good. Patriotic love, self-devotion and yielding to fate or to God in drama, raptures of lovers in. a novel, voluptuousness in a picture, courage in triumphal marches, merriment in a dance, and humor in a funny story are all different forms of art. If the feelings of the author are transmitted to the spectators, they are deemed to be rightly infected.

Art is certainly a human activity that consists of external signs hands-on to other’s feelings he had lived through and that others are infected by these feelings and also experience them. The metaphysicians state that art is not the manifestation of some, mysterious idea of the beauty of God. Physiologists view a game in which man lets off his excess stored-up energy, is not man s expression of emotion by external signs. It is neither pleasure nor the production of pleasing objects.

Analytical outlines of Part Two.

• The artist’s transmission of feelings to others is varied and many.
• Some are very strong and others are very weak.
• Some are significant and others are insignificant.
• Even some are very good and others are very bad.
• There are different forms of art.
• Patriotic love and self-devotion are the same.
• Raptures of lovers in a novel, and voluptuousness in a picture are others.
• Also, courage in a triumphal march, merriment in a dance, and humor in a tunny spry are still others.
• If the feelings of the writer are transmitted to the audience, they are deemed to be rightly infected.

• Art is, certainly, a human activity.
• It consists of external signs hands-on to other’s feelings.
• Others are infected by these feelings.
• They also experience these feelings.
• The metaphysicians opine that art is not the manifestation of some mysterious idea of the beauty of God.
• Physiologists view it is a game in which man gets off his excess stored up energy.
• It is not the expression of man’s emotion by external signs.
• It is neither pleasure nor the production of pleasing objects.
• It is a means of union among men joining them together in the same feelings.

Odisha State Board BSE Odisha 8th Class English Solutions Lesson 1 The Missing Ring Textbook Exercise Questions and Answers.

BSE Odisha Class 8 English Solutions Lesson 1 The Missing Ring

Session – 1
I. Pre-Reading (ପ୍ରାକ୍-ପଠନ):

→  You must have heard or read the stories about Birbal, a wise and clever man. He had uncommon wit and intelligence. He was a minister in Emperor Akbar’s court. Here is a story in which Birbal helps Akbar in finding a thief.
(ତୁମେମାନେ ନିଶ୍ଚୟ ଚତୁରମନ୍ତ୍ରୀ ବୀରବଲ୍ଲଙ୍କ ଉପରେ ଗଳ୍ପସବୁ ଶୁଣିଥ‌ିବ । ସେ ଅସାଧାରଣ ଚତୁର ଏବଂ ବୁଦ୍ଧିମାନ ଥିଲେ । ସେ ଆକବରଙ୍କ ରାଜଦରବାରରେ ଜଣେ ମନ୍ତ୍ରୀ ଥିଲେ । ଏଠାରେ ସେହିପରି ଏକ ଗଳ୍ପର ଅବତାରଣା କରାଯାଉଛି, ଯେଉଁଥରେ ସମ୍ରାଟ ଆକବରଙ୍କୁ ତାଙ୍କର ହଜିଥ‌ିବା ମୁଦି ପାଇବାରେ ସାହାଯ୍ୟ କରିଥିଲେ ।)

II. While Reading (ପଢ଼ିବା ସମୟରେ)

• SGP – 1
• Read paragraphs 1-3 silently and answer the questions that follow.
  (ପ୍ରଥମ ଅନୁଚ୍ଛେଦରୁ ତୃତୀୟ ଅନୁଚ୍ଛେଦପର୍ଯ୍ୟନ୍ତ ପାଠକୁ ନୀରବରେ ପଢ଼ ଏବଂ ସେଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ)

1. Akbar was ………………………………….. from the queen.
ଓଡ଼ିଆ ଅନୁବାଦ | ସାରାଂଶ :
ସମ୍ରାଟ ଆକବର ଖୁବ୍ ଅଳଙ୍କାରପ୍ରିୟ ଥିଲେ । ହୀରା, ନୀଳା ଭଳି ବହୁ ମୂଲ୍ୟବାନ ପଥର ଖଚିତ ଶହ ଶହ ମୁଦି ତାଙ୍କର ଥିଲା । ମାତ୍ର ସେଗୁଡ଼ିକ ମଧ୍ୟରୁ ଗୋଟିଏ ମୁଦି ତାଙ୍କର ସବୁଠାରୁ ପ୍ରିୟ ଥିଲା । ଏହି ମୁଦିଟିରେ ଅସଂଖ୍ୟ ହୀରା ଓ ମୋତି ଖଚିତ ହୋଇ ରହିଥିଲା । ଏହି ମୁଦିଟିକୁ ତାଙ୍କୁ ତାଙ୍କ ରାଣୀ ଉପହାର ସ୍ବରୂପ ପ୍ରଦାନ କରିଥିଲେ ।

2. At Akbar ………………………………….. Emperor’s room.
ଓଡ଼ିଆ ଅନୁବାଦ | ସାରାଂଶ :
ସମ୍ରାଟ ଆକବରଙ୍କ ରାଜପ୍ରାସାଦରେ ତାଙ୍କର ପୋଷାକ ଏବଂ ଅଳଙ୍କାରର ଯନୂ ନେବାକୁ ୮ ଜଣ କର୍ମଚାରୀ ନିଯୁକ୍ତ ଥିଲେ । ପ୍ରତିଦିନ ଏମାନଙ୍କ ମଧ୍ୟରୁ ଜଣେ ଲେଖାଏଁ କର୍ମଚାରୀ ସମ୍ରାଟଙ୍କର ଦରବାର ଯିବାପୂର୍ବରୁ ତାଙ୍କୁ ଆବଶ୍ୟକ ପୋଷାକ ଓ ଅଳଙ୍କାରରେ ମଣ୍ଡିତ କରୁଥିଲା । ଏମାନଙ୍କ ବ୍ୟତୀତ ଅନ୍ୟ କୌଣସି ବ୍ୟକ୍ତି ବା କର୍ମଚାରୀଙ୍କୁ ସମ୍ରାଟଙ୍କ କୋଠରି ମଧ୍ୟକୁ ପ୍ରବେଶ କରିବାକୁ ଅନୁମତି ଦିଆଯାଉ ନଥିଲା ।

3. One day …………………………………… not to be found.
ଓଡ଼ିଆ ଅନୁବାଦ | ସାରାଂଶ :
ଦିନେ ସମ୍ରାଟ ଦରବାରକୁ ଯିବାକୁ ପ୍ରସ୍ତୁତ ହେଉଥ‌ିବାବେଳେ ତାଙ୍କର ଅତି ପ୍ରିୟ ମୁଦିଟିକୁ ପିନ୍ଧିବାକୁ ଇଚ୍ଛାପ୍ରକାଶ କଲେ । ସେ ତାଙ୍କର ୮ ଜଣ କର୍ମଚାରୀଙ୍କ ମଧ୍ୟରୁ ଜଣକୁ ଡାକି ଉକ୍ତ ମୁଦିଟିକୁ ନିର୍ଦ୍ଦିଷ୍ଟ ସ୍ଥାନରୁ ଆଣିବାକୁ ଆଦେଶ ଦେଲେ । ଏହି କର୍ମଚାରୀ ଜଣକ କିଛି ସମୟ ପରେ ଫେରିଆସି ମୁଦିଟିକୁ ପାଉନଥ‌ିବା ଜଣାଇଲା । ଆକବର ଉକ୍ତ ମୁଦିଟିକୁ ଖୋଜିବାକୁ କର୍ମଚାରୀମାନଙ୍କୁ ଆଦେଶ ଦେଲେ ମାତ୍ର ମୁଦିଟି କେଉଁଠି ମିଳିଲା ନାହିଁ ।

Word Meaning

fond of : to like something very much (ପ୍ରିୟ | ଭଲ ପାଇବା)
diamond : a very hard, bright, clear precious stone used in jewellery (ହୀରା, ଏକ ମୂଲ୍ୟବାନ ଧାତୁ)
pearls : a small, hard, round white object that grows inside the shell of an oyster (ଶାମୁକା) and used to make jewellery (ଅଳଙ୍କାର)
gem : any sort of valuable stone used as jewellery (ରତ୍ନ)
present : honour / reward / gift (ଉପହାର)
palace : place of kings (ରାଜପ୍ରାସାଦ)
look after : take care of (ଯତ୍ନ ନେବା)
get ready : to prepare (ପ୍ରସ୍ତୁତ ହେବା)
enter : to move inside (ପ୍ରବେଶ କରିବା)
favourite : something one likes the most / valued more (ଅତିପ୍ରିୟ)
ordered : to give instruction / to do something with authority (ଆଦେଶ ଦେବା)

Comprehension Questions and Answers : (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
What was Akbar fond of?
(ଆକବର କ’ଣ ଭଲ ପାଉଥିଲେ ?)
Answer:
Akbar was very fond of jewellery.

Question 2.
How many rings did he have?
(ତାଙ୍କର କେତୋଟି ମୁଦ୍ରିକା ଥିଲା ?)
Answer:
He had hundreds of rings.

Question 3.
Which ring did he like most? Why?
(କେଉଁ ମୁଦ୍ରିକା ସେ ଅଧ୍ଵ ଭଲ ପାଉଥିଲେ ? କାହିଁକି ?)
Answer:
He liked one of his rings most which was a large ring with a number of pearls and diamonds set in it.

Question 4.
Who looked after the Emperor’s clothes and jewelry?
(କିଏ ରାଜାଙ୍କ ପୋଷାକ ଓ ରତ୍ନଗୁଡ଼ିକର ରକ୍ଷଣାବେକ୍ଷଣ କରୁଥିଲେ ?)
Answer:
There were eight servants who looked after the Emperor’s clothes and jewelry.

Question 5.
What did they do every day?
(ସେମାନେ ପ୍ରତିଦିନ କ’ଣ କରୁଥିଲେ ?)
Answer:
Every day one of these eight servants used to help Akbar get ready to go to the court.

Question 6.
Were others allowed to the Emperor’s room? Why?
(ଅନ୍ୟମାନେ ରାଜାଙ୍କ କକ୍ଷକୁ ପ୍ରବେଶ କରିପାରୁଥିଲେ କି ? କାହିଁକି ?)
Answer:
No, none other than these eight servants could enter the Emperor’s room, because the king did not believe others other than these eight servants.

Question 7.
What did Akbar want to wear while getting ready to go to the court?
(ଆକବର ଦରବାରକୁ କେଉଁ ପୋଷାକରେ ଯିବାକୁ ପସନ୍ଦ କରୁଥିଲେ ?)
Answer:
One day Akbar wanted to wear his favorite ring in which a number of pearls and diamonds were set while getting ready to go to the court.

Question 8.
What was the servant’s report?
(ଭୃତ୍ୟମାନେ କ’ଣ କହିଲେ ?)
Answer:
The servant reported that he could not find the ring.

Question 9.
What was the Emperor’s order?
(ରାଜାଙ୍କ ଆଦେଶ କ’ଣ ଥିଲା ?)
Answer:
The Emperor’s order was to search for his favorite ring.

Question 10.
What was the result?
(ପରିଣାମ କ’ଣ ହେଲା ?)
Answer:
But the ring could not be found.

• SGP – 2
• Read paragraphs 4-5 silently and answer the questions that follow.

4. Akbar………………………………………….. the thief.
ଓଡ଼ିଆ ଅନୁବାଦ / ସାରାଂଶ :
ଆକବର ଭୀଷଣ ରାଗିଗଲେ । ସେ ଅନୁଭବ କରିଥିଲେ ଯେ ତାଙ୍କର ସେହି ପ୍ରିୟ ମୁଦିକୁ ତାଙ୍କ ୮ ଜଣ କର୍ମଚାରୀଙ୍କ ମଧ୍ୟରୁ କେହି ଜଣେ ଚୋରି କରିଥିଲା । ସେ ବୀରବଲ୍ଲଙ୍କୁ ଡକାଇଲେ । ବୀରବଲ୍ଲ ଆସିବାପରେ ସେ ତାଙ୍କୁ ସବୁକଥା ଜଣାଇଲେ ଏବଂ ପ୍ରକୃତ ଚୋରକୁ ଠାବକରିବାକୁ କହିଲେ ।

5. The following ……………………………………………….. that night.
ଓଡ଼ିଆ ଅନୁବାଦ / ସାରାଂଶ :
ତା’ପରଦିନ ବୀରବଲ୍ଲ ସମ୍ରାଟଙ୍କ କୋଠରି ଦାୟିତ୍ଵରେ ଥିବା ଆଠଜଣ କର୍ମଚାରୀଙ୍କୁ ଡକାଇଲେ । ସମସ୍ତ ଆଠଜଣଙ୍କୁ ଛୋଟ ଛୋଟ ଖଣ୍ଡିଏ ଲେଖାଏଁ ବାଡ଼ି ଦେଲେ ଏବଂ ସେ ବାଡ଼ିକୁ ଧରି ତା’ପରଦିନ ତାଙ୍କ ନିକଟକୁ ଆସିବାକୁ କହିଲେ । ସେ ସେମାନଙ୍କୁ ସୂଚନା ଦେଲେ ଯେ ତାଙ୍କ ମଧ୍ୟରୁ ଜଣେ କେହି ନିଶ୍ଚୟ ମୁଦିଟି ଚୋରିକରିଛି । ଚୋରି କରିଥିବା ବ୍ୟକ୍ତିର ବାଡ଼ିଟି ଆଜି ରାତି ମଧ୍ୟରେ ଆକାରରେ ଗୋଟିଏ ଆଙ୍ଗୁଠି ଲମ୍ବର ବଡ଼ ହୋଇ ହୋଇଯିବ ।

Word Meaning

angry : to get upset / enraged (କ୍ରୁଦ୍ଧ)
stolen : take without owner’s consent (ଚୋରି ହୋଇଥ‌ିବା/ଚୋରି କରିଥୁବା )
happen : befall / take place (ଘଟିବା / ସଂଘଟିତ ହେବା)
in charge of : taking the responsibility of something (ଦାୟିତ୍ଵରେ ଥ‌ିବା)
short : small (ଛୋଟ)
stick : a small thin branch of tree (ବାଡ଼ି, ଲାଠି / ଦଣ୍ଡ)

Comprehension Questions and Answers : (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
Akbar felt something about the missing ring, what was it? What did he ask Birbal to do?
(ଆକବର ତାଙ୍କର ହଜିଥ‌ିବା ମୁଦି ବିଷୟରେ କିଛି ଜାଣିପାରିଲେ ତାହା କ’ଣ ? ସେ ବୀରବଲ୍ଲଙ୍କୁ ଡାକି କ’ଣ କହିଲେ ?)
Answer:
Akbar felt something about his missing ring one of his eight servants had stolen the ring. Akbar asked Birbal to find the thief.

Question 2.
What did Birbal give to the servants?
(ବୀରବଲ୍ଲ କର୍ମଚାରୀମାନଙ୍କୁ କ’ଣ ଦେଲେ ?)
Answer:
Birbal gave each of the servants a short stick.

Question 3.
What did he tell about the thiefs stick?
(ଚୋରର ବାଡ଼ି ସମ୍ପର୍କରେ ସେ କ’ଣ କହିଲେ ?)
Answer:
He told the servants that he had known one of the eight servants had stolen the ring. The stick of the thief he gave would become longer as much as the length of a finger that night.

Session – 2

• SGP-3
• Read paragraphs 6-7 silently and answer the questions that follow.
  (୬ଷ୍ଠ ଓ ୭ମ ଅନୁଚ୍ଛେଦ ଦୁଇଟି ନୀରବରେ ପାଠ କରି ପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

6. The next morning …………………………………. by the Emperor.
ଓଡ଼ିଆ ଅନୁବାଦ | ସାରାଂଶ :
ପରଦିନ ସକାଳେ ଆଠଜଣଯାକ କର୍ମଚାରୀ ବୀରବଲ୍ଲଙ୍କ ନିକଟକୁ ସେମାନଙ୍କ ବାଡ଼ି ସହିତ ଆସି ପହଞ୍ଚିଲେ । ବୀରବଲ୍ଲ ସେମାନଙ୍କୁ ଗୋଟିଏ ଧାଡ଼ିରେ ଛିଡ଼ା ହେବାକୁ କହିଲେ ଏବଂ ଅତି ସନ୍ତର୍ପଣରେ ସେମାନଙ୍କ ବାଡ଼ିଗୁଡ଼ିକୁ ଲକ୍ଷ୍ୟ କଲେ । ହଠାତ୍ ଜନୈକ କର୍ମଚାରୀକୁ ଧରିପକାଇଲେ ଏବଂ ଆକବରଙ୍କ ନିକଟକୁ ନେଇଗଲେ । ସେ ଆକବରଙ୍କୁ କହିଲେ ଯେ ସେ ବ୍ୟକ୍ତିଟି ହିଁ ପ୍ରକୃତ ଚୋର । ଅଭିଯୁକ୍ତ କର୍ମଚାରୀଟି ଆକବରଙ୍କ ଗୋଡ଼ତଳେ ପଡ଼ିଯାଇ କହିଲା ଯେ ସେ ହିଁ ସେ ମୁଦିଟି ଚୋରି କରିଥିଲା । ଏହାପରେ ସମ୍ରାଟ କର୍ମଚାରୀଟିକୁ ଦଣ୍ଡିତ କରିଥିଲେ ।

7. How did …………………………………………. of others.
ଓଡ଼ିଆ ଅନୁବାଦ | ସାରାଂଶ :
ଆଠଜଣ କର୍ମଚାରୀଙ୍କ ମଧ୍ୟରୁ କେଉଁ ବ୍ୟକ୍ତିଟି ମୁଦି ଚୋରି କରିଥିଲା, ତାହା ବୀରବଲ୍ଲ ଜାଣିଲେ କିପରି ? ଚୋରି କରିଥିବା କର୍ମଚାରୀଜଣକ ବୀରବଲ୍ଲ ଙ୍କ କହିବା ଅନୁଯାୟୀ ବିଶ୍ଵାସ କରିଗଲା ଯେ ସେହି ରାତିରେ ତା’ବାଡ଼ିଟି ନିଶ୍ଚୟ ଆଙ୍ଗୁଳେ ବଢିଯିବ । ଏଣୁ ସେ ଘରକୁ ଯାଇ ଛୁରି ସାହାଯ୍ୟରେ ଗୋଟିଏ ଆଙ୍ଗୁଳି ଲମ୍ବର ଅଂଶ କାଟିଦେଇଥିଲା । ଏଣୁ ସେ ଲୋକଟିର ବାଡ଼ିଖଣ୍ଡିକ ଅନ୍ୟମାନଙ୍କ ବାଡ଼ିଠାରୁ ଛୋଟ ହୋଇଯାଇଥିଲା ।

Word Meaning

line : a formation of people or things one behind other (ଧାଡ଼ି)
suddenly : quickly and unexpectedly (ହଠାତ୍ / ସହସା / ଆକସ୍ମିକ ଭାବେ)
caught hold of: to hold / prevent / find out (ଧରିବା / ଅଟକାଇବା)
punished : to make someone suffer due to bad act (ଡେରିରେ ଆସିଥିବା ପିଲା)
length : extent from end to end (ଦୈର୍ଘ୍ୟ)
noticed : to mark I observe / find (ଲକ୍ଷ୍ୟକରିବା / ଦେଖିବା)

Session – 3
III. Post-Reading (ପଢ଼ିବା ପରେ)

1. Visual Memory Development Technique (VMDT):
Whole Text : the emperor’s favourite ring – (ସମ୍ରାଟଙ୍କ ପ୍ରିୟ ମୁଦି)
The ring was found missing – (ମୁଦି ହଜିଯିବା)
Birbal’s plan to catch the thief – (ବୀରବଲ୍ଲଙ୍କର ଚୋର ଧରିବା ଯୋଜନା)
The thief was caught. – ( ଚୋର ଧରାପଡ଼ିଲା )
Part : Paragraph 1 – Akbar fond of jewellery, his favourite ring.

2 Comprehension Activity : (ବୋଧ ପରିମାପକ କାର୍ଯ୍ୟାବଳୀ)

(a) MCQs: Choose the correct alternatives and complete each sentence.
(ଠିକ୍ ବିକଳ୍ପ ଉତ୍ତରଗୁଡ଼ିକ ବାଛ ଏବଂ ପ୍ରତ୍ୟେକ ବାକ୍ୟ ପୂର୍ଣ୍ଣ କରି ଲେଖ ।)

Question 1.
Akbar liked one of his rings the most because ___________.
(A) a number of pearls and diamonds were set on it.
(B) it was the most precious of all the rings he had.
(C) It was made with special designs.
(D) it was a present to him from the queen.
Answer:
(A) a number of pearls and diamonds were set on it.

Question 2.
The ring was stolen by ___________.
(A) one of the eight servants
(B) all the eight servants
(C) Birbal and one of the servants
(D) Birbal and all the servants
Answer:
(A) one of the eight servants

Question 3.
Akbar asked Birbal ___________.
(A) to get a new ring made
(B) to search the lost ring
(C) to find the thief
(D) to entertain him with a story
Answer:
(C) to find the thief

Question 4.
Birbal told that the thief’s stick would ___________.
(A) become shorter
(B) become longer
(C) become gold
(D) be broken into pieces
Answer:
(B) become longer

Question 5.
Birbal was ___________ to catch the thief.
(A) kind
(B) cruel
(C) clever
(D) foolish
Answer:
(C) clever

(b) The following sentences are about the story you read. But they are not in order. Write their serial numbers in the boxes.

[ ] All the eight servants came back to Birbal with their sticks. Birbal called all the servants to his presence.
[ ] He said the thief’s stick would be longer that night.
[ ] One day Akbar lost his favorite ring.
[ ] He asked Birbal to find the thief.
[ ] The thief cut his stick to make it shorter.
[ ] He gave a stick to each servant.
[ ] The thief was caught at last.

Answer:
[ 6 ] All the eight servants came back to Birbal with their sticks.
[ 3 ] Birbal called all the servants to his presence.
[ 5 ] He said the thief’s stick would be longer that night.
[ 1 ] One day Akbar lost his favorite ring.
[ 2 ] He asked Birbal to find the thief.
[ 7 ] The thief cut his stick to make it shorter.
[ 4 ] He gave a stick to each servant.
[ 8 ] The thief was caught at last.

Session – 4

3. Listening : (ଶ୍ରବଣ)
Your teacher will read out the following statements. S/he will not read them in order. Listen to him/her and write the serial number in boxes of each as read out. (ତୁମ ଶିକ୍ଷକ ଏହି ତଥ୍ୟସମ୍ବଳିତ ବାକ୍ୟଗୁଡ଼ିକ ପାଠକରିବେ । ସେ ଏହାକୁ କ୍ରମ ଅନୁସାରେ ପଢ଼ିବେ ନାହିଁ । ତାଙ୍କଠାରୁ ଶୁଣି ତୁମେ ସେଗୁଡ଼ିକର ଠିକ୍ କ୍ରମିକ ନମ୍ବର ତୁମେ ଲେଖ ।)

(i) Akbar was very fond of jewelry. [ i ]
(ii) The queen presented him with a diamond ring. [ ii ]
(iii) The ring was his favorite. [ iii ]
(iv) He put on the ring when he went to the royal assembly. [ iv ]
(v) One day the ring was stolen. [ v ]
(vi) Clever Birbal found the thief. [ vi ]
(vii) Akbar was happy and praised Birbal. [ vii ]

4. Speaking : (କଥନ)
The jumbled up sentences of the story in Activity 2(b) have already been put in order. Teacher will ask the students to read out the sentences serially in a chain in the class. For example :
(ଏପଟସେପଟ ହୋଇ ରହିଥ‌ିବା ବାକ୍ୟଗୁଡ଼ିକ ତୁମ୍ଭେମାନେ ସଜାଇ ଲେଖୁଛି । ଶିକ୍ଷକ ଛାତ୍ରଛାତ୍ରୀମାନଙ୍କୁ ସେଗୁଡ଼ିକୁ ପଢ଼ି ଶୁଣାଇ ଛାତ୍ରଛାତ୍ରୀମାନଙ୍କୁ କହିବେ)

Student 1: One day Akbar lost his favourite ring.
Student 2: He asked Birbal to find out the thief.
Student 3: …………………………………………………………………..
…………………………………………………………………………………….

Session – 5
5. Vocabulary : (ଶବ୍ଦଜ୍ଞାନ)

Find new words hiding in the word.
Example:
Servant: Van, ant
catch: — — t
palace: — a —e
jewelry: j — — —l
ready: r — — d
missing: m— — s, s — n —
become: — o — e
punished: — — , s — e, sh — —
pearl : p — a — , e — —

Answer:
catch: c a t
palace: l a c e
jewelry: j e w e l
ready: r e a d
missing: m i s s, s i n g
become: c o m e
punished: i s , s h e, s h e d
pearl : p e a r, e a r

6. Writing : (ଲିଖନ)
a. You have put the story sentences in order in 2. b.
(2 (b) ରେ ଥ‌ିବା ଗଳ୍ପର ବାକ୍ୟଗୁଡ଼ିକୁ ତୁମେ ସଜାଇଲେଖୁଛି । ବର୍ତ୍ତମାନ ସେଗୁଡ଼ିକୁ ପ୍ରୟୋଗ କରି ଗଳ୍ପଟି ଲେଖ)

Now use the sentences and write the story ‘The Missing Ring’.
One day Akbar lost his favorite ring. _________________
___________________________________________
___________________________________________
___________________________________________

Answer:
(6) କୁ ଆଉଥରେ ଏଠାରେ ଲେଖ ।

Session  – 6

(b) Write answers to the following questions :

Question (i)
Which ring was Akbar’s favorite?
(କେଉଁ ମୁଦିଟି ଆକବରଙ୍କ ଅତିପ୍ରିୟ ଥିଲା ?)
Answer:
Akbar’s favorite ring was one large ring in which diamonds and many other gems had been set.

Question (ii)
What did the eight servants do every day?
(ଆଠଜଣ ଯାକ କର୍ମଚାରୀ ପ୍ରତିଦିନ କ’ଣ କରୁଥ‌ିଲେ ?)
Answer:
The eight servants looked after the -Emperor’s clothes and jewelry every day.

Question (iii)
What did Akbar feel about the missing ring? Why did he think so?
(ଆକବର ହଜିଥ‌ିବା ମୁଦି ସମ୍ପର୍କରେ କ’ଣ ଅନୁଭବ କଲେ; କାହିଁକି ଏପରି ସେ ଭାବିଲେ ?)
Answer:
Akbar felt that one of his servants had stolen his missing ring. He thought so because none other than those eight servants was allowed to enter the Emperor’s room.

Question (iv)
Who did he ask to find out the thief?
(ସେ କାହାକୁ ମୁଦିଟିକୁ ଖୋଜିବାକୁ କହିଲେ ?)
Answer:
He asked his clever minister Birbal to find out the thief.

Question (v)
What did Birbal say about the thief’s stick?
(ବୀରବଲ୍ଲ ଚୋରର କାଠି ସମ୍ବନ୍ଧରେ କ’ଣ କହିଲେ ?)
Answer:
Birbal said about the thief’s stick that the same stick would be longer in size by one finger’s length that night.

Question (vi)
Why did the thief cut his stick?
(ଚୋର କାହିଁକି ତା’ର ବାଡ଼ିଟିକୁ କାଟିଦେଲା ?)
Answer:
The thief believed the story of Birbal and cut his stick so that his stick would not be shown longer than other’s sticks the next day.

Session – 7

7. Mental Talk : (ମାନସିକ ଆଳାପ)

• You cannot hide wrongdoing that leads to another.
• It is the wise man who watches the fool.

Tail-Piece

You enjoyed how cleverly Birbal caught the thief. Read the story to learn an interesting way to find the guilty.
(ବୀରବଲ୍ଲ କିପରି ଚତୁରତାର ସହ ଚୋରଟିକୁ ଧରିଲେ । ତାହା ତୁମେ ପାଠକରି ଆନନ୍ଦ ଲାଭ କର । ଦୋଷୀକୁ ଧରିବା ଉପରେ ଆଉ ଗୋଟିଏ ଗପ ଏବେ ଶୁଣ ।)

FINDING THE THIEF (ଚୋର ଧରିବା)

One night …………………………………….. that he has taken.

ଗୋଟିଏ ରାତିରେ ଜଣେ ବୃଦ୍ଧବ୍ୟକ୍ତି ଗୋଟିଏ କ୍ଷୁଦ୍ର ଗ୍ରାମରେ ପହଞ୍ଚି ଛୋଟିଆ ହୋଟେଲରେ ରହିଲେ । ସେ ସେଠାରେ ତାଙ୍କ ରାତ୍ରିଭୋଜନ ସମାପ୍ତ କରି ଶୋଇବା ପାଇଁ ଉପର ମହଲାରେ ଥିବା ତାଙ୍କ କୋଠରିକୁ ଗଲେ । ସେହି ମୁହୂର୍ଭରେ ହୋଟେଲର ପରିଚାଳକ ସେଠାରେ ପହଞ୍ଚି ତାଙ୍କୁ ତୁରନ୍ତ ତଳ ମହଲାକୁ ଆସି ତାଙ୍କୁ କୌଣସି କାର୍ଯ୍ୟରେ ସାହାଯ୍ୟ କରିବାକୁ ନିବେଦନ କଲେ । ବୃଦ୍ଧଲୋକଟି ଉପରକୁ ଉଠିଲେ ଏବଂ ସଙ୍ଗେ ସଙ୍ଗେ ପୁନର୍ବାର ତଳକୁ ଚାଲିଲେ ଏବଂ ଘଟଣା କ’ଣ ବୋଲି ସେ ପଚାରିଲେ । ପରିଚାଳକ କହିଲେ, ‘ଦେଖନ୍ତୁ ଏଇ ପାଞ୍ଚଜଣ ଏଠାକୁ ଗତକାଲି ରାତିରେ ଆସିଥିଲେ । ବର୍ତ୍ତମାନ ସେମାନଙ୍କ ମଧ୍ୟରୁ ଜଣେ କହୁଛି ଯେ ତା’ର ଟଙ୍କା ହଜିଛି ଏବଂ ସେହି ସମୟରୁ ଏମାନଙ୍କ ମଧ୍ୟରୁ କୌଣସି ବ୍ୟକ୍ତି ବାହାରକୁ ଯାଇ ନାହାନ୍ତି ।

ଆମର ସମସ୍ତଙ୍କର ଆପଣଙ୍କ ଉପରେ ବିଶ୍ୱାସ ଅଛି । ଦୟାକରି ଯଦି ପାରନ୍ତି, କିଏ ଟଙ୍କା ଚୋରିକରିଛି ତାକୁ ବାହାରକରନ୍ତୁ । ବୃଦ୍ଧଜଣକ କହିଲେ ‘ମୋତେ ଗୋଟିଏ କୁକୁଡ଼ା ଦିଅ ।’’ ପରିଚାଳକ ବାହାରକୁ ଗଲେ ଏବଂ ଗୋଟିଏ କୁକୁଡ଼ା ଆଣିଦେଲେ । ଉପସ୍ଥିତ ବ୍ୟକ୍ତିମାନେ ବୃଦ୍ଧ କ’ଣ କରିବାକୁ ଯାଉଛନ୍ତି, ତାହା ଜାଣିବାପାଇଁ ଆଶ୍ଚର୍ଯ୍ୟ ହେଉଥିଲେ । ଏହାପରେ ବୃଦ୍ଧଜଣକ ଗୋଟିଏ ପାତ୍ର ଆଣିବାକୁ କହିଲେ ଏବଂ ପରିଚାଳକ ରୋଷେଇଶାଳ ଯାଇ ଚୁଲି ନିଆଁରେ ଥ‌ିବା ଗୋଟିଏ ବଡ଼ ପାତ୍ର ଆଣି ତାଙ୍କୁ ଦେଲେ । ଏହାପରେ ସେ କହିଲେ, ମୁଁ ବର୍ତ୍ତମାନ ଏହି ପାତ୍ରଟିକୁ କୁକୁଡ଼ା ଉପରେ ଉଗୁଡ଼ାଇଦେବି । ସେ କୁକୁଡ଼ାକୁ ଟେବୁଲ ଉପରେ ରଖ୍ ତା’ଉପରେ ଉତ୍ତପ୍ତ ପାତ୍ରଟିକୁ ଉଗୁଡ଼ାଇଦେଲେ । ଏହାପରେ ସେ ଘରର ଦ୍ବାର ବନ୍ଦ କରି ଘରର ଚାବିଟି ତାଙ୍କ ନିକଟରେ ରଖିଲେ ।

ଏହାପରେ ସେ ଲୋକମାନଙ୍କୁ ଘରକୁ ଅନ୍ଧାର କରିଦେବାକୁ କହି ନିଆଁ ଲିଭାଇଦେବାକୁ କହିଲେ । ସେ ଅନୁଯାୟୀ ସମସ୍ତେ ଘରଟିକୁ ସମ୍ପୂର୍ଣ୍ଣ ଅନ୍ଧାର କରିଦେଲେ । ତା’ପରେ ବୃଦ୍ଧ ବ୍ୟକ୍ତିଜଣକ କହିଲେ, ‘ବର୍ତ୍ତମାନ ତୁମେମାନେ ଗୋଟିଏ ଧାଡ଼ିରେ ଛିଡ଼ାହୋଇ ପାତ୍ରଟିକୁ ଜଣକ ପରେ ଜଣେ ଘଷ; ମାତ୍ର ଚୋରଟି ପାତ୍ରକୁ ଘଷିବା ସମୟରେ କୁକୁଡ଼ାଟି ତିନିଥର ରାବକରିବ । ସମସ୍ତ ଲୋକ ଜଣକ ପରେ ଜଣେ ଯାଇ ପାତ୍ରଟିକୁ ଘଷିଲେ । କୌଣସି ଶବ୍ଦ ହେଲା ନାହିଁ । କୁକୁଡ଼ାଟି ମଧ୍ୟ ରାବ କଲା ନାହିଁ । ଏହାପରେ ବୃଦ୍ଧବ୍ୟକ୍ତି ପଚାରିଲେ, ‘ତୁମମାନଙ୍କ ମଧ୍ୟରୁ କେହି ଏ ପାତ୍ର ନ ଘଷି ରହିଗଲ କି ?’’ କେହି କିଛି କହିଲେ ନାହିଁ । ବର୍ତ୍ତମାନ ଘରକୁ ଆଲୋକିତ କରିବାକୁ ସେ କହିଲେ । ଲୋକମାନେ ବତିଗୁଡ଼ିକ ପୁଣି ଜଳାଇଲେ ।

ଆଲୋକରେ ଘରଟି ପୁନର୍ବାର ଉଜ୍ଜ୍ଵଳ ଆଲୋକରେ ଉଦ୍ଭାସିତ ହେଲା । ତା’ପରେ ବୃଦ୍ଧଜଣକ ଲୋକମାନଙ୍କୁ କହିଲେ, ‘ବର୍ତ୍ତମାନ ତୁମେ ପୁଣିଥରେ ଧାଡ଼ିରେ ଛିଡ଼ାହୁଅ ଏବଂ ପ୍ରୟୋଗ ମୋ ଆଡ଼କୁ ହାତ ବଢ଼ାଅ । ସେମାନେ ତାହା ହିଁ କଲେ । ଏଥର ବୃଦ୍ଧବ୍ୟକ୍ତିଟି ସମସ୍ତ ହାତଗୁଡ଼ିକୁ ସତର୍କତାର ସହିତ ନିରୀକ୍ଷଣ କଲେ । ଲକ୍ଷ୍ୟକଲେ ସେଠାରେ ଥ‌ିବା ପାଞ୍ଚଜଣ ଲୋକମାନଙ୍କ ମଧ୍ୟରୁ ଜଣକର ହାତ ପାତ୍ରର କଳାରେ ଆଦୌ କଳା ହୋଇ ନଥିଲା । ସେଇ ଲୋକଟିକୁ ହିଁ ସେ ପ୍ରକୃତ ଚୋର ବୋଲି ଜାଣିପାରିଲେ ଏବଂ କହିଲେ ‘ଏଇ ବ୍ୟକ୍ତି ହିଁ ପ୍ରକୃତ ଚୋର । ତାଙ୍କୁ ଧର ଏବଂ ତାଙ୍କୁ ଚୋରି ଟଙ୍କା ଫେରାଇବାକୁ କୁହ ।

Word Meaning

crow (v): to make a loud noise by cock.
downstairs: (opposite: upstairs), on or to the ground floor.
missing : lost (ହଜିଯାଇଥ)
rub: to move your hand, a cloth, etc. backward and forwards over a surface, ଘଷିବା, ମାଜିବା
turn upside down: move the top part to the bottom
soot: a black powder that is produced when wood, coal, etc. is burnt

Odisha State Board BSE Odisha 8th Class English Solutions Lesson 2 Math-Magic Textbook Exercise Questions and Answers.

BSE Odisha Class 8 English Solutions Lesson 2 Math-Magic

Session – 1

I. Pre-Reading (ପ୍ରାକ୍-ପଠନ):

→ There are 35 chocolates. You have to divide them into three. The first one will get 1/2 of the total chocolates. The second will get l/3rd and the third one will get l/9th. Can you divide the chocolates among the three without breaking the chocolates according to the plan? If yes, you know math and you are intelligent.
୩୫ଟି ଚକୋଲେଟ୍ ଅଛି । ତିନିଜଣଙ୍କ ଭିତରେ ବାଣ୍ଟିବାକୁ ହେବ । ପ୍ରଥମ ଜଣକ ସମୁଦାୟ ଚକୋଲେଟ୍‌ର ଅର୍ଦ୍ଧେକ ପାଇବ । ଦ୍ବିତୀୟ ଜଣକ ସମୁଦାୟ ଚକୋଲେଟ୍‌ର ଏକ ତୃତୀୟାଂଶ ଏବଂ ତୃତୀୟ ଜଣକ ସମୁଦାୟର ଏକ ନବମାଂଶ ପାଇବ । କୌଣସି ଚକୋଲେଟ୍କୁ ନ ଭାଙ୍ଗି ଏହି ଅନୁଯାୟୀ ତୁମେ ତିନିଜଣଙ୍କ ମଧ୍ୟରେ ଚକୋଲେଟ୍‌କୁ ବାଣ୍ଟି ପାରିବ କି ? ଯଦି ପାରିବ, ତାହାହେଲେ ତୁମେ ଗଣିତ ଜାଣ ଏବଂ ତୁମେ ବୁଦ୍ଧିମାନ ବୋଲି ପରିଗଣିତ ହେବ ।

→ Read this story to see how this is to be done.
ଗପଟି ପଢ଼ ଏବଂ ଏହା କିପରି ହୋଇପାରେ ଦେଖ ।

II. While Reading (ପଢ଼ିବା ସମୟରେ)

Text

• SGP – 1
• Read paragraphs 1-2 silently and answer the questions that follow.

1. Once upon ………………………………….. on sand.
ଏକଦା ଆରବ ଭୂମିରେ ଜଣେ ଯୁବକ ରହୁଥିଲେ । ଦିନେ ସେ ଗୋଟିଏ ବାଲୁକା-ପାହାଡ଼ ଉପରେ ବସି ଆକାଶରେ ଉଡୁଥ‌ିବା ପକ୍ଷୀଦଳଙ୍କୁ ଗଣନା କରୁଥିଲା ଏବଂ ତାକୁ ବାଲିରେ ଲେଖୁଥିଲା ।

2. A man riding ………………………………………….. man.
ଜଣେ ଲୋକ ଓଟରେ ଚଢ଼ି ସେଇବାଟ ଦେଇ ଅତିକ୍ରମ କରୁଥିଲା । ସେ ଯୁବକଟିର ସଠିକ୍ ଉଡ଼ନ୍ତା ପକ୍ଷୀ ଗଣନା ଦୃଶ୍ୟ ଦେଖ୍ ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ । ସେ ମନେ ମନେ ଭାବିଲେ, ବାଳକଟି କିପରି ଆକାଶରେ ଉଡୁଥ୍‌ ଅଗଣିତ ପକ୍ଷୀମାନଙ୍କୁ ଠିକ୍‌ରୁପେ ଗଣନା କରିପାରୁଛି ? ସେ ଓଟ ପିଠିରୁ ଓହ୍ଲାଇ ସେ ବାଳକ ଆଡ଼େ ଆସିଲା ଏବଂ ସେ ଏହା କିପରି ଉପାୟରେ କରିପାରୁଛି ବୋଲି ପ୍ରଶ୍ନ କଲେ । ଯୁବକଟି କହିଲା, ‘ମୁଁ ଗୋଟିଏ ଧନୀଲୋକର ମେଣ୍ଢା ପାଳକ ଭାବରେ କାର୍ଯ୍ୟ କରୁଥିଲି । ମୁଁ ସେଠାରେ ପ୍ରତ୍ୟେକ ଦିନ ମେଣ୍ଢାମାନଙ୍କୁ ଗଣନା କରୁଥୁଲି । ଆଗନ୍ତୁକ ବ୍ୟକ୍ତି ଜଣକ କହିଲେ, ‘ତାହା ହେଲେ ତୁମେ ଗୋଟିଏ ଭଲ କାମ ପାଇପାରିବ । ମୋ ସାଙ୍ଗରେ ଆସ ।’’

Word Meaning

once upon a time : a time in the past (ଏକଦା)
sand-hill : hill of sand by the sea or in the desert (ବାଲି ସ୍ତୁପ, ବାଲିହୁଙ୍କା, ବାଲିର ପାହାଡ଼)
counting : the act of counting / reciting numbers in ascending order (ଗଣିବା)
flocks : a group of birds (ଦଳେ ପକ୍ଷୀ)
sky : the atmosphere and outer space as viewed from the earth (ଆକାଶ)
passing by : moving fast (ଅତିକ୍ରମ କରିବା)
suprised : to be amazed (ଆଶ୍ଚର୍ଯ୍ୟ ହେବା, ଆଶ୍ଚର୍ଯ୍ୟଚକିତ)
correctly : in a correct way / in an accurate way (ଠିକ୍ ଭାବରେ)
groups : any number of members considered as a unit (ଦଳ ଭାବରେ)
shepherd : one who takes care of sheep (ମେଷପାଳକ)
job : a specific piece of work required to be done as a duty or for a specific fee (ଚାକିରି)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
What was the young man doing?
(ଯୁବକଟି କ’ଣ କରୁଥିଲା ?)
Answer:
The young man was counting the flocks of birds flying in the sky.

Question 2.
Who saw him?
(ତାକୁ କିଏ ଦେଖିଲା ?)
Answer:
A man riding a camel saw him.

Question 3.
Why was he surprised?
(ସେ କାହିଁକି ଆଶ୍ଚର୍ଯ୍ୟ ହେଲା ?)
Answer:
He was surprised to see the boy counting the flying birds correctly.

Question 4.
How did the young man learn counting?
(କିପରି ଯୁବକଟି ଗଣିବା ଶିଖ୍ ?)
Answer:
The young man learnt counting when he was working as a shepherd boy with a rich man.

Question 5.
What did the man say to praise the boy?
(ଲୋକଟି ବାଳକଟିକୁ ପ୍ରଶଂସା କରିବାକୁ ଯାଇ କ’ଣ କହିଲା ?)
Answer:
To praise the boy he said he could get a good job.

• SGP – 2
• Read paragraphs 3-4 silently and answer the questions that follow.

3. The man ……………………………………………… among themselves.
ଲୋକଟି ତାଙ୍କୁ ତା ଓଟରେ ବସାଇ ଚାଲିଲା । ସେ ରାଜାଙ୍କ ଉଆସରେ ଗୋଟିଏ ଚାକିରି ଦେବାକୁ ତାକୁ ଆଶ୍ୱାସନା ଦେଲା । ବାଟରେ ତିନିଜଣ ବ୍ୟକ୍ତି କଳି କରୁଥିବା ଏବଂ ନିକଟରେ ୩୫ଟି ଓଟ ଛିଡ଼ା ହେବା ସେମାନେ ଦେଖ‌ିଲେ । ସେମାନଙ୍କ ମଧ୍ୟରେ ୩୫ଟି ଓଟ କିପରି ବଣ୍ଟା ହେବ ସେଥ‌ିପାଇଁ କଳି ହେଉଥିଲା । ସେମାନଙ୍କର ପିତା ମୃତ୍ୟୁବରଣ କରିଥିଲେ । ତାଙ୍କ ମୃତ୍ୟୁ ପୂର୍ବରୁ ସେ ଗୋଟିଏ ଇଚ୍ଛାପତ୍ର ଲେଖୁଥିଲେ । ତାଙ୍କ ଇଚ୍ଛାପତ୍ର ଅନୁଯାୟୀ ୩୫ଟି ଓଟ ମଧ୍ୟରୁ ବଡ଼ପୁଅ ସମୁଦାୟ ଓଟ ସଂଖ୍ୟାର ଅର୍ଦ୍ଧେକ, ଦ୍ୱିତୀୟ ପୁତ୍ର ଏକ ତୃତୀୟାଂଶ ଏବଂ ସାନପୁଅ ଏକ ନବମାଂଶ ପାଇବ ।

4. This was ……………………………………. 4 camels.
ଏହା ବଡ଼ କଠିନ ବ୍ୟାପାର ଥିଲା । ପିତାଙ୍କ ଇଚ୍ଛାପତ୍ର ଅନୁଯାୟୀ ବଡ଼ପୁଅ ୧୭ଟି, ଦ୍ୱିତୀୟ ପୁତ୍ର ୧୨ଟିରୁ ସାମାନ୍ୟ କିଛି କମ ଏବଂ ସାନପୁଅ ୪ଟିରୁ କିଛି କମ୍ ପାଇବେ ।

Word Meaning

promised : to make a commitment to do something in future (ପ୍ରତିଜ୍ଞା କରିବା)
quarrel : to make an argument / an angry dispute (କଳି କରିବା /ଯୁକ୍ତି କରିବା)
nearby : very near / close to (ଖୁବ୍ ନିକଟରେ)
divide : to separate / divide (ବାଣ୍ଟି ଦେବା | ଭାଗ କରିବା)
will : legal document about what will happen to
somebody’ s money and property after he dies (ଆଇନଗତ କାଗଜ ଯାହା ମୃତ୍ୟୁ ପରେ ଜଣକର ଟଙ୍କା ଓ ସମ୍ପତ୍ତି ପାଇବାର ଲୋକଙ୍କ ବିଷୟରେ ସୂଚନା ଥାଏ)
according to : a stated / reported by somebody (ଅନୁସାରେ)
one third : one out of three equal parts (ଏକ-ତୃତୀୟାଂଶ)
one ninth : one out of nine equal parts
less : not as much (କମ୍)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
What did the man promise the boy?
(ଲୋକଟି ବାଳକ ପାଖରେ କ’ଣ ପ୍ରତିଜ୍ଞା କଲା ?)
Answer:
The man promised the boy to give a good job.

Question 2.
What did they see on their way?
(ବାଟରେ ସେମାନେ କ’ଣ ଦେଖ‌ିଲେ ?)
Answer:
They saw three persons quarreling and thirty-five camels standing nearby.

Question 3.
What was the father’s wish?
(ବାପାଙ୍କର ଇଚ୍ଛା କ’ଣ ଥିଲା ?)
Answer:
The father’s wish was to give half of the whole camel to the eldest son, one-third to the second son, and one-ninth to the third son.

Question 4.
‘This was difficult’. What was difficult ? (para – 4)
(‘ଏହା କଷ୍ଟକର ଥିଲା ।’ ଏଠାରେ କ’ଣ କଷ୍ଟକର ଥିଲା ?)
Answer:
This was difficult because according to the father’s will the eldest son would get 17 1/2 camels. The second son would get one-third or a little less than 12 and the third son would get a little less than 4 camels.

Session – 2

• SGP – 3
• Read paragraphs 5-7 silently and answer the questions that follow.

5. The young ………………………………………… happy too.
ସେମାନଙ୍କ କଥା ଶୁଣି ଯୁବକଟି କହିଲା, ‘‘ଏହା ଖୁବ୍ ସହଜ ।’’ ଏହା କହି ୩୫ଟି ଓଟ ସହିତ ସେମାନେ ଆସିଥିବା ଓଟଟିକୁ ସେଥ୍ରେ ଯୋଗ କରିଦେଲା । ବର୍ତ୍ତମାନ ସମୁଦାୟ ଓଟଙ୍କ ସଂଖ୍ୟା ୩୬ ହୋଇଗଲା । ବଡ଼ପୁଅଟି ଯିଏ ୧୭– ଓଟ ପାଇବା କଥା ସେ ୧୮ଟି ଓଟ ପାଇଲା । ସେ ଖୁସି ହୋଇଗଲା । ଦ୍ୱିତୀୟ ପୁଅଟି ଯେ ୧୨ଟିରୁ କମ୍ ପାଇବାର ଥିଲା ସେ ୧୨ଟି ଓଟ ପାଇଲା । ସେ ମଧ୍ୟ ଖୁସି ହୋଇଗଲା । ସବା ସାନପୁଅ ଯେ ୪ଟିରୁ କମ୍ ପାଇବାର ଥିଲା, ସେ ୪ଟି ଓଟ ପାଇଲା । ସେ ମଧ୍ୟ ଖୁସି ହେଲା ।

6. But the ………………………………………………. Think how?
କିନ୍ତୁ ସବୁଠାରୁ ଅଧିକ ଖୁସି ଥିଲା ଗଣିତ ଦକ୍ଷ ବାଳକଟି । ସେମାନଙ୍କ ସମସ୍ୟା ସମାଧାନ ହୋଇଥ‌ିବାରୁ ତାକୁ ଗୋଟିଏ ଓଟ ମିଳିଗଲା । ତିନି ଭାଇଙ୍କର ୩୫ଟି ଓଟ ଥିଲା । କିନ୍ତୁ ସେମାନେ (୧୮ + ୧୨ + ୪) ହିସାବରେ ଓଟ ପାଇଲେ । ସବୁ ଭାଇ ଗୋଟିଏ ଲେଖାଏଁ ଓଟ ଅଧ୍ଵ ପାଇଲେ । ଗଣିତ ଦକ୍ଷ ଯୁବକଟିର କୌଣସି ଓଟ ନଥିଲେ ମଧ୍ୟ ସେ ଗୋଟିଏ ପାଇଲା କିପରି ଏହା ହେଲା ଭାବ ।

7. This ……………………………………… a job.
ଏହା ହିଁ ଗଣିତ ଜ୍ଞାନର କୁହୁକ ଏବଂ ଦୁଇ ଆରବୀୟ ଏହାପରେ ରାଜାଙ୍କ ପ୍ରାସାଦ ଆଡେ ଚାକିରି ଉଦ୍ଦେଶ୍ୟରେ ଚାଲିଲେ ।

Word Meaning

listen : to hear I pay attention (ଶୁଣିବା)
add : to join (ଯୋଗ କରିବା)
herd : a group of animals of the same type
eldest : first in order of birth (ବୟସରେ ସବୁଠାରୁ ବଡ଼)
solving : finding a solution to a problem (ସମାଧାନ କରିବା)
gain : to get some profit / to benefit (ଲାଭ କରିବା | ପାଇବା )
magic : tricks (ଯାଦୁ । କୌଶଳ)
seek : to search / want to get (ଖୋଜିବା | ପାଇବାକୁ ଚାହିଁବା)

Comprehension Questions and Answers: (ସଂକ୍ଷିପ୍ତ ପ୍ରଶ୍ନୋତ୍ତର)

Question 1.
“It’s very easy.” What was very easy?
(‘ଏହା ଭାରି ସହଜ ।’ କେଉଁଟି ଭାରି ସହଜ ?)
Answer:
“It is very easy”. This “very easy” was to distribute 35 camels among three brothers according to the will of their father.

Question 2.
What did the young Arab do to divide the camel?
(ଯୁବ ଆରବୀୟ ଲୋକଟି ଓଟପଲକୁ ଭାଗ ଭାଗ କରିବା ପାଇଁ କ’ଣ କଲା ?)
Answer:
The young Arab added the camel they were riding to divide the camels.

Question 3.
How many camels did each brother get?
(ପ୍ରତ୍ୟେକ ଭାଇ କେତେକ ଓଟ ପାଇଲେ ?)
Answer:
The eldest brother got 18 camels which were half of the total camels, the camels, the second brother got 12 which was 1/3 of the total camels and the youngest brother got 4. which was 1/9 of the total camels.

Question 4.
Why were they happy?
(ସେମାନେ କାହିଁକି ଖୁସି ହେଲେ ?)
Answer:
They all were happy because they got one more camel than they expected.

Question 5.
Who was the happiest of all the persons present? Why?
(ଉପସ୍ଥିତ ଲୋକମାନଙ୍କ ମଧ୍ୟରୁ କିଏ ଅଧିକ ଖୁସି ହେଲା ? କାହିଁକି ?)
Answer:
The young math man was the happiest of all persons present. Because he gained one though he had no camel of his own.

Question 6.
Every brother thought he had gained. But did they really gain or lose?
(ପ୍ରତ୍ୟେକ ଭାଇ ଅଧିକ ପାଇଲେ ବୋଲି ଭାବିଲେ । କିନ୍ତୁ ସେମାନେ ପ୍ରକୃତରେ ଲାଭ ବା କ୍ଷତିରେ ରହିଲେ କି ?)
Answer:
Every brother thought he had gained. But really they did not gain. Out of the total channels they lost one.

Question 7.
Do you think the young man will get a job in the king’s court?
(ଯୁବକଟି ରାଜାଙ୍କ ଦରବାରରେ ଚାକିରି ପାଇଲା ବୋଲି ତୁମେ ଭାବୁଛ କି ?)
Answer:
Yes, we think the young man will get a job in the king’s court.

Session – 3

III. Post-Reading (ପଢ଼ିବା ପରେ)

(The math man solved the problem with the camels. Try to solve your chocolate problem. See, whether this is possible or not. One who solves the problem gets one chocolate as a reward).

1. Visual Memory Development Technique (VMDT)
Whole Text: The young man was sitting on a sand hill and counting the birds. three brothers quarreling, The young man got a horse by solving their problem.
Part: Paragraph 5 – got 18, thirty-five, It’s very easy.

2. Comprehension Activities : (ବୁଦ୍ଧିପରିମାପକ କାର୍ଯ୍ୟାବଳୀ)

MCQs:
Choose the correct alternatives and fill in the blanks.

Question 1.
The young man counted the ___________birds.
(A) swimming
(B) sleeping
(C) sitting
(D) flying
Answer:
(D) flying

Question 2.
The three persons were quarreling because they could not divide ___________.
(A) their cows
(B) their goats
(C) their buffaloes
(D) their camels
Answer:
(D) their camels

Question 3.
The young man was good at ___________.
(A) Mathematics
(B) Language
(C) Science
(D) Geography
Answer:
(A) Mathematics

Question 4.
In the end, ___________ lost a camel.
(A) the eldest brother
(B) the second brother
(C) the youngest brother
(D) all the brothers
Answer:
(D) all the brothers

Session – 4

3. Listening : (ଶ୍ରବଣ)
(a) Your teacher read out the following lines from paragraph 3. Listen to him/her carefully and fill in the blanks with the words missing.
On their way they saw three persons ___________ and thirty-five ___________ standing ___________. Their ___________ was on how to ___________ the ___________ among them.

Answer:
On their way, they saw three persons quarreling and thirty-five camels standing nearby. Their quarrel was on how to divide the camels among them.

4. Speaking : (କଥନ)
→  Practise these dialogues.
→  Steps:
1. Rehearsal-teacher reads aloud, and Students listen. The teacher reads aloud and students repeat after him/her dialogue by dialogue.
2. Teacher vs Students
3. Students vs Students (in two groups)
(They do this reading from the text).

Man: Hey young man, what are you doing here?
Young man: Counting the birds.
Man: Counting the flying birds?
Young man: Yes.
Man: Such a large number?
Young man: I can count them correctly.
Man: How did you learn to count?
Young man: I worked as a shepherd boy. Every day I counted the herd of my sheep.
Man: You are very good at counting. You’ll get a good job.
Young man: Shall I ? Where? When?
Man: Come with me. I’ll take you to the king.

Session – 5

5. Vocabulary : (ଶବ୍ଦଜ୍ଞାନ)
Put the letters in the right order and make words. Clues/hints will help you.

• ands: — — — —
  (You can see it on deserts, seashores,s or river banks)
• imeca: — — — — —
  (This animal is called the ship of the desert)
• aing: — — — —
  (The opposite word of ‘loss’)
• scimatheMta: — — — — — — — — — — —
  (a subject of study)
• herdshep: — — — — — — — —
  (a person whose job is to take care of sheep)
• barA: — — — —
  (a man born in Arab)

Answer:
sand
camel
gain
Mathematics
shepherd
Arab

6. Writing : (ଲିଖନ)

(a). Put the following sentences in the right order and get the story. Put the serial number in the box given against each sentence.
(i) Everybody was happy. [ ]
(ii) The man took him on his camel and went away. [ ]
(iii) He was surprised to see the boy good at counting. [ ]
(iv) Once a young man was sitting on a sand hill. [ ]
(v) The boy divided the camels among them. [ ]
(vi) A man sitting on a camel was passing by. [ ]
(vii) On the way they saw three men quarreling over dividing thirty-five camels among themselves. [ ]
(viii)He promised him a job with the king. [ ]
(ix) The men got thirty-four camels and the boy got one. [ ]
(x) He was counting a large number of flying birds. [ ]

Answer:
Answer:
(i) Everybody was happy. [ 10 ]
(ii) The man took him on his camel and went away. [ 5 ]
(iii) He was surprised to see the boy good at counting. [ 4 ]
(iv) Once a young man was sitting on a sand hill. [ 1 ]
(v) The boy divided the camels among them. [ 8 ]
(vi) A man sitting on a camel was passing by. [ 3 ]
(vii) On the way they saw three men quarreling over dividing thirty-five camels among themselves. [ 7 ]
(viii)He promised him a job with the king. [ 6 ]
(ix) The men got thirty-four camels and the boy got one. [ 9 ]
(x) He was counting a large number of flying birds. [ 2 ]

Session – 6

(b). Use the sentences you have arranged and rewrite the story in the space below.
Math Magic
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
Answer:
Once a young man was sitting on a sand hill. He was counting a large number of flying birds. A man sitting on a camel was passing by. He was surprised to see the boy good at counting. The man took him on his camel and Went away. He promised him a job with the Icing. On the way, they saw three men quarreling over dividing thirty-five camels among themselves. The boy divided the camels among them. The men got thirty-four camels and the boy got one. Everybody was happy.

(c). Write answers to the following questions.

Question (i)
Where was the young man sitting? What was he doing there?
Answer:
The young man was sitting on a sand hill. He was counting a large number of flying birds in the sky.

Question (ii)
How did he learn counting?
Answer:
He learnt counting when he was working as a shepherd boy with a rich man.

Question (iii)
Who was quarreling?
Answer:
The three brothers were quarreling.

Question (iv)
Who decided their quarrel?
Answer:
The young math man decided their quarrel.

Question (v)
The three brothers and the young men were happy. What made them happy?
Answer:
The three brothers and the young men were happy because the three brothers got one more camel each they expected. The young man gained a camel though he had no camel of his own.

Session – 7

7. Mental Talk : (ମାନସିକ ଆଳାପ)
One who knows math can solve many problems.

8. Let’s think : (ଆସ ଚିନ୍ତା କରିବା)
Our brain is powerful. We should make good use of it.

Tail-Piece

Wise men solve other’s problems. But who will help them when they are in a fix (in trouble) ? Here follows a story for you to read and get the answer.

SIX WISE MEN (ଛଅଜଣ ବିଜ୍ଞ ଲୋକ)

Once ……………………………………………………….. on our journey.
ଥରେ ଛ’ଜଣ ବୁଦ୍ଧିମାନ ବ୍ୟକ୍ତି ଏକତ୍ର ଏକ ଯାତ୍ରା ଆରମ୍ଭ କଲେ । ବାଟରେ ଗୋଟିଏ ଗଭୀର ନଦୀ ପଡ଼ିଲା । ସେଠାରେ କୌଣସି ଡଙ୍ଗା ନଥିଲା । ତେଣୁ ସେମାନେ ନଦୀକୁ ପହଁରି ଅତିକ୍ରମ କଲେ ଏବଂ ନଦୀ ଆରପଟେ ନିରାପଦରେ
ପହଞ୍ଚିଲେ ।

‘‘ଆମେ ସମସ୍ତେ ନିରାପଦ କି ?’’ ସେମାନଙ୍କ ମଧ୍ୟରୁ ଜଣେ ବିଜ୍ଞ ବ୍ୟକ୍ତି ପ୍ରଶ୍ନ କଲେ ।
“ ଆସ ଏ ସମ୍ପର୍କରେ ନିଶ୍ଚିତ ହେବା ।’’ ଅନ୍ୟ ସମସ୍ତ ବିଜ୍ଞ ବ୍ୟକ୍ତି କହିଲେ ।
ପ୍ରଥମ ବ୍ୟକ୍ତିଟି ଅନ୍ୟମାନଙ୍କୁ ଗଣିଲା ।
ଏକ, ଦୁଇ, ତିନି, ଚାରି, ପାଞ୍ଚ ଏବଂ କହିଲା,
‘‘ଦେଖ ଆମ ଭିତରୁ ଜଣେ କେହି ନାହିଁ । ଆମେ
ମାତ୍ର୫ ଜଣ ଅଛୁ ।’’
‘ତୁମେ ଗୋଟିଏ ବାଉଳା ଲୋକ ।’’
ଦେଖ, ମୁଁ ଗଣୁଛି । ଏକ, ଦୁଇ, ତିନି, ଚାରି,
ପାଞ୍ଚ…….. ‘ ଆରେ ! ସତେତ ଆମେ ପାଞ୍ଚ
ଅଛୁ । ହେ ଭଗବାନ ! ଆମ ଭିତରୁ ଜଣେ
ହଜିଛି ।’’

ତୃତୀୟ ବ୍ୟକ୍ତିଟି ଗଣିଲା ଏବଂ ଚତୁର୍ଥ
ବ୍ୟକ୍ତିଟି ମଧ୍ୟ । ହଁ, ପ୍ରକୃତରେ ସେମାନେ ପାଞ୍ଚଜଣ ?
ସେମାନେ ସମସ୍ତେ ବସି ପଡ଼ିଲେ ଏବଂ କ୍ରନ୍ଦନ କଲେ, ‘ଆମ ଭିତରୁ ଜଣେ ବୁଡ଼ିଯାଇଛି । କି ଦୁଃସମ୍ବାଦ ! ଆମର ପ୍ରିୟ ବନ୍ଧୁ ଜଣେ ବୁଡ଼ିଯାଇଛି । ଆମେ ବର୍ତ୍ତମାନ କ’ଣ କରିବା ?
କିଛି ସମୟ ପରେ ଦଳର ମୁଖ୍ୟା କହିଲେ, ‘ଆମର ଯାତ୍ରାର ଅନୁକୂଳ ଖରାପ ଥିଲା । ଏଣୁ ଆମେ ଜଣକୁ ହରେଇଲେ । ଆମେ ଆଉ ଯିବା ନାହିଁ । ଆସ ଘରକୁ ଫେରିଯିବା । ସମସ୍ତେ ରାଜି ହେଲେ ।
ବୁଦ୍ଧିମାନ ବ୍ୟକ୍ତିମାନେ ନଦୀକୁ ଅତିକ୍ରମ କରି ଆରପାରିରେ ପହଞ୍ଚଲେ । ତାଙ୍କ ଗାଁର ଜଣେ ଗରିବ ଅପାଠୁଆ ବ୍ୟକ୍ତି ଗାଁର ସେମାନଙ୍କୁ ଦେଖୁଲା ଏବଂ କହିଲା, ‘ଆପଣମାନେ ଆଉ ଭ୍ରମଣରେ ଗଲେ ନାହିଁକି ?’’
ନା ଆମେ ଯାଉନାହୁଁ । ‘ଦଳର ମୁଖ୍ୟା କହିଲେ, ‘ଆମେ ଆମ ଭିତରୁ ଜଣକୁ ହରାଇଲୁ । ଏଣୁ ଘରକୁ ଫେରି ଯାଉଛୁ ।’’
ଅପାଠୁଆ ଲୋକଟି ପଚାରିଲା, ‘‘ଆପଣଙ୍କ ଭିତରୁ କିଏ ହଜିଲା ?’’
‘‘ଆମେ ଜାଣୁନା । ଆମେ ଛ’ଜଣ ଥିଲୁ । ବର୍ତ୍ତମାନ ପାଞ୍ଚ ଅଛୁ ।’’ ଅପାଠୁଆ ଗାଁ ବାଲାଟି ସେମାନଙ୍କୁ ଗଣିଲା । ଦେଖିଲା, ସେମାନେ ଛ’ଜଣ ଅଛନ୍ତି । ପାଞ୍ଚଜଣ ନୁହଁନ୍ତି । ସେ ଚିନ୍ତାକଲା । ‘ଏ ଲୋକଗୁଡ଼ା ପ୍ରକୃତରେ ବୋକା । ସେମାନେ ପ୍ରତ୍ୟେକ ଅନ୍ୟମାନଙ୍କୁ ଗଣି ନିଜକୁ ଛାଡ଼ି ଦେଇଛନ୍ତି । ବର୍ତ୍ତମାନ କେମିତି ଗଣନା ହୁଏ, ତାଙ୍କୁ ମୁଁ ଶିଖେଇବି ।’’
‘‘ତୁମେମାନେ ଧାଡ଼ିକରି ଛିଡ଼ା ହୁଅ । ମୁଁ ଗଣିବି ।’’ ଗାଉଁଲି ଲୋକଟି ପ୍ରଥମ ବ୍ୟକ୍ତି ପାଖକୁ ଗଲା ତା’ମୁଣ୍ଡରେ ଗୋଟିଏ ଦାଗ ଦେଇ ଗଣିଲା ‘ଏକ’। ସେ ତା’ପରେ ଦ୍ବିତୀୟ ବ୍ୟକ୍ତିର ମୁଣ୍ଡରେ ଦାଗ ଦେଇ କହିଲା ‘ଦୁଇ’ । ଏହିପରି ଗଣି ଗଣି ଗଲା । ଷଷ୍ଠ ବ୍ୟକ୍ତି ମୁଣ୍ଡରେ ଦାଗ ଦେଇ ଗଣିଲା ‘ଛଅ’ । ବର୍ତ୍ତମାନ ତୁମେ ଦେଖ । ତୁମେମାନେ ‘ଛଅ’, ପାଞ୍ଚ ନୁହଁ । ‘ହେ, ପ୍ରକୃତରେ ତୁମେ ଠିକ୍ । ଆମେ ବର୍ତ୍ତମାନ ଛଅଜଣ ଅଛୁ ।’’ ସେମାନଙ୍କ ମଧ୍ୟରୁ ଜଣେ ବୁଦ୍ଧିମାନ ବ୍ୟକ୍ତି କହିଲା । ବୁଦ୍ଧିମାନ ବ୍ୟକ୍ତିଙ୍କ ମୁଖୁଆ କହିଲେ,’’ ଯାହାହେଉ, ଭଲହେଲା । ହଜିଥ‌ିବା ବ୍ୟକ୍ତିଟି ଫେରିଆସିଲା । ବର୍ତ୍ତମାନ ଆସ ଯାତ୍ରା ଆରମ୍ଭ କରିବା ।’’

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 9 Binomial Theorem will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 9 Binomial Theorem

Binomial Theorem For Positive Integral Index:
For any a,b ∈ R, and n ∈ N
(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1b + ….. ⁿC_n bⁿ

Note:

(a) (a + b)ⁿ = aⁿ + na^n-1 b + \(\frac{n(n-1)}{2 !}\) a^n-2b² ….. + bⁿ
(b) (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ….. + ⁿC_n xⁿ
(c) (a – b)ⁿ = ⁿC₀ aⁿ – ⁿC₁ a^n-1 b + ⁿC₂ a^n-2b² ….. + (-1)ⁿ bⁿ
(d) (1 – x)ⁿ = ⁿC₀ – ⁿC₁ x + ⁿC₂ x² ….. + (-1)ⁿ xⁿ

Some conclusions from the Binomial theorem:

• There are (n + 1) terms in the expansion of (a + b)ⁿ
• We can write (a + b)ⁿ = \(\sum_{r=0}^n{ }^n \mathrm{C}_r a^{n-r} b^r\) and (a – b)ⁿ = \(\sum_{r=0}^n(-1)^r{ }^n \mathrm{C}_r a^{n-r} b^r\)
• The sum of powers of a and b in each term = n
• As ⁿC_r = ⁿC_n-r (The coefficient of terms equidistant from the beginning and the end are equal).
• (r + 1)^th term (General term)
  = t_r+1 = ⁿC_r a^n-rb^r
• (a + b)ⁿ + (a – b)ⁿ = 2[ⁿC₀aⁿ + ⁿC₂ a^n-2b² + ….]
• (a + b)ⁿ – (a – b)ⁿ = 2[ⁿC₁ a^n-1b + ⁿC₃ a^n-3b³ + ….]
• (middle terms):
  ⇒ If n is even then the middle term = \(t_{\left(\frac{n+2}{2}\right)}=t_{\left(\frac{n}{2}+1\right)}\)
  ⇒ If n is odd there are two middle terms. They are = \(t_{\left(\frac{n+1}{2}\right)} \text { and } t_{\left(\frac{n+3}{2}\right)}\)
• t_r+1 from the end in the expansion of (a + b)ⁿ = t_r+1 from the beginning in the expansion of (b + a)ⁿ.

Binomial Theorem For Any Rational Index:
If n ∈ Q and x ∈ R such that |x| < 1 then (1 + x)ⁿ = 1 + nx + \(\frac{n(n-1)}{2 !} x^2\) + \(\frac{n(n-1)(n-2)}{3 !} x^3+\ldots .\)

Note:

(1) (1 + x)^-1 = 1 – x + x² – x³ + …..
(2) (1 – x)^-1 = 1 + x + x² + …..
(3) (1 + x)^-2 = 1 – 2x + 3x² – 4x³ + …..
(4) (1 – x)^-2 = 1 + 2x + 3x² + 4x³ + …..

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 8 Permutations And Combinations

Fundamental Principle Of Counting:
(a) Fundamental principle of Multiplication:
If we choose an element from set A with m element and then one element from set B  with n elements, then are total number of ways we can make a choice is exactly mn.

OR

If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of ways in which both the events can occur in succession in mn ways.

(b) Fundamental Principle of addition: If there are two events such that they can be performed independently in m and n different ways respectively, then either of two events can be performed in (m + n) ways.

Note:
(a) Use the multiplication principle if by doing one part of the job, the job remains incomplete.
(b) Use the addition principle if by doing one part of the job, the job is completed.

Factorial Notation:
If n ∈ N then the factorial of n, denoted by n! or ∠n is defined as
n! = n (n – 1). (n – 2) … 3.2.1.

Note:
0! = 1

Properties of Factorial:
(1) Factorial of negative integers is not defined
(2) n! = n(n – 1)!
= n(n – 1) (n – 2)!
= n(n – 1) (n – 2) (n – 3)!
(3) \(\frac{n !}{r !}\) = n(n – 1) (n – 2) ….. (r + 1)
(4) Exponent of a prime number p in n! denoted by
\(\mathrm{E}_p(n !)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\ldots \ldots\)

Permutation:
Each of the arrangements which can be made by taking some or all objects or things at a time is called a permutation.

(a) Permutation of n different objects:

• Number of permutations of n different objects have taken all at a time = \({ }^n \mathrm{P}_n\) = n!.
• Number of permutations of n different objects taken none at a time = \({ }^n \mathrm{P}_0\) = 1
• Number of permutations of n different objects taken r at a time = \({ }^n \mathrm{P}_r\) = P(n, r) = \(\frac{n !}{(n-r) !}\)

(b) Permutation ofnon-distinct objects:
(1) Number of permutations of n objects taken all at a time of which p objects are of same kind and others are distinct = \(\frac{n !}{p !}\)
(2) Number of permutations of n objects taken all at a time of which p objects are of one kind, q objects are of a second kind and other are distinct = \(\frac{n !}{p ! q !}\)
(3) Number of permutations of n objects taken all at a time in which p₁ objects are of one kind, p₂ are of second kind, p₃ are 3rd kind ….. and
p_n are of nth kind and other are distinct. = \(\frac{n !}{p_{1} ! p_{2} ! \cdots p_{n} !}\)

(c) Restricted permutations:

• Permutation of distinct objects with repetition: The number of permutations of n different things taken r at a time when each thing may be repeated any number of times = n^r
• Number of permutations of n different things taken r at a time when a particular thing is to be always included in each arrangement = r. ^n-1P_r-1.
• Number of permutations of n different things, taken r at the time when p particular are to be always included in each arrangement = P(r – (p – 1) ^n-pP_r-p.
• Number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement = ^n-1P_r.
• Number of permutations of n different things taken r at a time, when p particular things never taken in each arrangement = ^n-pP_r.

(d) Circular permutation:
(1) When we do an arrangement of objects along a closed curve we call it the circular permutation.
(2) Number of circular permutations of n distinct objects taken all at a time = (n – 1)!, where clockwise and anti-clockwise orders are taken as different, as arrangements round a table.
(3) Number of circular permutations of n distinct objects taken all at a time, where clockwise and anti-clockwise orders make no difference as beads or flowers in a necklace or garland.
= \(\frac{(n-1) !}{2}\)
(4) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders are different = \(\frac{\left({ }^n \mathrm{P}_r\right)}{r}\)
(5) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders make no difference = \(\frac{\left({ }^n \mathrm{P}_r\right)}{2 r}\)

(e) Some more restricted permutations:

• Number of permutations of n different things taken all at a time, when m specified things come together = m!(n – m + 1)!.
• Number of permutations of n different things taken all at a time when m specified things never come together = n!  – m!(n – m + 1)!.

Combinations:
Each of the different selections made by taking some or all objects at a time irrespective of any order is called a combination.

(a) Difference between permutation and combination:

• A combination is a selection but a permutation is not a selection but an arrangement.
• In combination the order of appearance of objects is immaterial, whereas in a permutation the ordering is essential.
• Practically to find permutations of n different objects taken r at a time, we first select objects then we arrange them.
• One combination corresponds to many permutations.

(b) Combinations of n different things taken r at a time:
The number of combinations of n different things have taken r at a time ⁿc_r = C(n, r) = \(\left(\begin{array}{l}
n \\
r
\end{array}\right)=\frac{n !}{r !(n-r) !}\)

(c) Properties of ⁿc_r :
(1) ⁿc_r = ⁿC₀ = 1, ⁿC₁ = n
(2) ⁿC_r = ⁿC_n-r
(3) ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r (Euler’s formula)
(4) ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
(5) n. ^n-1C_r-1 = (n – r + 1) ⁿC_r-1
(6) ⁿC_r = \(\frac{n}{r}{ }^{n-1} \mathrm{C}_{r-1}\)
(7) \(\frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r-1}}=\frac{n-r+1}{r}\)
(8) If n is even then the greatest value of ⁿC_r is ⁿC_n/2.
⇒ If n is odd then the greatest value of ⁿC_r is \({ }^n \mathrm{C}_{\left(\frac{n+1}{2}\right)} \text { or }{ }^n \mathrm{C}_{\left(\frac{n-1}{2}\right)}\)

(d) Number of combinations of n different things taken r at a time, when k particular things always occur = ^n-kC_r-k

(e) The number of combinations of n different things, taken r at a time where k particular things never occur = ^n-kC_r

(f) The total number of combinations of n different things taken one or more at a time (or the number of ways of n different things selecting at least one of them) = ⁿC₁ + ⁿC₂ + ⁿC₃ + ….. + ⁿC_n = 2ⁿ -1

(g) The number of combinations of n identical things taken r at a time = 1.

(h) Number of ways of selecting r things out of n alike things where r = 0, 1, 2, 3 ….. n is (n+ 1).

(i) Division into groups:

• The number of ways in which (m + n) different things can be divided into two groups which contain m and n things respectively = \(\frac{(m+n) !}{m ! n !}\) for m ≠ n.
• If m-n then the groups are of equal size. Thus, division can be done in two ways as:
  ⇒ If order of groups is not important: In this case the number of ways = \(\frac{(2 n) !}{2 !(n !)^2}\)
  ⇒ If order of groups is important: In this case the number of ways = \(\frac{(2 n) !}{(n !)^2}\)

(j) Arrangement in groups:

1. The number of ways in which n different things can be arranged into r different groups = ^n+r-1P_n or n! ^n-1C_r-1
2. The number of ways in which n different things can be distributed into r different groups = rⁿ – ^rC₁(r – 1)ⁿ + ^rC₂(r – 2)ⁿ ….. + (-1)^r-1 . ^rC_r-1. (Blank groups are not allowed)
3. The number of ways in which n identical things can be distributed into r different groups where blank groups are allowed
   = ^(n+r-1)C_(r-1)
   = ^(n+r-1)C_n
4. Number of ways in which n identical things can be distributed into r different groups where blank groups are not allowed (each group receives at least one item) = ^n-1C_r-1

(k) Number of divisors:

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text A: That Lean, Hungry Look School Textbook Activity Questions and Answers.

CHSE Odisha 11th Class Alternative English Solutions Unit 4 Text A: That Lean, Hungry Look

Activity – 1
Purpose And Attitude And The Text Type

Question 1.
Which of the following described the writer’s attitude to thin people?
(i) impressed
(ii) complementary
(iii) disapproving
(iv) condemning
(v) approving
(vi) noncommittal
Answer:
(iv) condemning.

Question 2.
Which of the following phrases best expresses the writer’s purpose?
(i) to present objective information
(ii) to present both sides of a controversial issue
(iii) to shock the reader with an unconventional point of view
(iv) to persuade the reader that fat people are better than thin people.
(v) to express his dislike of thin people.
Answer:
(iv) to persuade the reader that fat people are better than people.

Question 3.
Which of the following categories of text type would you say the article belongs to?
(i) informative
(ii) imaginative
(iii) expressing an opinion
(iv) descriptive
(v) narrative
Answer:
(iii) expressing an opinion.

Question 4.
What is the general tone of the article?
(i) ironic
(ii) humorous
(iii) matter of fact
(iv) Passionate
(v) serious
Answer:
(iii) matter of fact.

Activity-3
Getting The Main Ideas Of Paragraphs

Match the paragraph in column A with the titles in column B and then say whether a title refers to thin people or fat people.

Answer:

A	B
1.	(vi) dangerous people.
2.	(v) no absolute truth.
3.	(ix) speedy metabolism
4	(vii) relaxed and fun-loving
5.	(xi) seeing all sides
6.	(xii) life is illogical and unfair.
7.	(i) a long list of logical things
8.	(viii) happiness is elusive.
9.	(iv) muddling through rather than saving time.
10.	(x) not enough time for work.
11.	(iii) love of math and morality.
12.	(ii) loving and accepting

Activity 4
Understanding Patterns Of Comparison And Contrast

There are two important ways of developing a comparison and contrast text, namely (i) the block method and the point-by-point method. In the Block method, you single out one basic way in which the two objects are alike or different. For example, if you are comparing two people at work, the introductory paragraph would tell the reader what your article would be about. The first body paragraph of the article would show something about one person’s approach to work, the following body paragraph would focus on the other person’s approach.

And in the concluding paragraph, you would briefly summarise the topic and give a dominant impression about the similarities and/or differences in the two worker’s approaches to their jobs. However, instead of deciding to compare and contrast the two objects one after another, you may decide not to separate the two objects you are discussing. You may then adopt the point-by-point method and treat’ both objects together as you present each point of comparison or contrast. You may have discovered that both of these methods have been employed in Text A.

a) Which patterns of comparison and Contrast does the writer use in paragraphs 2-5 and paragraphs 8-14?
Answer:
Point-by-point method.

b) Which pattern does she use in paragraphs 6,7,11 and 12?
Answer:
Block method.

c) Which of these two patterns do you find more effective and why?
Answer:
Both these patterns are effective in dealing with a problem. However, the point-by-point pattern is more effective because the comparison and contrast will be clear in the treatment and approach in this pattern.

d) Does the writer state the thesis explicitly? If so, where does she state it?
Answer:
The writer states the thesis of this work of art explicitly. It appears in paragraph -1.

e) How does the conclusion support the thesis? Write a few words on the appropriateness or otherwise of the conclusion.
Answer:
The conclusion almost sums up the nature and pursuit of the thin and fat people described in the previous paragraphs. The concluding paragraph is eloquent of the strong contrasts between fat and thin people.

Section – A
The paragraph below is the beginning paragraph of Text-A. Read it quickly and try to guess what Text-A is about.

Ceasar was right. Thin people need watching. I have been watching them for most of my adult life and 1 do not like what I see. When these narrow fellows spring at me, quiver to my toes. Thin people come in all personalities, most of them menacing. You have got your ‘together’ thin person, your mechanical thin person. Your condescending thin person, your efficiency expert thin person. All of them are dangerous.
Now read Text-A, which is adopt?  from an article in news week in the year 197. in order to check whether your prediction made above is right.

That Lean, Hungry Look Summary in English

According to Ceasar, thin people need watching. The writer has been watching such people for most of his adult life and never likes what he sees. He says when these thin fellows spring at him he trembles to his toes. Thin people come in all personalities and most if they are dangerous. Thin people in the first place are not fun. They have always got to be going something. They make others tired. They get speedily little metabolisms that cause them to burtle briskly. Sluggish, inert, easy-going fat people are preferable to thin ones.

Fat people don’t chattel all day long. Thin people turn mean and hard at a young age because they never learn the value of a hot fudge Sunday for easing tension. They are firm and fresh and dull like carrots. Thin people believe in logic, fat people see all sides. Fat people realize that life is illogical and unfair. They know well that God is not in heaven and all is not right with the world. If God was up there, fat people could leave two doughnuts and a big orange drink the time they wanted it.

Thin people have a long list of logical things they are always spouting off to me. They hold up one finger at a time as they reel off these things. They speak slowly as if to a young child. The list is long and full of holes. They also think these 2,000-point plans lead to happiness. Fat people know happiness is elusive at best and even if ey could get the kind of thin people to talk about, they wouldn’t want it. Fat people see that such programs are too dull, too hard, and too off the mark. They are never better than a whole cheesecake.

However, fat people know all about the mystery of life. They get acquainted with the night, luck, and fate, and play them by ears. The main problem with people is that they oppress. Thin people are downers. They like math and morality and reasoned evaluation of the limitations of human beings. They expound prognosis, probe, and prick. Fat people are friendly and cheerful. Fat people will talk continuously, trade quickly, laugh loudly, gyrate, and gossip. They are generous, giving, and gallant. They are gluttonous, goodly, and great.

Analytical outlines:

• According to Ceasar, thin people need to be watched minutely.
• He has been watching such people for most of his adult life.
• He calls them as narrow fellows.
• When they spring at him, he trembles to his toes.
• They appear in all personalities.
• Most of them are dangerous.
• Thin people in the first place are not having fun.
• They have always got to be doing something.
• Give them a coffee break.
• They will job around the block.
• They make others tired.
• They have got a speedily little metabolism.
• It makes them to burtle briskly.

• They have forever been rubbing their bony hands together.
• They have also been eying new problems to tackle.
• But the fat people are sluggish, inert, easy going.
• So, they are preferable to the thin ones.
• Fat people don’t chartle all day long.
• Thin people turn mean and hard at a young age.
• Because they never learn the value of a hot fudge Sunday for casing tension.
• They are firm and fresh and dull like carrots.
• They go straight to the heart of the matter.
• But fat people let things stay all blurry, hazy, and vague.
• They want to race the truth.
• Fat people know there is no truth.
• Thin people believe in logic.
• Fat people see all sides.
• Fat people realize that life is illogical and unfair.
• They know very well that God is not in heaven.
• They consider all is not right with the world.
• If God was up there, fat people could have two doughnuts and a big orange drink any time they wanted it.
• Thin people have a long list of logical things.
• They are always spouting off to him.
• They hold up one finger. at a time as they reel off these things.
• They speak slowly as if to a young child.
• Their list is long and full of holes.
• They think about 2000-point plans.
• They think it must lead them to happiness.
• Fat people know happiness is elusive at best.

• They don’t want as the thin people talk about it.
• To fat people, such programs are too dull, too hard, and too off the mark.
• They are never better than a whole cheesecake.
• Fat people know all about the mystery of life.
• They get acquainted with might, luck, and fate, and playing them by ears.
• But the main problem with the thin men is that they oppress.
• Their good intentions, bony torsos, tight, ships, neat corners, cerebral machinations, and pet solutions loom like dark clouds over the
• loose, comfortable, spread out, soft world of the fat.
• Thin people are downers.
• They like math and morality.
• They also like reasoned evaluation of the limitations of human beings.
• They have their skinny little acts together.
• They expound prognoses, probes, and prick.
• Fat people are convivial that is jovial.
• They even like irregular people.
• They will come up with a good reason.
• Fat people are generous, giving, and gallant.
• They are also gluttonous, goodly, and great.
• They are friendly and cheerful.
• Fat people will gab, giggle, guffa, gyrate, and gossip.
• They have plenty of room to be free and frank.

Meaning of difficult words:

to goof off – to make a trivial mistake.
to burtle – to move around quickly.
sluggish – moving or reacting more slowly than normal.
chartling – bulging out of amusement.
wizened – small and thin and wrinkled.
shrivel led – dried up and bent, became small.
gooey – sticky, soft, and sweet.
not-fudge sonde- a hot and soft creamy light
brown sweet dish made from the ice-cream, fruits, and nuts.
crunchy – firm and fresh.
nebulous – not clear or exact, fainted.
doughnuts – small and cakes.
elusive – difficult to achieve.
muscled – covered the ground with

decaying leaves to improve its quality.
double-fudged – with two layers of chocolate or cream dressing.
cerebral machination – secret and clever plans made by the brain.
rutabagas – a king of roots.
punch line – the last few words of a joke or story.
dovners – a person who stops your
feeling cheerful or happy.
convivial – friendly and cheerful.
gab – talk continuously.
guffaw – laugh loudly.
gyrate – turn around fast in circles.
giggle – moving from side to side with quick short movements.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Unit imaginary number ‘i’.
The unit imaginary number i = √-1
i² = -1
i³ = -i
i⁴ = 1
In general (i)⁴ⁿ = 1, (i)⁴ⁿ⁺¹ = i, (i)⁴ⁿ⁺² = -1, and (i)⁴ⁿ⁺³ = -i.
⇒ If a and b are positive real numbers then
√-a × √-b = -√ab
√a × √b = √ab

Complex Number
General form: = z = a +ib

• a = Real part of (z)  = Re (z)
• b = Imaginary part of (z) = I_m(z)
• a + i0 is purely real and 0 + ib is purely imaginary .
• a + ib = c + id iff a = c and b = d

Complex Algebra
(a) Addition of complex numbers
If z₁ = a + ib and z₂ = c + id then z₁ + z₂ = (a + c) + i(b + d)

Properties:

• Addition is commutative: z₁ + z₂ = z₂ + z₁
• Addition is associative: (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
• 0 + i0 is the additive identity.
• -z is the additive inverse of z.

(b) Subtraction of complex numbers:
z₁ = a + ib and z₂ = c + id then z₁ – z₂ = (a – c) + i(b – d)

(c) Multiplication of complex numbers:
z₁ = a + ib and z₂ = c + id then z₁z₂ = (ac – bd) + i(bc + ad)

Properties:

• Multiplication is commutative: z₁z₂ = z₂z₁
• Multiplication is associative: z₁(z₂z₃) = z₁z₂(z₃)
• 1 = 1 + i0 is the multiplicative identity.
• If z = a + ib then the inverse of z.
  z^-1 = \(\frac{1}{a+i b}=\frac{a-i b}{(a+i b)(a-i b)}\)
  = \(\frac{a-i b}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{i b}{a^2+b^2}\)
• Multiplication is distributive over addition. z₁(z₂ + z₃) = z₁z₂ + z₁z₃

Conjugate and modulus of a complex number:
If  z = a + ib the conjugate of z is \(\bar{Z}\) = a – ib.
⇒ We get conjugate by replacing i by (-i) Modulus of z = a + ib is denoted by |z| and |z| = \(\sqrt{a^2+b^2}\)

Properties Of Conjugate:
(i) \((\overline{\bar{z}})\) = z
(ii) z + \(\bar{z}\) = 2 Re (z)
(iii) z – \(\bar{z}\) = 2i m̂ (z)
(iv) z – \(\bar{z}\) ⇔ z is purely real
(v) Conjugate of real number is itself.
(vi) z + \(\bar{z}\) = 0 ⇒ z is purely imaginary.
(vii) z. \(\bar{z}\) = [Re(z)]² + [m̂(z)]²
= a² + b²
= |z|²
(viii) \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\)
(ix) \(\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\)
(x) \(\overline{z_1z_2}=\overline{z_1}\overline{z_2}\)
(xi) \(\left(\overline{\frac{z_1}{z_2}}\right)=\frac{\overline{z_1}}{\overline{z_2}}\)

Properties of modulus:
(1) Order relations are not defined for complex numbers. i,e,. z₁ > z₂ or z₁ < z₂ has no meaning but |z₁| < |z₂| or |z₁| > |z₂| is meaningful because |z₁| and |z₂| are real numbers.
(2) |z|  = 0 ⇔ z = 0
(3) |z| = |\(\bar{z}\)| = |-z|
(4) |z| ≤ Re (z) ≤ |z| and -|z| ≤ m̂ (z) ≤ |z|
(5) |z₁z₂| = |z₁| |z₂|
(6) \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\)
(7) |z₁ ± z₂|² = |z₁|² + |z₂|² ± 2 Re (z₁\(\bar{z}_2\))
(8) |z₁ + z₂|² = |z₁ – z₂|² = 2(|z₁|² + |z₂|²)
(9) |z₁ + z₂|² ≤ |z₁| + |z₂|

Square Root Of Complex Number:
Let z = a + ib
Let √z = x + iy

If b > 0 then x and y are taken as same sign.
If b < 0 then x and y are of opposite sign.

Representation of a complex number:
We represent a complex number in different forms like
(i) Geometrical form
(ii) Vector form
(iii) Polar form
(iv) Eulerian form or Exponential form

(i) Geometrical form:
Geometrically z = x + iy = (x, y) represents a point in a coordinate plane known as Argand plane or Gaussian plane.

(ii) Vector form:
In vector form a complex number z = x + iy is the vector \(\overrightarrow{\mathrm{OP}}\) where p(x, y) is the point in the cartesian plane.

(iii) Polar form:
A complex number z = x + iy  in polar form can be written as z = r(cos θ + i sin θ) where r = \(\sqrt{x^2+y^2}\) = |z| and θ is called the argument and -π < θ ≤ π. Technique to write z = x + iy in polar form.
Step – 1: Find r = |z| = \(\sqrt{x^2+y^2}\)
Step – 2: Find α = tan^-1 \(\left|\frac{y}{x}\right|\)
Step – 3:
θ = α for x > 0, y > 0
θ = π – α for x > 0, y > 0
θ = -π + α for x > 0, y > 0
θ = -α for x > 0, y > 0
Step – 4: Write z = r(cos θ + i sin θ)

(iv) Eulerian form or Exponential form z = r e^iθ, because e^iθ = cos θ + i sin θ where θ is the argument and r is the modulus if z.

Note:
(1) |z₁ z₂ z₃ ….. z_n| = |z₁||z₂| …. |z_n|
(2) arg (z₁z₂ …. Z_n) = arg (z₁) + arg (z₂) + ….. + arg (z_n)
(3) arg \(\left(\frac{z_1}{z_2}\right)\) = arg (z₁) – arg (z₂)
(4) arg \((\bar{z})\) = -arg (z)

Cube Roots Of Unity:
Cube roots of unity are 1, ω, ω² where ω = \(\frac{-1 \pm i \sqrt{3}}{2}\)

Properties of Cube roots of unity:
(i) Cube roots of unity lie on unit circle |z| = 1
(ii) 1 + ω + ω² = 0
(iii) Cube roots of -1 are -1, -ω, -ω²
(iv) 1 + ωⁿ + ω²ⁿ \(=\left\{\begin{array}{l}
0 \text { if } n \text { is not a multiple of } 3 \\
3 \text { if } n \text { is a multiple of } 3
\end{array}\right.\)
(v) z³ + 1 = (z + 1) (z + ω) (z + ω²)
(vi) -ω and -ω² are roots of z² – z + 1  = 0.

De-moivre’s theorem:
(a) (De-moivre’s theorem for integral index)
(cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)

(b) (De-moivre’s theorem for rational index)
cos (nθ) + i sin (nθ) is one of the values of (cos θ + i sin θ)ⁿ

(c) nth roots of unity
nth roots of unity are 1, α, α², α³ …..α^n-1. where α = e^{i\(\frac{2 \pi}{n}\)} = cos \(\frac{2 \pi}{n}\) + i sin \(\frac{2 \pi}{n}\)

Properties:

• 1 + α + α² ….. + α^n-1 = 0
• 1 + α^p + α^2p + ….. + α^(n-1)p \(= \begin{cases}0 & \text { if } p \text { is not a multiple of } n \\ n & \text { if } p \text { is a multiple of } n\end{cases}\)
• 1. α. α² ….. α^n-1 = (-1)^n-1
• zⁿ – 1 = (z – 1) (z – α) (z – α²) …..(z – α^n-1)

Quadratic Equations:
The general form: ax² + bx + c = 0  …(i)
Solutions of quadratic equation(1) are
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
D = b² – 4ac is called the discrimination of a quadratic equation.
D > 0 ⇒ The equation has real and distinct roots.
D = 0 ⇒ The equation has real and equal roots.
D < 0 ⇒ The equation has complex roots.

Note:
In a quadratic equation with real coefficients, the complex roots occur in conjugate pairs.