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Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.
1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)
Solution:
Let p_n be the given statement

Question 2.
1² + 2² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p_n be the given statement
when n = 1
1² =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)
P₁ is true
Let P_k be true.
i.e. 1² + 2² + … + k² = \(\frac{k(k+1)(2 k+1)}{6}\)
we shall prove P_k + 1 is true i.e., 1² + 2² + … + k² + (k + 1)²
\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)

∴ P_k+1 is true
∴ P_n is true of all values of n∈N.

Question 3.
1 + r + r²+ …. + rⁿ = \(\frac{r^{n+1}-1}{r-1}\)
Solution:
Let P_n be the given statement

Question 4.
5ⁿ – 1 is divisible by 4.
Solution:
Let P_n = 5ⁿ – 1
When n = 1,
5¹ – 1 = 4 is divisible by 4.
∴ P₁, is true.
Let P_k be true i.e.,
5^k – 1 is divisible by 4.
Let 5^k+1 – 1 = 4m, me Z
Now 5k + 1 – 1 = 5^k. 5 – 5 + 4
= 5 (5^k – 1) + 4
= 5 × 4m + 4 = 4 (5m + 1)
which is divisible by 4.
∴ P_k+1 is true.
∴ P_n is true for all values of n ∈ N

Question 5.
7²ⁿ + 2^3n-3  3^n-1 is divisible by 25 for any natural number n > 1.
Solution:
Let 7²ⁿ + 2^3n-3 . 3^n-1
when n = 1, 7¹ + 2⁰ . 3⁰ ⇒ 49 + 1 = 50
which is divisible by 25.
∴ P₁ is true. Let P_k be true.
i.e., 72^k + 2^3k-3 3^k-1 is divisible by 35
Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)
=7^2k+2 + 2^3k. 3^k
= 7^2k. 7² + 2^3k-3 2³. 3^k-1. 3¹
= 7^2k. 49 + 2^3k-3. 3^k-1. 24
= 7^2k (25 + 24) + 24. 2^3k-3. 3^k-1
= 7^2k. 25 + 24 (7^2k + 2^3k-3. 3^k-1)
= 7^2k. 25 + 24 × 25m
Which is divisible by 25 (∵ P_k is true)
∴ P_k+1 is true
∴ P_n is true for all values of n > 1.

Question 6.
7. 5^2n-1 + 23ⁿ⁺¹ is divisible by 17 for every natural number n ≥ 1.
Solution:
Let P_n = 7. 5²ⁿ – 1 + 2³ⁿ⁺¹.
When n = 1, 7.5 + 2⁴ = 35 + 16 = 51
Which is divisible by 17.
P₁, is true.
Let P_k be true i.e., 7.5^2k-1 + 2^3k+1  is divisible by 17.
Let 7.5^2k-1 + 2^3k+1 = 17 m, m ∈ Z
Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)
= 7.5^2k-1 + 2^3k+4
= 7.5^2k-1 . 5² + 2^3k+1 . 2³
= 25. 7. 5^2k-1 + 8. 2^3k+1
= (17 + 8) 7.5^2k-1 + 8. 2^3k+1
= 17. 7. 5^2k-1 + 8 (7. 5^2k-1 + 2^3k+1)
= 17 × 7 × 5^2k-1 + 8 × 17m
Which is divisible by 17.
Hence P_k+1. is true.
∴ P_n is true for all values of n ≥ 1.

Question 7.
4ⁿ⁺¹ + 15n + 14 is divisible by 9 for every natural number n ≥ 0.
Solution:
Let P_n = 4ⁿ⁺¹+ 15 n + 14
when n = 1, 4² + 15 + 14 = 45 is divisible by 9.
∴ P₁ is true. Let P_k be true.
i.e., 4^k+1 + 15k + 14 is divisible by 9.
Now, 4^k+1+1 + 15 (k + 1) + 14
= 4^k+2 + 15k + 29
= 4^k+1. 4 + 60k + 56 – 45k – 27
= 4 (4^k+1 + 15k + 14) – 9 (5k + 3)
Which is divisible by 9.
∴ P_k+1, is true.
∴ P_n is true for all values of n ≥ 0.

Question 8.
3^(2n-1) + 7 is divisible by 9 for every natural number n ≥ 2.
Solution:
Let P_n = 3^2(n-1) + 7
When n = 2. 3² + 7 = 16 is divisible by 8.
∴ P₂ is true.
Let P_k be true.
i.e., 5^2(k-1) + 7 is divisible by 8.
Let 3^2k-2 + 7 = 8m. m ∈ Z.
Now \(3^{2(\overline{k+1}-1)}\) + 7 = 3^2k + 7
= 3^2k-2. 3² + 63 – 56
= 9(3^2k-2 + 7) – 56
= 9 × 8m – 56 = 8 (9m – 7)
Which is divisible by 8.
P_k+1 is true.
P_n is true for all values on n ≥ 2.

Question 9.
5^(2n-4)  – 6n + 32 is divisible by 9 for every natural number n ≥ 5.
Solution:
Let P = 5^2(n-4) – 6n + 32
For n = 5, P₅ = 5² – 6. 5 + 32
= 25 – 30 + 32 = 27
Which is divisible by 9.
Hence P₅ is true.
Let P_k is true.
Let P_k is divisible by 9.
Let P_k = 5^2(k-4) – 6k + 32 = 9., m ≥ Z
5^2k+2 = 576 m + 24k  + 25 … (1)
we shall prove that P_k+1 is true.
Now 5^2(K+1)+2 – 24(k+1) – 25
= 5² (5^2k+2) – 24k – 24 – 25
= 5²[576m + 24k + 25] – 24k – 24 – 25
= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25
= 25 × 576 m + 576 k + 576
= 576 [25 m + k + 1]
which is divisible by 576
∴ P_k+1 is true.
So by the method of induction P_n is true for all n.
i.e., 5²ⁿ⁺²  – 24n – 25 is divisible by 576 for all n ∈ N.
Hence P_k+1 is true.
So by methods of induction P_n is true.
i.e., 5²ⁿ⁺² – 24n – 25 is divisible by 576 for all n.

Question 10.
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
Solution:
when n = 1,

Question 11.
1.3 + 2.4 + 3.5 + …….. + n(n + 2) = \(\frac{n(n+1)(2 n+7)}{6}\)
Solution:
when n = 1,
we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)
\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)

Question 12.
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R [Hint : Write xⁿ⁺¹ – yⁿ⁺¹ = x(xⁿ – yⁿ) + yⁿ(x – y)]
Solution:
Let p(n) is
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R
Step – 1:
For n = 2
x² – y² = (x – y) (x + y) (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., x^k – y^k = (x – y)(x^k-1 + x^k-2y + … +xy^k-2 + y^k-1)
Step – 3:
Let us prove P_k+1 is true.
i.e., x^k+1 – y^k+1 = (x – y) (x^k + x^k-1y + … (xy^k-1 + y^k)
L.H.S. = x^k+1 – y^k+1
= x^k+1 – xy^k + xy^k – y^k+1
= x(x^k – y^k) + y^k (x – y)
= x(x – y)(x^k-1 + x^k-2 y + … + xy^k-2 + y^k-1) + y^k(x – y) [by (1)]
= (x – y) [x^k + x^k-1 y + … + xy^k-2 + xy^k-1 + y^k]
= R.H.S.
∴ P_(k+1) is true.
Step – 4:
By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.
1 + 3 + 5 + ……. +(2n – 1) = n²
Solution:
Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n².
Step – 1:
For n = 2
L.H.S. = 1 + 3 = 4 = 2² (R.H.S)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., 1 + 3 + 5 … + (2k – 1) = k² …(1)
Step – 3:
We will prove that P(k + 1) is true
i.e., we want to prove.
1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)²
L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k² + 2k + 1          [By – (1)]
= (k + 1)² = R.H.S.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 1 + 3 + 5 ….+ (2n – 1) = n²

Question 14.
n > n; n is a natural number.
Solution:
Let P(n) is 2ⁿ > n
Step – 1:
2¹ > 1 (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
⇒ 2^k > k
Step – 3:
We shall prove that P(k + 1) is true
i.e., 2^k+1 > k + 1
Now 2^k+1 = 2.2k > 2k ≥ k + 1 for k ∈ N.
∴ 2^k+1 > k + 1
⇒ P(k + 1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 2ⁿ > n for n ∈ N

Question 15.
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Solution:
Let P(n) is
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Step – 1:
For n = 4
(1. 2. 3. 4)³ = 24³ = 13824
8(1³+ 2³ + 3³ + 4³) = 808
∴ (1. 2 . 3 . 4)³ > 8(1³ + 2³ + 3³ + 4³)
∴ P(4) is true.
Step- 2:
Let P(k) is true.
(1. 2. 3…….k)³ > 8(1³ + 2³ + 3³ + …… + k³)
Step – 3:
We shall prove that P_(k+1) is true.
i.e., (1. 2. 3. …….. k(k+1))³ > 8(1³ + 2³ + … + k³ + (k + 1)³)
Now (1. 2. 3. …….. k(k + 1)³)
= (1. 2. 3 … k)³ (k + 1)³
> 8 (1³ + 2³ + … k³) (k + 1)³
> 8 (1³ + 2³ + … k³) + 8(k + 1)³
= 8 (1³ + 2³ + … + k³ + (k + 1)³)
P_(k+1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.
Solution:
Let P(n) is
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)



Step-4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 1 I Like Bats Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 1 I Like Bats

BSE Odisha 6th Class English Follow-Up Lesson 1 I Like Bats Text Book Questions and Answers

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

→ Socialisation (ସାମାଜିକୀକରଣ):
→ The teacher thinks of a pre-reading activity.
(ଶିକ୍ଷକ ଏକ ପଢ଼ିବା ପୂର୍ବବର୍ତ୍ତୀ କାର୍ଯ୍ୟ ବିଷୟରେ ଚିନ୍ତା କରନ୍ତୁ ।)

→ You may use pictures. You may also link the poem with the poem of the main lesson.
(ତୁମେ ଛବିଗୁଡ଼ିକୁ ବ୍ୟବହାର କରିପାର । ତୁମେ କବିତାଟିକୁ ମୁଖ୍ୟପାଠର କବିତା ସହ ଯୋଡ଼ିପାର ।)

→ What are there in the picture? What do they look like? That’s how bats sleep and rest hanging upside down. What an interesting way of resting and relaxing! Do you like to rest like bats hanging upside down?
(ଛବିରେ କ’ଣ ଅଛି ? ସେମାନେ କିପରି ଦେଖାଯାଉଛନ୍ତି ? ସେହିପରି ଭାବରେ ବାଦୁଡ଼ିମାନେ ଉପରୁ ତଳକୁ ଝୁଲିରହି ଶୁଅନ୍ତି ଏବଂ ବିଶ୍ରାମ ନିଅନ୍ତି । ବିଶ୍ରାମ ନେବା ଓ ନିଦ୍ରାଯିବାର କି କୌତୂହଳଜନକ ଉପାୟ ! ତୁମେ ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରୁ ତଳକୁ ଝୁଲିରହି ବିଶ୍ରାମ କରିବାକୁ ଭଲ ପାଅ କି ?)

→ In the poem ‘Mice’ the poet likes mice. Let’s read this poem to see if the poet likes bats.
(‘Mice” କବିତାରେ, କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଇଛନ୍ତି । ଆସ ଆମେ ଏହି କବିତାଟି ପଢ଼ିବା ଏବଂ କବି ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କି ନାହିଁ ଦେଖୁବା ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ ):
Follow the steps of the main lesson.
(ମୁଖ୍ୟ ବିଷୟର ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କର ।)
TEXT (ବିଷୟବସ୍ତୁ):
Read the poem silently and answer the questions that follow.
(କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

I like bats
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like a bat
A way to find
After a day’s work
To relax and rewind
Upside down
Hang like bats.
I like bats
Hanging upside down
Like rats.

ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅପସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ମ୍ୟାଟ୍‌ସ୍ । ଲାଇକ୍ ସିଲ୍‌କ୍-କଟନ୍ ଫୁସ୍
ସୁଇଙ୍ଗିଙ୍ଗ୍ ଇନ୍ ଉଇଣ୍ଡ୍
ଦ୍ଵାଟ୍ ଏ ୱେ ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍ ।
ଆଇ ଉଇସ୍ ଆଇ କୁଡ଼୍
ଡୁ ଦ୍ଯାଟ୍
ଲାଇକ୍ ଏ ବ୍ୟାଟ୍
ଏ ୱେ ଟୁ ଫାଇଣ୍ଡ୍
ଆଫ୍‌ଟର୍ ଏ ଡେ’ଜ୍ ୱାର୍କ
ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍
ଅପ୍‌ସାଇଡ୍ ଡାଉନ୍
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍ ।
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅଫ୍‌ସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ଗ୍ୟାସ୍ ।

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):

like – ସଦୃଶ
bats – ବାଦୁଡି
hanging – ଫାଶୀ
upside – ଓଲଟା |
down – ତଳକୁ
like – ପରି
rats – ମୂଷା
silk – cotton – ଶିମିଳି-ତୁଳା
fruits – ଫଳ
swinging – ସୁଇଙ୍ଗ୍
in wind — ପବନରେ
could — କରିପାରନ୍ତି |
What a way — କି ଉପାୟ |
find — ଖୋଜିବା
relax — ବିଶ୍ରାମ କରିବା
day’s work — ଦିନର କାମ
rewind – ରିଭାଇଣ୍ଡ୍ |
wish — ଇଚ୍ଛା

ସାରକଥା | ଓଡ଼ିଆ ଅନୁବାଦ:
ମୁଁ ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲପାଏ
ସେମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ସଦୃଶ । | ସେମାନେ ପବନରେ ଦୋଳି ଖେଳୁଥା’ନ୍ତି ଶିମିଳି ତୁଳା ଫଳଗୁଡ଼ିକ ପରି । | କି ସୁନ୍ଦର ଉପାୟ ବିଶ୍ରାମ କରିବାର ଏବଂ ପବନରେ ଦୋହଲିବାର । | ମୁଁ ଭାବୁଛି ମୁଁ ସେହିପରି କରି ପାରିଥା’ନ୍ତି ଏକ ବାଦୁଡ଼ି ପରି । | ଗୋଟିଏ ଦିନକର କାମ ପରେ ଏକ ଉପାୟ ଖୋଜି ପାଇବାକୁ ବିଶ୍ରାମ କରିବାକୁ ଏବଂ ପବନରେ ଦୋହଲିବାକୁ । | ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ବାଦୁଡ଼ିମାନଙ୍କ ସଦୃଶ । ମୁଁ ଭଲପାଏ ବାଦୁଡ଼ିମାନଙ୍କୁ ଯେଉଁମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ଭଳି ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

The teacher is to try to frame his/her own questions. Here are some for him/her.
(ଶିକ୍ଷକ ତାଙ୍କର ନିଜର ପ୍ରଶ୍ନଗୁଡ଼ିକ ତିଆରି କରିବାକୁ ଚେଷ୍ଟା କରିବେ । ଏଠାରେ କେତେକ ତାଙ୍କ (ପୁ/ସ୍ତ୍ରୀ) ପାଇଁ ଅଛି ।)

Question 1.
How do bats hang?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲୁଥା’ନ୍ତି ?)
Answer:
Bats hang upside down.

Question 2.
What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହିତ ତୁଳନା କରାଯାଇଛି ?)
Answer:
Bats are compared to rats.

Question 3.
Have you seen bats hanging upside down on trees in great numbers?
(ଗଛଗୁଡ଼ିକରେ ବାଦୁଡ଼ିମାନେ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲି ରହିଥ‌ିବାର ତୁମେ ଦେଖୁଛ କି ?)
Answer:
Yes, we have seen bats hanging upside down on trees in great numbers.

Question 4.
Have you seen silk-cotton fruit hanging in great numbers? Do they look alike?
(ଶିମିଳି-ତୁଳା ଫଳ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲୁଥ‌ିବା ତୁମେ ଦେଖୁଛ କି ? ସେଗୁଡ଼ିକ ଏକାପରି ଦେଖାଯାଆନ୍ତି କି ?)
Answer:
Yes, we have seen silk-cotton fruit hanging in great numbers. Really, they look alike.

Question 5.
What is the meaning of the word ‘rewind’? See the dictionary at the end of this lesson.
(‘ରିୱାଇଣ୍ଡ୍’ ଶବ୍ଦର ଅର୍ଥ କ’ଣ ? ଏହି ଅଧ୍ୟାୟର ଶେଷରେ ଥ‌ିବା ଅଭିଧାନ ବା ଶବ୍ଦାର୍ଥ ଦେଖ ।)
Answer:
The meaning of the word ‘rewind’ is taking a rest with occasional backward movement.
Teacher is to frame some questions from the second stanza.
(ଶିକ୍ଷକ ଦ୍ଵିତୀୟ ପଦରୁ କିଛି ପ୍ରଶ୍ନ ତିଆରି କରିବେ ।)

Session – 2 (ସୋପାନ – ୨)
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):
6. Writing (ଲେଖିବା ):

(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) How do bats hang ?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲନ୍ତି ?)
Bats hang ______________________________________.
Answer:
Bats hang upside down.

(ii) What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହ ତୁଳନା କରାଯାଇଛି ?)
Bats are ______________________________________.
Answer:
Bats are compared to rats.

(iii) What swings in the wind?
(ପବନରେ କ’ଣ ଝୁଲିଥାଏ ?)
The silk – ______________________________________.
Answer:
The silk-cotton fruits swing in the wind.

(iv) Why does the poet like to hang like bats upside down?
(କବି କାହିଁକି ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରପାଖ ତଳକୁ କରି ଝୁଲିବାକୁ ଭଲପାଆନ୍ତି ?)
______________________________________to relax and ____________________________.
Answer:
The poet likes to hang like a bats upside down because it is a nice way to relax and rewind.

(b) Let us summarise the poem. Fill in the gaps.
(ଆସ ଆମେ କବିତାର ସାରାଂଶ ବାହାର କରିବା । ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକୁ ପୂରଣ କର ।) (Question with Answer)

The poet _____________________to see bats. _____________________. Bats hang like _____________. This is a good way to relax and ____________. The poet wants to_____________ _____________bats to relax ________________ ________________after the day’s ________________.
Answer:
The poet likes to see bats. They are hanging upside down. Bats hang like rats and like silk-cotton fruits swinging in wind. This is a good way to relax and rewind. The poet wants to hang upside down like a bat to relax and rewind after the day’s work.

(c) Think of writing a poem. Start with replacing ‘bats’ with some fruits and make minimum changes in the poem. Change the title accordingly.
(ଗୋଟିଏ କବିତା ଲେଖିବା କଥା ଭାବ । ‘ବାଦୁଡ଼ିମାନଙ୍କ’’ ବଦଳରେ କେତେକ ଫଳକୁ ନେଇ ଏବଂ କବିତାରେ ସ୍ଵଳ୍ପ ପରିବର୍ତ୍ତନ କରି ଆରମ୍ଭ କର । ସେହି ଅନୁସାରେ କବିତାର ଶିରୋନାମା ପରିବର୍ତ୍ତନ କର ।)

APPLES
Answer:
I like apples
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like an apple
A way to find
After a day’s work
To relax and rewind
Upside down Hang like apples.
I like apples
Hanging upside down
Like rats.

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):
(The words/phrases have been defined mostly on contextual meanings.)

Nibble – gentle and playful bite of a mouse. ମୂଷାର କୁଟ୍ କୁଟ୍ କରି କାଟି ଖାଇବା
Pink – (colour) pale red, ଫିକା ନାଲି |
Swinging – hanging and moving (bats have)
relax and rewind – taking rest (with occassional backward movement) ଆରାମରେ ବିଶ୍ରାମ ନେବା
upside down – legs upward and head downward. ଗୋଡ଼ ଉପରକୁ ଓ ମୁଣ୍ଡ ତଳକୁ କରି ଓଲଟା ରହିବା ।

Odisha State Board BSE Odisha 6th Class English Solutions Lesson 1 Mice Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Lesson 1 Mice

BSE Odisha 6th Class English Lesson 1 Mice Text Book Questions and Answers

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

→ Your teacher will introduce the poem in the following way.
S/he will ask the following questions:
Don’t you like a mouse? Why / Why not? Give reasons. Here is a poem on mice. Let us read, enjoy and see whether the poet likes or dislikes mice.
(ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ନିମ୍ନଲିଖ୍ ଭାବରେ ଉପସ୍ଥାପନ କରିବେ । ସେ (ପୁ | ସ୍ତ୍ରୀ) ନିମ୍ନଲିଖିତ ପ୍ରଶ୍ନଗୁଡ଼ିକ ପଚାରିବେ : ତୁମେ ଗୋଟିଏ ମୂଷାକୁ ଭଲ ପାଅ ନାହିଁ କି ? କାହିଁକି | କାହିଁକି ନୁହେଁ ? କାରଣ ଦର୍ଶାଅ । ଏଠାରେ ମୂଷାମାନଙ୍କ ବିଷୟରେ ଗୋଟିଏ କବିତା ଅଛି । ଆସ ଆମେ ପଢ଼ିବା, ଉପଭୋଗ କରିବା ଏବଂ ଦେଖିବା କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କିମ୍ବା ଭଲପାଆନ୍ତି ନାହିଁ ।)

II. While-Reading (ପଠନକାଳୀନ ):

1. Your teacher will read the poem aloud, and you will listen to him/her without opening your books.
(ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ବଡ଼ପାଟିରେ ପଢ଼ିବେ, ତୁମେ ତୁମ ବହି ନ ଖୋଲି ତାଙ୍କୁ (ପୁ | ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣିବ ।)
2. S/he will read the poem aloud for the second time and you will listen to him / her following the poem in your books.
(ସେ (ପୁ | ସ୍ତ୍ରୀ) କବିତାଟିକୁ ଦ୍ଵିତୀୟ ଥର ପାଇଁ ବଡ଼ପାଟିରେ ପଢ଼ିବେ ଏବଂ ତୁମେ ତୁମ ବହିରେ କବିତାଟିକୁ ଅନୁସରଣ କରି ତାଙ୍କୁ (ପୁ | ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣିବ ।)
3. Read the poem silently and try to answer the questions asked by your teacher.
(କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ତୁମ ଶିକ୍ଷକ ପଚାରିଥିବା ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦେବାକୁ ଚେଷ୍ଟା କର ।)

Text (ପାଠ୍ୟବସ୍ତୁ):
I think mice
Are rather nice.
Their tails are long
Their faces small,
They haven’t any
Skins at all.
Their ears are pink,
Their teeth are white,
They run about
The house at night
They nibble things
They shouldn’t touch
And no one seems
To like them much.
But I think mice
Are nice.

ଓଡ଼ିଆ ଉଚ୍ଚାରଣ | କବିତା ପଠନ :
ଆଇ ଥ୍କ୍ ମାଇସ୍
ଆର୍ ରାଦର୍ ନାଇସ୍ ।
ଦେଆର୍ ଟେଲସ୍ ଆର୍ ଲଙ୍ଗ୍
ଦେଆର୍ ଫେସେସ୍ ସ୍କୁଲ୍,
ଦେ ହାଭଣ୍ଟ୍ ଏନି
ସ୍କିନସ୍ ଆଟ୍ ଅଲ୍ ।
ଦେଆର୍ ଇଅରସ୍ ଆର୍ ପିକ୍,
ଦେଆର୍ ଟିଥ୍ ଆର୍ ଦ୍ଵାଇଟ୍,
ଦେ ରନ୍ ଏବାଉଟ୍
ଦି ହାଉସ୍ ଆଟ୍ ନାଇଟ୍
ଦେ ନିବଲ୍ ଥ୍ଙ୍ଗସ୍
ଦେ ସୁଡ଼ଣ୍ଟ ଟଚ୍
ଆଣ୍ଡ୍ ନୋ ୱାନ୍ ସିମ୍‌
ଟୁ ଲାଇକ୍ ଦେମ୍ ମଚ୍ ।
ବଟ୍ ଆଇ ଥ୍ ମାଇସ୍
ଆର୍ ନାଇସ୍ ।

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):

think – କରିବା
mice – ମୂଷା
rather – ବରଂ
nice – ସୁନ୍ଦର
tails – ଲାଙ୍ଗୁଡ଼
long – ଲମ୍ବା
faces – ଚେହେରା
small – ଛୋଟ
any – କୌଣସି
skins – ଚର୍ମ
at all – ଆଦୌ
ears – କାନ
pink – ଗୋଲାପୀ |
teeth – ଦାନ୍ତ
white – ଧଳା
run about – ଚଲାନ୍ତୁ
house – ଘର
at night – ରାତିରେ
nibble – ନିବଲ୍
things – ଜିନିଷ
shouldn’t – ଉଚିତ ନୁହେଁ |
touch – ସ୍ପର୍ଶ କରନ୍ତୁ |
seems – ମନେହୁଏ |
like – ପରି
much – ବହୁତ

କବିତାର ସାରକଥା | ଓଡ଼ିଆ ଅନୁବାଦ:
ମୁଁ ଭାବୁଛି ମୂଷାମାନେ ବରଂ ସୁନ୍ଦର । ସେମାନଙ୍କର ଲାଞ୍ଜଗୁଡ଼ିକ ଲମ୍ବା । ସେମାନଙ୍କର ମୁହଁସବୁ ଛୋଟ । ସେମାନଙ୍କର ଆଦୌ ଚର୍ମ ନାହିଁ । ସେମାନଙ୍କର କାନଗୁଡ଼ିକ ପାଟଳ ରଙ୍ଗ । ସେମାନେ ରାତିରେ ଘରର ଏଣେତେଣେ ଘୂରିବୁଲନ୍ତି । ସେମାନେ ଜିନିଷଗୁଡ଼ିକୁ ଟିକେ ଟିକେ କାମୁଡ଼ନ୍ତି ଯେଉଁଗୁଡ଼ିକ ସେମାନେ ଛୁଇଁବା ଉଚିତ ନୁହେଁ । କେହି ବୋଧହୁଏ ସେମାନଙ୍କୁ ବେଶୀ ଭଲ ପାଆନ୍ତି ନାହିଁ ।
କିନ୍ତୁ ମୁଁ ଭାବୁଛି ମୂଷାମାନେ ସୁନ୍ଦର ଅଟନ୍ତି ।

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ):

Question 1.
What is the poem about ?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଆଧାରିତ ?)
Answer:
The poem is about mice.

Question 2.
Who is “I” in the first line of the poem?
(କବିତାର ପ୍ରଥମ ଧାଡ଼ିରେ ‘ମୁଁ’ (I) କିଏ ଅଟେ ?)
Answer:
In the first line of the poem ‘I’ means the poet.
Or, The poet is I in the first line of the poem.

Question 3.
What are the tails of mice like?
(ମୂଷାମାନଙ୍କର ଲାଞ୍ଜସବୁ କିପରି ଅଟେ ?)
Answer:
The tails of mice are long.

Question 4.
“They have no skin at all”. Does this mean that they have no skins or do they have very thin skins?
(‘‘ସେମାନଙ୍କର ପ୍ରାୟ ଆଦୌ ଚର୍ମ ନାହିଁ ।’’ ଏହା କ’ଣ ସେମାନଙ୍କର ଚର୍ମ ନଥିବା କିମ୍ବା ପତଳା ଚର୍ମ ଥ‌ିବାକୁ ବୁଝାଏ ?)
Answer:
They have no skin at all.’ It means that they have very thin skins.

Question 5.
What is the color of their ears?
(ସେମାନଙ୍କ କାନର ରଙ୍ଗ କିପରି ଅଟେ ?)
Answer:
The colour of their ears is pink.

Question 6.
Where do they run about at night ?
(ସେମାନେ ରାତିରେ କେଉଁଠି ଘୂରି ବୁଲନ୍ତି ?)
Answer:
They run about the house at night.

Question 7.
Which things do they nibble ?
(କେଉଁ ଜିନିଷଗୁଡ଼ିକୁ ସେମାନେ ଟିକିଟିକି କରି କାମୁଡ଼ି ଦିଅନ୍ତି ?)
Answer:
They nibble everything they find in the house. Those are cakes, vegetables, fruits, clothes, etc.

Question 8.
Do most people like mice? Why?
(ଅଧିକାଂଶ ଲୋକ ମୂଷାମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କି ? କାହିଁକି ?)
Answer:
No, most people do not like mice. Because they destroy many valuable things.

Question 9.
Does the poet like mice?
(କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କି ?)
Answer:
Yes, the poet likes mice.

Question 10.
Which line tells you so?
(କେଉଁ ଧାଡ଼ିଟି ତୁମକୁ ସେପରି କହେ ?)
Answer:
The last line, “But I think mice are nice” tells us so.

Question 11.
Which lines are repeated in this poem?
(ଏହି କବିତାରେ କେଉଁ ଧାଡ଼ିଗୁଡ଼ିକ ପୁନରାବୃତ୍ତି ହୋଇଛି ?)
Answer:
The first and last lines in this poem are repeated. They are “I think mice are rather nice.”

Session – 2 (ଦ୍ବିତୀୟ ପର୍ଯ୍ୟାୟ):
III. Post-Reading (ପଠନ ପରବର୍ତ୍ତୀ):

1. Visual Memory Development Technique (VMDT) :
(ଦୃଶ୍ୟ ସ୍ମୃତି ଉନ୍ନୟନ କୌଶଳ (VMDT))
Take a photograph of the poem in your eye cameras. Then put your index finger on the words/phrases your teacher says. Open your eyes and see whether your finger is on the right place. Repeat this activity for other words/phrases your teacher says.
(ତୁମ ଆଖ୍ ରୂପକ କ୍ୟାମେରାରେ କବିତାଟିର ଫଟୋଗ୍ରାଫ୍ ଉଠାଇନିଅ । ତା’ପରେ ତୁମ ବିଶି ଆଙ୍ଗୁଠିଟିକୁ ତୁମ ଶିକ୍ଷକ କହିଥିବା ଶବ୍ଦ | ଖଣ୍ଡବାକ୍ୟ ଉପରେ ରଖ । ତୁମ ଆଖ୍ ଖୋଲ ଏବଂ ଦେଖ ତୁମର ଆଙ୍ଗୁଠି ଠିକ୍ ସ୍ଥାନ ଉପରେ ଅଛି କି ନାହିଁ । ତୁମ ଶିକ୍ଷକ କହିଥିବା ଅନ୍ୟ ଶବ୍ଦ ବା ଖଣ୍ଡବାକ୍ୟ ପାଇଁ ଏହି କାର୍ଯ୍ୟ ପୁନରାବୃତ୍ତି କର ।)
Whole Text: Pink, nibble things, faces small, like them much, The lines repeated.

2. Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ) :

(a) MCQs: Choose the correct alternatives. (ସଠିକ୍ ବିକଳ୍ପଟିକୁ ବାଛ ।)
i. Mice don’t have ___________.
(a) thick skins.
(b) long tails.
(c) small faces.
(d) pink ears.
Answer:
(a) thick skins.

ii. Which of the following is not true?
(ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ଠିକ୍ ନୁହେଁ ?)
(a) The poet likes mice.
(b) very thin skins.
(c) No one likes mice.
(d) black skins.
Answer:
(b) Everyone likes mice.

iii. ‘They haven’t any skins at all’ means they have
(‘ସେମାନଙ୍କର ଚର୍ମ ଆଦୌ ନଥାଏ’ ବୁଝାଏ ଯେ ସେମାନଙ୍କର ______________ ଥାଏ ।)
(a) thick sinks.
(b) very thin skins.
(c) no skins at all.
(d) black skins.
Answer:
(b) very thin skins.

(b) Match items under column A with items under column B. Join them with lines. The first one is done for you.
(ସ୍ତମ୍ଭ ‘A’ ତଳେ ଥିବା ଶବ୍ଦ ସହ ସ୍ତମ୍ଭ ‘B’ ତଳେ ଥ‌ିବା ଶବ୍ଦକୁ ମିଳାଅ । ସେଗୁଡ଼ିକୁ ଗାର ଟାଣି ଯୋଗ କର । ପ୍ରଥମଟି ତୁମପାଇଁ କରିଦିଆଯାଇଛି ।)
(Question with Answer)

Answer:

Session – 3 (ତୃତୀୟ ପର୍ଯ୍ୟାୟ)
3. Listening (ମନଯୋଗ ସହକାରେ ଶୁଣିବା):
(a) Tick the words / phrases your teacher reads aloud.
(ତୁମ ଶିକ୍ଷକ ବଡ଼ ପାଟିରେ ପଢ଼ୁଥ‌ିବା ଶବ୍ଦ ବା ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକୁ ଟିକ୍ ( ✓) ଚିହ୍ନ ଦିଅ ।)
mice, tails, skins, ears, pink, teeth, white, nibble, nice (Listen to your teacher and tick the words/phrases.)

(b) Your teacher reads aloud the first six lines of the poem. Fill in the gaps while listening to him/her.
(ତୁମ ଶିକ୍ଷକ କବିତାର ପ୍ରଥମ ଛଅ ଧାଡ଼ିକୁ ବଡ଼ପାଟିରେ ପଢ଼ିବେ । ତାଙ୍କୁ ଶୁଣିବା ସମୟରେ ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକୁ ପୂରଣ କର ।)
(Question with Answer)
I think _________.
Are rather _________
Their tails are _________
Their _________ small
They haven’t any
_________ at all.
Answer:
I think mice
Are rather nice
Their tails are long
Their faces small
They haven’t any
skins at all.

Session – 4 (ସୋପାନ — ୪)
4. Speaking (କହିବା):

• Your teacher will read aloud one line, and you will repeat it after him/her. This is done twice.
  (ତୁମ ଶିକ୍ଷକ ଗୋଟିଏ ଧାଡ଼ି ପଢ଼ିବେ, ତୁମେ ତାଙ୍କ (ପୁ/ସ୍ତ୍ରୀ) ପଛରେ ପୁନରାବୃତ୍ତି କରିବ । ଏହିପରି ଦୁଇଥର କରାଯିବ ।)
• Your teacher will read one line, you will read the next line, and so on.
  (ତୁମ ଶିକ୍ଷକ ଗୋଟିଏ ଧାଡ଼ି ପଢ଼ିବେ, ତୁମେ ତା’ପର ଧାଡ଼ି ପଢ଼ିବ ଏବଂ ଏହିପରି ଚାଲୁରହିବ ।)
• Students sitting on the right side will read one line and students sitting on the left will read the next line and so on.
  (ଡାହାଣ ପାର୍ଶ୍ଵରେ ବସିଥିବା ଶିକ୍ଷାର୍ଥୀମାନେ ଗୋଟିଏ ଧାଡ଼ି ପଢ଼ିବେ ଏବଂ ବାମପାର୍ଶ୍ଵରେ ବସିଥିବା ଶିକ୍ଷାର୍ଥୀମାନେ ତା’ର ପରବର୍ତ୍ତୀ ଧାଡ଼ିଟି ପଢ଼ିବେ ଏବଂ ଏହିପରି ଚାଲୁ ରହିବ ।)

(b) Chaindrill: “I think mice are nice.”
(ଶୃଙ୍ଖଳ-ଅଭ୍ୟାସ : ‘‘ମୁଁ ଭାବୁଛି ମୂଷାମାନେ ସୁନ୍ଦର ।’’)

5.Vocabulary (ଶବ୍ଦସମ୍ଭାର):
Match the words/phrases under A with their opposite words under B. Join them with lines. The first one is done for you.
(ସ୍ତମ୍ଭ ‘A’ ତଳେ ଥିବା ଶବ୍ଦ | ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ସହ ସ୍ତମ୍ଭ ‘B’ ତଳେ ଥ‌ିବା ସେମାନଙ୍କର ବିପରୀତ ଅର୍ଥବୋଧକ ଶବ୍ଦଗୁଡ଼ିକ ସହ ମିଳାଅ । ସେମାନଙ୍କୁ ଗାର ଟାଣି ଯୋଗ କର । ତୁମ ପାଇଁ ପ୍ରଥମଟି କରି ଦିଆଯାଇଛି ।)

Answer:

Session – 5 (ପଞ୍ଚମ ପର୍ଯ୍ୟାୟ)
6. Usage (ପ୍ରଚଳିତ ପ୍ରୟୋଗ):
(a) When we talk about one person, it is singular. But when we talk about more than one person or thing, it is plural. A thing or a person is made plural in the following ways :
(ଯେତେବେଳେ ଆମେ ଜଣେ ବ୍ୟକ୍ତି ବିଷୟରେ କହୁ, ଏହା ଏକବଚନ ଅଟେ । କିନ୍ତୁ ଯେତେବେଳେ ଆମେ ଏକାଧ୍ଵକ ବ୍ୟକ୍ତି କିମ୍ବା ଜିନିଷଗୁଡ଼ିକ ବିଷୟରେ କହୁ, ଏହା ବହୁବଚନ ଅଟେ । ଗୋଟିଏ ଜିନିଷ ବା ଜଣେ ବ୍ୟକ୍ତି ନିମ୍ନଲିଖିତ ଭାବରେ ବହୁବଚନରେ ପରିଣତ କରାଯାଏ ।)

(i) By adding “s” to the word: (ଶବ୍ଦରେ ‘s’ ଯୋଗକରି )
Singular      Plural
( ଏକବଚନ )   (ବହୁବଚନ)
cat (ବିରାଡ଼ି)   cats (ବିଲେଇମାନେ)
dog (କୁକୁର)  dogs (କୁକୁରମାନେ)

(ii) By adding “es” to the word: (ଶବ୍ଦରେ ‘es’ ଯୋଗକରି)
bus (ଯାତ୍ରୀଗାଡ଼ି)  buses (ଯାତ୍ରୀଗାଡ଼ିସବୁ)
bush (ବୁଦା)       bushes (ବୁଦାଗୁଡ଼ିକ)
mango (ଆମ୍ବ)   mangoes (ଆମ୍ବଗୁଡ଼ିକ)
box (ବାକ୍ସ)         boxes (ବାକ୍ସଗୁଡ଼ିକ)

(iii) By changing the spelling of the word :(ଶବ୍ଦର ବନାନରେ ପରିବର୍ତ୍ତନ କରି)
mouse (ମୂଷା)    mice (ମୂଷାମାନେ)
tooth (ଦାନ୍ତ)      teeth (ଦାନ୍ତଗୁଡ଼ିକ)

(iv) By not changing the word : (ଶବ୍ଦକୁ ପରିବର୍ତ୍ତନ ନକରି)
deer (ହରିଣ)    deer (ହରିଣମାନେ)
sheep (ମେଣ୍ଢା) sheep (ମେଣ୍ଢାଗୁଡ଼ିକ)

(b) Given below are some singular forms of some words used in this poem. Find out the plural forms of these words from the poem and write them against each of the words. (ନିମ୍ନରେ ଏହି କବିତାରେ ବ୍ୟବହାର କରାଯାଇଥିବା କେତେକ ଶବ୍ଦର ଏକବଚନ ରୂପଗୁଡ଼ିକ ଦିଆଯାଇଛି । କବିତାରୁ ଏହି ଶବ୍ଦଗୁଡ଼ିକର ବହୁବଚନ ରୂପଗୁଡ଼ିକୁ ଖୋଜି ବାହାର କର ଏବଂ ସେଗୁଡ଼ିକୁ ପ୍ରତ୍ୟେକ ଶବ୍ଦ ପାର୍ଶ୍ଵରେ ଲେଖ ।)

Answer:

Session – 6 (ଷଷ୍ଠ ପର୍ଯ୍ୟାୟ)
7. Writing (ଲିଖନାତ୍ମକ):
(a) Based on your matching above in-2 (b), Comprehension Activities, write some sentences. One is done for you. (ଉପରିସ୍ଥ ବୋଧମୂଳକ କାର୍ଯ୍ୟାବଳୀର 2 (b) ବିଭାଗରେ ତୁମର ମେଳକକୁ ଭିଭିକରି କେତେକ ବାକ୍ୟ ଲେଖ । ଗୋଟିଏ ତୁମ ପାଇଁ କରି ଦିଆଯାଇଛି ।)

Tails …………………………………… long
Tails of mice are long. (Question with Answer)
Face __________________________
Skin __________________________
Ear __________________________
Teeth __________________________.
Run about __________________________
Nibble things __________________________

Answer:
Face Faces of mice are small.
Skin Mice haven’t any skins at all.
Ear Ears of mice are pink.
Teeth  Teeth of mice are white.
Run about They run about the house at night.
Nibble things They nibble things they shouldn’t touch.

(b) Fill in the gaps using the words given. The first one is done for you. (ଦିଆଯାଇଥ‌ିବା ଶବ୍ଦଗୁଡ଼ିକୁ ବ୍ୟବହାର କରି ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକୁ ପୂରଣ କର । ପ୍ରଥମଟି ତୁମ ପାଇଁ କରି ଦିଆଯାଇଛି ।)
(Do not see the poem while doing this task)
(ଯେତେବେଳେ ଏହି କାର୍ଯ୍ୟଟି କରୁଛ, କବିତାଟିକୁ ଦେଖ ନାହିଁ ।)

Answer:

Session – 7 (ସୋପାନ – ୭):
(c) The poet likes mice. However, you don’t like them. Rewrite the poem expressing your dislikes. The first two lines are done for you.
(କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି । ଯାହା ହେଉନା କାହିଁକି, ତୁମେ ସେମାନଙ୍କୁ ପସନ୍ଦ କରନାହିଁ । ତୁମର ନାପସନ୍ଦଗୁଡ଼ିକୁ ପ୍ରକାଶ କରି କବିତାଟିକୁ ଆଉଥରେ ଲେଖ । ପ୍ରଥମ ଦୁଇଟି ଧାଡ଼ିକୁ ତୁମ ପାଇଁ କରି ଦିଆଯାଇଛି ।)
I think mice
Are not nice.
_____________________________________________
_____________________________________________
_____________________________________________
_____________________________________________
_____________________________________________
_____________________________________________
_____________________________________________
_____________________________________________
Answer:
Their tails are long.
Their faces small.
They haven’t any skins at all.
Their ears are pink.
Their teeth are white.
They run about
The house at night
They nibble things
They shouldn’t touch
And no one seems
To like them much.
And I also think
mice Are not nice.

Session – 8 (ସୋପାନ – ୮):
(d) Write the answers to the following questions.
(ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଲେଖ ।)

(i) What is the poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଆଧାରିତ ? )
Answer:
The poem is about mice.

(ii) Who is “I” in the first line of the poem?
(କବିତାଟି ପ୍ରଥମ ଧାଡ଼ିରେ ‘ମୁଁ’ (I) କିଏ ?)
Answer:
In the first line of the poem, the word T is the poet.

(iii) What does the tail of a mouse look like?
(ଗୋଟିଏ ମୂଷାର ଲାଞ୍ଜଟି ଦେଖିବାକୁ କିପରି ଅଟେ ?)
Answer:
The tail of a mouse looks long.

(iv) Where do they run about at night?
(କେଉଁଠାରେ ସେମାନେ ରାତିରେ ଧାଆଁନ୍ତି ?)
Answer:
They run about the house at night.

(v) Which things do they nibble?
(କେଉଁ ଜିନିଷଗୁଡ଼ିକୁ ସେମାନେ ଟିକିଟିକି କରି କାମୁଡ଼ନ୍ତି ?)
Answer:
They nibble everything they find in the house. Those are cakes, vegetables, fruits, clothes, and books.

8. Mental Talk (ମାନସିକ କଥୋପକଥନ):
Mentally repeat the following line of the poem :
(ମନେ ମନେ କବିତାଟିର ନିମ୍ନଲିଖିତ ଧାଡ଼ିଟିକୁ ପୁନରାବୃତ୍ତି କର ।)
“I think mice are rather nice.”

9. Let Us Think (ଆସ ଆମେ ଭାବିବା ):
The poet likes mice. Do we like his / her attitude? Why?
(କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି । ଆମେ ତାଙ୍କର (ପୁ/ସ୍ତ୍ରୀ) ମନୋଭାବକୁ ପସନ୍ଦ କରୁ କି ? କାହିଁକି ?)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(a)

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(a)

Question 1.
Which of the following in a sequence?
(i) f(x) = [x], x ∈ R
(ii) f(x) = |x|, x ∈ R
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N
Solution:
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.
Determine if (t_n) is an arithmetic sequence if :
(i) t_n = an² + bn
Solution:
t_n = an² + bn
⇒ t_n+1 = a(n + 1)² + b(n – 1)
⇒ t_n+1 – t_n = a{(n + 1)² – n²} + b{n + 1 – n}
= a(2n + 1) + b
Which is not independent of n.
∴ (t_n) is not an A.P.

(ii) t_n = an + b
Solution:
t_n = an + b
⇒ t_n+1 = a(n + 1) + b
Now t_n+1 – t_n
= {a(n + 1) + b} – {an + b}
= a (constant)
∴ (t_n) is an arithmetic sequence.

(iii) t_n = an² + b
Solution:
t_n = an² + b
⇒ t_n+1 = a(n + 1)² + b
∴ t_n+1 – t_n = a[(n + 1)² – n²] + b – b
= a(2n + 1)
(does not independent of n)
∴ (t_n) is not an arithmetic sequence.

Question 3.
If a geometric series converges which of the following is true about its common ratio r?
(i) r > 1
(ii) -1 < r < 1
(iii) r > 0
Solution:
(ii) -1 < r < 1

Question 4.
If an arithmetic series ∑tn converges, which of the following is true about t_n?
(i) t_n < 1
(ii) |t_n| < 1
(iii) t_n = 0
(iv) t_n → 0
Solution:
(iii) t_n = 0

Question 5.
Which of the following is an arithmetic-geometric series?
(i) 1 + 3x + 7x² + 15x³+ ….
(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x² + ….
(iii) x + (1 + 2)x² + (1 + 2 + 3)x³ +…
(iv) x + 3x² + 5x³ + 7x⁴ + …
Solution:
(iv) x + 3x² + 5x³ + 7x⁴ + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.
For an arithmetic sequence (t_n) t_p = q, t_q = p, (p ≠ q), find t_n.
Solution:
t_p = q ⇒ a + (p – 1)d = q    ……(1)
t_q = p ⇒ a + (q – 1)d = p    ……(2)
From (1) and (2) we have (p – q)d = q – p
⇒ d = (-1)
Putting d = (-1) in (1)
we have a = p + q – 1
∴ t_n = a + (n – 1)d
= (P + q – 1) + (n – 1) (-1)
= p + q – n

Question 7.
For an arithmetic series, ∑a_n S_p = q and S_q = p (p ≠ q) find S_p+q
Solution:
S_p = q and S_q = p

Question 8.
The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.
Solution:

Question 9.
The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.
Solution:


∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.
Find the sum as directed
(i) 1 + 2a + 3a² + 4a³ + …..(first n terms(a ≠ 1))
Solution:

(ii) 1 + (1 + x)y + (1 + x + x²)y² + …..(to infinity)
Solution:
Let S_∞ = 1 + (1 + x)y + (1x + x²)y² + …
⇒ S_n = 1 + (1 + 1 + x)y + (1 + x + x²)y² + ……+(1 + x + …. x^n-1)y^n-1

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)
Solution:

(iv) 1 + 4x + 8x² + 13x³ + 19x⁴ + …..(to infinity). Assuming that the series has a sum for |x| < 1.
Solution:

(v) 3.2 + 5.2² + 7.2³ + …..(first n terms)
Solution:
Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.
Find the sum of the infinite series.
(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)
Solution:

= 1 – \(\frac{1}{n+1}\)
∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)
Solution:
\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)
Solution:
Here t_n = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)
The denominator of t_n is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have
\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_n = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]
Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)
Solution:

Question 12.
Find S_n for the series.
(i) 1.2 + 2.3 + 3.4 + ….
Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …
Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…
Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …
[Hint : t_n = (3n – 1) (3n + 2)(3n + 5)]
Solution:

(v) 1.5 + 2.6 + 3.7 + …
[Hint: t_n = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_n = n² + 4n and find S_n = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]
Solution:

(vi) 2.3 + 3.6 + 4.11 + …
[Hint : Take t_n = (n + 1) (n² + 2)]
Solution:

(vii) 1.3² + 2.5² + 3.7² + ….
Solution:

Question 13.
Find the sum of the first n terms of the series:
(i) 5 + 6 + 8 + 12 + 20 + …
Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …
Solution:

Question 14.
(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]
Solution:
We know that
(x₁ + x₂ + x₃ + …. + x_n)²
= (x²₁ + xⁿ₂ + … + x²_n) + 2 (Sum of all possible Products taken two at a time)
∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)
= (1 + 2 + 3 + … 20)² – (1² + 2² + … + 20²)
\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)
= (210)² – 70 × 41
= 44100 – 2870 = 41230
∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.
Solution:
Here the required sum

Question 15.
If a = 1 + x + x² + ….. and b = 1 + y + y² + ….|x| <  1 and |y| <  1, then prove that 1 – xy + x²y² + x³y³ + ….. =  \(\frac{a b}{a+b-1}\)
Solution:

Question 16.
If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0
Solution:
Let the first term of an A.P. = A and the common difference = D.
According to the question
t_P = a, t_q = b, t_r = c
⇒ A + (p – q)D = a      …..(1)
A + (q – 1)D = b           …..(2)
A + (r – 1)D = c            …..(3)
L.H.S = a(q – r) + b (r – p) + c (p – q)
= (A + (p – 1)D) (q – r) + (A + (q – 1)D)
(r- p) + (A + (r – 1)D) (p – q)
= A (q – r + r – p + p – q) + D [(p – 1)
(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]
= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.
If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.
Solution:

Question 18.
If a², b², c² are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.
Solution:

Question 19.
If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0
Solution:

Question 20.
If (b – c)², (c – a)², (a – b)² are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.
Solution:

Question 21.
If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0
Solution:

Question 22.
If a, b, c,d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let a, b, c, and d are in G.P.
Let the common ratio = r
⇒ b = ar, c = ar², d = ar³
LHS = (a² + b² + c²) (b² + c² + d²)
= (a² + a²r² + a²r⁴) (a²r² + a²r⁴ + a²r⁶)
= a⁴r² (1 + r² + r⁴)²
= (a²r + a²r³ + a²r⁵)²
= (a.ar + ar.ar² + ar².ar³)²
= (ab + bc + cd)² = R.H.S. (proved)

 

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.
Compute the following :
(i) ¹²C₃
Solution:
¹²C₃ = \(\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}\) = 220

(ii) ¹⁵C₁₂
Solution:
¹⁵C₁₂ = \(\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}\)
= 5.7.13 = 455

(iii) ⁹C₄ + ⁹C₅
Solution:
⁹C₄ + ⁹C₅ = \(\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}\)
\(=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}\) × 2 = 252

(iv) ⁷C₃ + ⁶C₄ + ⁶C₃
Solution:
⁷C₃ + ⁶C₄ + ⁶C₃ = ⁷C₃ + ⁶C₄ + ⁶C4-1
= ⁷C₃ + ⁶⁺¹C₄ = ⁷C₃ + ⁷C₄
(∴ ⁿc_r + ⁿC_r-1– = ⁿ⁺¹c_r)
= ⁷C₄ + ⁷C_4-1 = ⁷⁺¹C₄
= ⁸C₄ = \(\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}\) = 70

(v) ⁸C₀ + ⁸C₁ + …….. + ⁸c₈
Solution:
⁸C₀ + ⁸C₁ + …….. + ⁸c₈ = 2⁸ = 256

Question 2.
Solve :
(i) ⁿC₄ = ⁿC₁₁ ;
Solution:
ⁿC₄ = ⁿC₁₁ ;  (∴ n = 4 + 11 = 15)

(ii) ²ⁿC₃ : ⁿC₃ = 44: 5
Solution:
²ⁿC₃ : ⁿC₃ = \(\frac{44}{5}\)
⇒ \(\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}\)

Question 3.
Find n and r if ⁿP_r = 1680, ⁿC_r = 70.
Solution:
ⁿP_r = 1680, ⁿC_r = 70
∴ \(\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}\)
or, r ! = 24 = 4!
∴ r = 4
Again, ⁿC_r = 70 or ⁿC₄ = 70
or, \(\frac{n !}{4 !(n-4) !}=70\)
or, n(n – 1) (n – 2) (n – 3)
= 70 × 4! = 7 × 10 × 4 × 3 × 2
= 8 × 7 × 6 × 5
or, n(n – 1) (n – 2) (n – 3)
or, 8(8 – 1) (8 – 2) (8 – 3)
∴ n = 8

Question 4.
How many diagonals can an n-gon(a polygon with n sides) have?
Solution:
A polygon of n – sides has n vertices.
∴ The number of st. lines joining the n-vertices is ⁿC₂.
∴ The number of diagonals is ⁿC₂ – n

Question 5.
If a set A has n elements and another set B has m elements, what is the number of relations from A to B?
Solution:
If |A| = n, |B| = m
then |A × B| = mn
∴ The number of possible subsets of
A × B = 2^mn
∴ The number of relations from A to B is 2^mn.

Question 6.
From five consonants and four vowels, how many words consist of three consonants and two vowels?
Solution:
Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.
∴ The number of ways = ⁵C₃ × ⁴C₂
Again, 5 letters can be arranged among themselves in 5! ways.
∴ The total number of ways
= ⁵C₃ × ⁴C₂ × 5! = 10 × 6 × 120 = 7200.

Question 7.
In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?
Solution:
A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.
∴ The number of ways in which the committee can be formed.
⁷C₄ × ⁶C₃ = \(\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}\) = 700

Question 8.
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.
Solution:
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.
∴ The number of ways in which balls are drawn \({ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\) = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4)	White(5)
3	3
4	2

The number of ways in which at least 3 black balls are drawn
= (⁴C₃ × ⁵C₃) + (⁴C₄ × ⁵C₂)
= (4 × 10) + (1 × 10) = 50

Question 9.
How many triangles can be drawn by joining the vertices of a decagon?
Solution:
A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.
∴ The number of triangles formed by the joining of the vertices of a decagon is
¹⁰C₃ = \(\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}\) = 120

Question 9.
How many triangles can be drawn by joining the vertices and the center of a regular hexagon?
Solution:
A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.
⁷C₃ = \(\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35
As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.
∴ The required number of triangles 35 – 3 = 32

Question 11.
Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?
Solution:
Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is
⁶⁰C₂ = \(\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}\) = 1770

Question 12.
In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?
Solution:
10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in ¹¹C₁₀ ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.
∴ The total number of ways = ¹¹C₁₀ × 10! × 10! = (11)! × (10)!

Question 13.
In how many ways can six men and seven girls sit in a row so that the girls always sit together?
Solution:
Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.
∴ The total number of arrangements
= 7! × 7!
= (7!)²

Question 14.
How many factors does 1155 have that are divisible by 3?
Solution:

∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.
∴ The number of ways = ³C₁ + ³C₂ + ³C₀ = 2³ – 1 = 7
∴ The number of factors is 7 excluding 1155 itself.

Question 15.
How many factors does 210 have?
Solution:

∴ We can choose at least one 2, 3, 5, or 7  to be a factor of 210.
∴ The number of factors.
= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ – 1 = 15
∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.
If n is a product of k distinct primes what is the total number of factors of n?
Solution:
n is a product of k distinct primes.
∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.
∴ The number of ways = ^kC₁ + ^kC₂ + ……… ^kC_k-1 = 2^k  – 1 – 1
∴ The number of factors of n is 2^k – 2.
(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.
If m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn, what is the total number of factors of m (excluding 1)?
Solution:
m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn,
∴ m = P₁^r1, P₂^r2 ….. P_n^rn,
P₁ is a factor of m which occurs r₁ times. Each of the factors P₁^r1 will give rise to (r₁ + 1) factors.
Similarly
P₂^r2 gives (r₂ + 1) factors and so on.
∴ The total number of factors (r₁ + 1) (r₂ + 1) ….. (r_n + 1) – 1 (including m).

Question 18.
If 20! were multiplied out, how many consecutive zeros would it have on the right?
Solution:
If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.
How many factors of 10,000 end with a 5 on the right?
Answer:
We have 1000 = 2⁴ × 5⁴ The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125
5 × 5 × 5 × 5 = 625
∴ There are 4 factors ending With 5.

Question 20.
A man has 6 friends. In how many ways can he invite two or more to a dinner party?
Solution:
A man has 6 friends. He can invite 2 or more of his friends to a dinner party.
∴ He can invite 2, 3, 4, 5, or 6 of his friends in
⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 2⁶ – ⁶C₀ – ⁶C₁
= 64 – 7 = 57 ways.

Question 21.
In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?
Solution:
A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in ⁷C₃ = 35 ways.

Question 22.
In how many ways can a student choose five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together?
Solution:
A student chooses five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together.
∴ He is to choose 3 courses out of C₃, C₄, ………… ,C₈, C₉.
Without taking any restrictions 3 courses out of C₃, C₄, …… C₉, i.e. from 7 courses in ⁷C₃ ways. If C₆, C₈ are taken together then one course only to be choosen from C₃, C₄, C₅, C₇, C₉ by ⁵C₁ ways. Hence required number of ways.
= ⁷C₃ – ⁵C₁
= \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) – 5
= 35 – 5 = 30 ways.

Question 23.
A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?
Solution:
A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.
The selection can be made as follows :

Batsmen(8)	Bowlers(5)
8	3
7	4
6	5

The number of selections is
(⁸C₈ × ⁵C₃) + (⁸C₇ × ⁵C₄) + (⁸C₆ × ⁵C₅)
= (1 × 10) + (8 × 5) + (28 × 1)
= 10 + 40 + 28 = 78

Question 24.
There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?
Solution:
There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.
∴ We can form a straight line by joining any two points.
n-collinear points form one line and r-non-collinear points form ^rC₂ lines.
Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.
∴ The number of such limes is r x n.
∴ The total number of lines

Question 25.
There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?
Solution:
There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.
∴ The total number of arrangements
= 5! x 6!

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(B) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(B)

BSE Odisha 6th Class English Test-1(B) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Write the following Odia names of the persons in English. (Teacher will give the names of six persons in Odia.) [06]
Answer:
ଲାଲା ଲଜପାଟ ରୟ – Lala Lajpat Roy
B.R. ଆମ୍ବେଦକର | – B.R. Ambedkar
ବାଙ୍କିମ୍ ଚନ୍ଦ୍ର ପାଲ୍ – Bankim Chandra Pal.
ବୁକ୍ସି ଜଗବନ୍ଧୁ – Buxi Jagabandhu
ଫାକିର ମୋହନ ସେନାପତି – Fakir Mohan Senapati
ରାଧନାଥା ରାଧା – Radhanatha Ratha

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ଶ୍ରୀନଗର – Srinagar
ହିମାଚଳ ପ୍ରଦେଶ – Himachal Pradesh
ପୁଡୁଚେରୀ – Puducheri
ଦ୍ୱାରିକା – Dwarika
ଚିଲିକା ହ୍ରଦ – Chilika lake
ନନ୍ଦନକାନନ୍ ପ୍ରାଣୀ ଉଦ୍ୟାନ – Nandankanan Zoo

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer :
Jagannatha     Moon
Saraswati        Capital
Ganesha         Chief-Minister
Earth /World  President
Sun                 Garden

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
temper, catch, squirrel, little, agile, climbed, thrash, burst, bamboo, curry, foolish.
[Listen to your teacher and tick those words s/he reads.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. (Question with Answer) [08]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. Do you know how much tea he took?

6. Match the words which sound alike at the end. (Question with Answer)

7. Read the poem and answer the questions. [10]
Their ears are pink,
Their teeth are white,
They run about
The house at night
They nibble things
They shouldn’t touch
And no one seems
To like them much.

Question (i).
What is the color of their ears?
Answer:
The color of their ears is pink.

Question (ii).
What is the color of their teeth?
Answer:
The color of their teeth is white.

Question (iii).
Where do they run about at night?
Answer:
They run about the house at night.

Question (iv).
Which things do they nibble?
Answer:
They nibble things that they shouldn’t touch.

Question (v).
Do most people like mice?
Answer:
No, no one seems to like mice.

8. Read the following paragraph and answer the questions in complete sentences. [20]
After eating the dog, the man walked, walked, and walked till he met a little squirrel. The little squirrel asked the old man, “Old man, old man, what makes you so fat? The old man said, “I’ve taken a very heavy breakfast. In my breakfast, I took two mugs of tea, two liters of milk, three tins of biscuits, and five big pieces of cakes.” Then I ate a little boy and a small dog. I’ll also eat you up if I can catch you”.

“But you cannot catch me, old man,” said the active, agile, little squirrel. Then the squirrel jumped up the tree, the old man also climbed up the tree. The little squirrel jumped up to the main branch of the tree. The old man also climbed up to the main branch of the tree. Next, the little squirrel jumped up to a thin branch. The old man also climbed up to the thin branch. But thrash ! the small branch broke and the old man fell to the ground. His big belly burst out.

Question (i).
Whom did the old man meet in this section?
Answer:
In this section, the old man met a little squirrel.

Question (ii).
What did the squirrel ask the old man?
Answer:
The squirrel asked the old man, “What makes you so fat?”

Question (iii).
How many mugs of tea did the old man take?
Answer:
The old man took two mugs of tea.

Question (iv).
How much milk did he take?
Answer:
He took two liters of milk.

Question (v).
How many tins of biscuits did he take in his breakfast?
Answer:
For his breakfast, he took three tins of biscuits.

Question (vi).
How many pieces of cake did he take?
Answer:
He took five big pieces of cake.

Question (vii).
Where did the squirrel jump up first?
Answer:
The squirrel jumped up first to the tree.

Question (viii).
What did the old man do?
Answer:
The old man also climbed up the tree.

Question (ix).
Where did the squirrel jump up then?
Answer:
Then the squirrel jumped up to the main branch of the tree.

Question (x).
What happened to the old man when he fell down?
Answer:
When the old man fell down, his big belly burst out.

9. Read the following poem and answer the questions in complete sentences. [10]
“My story is said
The flowering plant is dead.
O flowering plant! Why did you die?
The black cow ate me up and made me lie.
O black cow! Why did the plant you eat?
Because the cowherd did not do well treat.
O, cowherd! Why didn’t you well the cow treat to eat?
The daughter-in-law did not give me food.”

Question (i).
Why is the flowering plant dead?
Answer:
The flowering plant is dead because the black cow ate it up.

Question (ii).
Why did the cow eat up the flowering plant?
Answer:
The cow ate up the flowering plant because the cowherd did not well her treat.

Question (iii).
Why didn’t the cowherd well the cow treat to eat?
Answer:
The cowherd didn’t well the cow treat to eat because the daughter-in-law did not give him food.

Question (iv).
What is said?
Answer:
Your story is said.

Question (v).
What type of plant is dead?
Answer:
The flowering plant is dead.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
Once a foolish Santal son-in-law went to his in-law’s house. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from? Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” Yes, son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”. The next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Question (i).
Who went to his father-in-law’s house?
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question (ii).
What did his mother-in-law cook for him?
Answer:
His mother-in-law cooked delicious dishes for him.

Question (iii).
What was one of the dishes made?
Answer:
One of the dishes was a curry made out of the bamboo shoot.

Question (iv).
How was the curry?
Answer:
The curry was extremely delicious.

Question (v).
What did the son-in-law ask his mother-in-law?
Answer:
The son-in-law asked his mother-in-law what the curry was made from.

Question (vi).
What did the mother-in-law answer?
Answer:
Instead of answering his question, she pointed at the bamboo door.

Question (vii).
Why is the curry called “Bamboo Curry”?
Answer:
The curry is called ‘Bamboo Curry’ because it is made from bamboo.

Question (viii).
When was the son-in-law returning to his home?
Answer:
The next day, the son-in-law was returning to his home.

Question (ix).
What did he think of doing?
Answer:
He thought of cooking bamboo curry at home.

Question (x).
What did not they have? What did he do for that?
Answer:
They did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(A) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(A)

BSE Odisha 6th Class English Test-1(A) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English. [06]
(Teacher will give the names of six persons in Odia.)
Answer:
ଲାଲ ବାହାଦୂର ଶାସ୍ତ୍ରୀ | – Lal Bahadur Shastri
ବାଲ ଗଙ୍ଗାଧର ତିଲକ | – Bal Gangadhar Tilak
ସର୍ଦ୍ଦାର ଭାଲାଭଭାଇ ପଟେଲ | – Sardar Ballavabhai Patel
ହରେକୃଷ୍ଣ ମହାତ୍ମା – Harekrushna Mahatab
ଭୀମା ଭୋଇ – Bhima Bhoi
ପ୍ରାଣକୃଷ୍ଣ ପରଜା | – Pranakrushna Parija

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ହିମାଳୟ – Himalaya
ପଞ୍ଜାବ – Punjab
ଅରୁଣାଚଳ ପ୍ରଦେଶ – Arunachal Pradesh
କୋଲକାତା – Kolkata
ବ୍ରହ୍ମପୁର – Berhampur
ବାଲାସୋର – Balasore

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer:
Grandfather      Ice-cream       Friend
Grandmother   Computer       Neighbour
Bread               Blackboard      Government
Mango

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
kite, paper, greedy, breakfast, tea, mugs, biscuits, cakes, conversation, heavey, squirrel.
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and nil in the gaps. (Question with Answer) [08]
Answer:
The old man said, “I’ve taken two mugs of tea, and two liters of milk. I also took three tins of biscuits and five kilograms of cakes. And if I can catch you, I’ll eat you up .’’Then the old man caught the thin little boy and ate him up.

6. Match the words which sound alike at the end. (Question with Answer) [10]

7. Read the poem and answer the questions. [10]
I think mice
Are rather nice.
Their tails are long
Their faces small,
They haven’t any
Skins at all.

Question (i).
What is this poem about?
Answer:
This poem is about mice.

Question (ii).
Who is ‘I’ at the beginning of the poem?
Answer:
‘I’ at the beginning of the poem is the poet himself.

Question (iii).
What are the tails of mice like?
Answer:
The tails of mice are long.

Question (iv).
How are their faces?
Answer:
Their faces are small.

Question (v).
They haven’t any skint at all”- What does this line mean?
Answer:
‘‘They haven’t any skins at all”- This line means that mice have very thin skins.

8. Read the following paragraph and answer the questions in complete sentences. [20]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. He took two mugs of tea and two liters of milk. Then he took three tins of biscuits and five big pieces of cake. After breakfast, he looked really very very fat.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a greedy fat old man.

Question (ii).
How was the old man?
Answer:
The old man was greedy and fat.

Question (iii).
When did he get up one day?
Answer:
One day he got up at 6 a.m.

Question (iv).
When did he brush his teeth?
Answer:
He brushed his teeth at 6.30 a.m.

Question (v).
What did he take at 7 a.m.?
Answer:
He took tea at 7 a.m.

Question (vi).
When did he take his breakfast?
Answer:
He took his breakfast at 8.30 a.m.

Question (vii).
How many mugs of tea did he take?
Answer:
He took two mugs of tea.

Question (viii).
How much milk did he take?
Answer:
He took two liters of milk.

Question (ix).
How many tins of biscuit did he take?
Answer:
He took three tins of biscuits.

Question (x).
How did he look like after breakfast?
Answer:
After breakfast, he looked really very very fat.

9. Read the following poem and answer the questions in complete sentences. [10]

For want of a nail,
the shoe was lost.
For want of a shoe,
the horse was lost.
For want of a horse,
the rider was lost.
For want of a rider,
the battle was lost.
For want of a battle,
the kingdom was lost.
And all for the want of a
horseshoe nail.

Question (i).
Why was the shoe lost?
Answer:
The shoe was lost for want of a nail.

Question (ii).
Why was the horse lost?
Answer:
The horse was lost for want of a shoe.

Question (iii).
Why was the rider lost?
Answer:
The rider was lost for want of a horse.

Question (iv).
Why was the battle lost?
Answer:
The battle was lost for want a rider.

Question (v).
Why was the kingdom lost?
Answer:
The kingdom was lost for want of a battle.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
A son-in-law, after his marriage, was planning to visit his father-in-law’s house for the first time. A man from his village gave him the advice, “Use big and high-sounding words in your father-in-law’s house. Always sit on a high place. First say ‘No’ to any food given to you.”
He, therefore, used very long and high-sounding words. He told his mother-in-law, “You are the sweetest, kindest, greatest, and gentlest lady.” The mother-in-law was very much pleased to hear this. She praised her son-in-law in front of her neighbors for using high-sounding words and calling her the kindest and greatest lady.

Question (i).
What was the son-in-law’s plan?
Answer:
The son-in-law’s plan was to visit his father-in-law’s house.

Question (ii).
When was the son-in-law planning to visit his father-in-law’s house?
Answer:
After his marriage, the son-in-law was planning to visit his father-in-law’s house.

Question (iii).
Did he visit his father-in-law’s house before?
Answer:
No, he did not visit his father-in-law’s house before.

Question (iv).
Who advised him to do something at his in-law’s house?
Answer:
A man from his village advised him to do something at his in-law’s house.

Question (v).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (vi).
What was the second advice for the son-in-law?
Answer:
The second piece of advice for the son-in-law was always to sit on a high place.

Question (vii).
What was the third piece of advice for the son-in-law?
Answer:
The third piece of advice for the son-in-law was first to say ‘No’ to any food given to him.

Question (viii).
What did he tell his mother-in-law?
Answer:
He told his mother-in-law that she was the sweetest, kindest, greatest and gentlest lady.

Question (ix).
How did the mother-in-law feel to hear his words?
Answer:
The mother-in-law was very much pleased to hear his words.

Question (x).
What did she do?
Answer:
She praised her son-in-law in front of her neighbors for calling her the kindest and greatest lady.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1 Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1

BSE Odisha 6th Class English Test -1 Text Book Questions and Answers

• The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English.
(Teacher will give the names of six persons in Odia.) [06]

Answer:
ମହାତ୍ମା ଗାନ୍ଧୀ – Mahatma Gandhi
ଜବାହରଲାଲ ନେହେରୁ | – Jawaharlal Nehru
ସୁଭାଷ ବୋଷ | – Subhas Bose
ଗୋପାବନ୍ଧୁ ଦାସ – Gopabandhu Das
ମଧୁସୂଦନ ଦାସ – Madhusudan Das
ବିଜୁ ପଟ୍ଟନାୟକ – Biju Patnaik

2. Write the following place names in English.
(Teacher will give names of six places in Odia.) [06]

Answer:
ଦିଲ୍ଲୀ – Delhi
କାଶ୍ମୀର – Kashmir
ଅୟୋଧ୍ୟା – Ayodhya
ଆସାମ | – Assam
ଭୁବନେଶ୍ୱର – Bhubaneswar
କଟକ | – Cuttack

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]

Answer :
School     Mother     Town
Book        House      Temple
Teacher    Village      God
Father

4. Given below are some words. Your teacher will read aloud five of them. Tick those which s/he reads aloud. [05]
house, gentle, great, night, cotton, lemon, nibble, thief, delicious, curry, kingdom, nail, battle, ground.
[Listen to your teacher and tick those words your teacher reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. [08]
(Question with Answer)
Next ____________the son-in-law was ___________to leave for his ______________ . The bamboo _____________came to his ___________________. He thought of_____________ bamboo curry at home. But ________________ did not have _______________________.
Answer:
Next day the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo.

6. Match the words which sound alike at the end. (Question with Answer) [10]
(i)

Answer:

(ii)

Answer:

7. Read the poem and answer the questions. [10]

A kite on the ground
Is just paper and string
But up in the air
Will dance and sing.
A kite in the air
Will dance and caper
But back on the ground
Is just string and paper.

Question (a).
What is this poem about?
Answer:
This poem is about a kite.

Question (b).
What does it do when up in the air?
Answer:
When Up in the air, it will dance and sing.

Question (c).
What is it when it is on the ground?
Answer:
When on the ground, it is just paper and string.

Question (d).
Do you like flying a kite?
Answer:
Yes, I like flying a kite very much.

Question (e).
What are kites normally made of?
Answer:
Normally, kites are made of paper and string.

8. Read the following paragraph and answer the questions in complete sentences. [20]
A son-in-law after his marriage was planning to visit his father-in-law for the first time. A man from his village gave him some advice, “Use big and high-sounding words in your father-in-law’s house. Always sit in a high place. First, say ‘no’ to any food given to you.’’

Question (a).
What is the paragraph about?
Answer:
The paragraph is about a son-in-law.

Question (b).
What was he planning?
Answer:
He was planning to visit his father-in-law’s house.

Question (c).
When was he planning?
Answer:
He was planning after his marriage.

Question (d).
Had he gone to his father-in-law’s house before?
Answer:
No, he had not gone to his father-in-law’s house before.

Question (e).
Who gave him some advice?
Answer:
A man from his village gave him some advice.

Question (f).
How many pieces of advice did he give?
Answer:
He gave three pieces of advice.

Question (g).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (h).
What was the second piece of advice?
Answer:
The second piece of advice was always to sit in a high place.

Question (i).
What was the third piece of advice?
Answer:
The third piece of advice was first to say ‘no’ to any food given to him.

Question (j).
Will he carry out the advice?
Answer:
Yes, he will carry out the advice.

9. Read the following poem and answer the questions in complete sentences. [10]

I woke up this morning
And I got out of bed,
Then I took a cup of tea
And ate a slice of bread.
1 went to the bus stop
And caught the bus to school,
On my way back it rained
And the weather was cool.

Question (i).
Who is T in the poem?
Answer:
T in the poem is the poet.

Question (ii).
What did s/he take after getting up?
Answer:
After getting up, s/he took a cup of tea.

Question (iii).
What did s/he eat?
Answer:
S/he ate a slice of bread.

Question (iv).
How did s/he go to school?
Answer:
S/he went to school by bus.

Question (v).
What happened on his / her way back?
Answer:
On his/her way back, it rained and the weather was cool.

10. Read the following paragraph and answer the questions in complete sentences. [20]
There was a crow and there was a cuckoo. They lived together happily. The crow was hard-working. But the cuckoo was very lazy. The crow brought food. The cuckoo only ate. The crow built a nest. She laid her eggs there. The cuckoo also wanted to lay eggs. But she had
not built a nest. One day the crow was not in her nest. The cuckoo threw away the eggs and laid her own eggs there. The crow did not know this. She sat over the eggs for some days. Young ones came out. She took care of them. But when the young ones grew? up, they sang like cuckoos. So she drove them away. From that day the crow always drives away the cuckoo.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a crow and a cuckoo.

Question (ii).
Who was lazy?
Answer:
The cuckoo was very lazy.

Question (iii).
Who was hard-working?
Answer:
The crow was hard-working.

Question (iv).
Who brought food?
Answer:
The crow brought food.

Question (v).
Who ate it?
Answer:
The cuckoo only ate it.

Question (vi).
Who built a nest?
Answer:
The crow built a nest.

Question (vii).
What did the cuckoo do with the eggs of the crow?
Answer:
The cuckoo threw away the eggs of the crow and laid her own eggs there.

Question (viii).
Whose young ones the crow was taking care of?
Answer:
The crow was taking care of the young ones of the cuckoo.

Question (ix).
When did she come to know about this?
Answer:
When the young ones grew up and sang like cuckoos, she (crow) came to know about this.

Question (x).
Why could not she recognize them before?
Answer:
She (the crow) could not recognize them before because in her absence the cuckoo had thrown away her eggs and laid her (cuckoo’s) own eggs there. Further, the young ones of the crow and the cuckoo look alike.

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(a) ବିମ୍ନ ଟିତ୍ର (a) 6ର L₁ || L₂ || L₃ ଏର୍ଚ T₁ ଓ T₂ ଛେଦ ଦା |
(i) AB = 2 6ସ.ର୍ମି., BC = 3 6ସ.ର୍ମି. ଓ DE = 3 6ସ.ର୍ମି. 6ହ6ଲ EF = ……….|
(ii) DE = 6 6ସ.ର୍ମି., EF = 8 6ସ.ର୍ମି. ଓ BC = 6 6ସ.ର୍ମି. 6ହ6ଲ AC = ……….|

(b) ଗପଭୋକ୍ତ ତିତ୍ର (b) 6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଛେଦକ |
(i) AB = 1.5 × BC ହେଲେ, \(\frac { EF }{ FD }\) = ……………..
(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟବିରୁ B 6ହ6ଲ, EF ର ……… ମୁଶ 6ହରଛି FD|
Solution:
(a) (i) L₁ || L₂ || L₃ ⇒ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac { 2 }{ 3 }\) = \(\frac { 3 }{ EF }\) ⇒ EF = \(\frac{3 \times 3}{2}\) 6 ସ.ମି. = 4.5 6ସ.ମି.

(ii) L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ରୁକଚି 6ଛଦକ |
∴ \(\frac { DE }{ EF }\) = \(\frac { AB }{ BC }\) ⇒ \(\frac { 6 }{ 8 }\) = \(\frac { AB }{ 6 }\)
⇒ AB = \(\frac{6 \times 6}{8}\) = 4.5 6ସ.ମି. |
AC = AB + BC = 4.5 6ସ.ମି. + 6ସ.ମି. = 10.5 6ସ.ମି. |

(b) (i) ଏଠ।6ର L₁ || L₂ || L₃ ଏରଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଚଦକ |
\(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { 3 }{ 2 }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac{3+2}{2}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) ⇒ \(\frac { DE }{ EF }\) = \(\frac { 5 }{ 2 }\) ⇒ \(\frac { EF }{ DF }\) = \(\frac { 2 }{ 5 }\)

(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟ ଦିନୁ B ଅର୍ଥ।ତ୍ AB = BC
\(\frac { EF }{ FD }\) = \(\frac { BC }{ AC }\) = \(\frac{\mathrm{BC}}{\mathrm{AB}+\mathrm{BC}}\) = \(\frac { BC }{ 2BC }\) = \(\frac { 1 }{ 2 }\) ⇒ 2EF = FD>
ଅର୍ଥାତ୍ EF ଭ 2 ଣ୍ଣଣ 6ଦ୍ରଣଛି FD |

Question 2.
ଟିତ୍ର6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଛଦକ | L₂ ଓ L₃ ରପ6ର ପ୍ଥଥ।କୁ6ର G ଓ H ଦିହୁ ରିଜିତ 6 ପ୍ପପତି BC = AD ଏବଂ CH = BE;

ପ୍ରମାଣ କର 6ଯ
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF
Solution:
ଦଉ : L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁକଟି 6ଚ୍ଛଦକ ଯଥାସ୍ତ6ମ L₁, L₂ ଓ L₃ କୁ A,B,C ଓ D,E,F ଦିନ୍ଦ6ର 6ଚ୍ଛଦକ6ର | L₂ ରପ6ର G ଏକ ଦିବ୍ର 6ସ୍ ପରି BG = AD ଏର୍ବ L₃ ରପ6ର H ଏକ ଦିନ୍ଦୁ 6ନ୍ଦ୍ ପରି CH = BE |
ପ୍ରାମାଣ୍ୟ :
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF

ଅକ୍ଳଜ : \(\overline{\mathrm{DG}}\) ଓ \(\overline{\mathrm{EH}}\) ଅଲ୍ଲବ ଦବାପାର |
ପ୍ତମାଣ : (i) AD = BG (ଦର) \((\overline{\mathrm{AD}} \| \overline{\mathrm{BG}})\), ∵ L₁ || L₂ (ଦର)
⇒ ABGD ଏକ ସାମାତ୍ରତିଦ ଚିତ୍ର | AB = DG
6ସଦୃପତି BEHC ଏକ ସାମାତ୍ର ଚିତ୍ର ଓ BC = EH
∴ \(\frac { DG }{ EH }\) = \(\frac { AB }{ BC }\) …(1)
ପୁନଣ୍ଡ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) (1ରୁ) (ପ୍ରମାଣିତ)

(ii) ପୁଫରୁ ପ୍ରମାଣିର \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) (Componendo ଦ୍ଦାରା )
⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac { DF }{ EF }\) (ପ୍ରମାଣିତ)

Question 3.
ନିମ୍ନ ଚିତ୍ରରେ L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । ଯଦି AB = BC ହୁଏ, ପ୍ରମାଣ କର ଯେ 2 BE = AD + CF |
Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଛେଦକ T₁ ଓ T₂ , L₁, L₂, ଓ L₃, କୁ ଯଥାକ୍ରମେ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦ କରେ ଏବଂ AB = BC |
ପ୍ରାମାଣ୍ୟ : 2BE = AD + CF
ଅକନ : E ଦିଦୁ ମଧ୍ୟ6ଦକ AC ସମାତ ତା6ଦ ଅଜିତ ବେଡା L₁ || L₃ କୁ ଯଥାକୁ6ମ X ଓ Y ଦିଦ6ର 6ଛଦକ୍ଳରୁ |

ଅଙ୍କନ : E ବିନ୍ଦୁ ମଧ୍ୟଦେଇ \(\overline{\mathrm{AC}}\)ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା L₁ ଓ L₃, କୁ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁରେ ଛେଦକରୁ ।

Question 4.
ଚିତ୍ରରେ L₁ || L₂ || L₃, ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₁, ଯଥାକ୍ରମେ A, B ଓ C ବିନ୍ଦୁରେ ଛେଦକରେ ଏବଂ L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₂, ଯଥାକ୍ରମେ D, E ଓ F ବିନ୍ଦୁରେ ଛେଦ କରେ । DE = EF ହେଲେ, ପ୍ରମାଣ କର ଯେ, CF – AD = 2 EB । (ସୂଚନା : AF ଅଙ୍କନ କର)

Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ ଯଥାକ୍ରମେ L₁, L₂, ଓ L₃, କୁ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : CF – AD = 2EB.
ଅଙ୍କନ : \(\overline{\mathrm{AF}}\) ଅଙ୍କନ କର । B ବିନ୍ଦୁରୁ \(\overline{\mathrm{AF}}\) ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା \(\overline{\mathrm{FC}}\) କୁ ଓ ବିନ୍ଦୁରେ ଛେଦକରୁ ।
\(\overline{\mathrm{AF}}\), L₂, କୁ H ବିନ୍ଦୁରେ ଛେଦକରୁ ।

ପ୍ରମାଣ : DE = EF (ଦଣ) ଓ \(\stackrel{\leftrightarrow}{\mathrm{BE}}\) || \(\overline{\mathrm{AD}}\) (∵ L₂ || L₁)
⇒ AH = HF,
ଦର୍ମାବ AH = HF ଓ \(\overline{\mathrm{BH}}\) || \(\overline{\mathrm{CF}}\) ⇒ AB = BC
\(\overline{\mathrm{BG}}\) || \(\overline{\mathrm{AF}}\) (ଅକo) ⇒ CG = GF ⇒ CF = 2GF △AFD 6ର H ଓ E ଯଥାକ୍6ମ \(\overline{\mathrm{AF}}\) ଓ \(\overline{\mathrm{DF}}\) ର ମଧ୍ୟଦିଦୁ
⇒ \(\overline{\mathrm{EH}}\) || \(\overline{\mathrm{AD}}\)
∴ ଦଘଣପଷ୍ଟ = CF – AD = 2GF – 2HE
= 2BH – 2HE (∵ GF = BH)
= 2(BH – HE) = 2BE = ଦାମପଖ (ପ୍ରମାଣିତ)

Question 5.

(i) ଭପରିସ୍ଥ ଚିତ୍ର (a) 6ର A – D – B ଏବଂ A – E – C | m∠DAE = 50°, m ∠AED = m∠ABC = 65° | AD = 3 6ସ.ମି., AE : EC = 2 : 1 6ଦୃ6କ, \(\overline{\mathrm{DB}}\) ଓ \(\overline{\mathrm{AB}}\) ର 6ଦିଘ୍ୟ ବ୍ଶଷଯ ଦର |
(ii) ଉପରିସ୍ଥ ବିତ୍ର (b) 6ର \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\), NR = \(\frac { 2 }{ 5 }\) PR ଏବଂ PQ = 10 6ସ. ମି. 6ଦୃ6କ, PM ଓ QM କିଣପ୍ର କର |
(iii) କପଟାସ୍ଟ ଚିତୃ (b) 6ର PM = \(\frac { 1 }{ 2 }\) PQ, NR = 1.2 6ସ.ମି. ଓ \(\overline{\mathbf{M N}} \| \overline{\mathbf{Q R}}\) 6 ଦୁ6 କ, PR ଷ୍ଟିର କର
Solution:
(i) △ ADE 6ର m∠AD = 180° – (50° + 65°) = 65°
∠ADE ≅ ∠ABC ⇒ \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\)

⇒ \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) = \(\frac { 2 }{ 1 }\)
⇒ \(\frac{3}{DB}\) = \(\frac { 2 }{ 1 }\) ⇒ DB = \(\frac { 3 }{ 2 }\) 6ପ.ମି. = 1.5 6ସ.ମି
AB = AD + DB = 3 6ସ.ମି. + 1.5 6ସ.ମି = 4.5 6ସ.ମି. |

(ii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) (ଦଉ)
\(\frac { PM }{ MQ }\) = \(\frac { PN }{ NR }\) ⇒ \(\frac{P M+M Q}{M Q}\) = \(\frac{\mathrm{PN}+\mathrm{NR}}{\mathrm{NR}}\)

⇒ \(\frac { PQ }{ MQ }\) = \(\frac { PR }{ NR }\) ⇒ \(\frac { 10 }{ MQ }\) = \(\frac{\mathrm{PR}}{\frac{2}{5} \mathrm{PR}}\)
⇒ \(\frac { 10 }{ MQ }\) = \(\frac { 5 }{ 2 }\)
⇒ MQ = \(\frac{10 \times 2}{5}\) 6ସ.ମି. = 4 6ସ.ମି. |
PM = PQ – MQ = 10 6ସ.ମି. – 4 6ସ.ମି. = 6 6ସ.ମି. |
∴ \(\overline{\mathrm{PM}}\) \(\overline{\mathrm{QM}}\) ର ଦେଶ ଯଥାଦୃ6ମ 6 6ସ.ମି. ଓ 4 6ସ.ମି.|

(iii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) , PM = \(\frac { 2 }{ 3 }\) PQ
⇒ MQ = PQ – PM = PQ – \(\frac { 2 }{ 3 }\) PQ = \(\frac { 1 }{ 3 }\) PQ

⇒ \(\frac { 1.2 }{ PR }\) = \(\frac{\frac{1}{3} \mathrm{PQ}}{\mathrm{PQ}}\) ⇒ \(\frac { 1.2 }{ PR }\) = \(\frac { 1 }{ 3 }\)
⇒ PR = 1.2 × 3 6ସ.ମି. = 3.6 ସେ.ମି. |
∴ \(\overline{\mathrm{PR}}\) ର 6ଦଣu = 3.6 ପେ.ପି. |

Question 6.
(i) △ABC 6ର, \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{AC}}\) ର ମଧତିହୁ ଯଥାକ୍ର6ମ X ଓ Y 6ହ6କ, ଦଶାଥ 6ଯ \(\overline{\mathbf{X Y}} \| \overline{\mathbf{B C}}\) |
(ii) ଏକ ତ୍ରିସ୍ତକର 6ଣାଟିଏ ତାଦୃତ ମଧ୍ୟଝିହୁ 6ଦକ ଅଠ୍ୟ ଏଜ ଚାହପ୍ତତି ଅଜାଡ ସମାତର 6ରଖା, ତୃତୀୟ ବlନୁ ସମକ୍ରିଖଶ୍ନ କ6ର |
(iii) 6ଣାଟିଏ ପମ6କାଣା ତ୍ରିହ୍ନର ଦଶାବ ମଧ୍ୟବିଦୁ ଅଠ୍ୟ ଏବ୍ଲ ଗ୍ଲାସ୍ଥଗି ସମସ୍ତିଖଣ୍ଡକ6ର , ପ୍ରମାଣ କର |
Solution:

Question 7.
△PQR ରେ, \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ବାହୁଦ୍ୱୟର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ M ଓ N । \(\overline{\mathbf{PR}}\) ଉପରିସ୍ଥ S ଯେକୌଣସି ଏକ ବିନ୍ଦୁ ହେଲେ, ପ୍ରମାଣ କର ଯେ \(\overline{\mathbf{MN}}\), \(\overline{\mathbf{QS}}\) କୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରିବ ।
Solution:
ଦତ : △PQR 6ର \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ର ମଧ୍ୟବିଦୁ ଯଥାଦୃ6ମ M ଓ N | \(\overline{\mathbf{PR}}\) ଭପରିସ୍ଥ ‘S’ ଏକ ବିଦୁ | \(\overline{\mathbf{QS}}\) ଓ \(\overline{\mathbf{MN}}\) ର 6ଛଦଦିଦୁ O |
ସ୍ତାମାଶ୍ୟ : OQ = OS

ପ୍ରମାଣ : \(\frac { QM }{ MP }\) = 1 (∵ QM = MP (ଦର) )
\(\frac { QN }{ NR }\) = 1 (∵ QN = NR (ଦର) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QN }{ NR }\) ⇒ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\)
\(\overline{\mathrm{MO}} \| \overline{\mathrm{PS}}\) (∵ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QO }{ OS }\) ⇒ 1 = \(\frac { QO }{ OS }\) (∵ \(\frac { QM }{ MP }\) = 1)
⇒ OQ = OS (ପ୍ତମାଣିତ)

Question 8.
ABCD ଣ୍ଠାପିଲିଷ୍ଟମ6ର \(\overline{\mathbf{AB}} \| \overline{\mathbf{CD}}\) | ଲଣ \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର 6ବ୍ଳଦଦିନ୍ଦି P 6ଦୃ6କ , ପ୍ତମାଶା କର 6ମ
(i) AP : PC = BP : PD
(ii) CP : AC = DP : BD |
Solution:

Question 9.
ABCD ଗ୍ରାପିଦିକ୍ଷ୍ମମ6ର \(\overline{\mathbf{AB}}\) || \(\overline{\mathbf{DC}}\) ଏର \(\overline{\mathbf{AD}}\) ର ମଧ୍ୟତିଦୁ P | \(\overline{\mathbf{AB}}\) ସଦୃ ସାପବ୍ଦର ଉ6ର ଅବିର \(\overleftrightarrow{\mathbf{P Q}}\) \(\overline{\mathbf{BC}}\) କ Q ଦିହି6ର 6ଛଦନ6କ , ପ୍ରମାଣ ଦ୍ଦର 6ଯ Q 6ଦୃରଚି \(\overline{\mathbf{BC}}\) ର ମଧ୍ୟନିନ୍ଦୁ |
Solution:

Question 10.
ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) ,ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
(a) ପ୍ରମାଣ କର ଯେ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
(b) ଉପରୋକ୍ତ ଚତୁର୍ଭୁଜ ABCD ର କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପର ପ୍ରତି ଲମ୍ବ ହେଲେ, ପ୍ରମାଣକର ଯେ PORS ଏକ ଆୟତ ଚିତ୍ର ।
Solution:
(a) ଦତ୍ତ : ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) , ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
ଅଙ୍କନ : \(\overline{\mathrm{BD}}\) ଅଙ୍କନ କରାଯାଉ ।
ପ୍ରମାଣ : △ABD ରେ \(\overline{\mathrm{AB}}\) ର ମଧ୍ୟବିନ୍ଦୁ P ଓ \(\overline{\mathrm{AD}}\) ର ମଧ୍ୟବିନ୍ଦୁ S ।
\(\frac { AP }{ BP }\) = 1 (∵ AP = BP (ଦତ୍ତ))
ସେହିପରି ଧୂ) \(\frac { AS }{ SD }\) = 1 (∵ AS = SD (ଦତ୍ତ))

∴\(\frac { AP }{ BP }\) = \(\frac { AS }{ SD }\) ⇒ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\)
ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରିବ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\) |
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\)
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{QR}}\)
\(\overline{\mathrm{AC}}\) ଅଙ୍କନ କରି ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରେ ଯେ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{SR}}\) |
∴ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର |

(b) ଦତ୍ତ : ABCD ଚତୁର୍ଭୁଜରେ \(\overline{\mathrm{AC}}\) ⊥ \(\overline{\mathrm{BD}}\) | \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CD}}\) ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ P, Q, R, S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ଆୟତଚିତ୍ର ।
ପ୍ରମାଣ : ମନେକର \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{BD}}\) ଛେଦବିନ୍ଦୁ ( ଏବଂ \(\overline{\mathrm{PS}}\) ଓ \(\overline{\mathrm{AC}}\) ର ଛେଦବିନ୍ଦୁ X ।

(a) ର ଅନୁରୂପ ପ୍ରମାଣ ଦ୍ଵାରା PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର 6ଚିତ୍ର ।
m∠AOB = 90° ⇒ m∠PXO = 90°
(∵ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\))
m∠XPXQ = 90° (∵ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\))
∴ PQRS ଏକ ସାମାନ୍ତରିକ |

Question 11.
ନିମ୍ନ ଚିତ୍ରରେ △ABC ର \(\overline{\mathrm{BA}}\) ବାହୁ ସହ \(\overline{\mathrm{CM}}\) ସମାନ୍ତର, \(\overline{\mathrm{AB}}\) ର ପ୍ରାମାବିନ୍ଦୁ P | \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\), \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\); ସାମାନ୍ତରିକ ମେ
\(\overline{\mathrm{PR}}\) || \(\overline{\mathrm{AM}}\) |
Solution:
ଦତ୍ତ : △ABC ରେ \(\overline{\mathrm{BA}}\) || \(\overline{\mathrm{CM}}\), AP = BP,
\(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\)
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathrm{PR}}\)||\(\overline{\mathrm{AM}}\)
ପ୍ରମାଣ : △ABC ରେ \(\overline{\mathrm{PQ}}\) ||\(\overline{\mathrm{AC}}\) (ଦତ୍ତ) ।

⇒ \(\frac { BP }{ AP }\) = \(\frac { BQ }{ QC }\)
△BCM ରେ \(\overline{\mathrm{QR}}\) ||\(\overline{\mathrm{CM}}\) (ଦତ୍ତ) ।
⇒ \(\frac { BQ }{ QC }\) = \(\frac { BR }{ RM }\) ……(i)
∴ (i) ଓ (ii)ରୁ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\)
△BCM ରେ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\) ⇒ \(\overline{\mathrm{PR}}\) ||\(\overline{\mathrm{AM}}\) (ପ୍ରମାଣିତ)
ତ୍ରିଭୁଜର କୋଣର ସମଦ୍ବିଖଣ୍ଡକ ସମ୍ବନ୍ଧୀୟ ଆଲୋଚନା

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c)

Question 1.
ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 137 ଓ ସେମାନଙ୍କର ବିୟୋଗଫଳ 43। ତେବେ ସଂଖ୍ୟାଦ୍ୱୟ ନିରୂପଣ କର ।
ସମାଧାନ:
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y ।
ପ୍ରଶ୍ନନୁସାରେ x + y = 137 ……..(i) ଏବଂ x – y = 43 …….(ii)
ସମୀକରଣ (i) ଓ (ii) କୁ ପ୍ରୟୋଗକଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, x + y = 137
⇒ 90 + y = 137 ⇒ y = 137 – 90 ⇒ y = 47
∴ ସଂଖ୍ୟାଦ୍ଵୟ 90 ଓ 47।

Question 2.
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁ ତ୍ରୟର ଦୈର୍ଘ୍ୟ x + 4 ସେ.ମି., 4x – y ସେ.ମି. ଓ y + 2 ସେ.ମି. ହେଲେ ବାହୁର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୂତ୍ରୟର ଦୈର୍ଘ୍ୟ ସମାନ । ଅର୍ଥାତ୍ x + 4 = 4x – y = y + 2
⇒ x + 4 = y + 2 = x – y = 2 – 4
⇒ x – y = -2 … (i)
ପୁନଶ୍ଚ 4x – y = y + 2 ⇒ 4x – y – y = 2 ⇒ 4x – 2y = 2
⇒ 2(2x – y) = 2 ⇒ 2x – y = 1 …… (ii)
ସମୀକରଣ (ii)ରୁ ସମୀକରଣ (i)କୁ ବିୟୋଗ କଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, 3 – y = – 2
⇒ – y = -2 – 3 = -5 ⇒ y = 5
∴ ସମବାହୁ ତ୍ରିଭୁଜର ଏକ ବାହୁର ଦୈର୍ଘ୍ୟ = y + 2 = 5 + 2 = 7 ସେ.ମି.

Question 3.
ABCD ଅ।ୟତକ୍ଷେତ୍ରର AB = 3x + y ସେ.ମି, BC = 3x + 2 ସେ.ମି, CD = 3y – 2x ସେ.ମି, ଓ DA = y + 3 ସେ.ମି. ହେଲେ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ନିରୂପଣ କର ।
ସମାଧାନ :
ଆୟତଚିତ୍ରର ବିପରୀତ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ପରସ୍ପର ସମାନ ।
ତେଣୁ ABCD ଆୟତଚିତ୍ରରେ AB = CD ଏବଂ BC = AD ହେବ ।
AB = CD ହେଲେ, 3x + y = 3y – 2x ⇒ 3x + 2x = 3y – y ⇒ 5x – 2y=0 (i)
ସେହିପରି BC = AD ହେଲେ, 3x + 2 = y + 3
⇒ 3x – y = 3 – 2 ⇒ 3x – y =1 …..(ii)

‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 2 – y = 1 ⇒ 6 – 1 = y ⇒ y = 5
∴ AB = 3x + y = 3 × 2 + 5 = 11 ସେ.ମି
BC = 3x + 2 = 3 × 2 + 2 = 8 ସେ.ମି
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = AB × BC = 11 × 8 = 88

Question 4.
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କଦ୍ଵୟର ଯୋଗଫଳର 4 ଗୁଣ । କିନ୍ତୁ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗ କଲେ ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳିଯାଏ । ତେବେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଟିର ଦଶକ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଦ୍ଵୟ ଯଥାକ୍ରମେ x ଏବଂ y l .:. ସଂଖ୍ୟାଟି = 10x + y
ପ୍ରଶ୍ନନୁସାରେ, ସଂଖ୍ୟାଟି ତାହାର ଅଙ୍କ ଦ୍ବୟର ଯୋଗଫଳର 4 ଗୁଣ ।
⇒ 10x + y = 4(x + y) ⇒ 10x + y = 4x + 4y ⇒ 10x – 4x + y – 4y = 0
⇒ 6x – 3y = 0 ⇒ 3(2x – y) = 0 … (i)
⇒2x – y = 0
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗକଲେ ଅଙ୍କ ଦୁଇଟିର ସ୍ଥାନ ବଦଳିଯାଏ ।
⇒ 10x + y + 18 = 10y + x ⇒ 10x – x + y – 10y = -18
⇒ 9(x- y) = -18 ⇒ x – y = \(\frac{-18}{9}=-2\)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 2x – y = 0
⇒ 4 – y = 0 ⇒ y=4
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 2 + 4 = 24

Question 5.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା ଓ ତାହାର ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଯେଉଁ ସଂଖ୍ୟା ମିଳିବ, ସେ ଦୁହିଁଙ୍କର ଯୋଗଫଳ 99 ଓ ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ହେଲେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଦଶକ ଓ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଦ୍ବୟ ଯଥାକ୍ରମେ x ଓ y
∴ ସଂଖ୍ୟାଟି = 10x + y
ଅଙ୍କଦ୍ୱୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ସଂଖ୍ୟାଟି ହେବ = 10y + x
ପ୍ରଶ୍ନନୁସାରେ, 10x + y + 10y + x = 99
⇒ 11x + 11y = 99 ⇒ x + y = \(\frac{99}{11}=9\) ………(i)
ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ।
ଅର୍ଥାତ୍ x – y = 3 ବା y – x = 3 … (ii)
ଯଦି x – y = 3 ହୁଏ,
∴ x = \(\frac{x+y+x-y}{2}=\frac{9+3}{2}=\frac{12}{2}=6\)
y = 9 – x = 9 – 6 = 3
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 6 + 3 = 63
ଯଦି ଆମେ ଦ୍ବିତୀୟ ସମୀକରଣକୁ y – x = 3 ନେବା ତେବେ ସଂଖ୍ୟାଟି 36 ହେବ ।
∴ ସଂଖ୍ୟାଟି 63 ବା 36 ।

Question 6.
ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି, ସେମାନଙ୍କ ବିୟୋଗଫଳର 4 ଗୁଣ ଏବଂ ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ 8 । ତେବେ ସଂଖ୍ୟା ଦୁଇଟି କେତେ ?
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y
ପ୍ରଶ୍ନନୁସାରେ, x + y = 4 (x – y)
⇒ 4x – 4y – x – y = 0 ⇒ 3x – 5y = 0 … (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ x + y = 8 …… (ii)

x ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ, x + y = 8
⇒ y = 8 – x = 8 – 5 = 3
∴ ସଂଖ୍ୟାଦ୍ୱୟ 5 ଓ 3 ।

Question 7.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଅଙ୍କମାନଙ୍କର ସମଷ୍ଟି 10; କିନ୍ତୁ ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି ମୂଳ ସଂଖ୍ୟାର ଦୁଇ ଗୁଣରୁ 1 ଊଣା ହୁଏ, ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଟିର ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଯଥାକ୍ରମେ x ଏବଂ y ।
∴ ସଂଖ୍ୟାଟି 10x + y ହେବ ।
ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳିଗଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି 10y + x ହେବ ।
ପ୍ରଶ୍ନନୁସାରେ, ଅଙ୍କଗୁଡ଼ିକର ଯୋଗଫଳ 10 । x + y = 10 … (i)
ପୁନଶ୍ଚ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି, ମୂଳ ସଂଖ୍ୟାର ଦୁଇଗୁଣରୁ 1 କମ୍ ।
ଅର୍ଥାତ୍ 10y + x = 2(10x + y) – 1 = 10y + x = 20x + 2y – 1
⇒ 8y – 19x = -1 ⇒ 19x – 8y = 1 … (ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 + y = 10 ⇒ y = 10 -3 = 7
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 3 + 7 = 37

Question 8.
ଦୁଇଟି ସଂଖ୍ୟା ମଧ୍ୟରୁ ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ 2 ହେବ ଏବଂ ଦ୍ଵିତୀୟଟିରେ 7 ଯୋଗ କଲେ ଯୋଗଫଳ ପ୍ରଥମଟିର 2 ଗୁଣ ହେବ। ସଂଖ୍ୟାଦ୍ଵୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପ୍ରଥମ ସଂଖ୍ୟା ଓ ଦ୍ୱିତୀୟ ସଂଖ୍ୟା ଯଥାକ୍ରମେ x ଏବଂ y ହେଉ ।
ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗକଲେ ବିୟୋଗଫଳ 2 ହେବ ।
ଅର୍ଥାତ୍ 3x – 2y = 2 …(i)
ସେହିପରି ଦ୍ବିତୀୟ ସଂଖ୍ୟାରେ 7 ଯୋଗକଲେ, ଯୋଗଫଳ ପ୍ରଥମ ସଂଖ୍ୟାର 2 ଗୁଣ ସଙ୍ଗେ ସମାନ ହେବ ।
ଅର୍ଥାତ୍ y + 7 = 2x = 2x – y = 7 …(ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 12 – 2y = 2 ⇒ 36 – 2y = 2
⇒ 2y = 34 ⇒ y = 17
∴ ପ୍ରଥମ ସଂଖ୍ୟାଟି 12 ଏବଂ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 17 ।

Question 9.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବ ଓ ହରରେ 2 ଯୋଗକଲେ ତାହା \(\frac{9}{11}\) ହୁଏ । ମାତ୍ର ଲବ ଓ ହରରେ 3 ଯୋଗକଲେ \(\frac{5}{6}\)
ହୁଏ । ତେବେ ଭଗ୍ନାଂଶଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{x+2}{y+2}=\frac{9}{11}\)
⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = – 4 (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+3}{y+3}=\frac{5}{6}\)
⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = – 3 … (ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 11 × 7 – 9y = -4
⇒77 – 9y = 4 ⇒ – 9y = – 4 – 77 = – 81 ⇒ y =\(\frac{81}{9}=9\)
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{7}{9}\)

Question 10.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବର 3 ଗୁଣ ଓ ହରରୁ 3 ବିୟୋଗ କଲେ ଭଗ୍ନାଂଶଟି \(\frac{18}{11}\) ହୁଏ । ମାତ୍ର ଲବରେ 8 ଯୋଗକଲେ ଓ ହରକୁ 2 ଗୁଣ କଲେ ତାହା \(\frac{2}{5}\) ହୁଏ । ତେବେ ଭଗ୍ନାଂଶ କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3x}{y-3}=\frac{18}{11}\)
⇒ 33x= 18y – 54 ⇒ 33x – 18y = – 54 ……..(i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+8}{2y}=\frac{2}{5}\)
⇒ 5x + 40 = 4y ⇒ 5x – 4y = -40 ……….(ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ବସାଇଲେ,
⇒ 33 × 12 – 18y = 54 ⇒ -18y = -54 – 396
⇒ 18y = 450 ⇒ y = \(\frac{450}{18}=25\)
∴ ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{12}{25}\)

Question 11.
5ଟି କଲମ ଓ ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି ୨ ଟଙ୍କା ଏବଂ 3ଟି କଲମ ଓ 2ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି 5 ଟଙ୍କା ହୁଏ । ତେବେ ଗୋଟିଏ କଲମ ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ କେତେ ?
ସମାଧାନ :
ମନେକର ଗୋଟିଏ କଲମର ଦାମ୍ x ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, 5x + 6y = 9 …… (i)
3x + 2y = 5 …….. (ii)

xର ମାନ ସମୀକରଣ (ii)ରେ ବସାଇଲେ, 3 × \(\frac{3}{2}\) + 2y = 5
⇒ 2y = 5 – \(\frac{9}{2}\) ⇒ 2y = \(\frac{10-9}{2}\)
⇒ y = \(\frac{1}{4}\) ଟଙ୍କା ।
∴ ଗୋଟିଏ କଲମର ଦାମ୍ \(\frac{3}{2}\) ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ \(\frac{1}{4}\) ଟଙ୍କା ।

Question 12.
ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 3 ଗୁଣ । 12 ବର୍ଷ ପରେ ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 2 ଗୁଣ ହେବ । ତେବେ ପିତା ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ କେତେ ?
ସମାଧାନ :
ମନେକର ପିତାଙ୍କର ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ y ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ, x = 3y … (i)
ପୁନଶ୍ଚ 12 ବର୍ଷ ପରେ ପିତାଙ୍କର ବୟସ (x + 12) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ (y + 12) ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ x + 12 = 2 (y + 12) ……..(ii)
ସମୀକରଣ (i)ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ,
3y +12 = 2 (y + 12) ⇒ 3y + 12 = 2y + 24
⇒3y – 2y = 24 – 12
⇒ y = 12 ବର୍ଷ |
ସମୀକରଣ (i)ରେ y = 12 ପ୍ରୟୋଗ କଲେ
x = 3y = 3 × 12 = 36 ବର୍ଷ |
∴ ପିତାଙ୍କର ବର୍ତ୍ତମାନ ବୟସ 36 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 12 ବର୍ଷ ।

Question 13.
ଏକ ଆୟତ କ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 5 ସେ.ମି. କମାଇ ପ୍ରସ୍ଥକୁ 3 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ଵାରା ଏହାର କ୍ଷେତ୍ରଫଳ 9 ବର୍ଗ ସେ.ମି. କମିଯାଏ । ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 3 ସେ.ମି. ଓ ପ୍ରସ୍ଥକୁ 2 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ୱାରା କ୍ଷେତ୍ରଫଳ 67 ବର୍ଗ ସେ.ମି. ବଢ଼ିଯାଏ । ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ x ସେ.ମି. ଓ ପ୍ରସ୍ଥ y ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) (y + 3) = xy – 9 ⇒ 3x – 5y – 6 = 0 … (i)
ପୁନଶ୍ଚ (x + 3) (y + 2) = xy + 67 = 2x +3y – 61 = 0 …….(ii)

⇒ \(\frac{x}{323}=\frac{y}{171}=\frac{1}{19}\)
⇒ x = \(\frac{323}{19}\) = 17 ଏବଂ y = \(\frac{171}{19}\) = 9
∴ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 17 ସେ.ମି. ଓ ପ୍ରସ୍ଥ 9 ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = 17 × 9 = 153 ବର୍ଗ ସେ.ମି. ।

Question 14.
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 5 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେହି କାର୍ଯ୍ୟକୁ 4 ଜଣ ପୁରୁଷ ଓ ୨ ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର 2 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ତେବେ ଜଣେ ସ୍ତ୍ରୀ ଲୋକ କିମ୍ବା ଜଣେ ପୁରୁଷ ସେହି କାର୍ଯ୍ୟକୁ କେତେ ଦିନରେ ଶେଷ କରିପାରବେ ?
ସମାଧାନ :
ମନେକର ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ ଗୋଟିଏ କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ ଶେଷ କରିପାରିବେ ।
∴ ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିବେ ।
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର = \(\frac{2}{x}\) + \(\frac{3}{y}\) ଅଂଶ କରିବେ ।
କିନ୍ତୁ 2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 5 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{5}\) ଅଂଶ ।
ପ୍ରଶାନୁସାରେ, \(\frac{2}{x}+\frac{3}{y}=\frac{1}{5}\) ……….(i)
ପୁନଶ୍ଚ, 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର କରିବେ = \(\frac{4}{x}\) + \(\frac{9}{y}\) ଅଂଶ ।
କିନ୍ତୁ 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 2 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{2}\) ଅଂଶ ।
ପୁନଶ୍ଚ, ପ୍ରଶ୍ନନୁସାରେ, \(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\)

⇒ \(– \frac{3}{y}=\frac{4-5}{10}=\frac{-1}{10}\) ⇒ \(\frac{3}{y}=\frac{1}{10}\)
⇒ y = 30
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{2}{x}+\frac{3}{30}=\frac{1}{5}\) ⇒ \(\frac{2}{x}+\frac{1}{10}=\frac{1}{5}\)
\(\frac{2}{x}+\frac{1}{5}-\frac{1}{10}\) ⇒ \(\frac{2}{x}=\frac{1}{10}\) ⇒ x = 20
∴ ଜଣେ ପୁରୁଷ କିମ୍ବା ଜଣେ ସ୍ତ୍ରୀ ସେହି କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ 20 ଦିନରେ କିମ୍ବା 30 ଦିନରେ କରିବେ ।

Question 15.
A ଓ B ଏକତ୍ର କାମ କରି ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 8 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେମାନେ ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରି 3 ଦିନ କାର୍ଯ୍ୟ କରିବା ପରେ A ଚାଲିଗଲା ଓ ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟକୁ B ଏକା ଆଉ 15 ଦିନରେ ଶେଷ କଲା । ପ୍ରତ୍ୟେକ ଏକାକୀ କାମ କଲେ କେତେ ଦିନରେ କାର୍ଯ୍ୟକୁ ଶେଷ କରିପାରିବେ ?
ସମାଧାନ :
ମନେକର A ଓ B କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ କରିପାରିବେ ।
A ଓ B 1 ଦିନରେ କାର୍ଯ୍ୟଟିର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
A ଓ B ମିଶି 1 ଦିନରେ କାର୍ଯ୍ୟଟିର \(\frac{1}{x}+\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
କିନ୍ତୁ ପ୍ରଶ୍ନ ଅଛି A ଓ B କାର୍ଯ୍ୟଟିକୁ 8 ଦିନରେ କରିପାରନ୍ତି ।
∴ 1 ଦିନରେ କରିବେ = \(\frac{1}{8}\) ଅଂଶ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{8}\) …….. (i)
A ଓ B ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର 3 ଦିନ ପରେ A ଚାଲିଗଲା ।
ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟଟିକୁ B ଆଉ 15 ଦିନରେ ଶେଷକଲା ।
ଅର୍ଥାତ୍ A, 3 ଦିନ ଓ B (15 + 3) = 18 ଦିନ କାର୍ଯ୍ୟ କଲାପରେ କାର୍ଯ୍ୟଟି ସମ୍ପୂର୍ଣ ହେଲା ।
∴ A, 3 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ = \(\frac{3}{x}\) ଅଂଶ । B, 18 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ \(\frac{18}{y}\) ଅଂଶ ।
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3}{x}+\frac{18}{y}=1\) ……….(ii)

⇒ \(– \frac{15}{y}=\frac{3-8}{8}\) ⇒ \(– \frac{15}{y}=\frac{-5}{8}\)
⇒ \(\frac{3}{y}=\frac{1}{8}\) ⇒ y = 24
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{1}{x}+\frac{1}{24}=\frac{1}{8}\) ⇒ \(\frac{1}{x}=\frac{1}{8}-\frac{1}{24}\)
\(\frac{1}{x}=\frac{3-1}{24}=\frac{2}{24}=\frac{1}{12}\) ⇒ x = 12
∴ A ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 12 ଦିନରେ ଓ B ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 24 ଦିନରେ ଶେଷ କରିପାରିବେ ।

Question 16.
A ଓ Bର ଆୟର ଅନୁପାତ 8 : 7 ଓ ବ୍ୟୟର ଅନୁପାତ 19 : 16 । ଯଦି ଉଭୟେ 1250 ଟଙ୍କା ସଞ୍ଚୟ କରିପାରନ୍ତି, ତେବେ ସେମାନଙ୍କର ଆୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
A ଓ B ର ଆୟର ଅନୁପାତ 8 : 7 ।
ମନେକର A ଓ B ର ଆୟ ଯଥାକ୍ରମେ 8x ଟଙ୍କା ଏବଂ 7x ଟଙ୍କା ।
ସେହିପରି A ଓ B ର ବ୍ୟୟର ଅନୁପାତ 19 : 16 |
ତେଣୁ ମନେକର A ଓ B ର ବ୍ୟୟ ଯଥାକ୍ରମେ 19y ଟଙ୍କା ଏବଂ 16y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ A ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 8x – 19y = 1250 ……..(i)
ସେହିପରି B ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 7x – 16y = 1250 ……….(ii)

∴ A = 8x = 8 × 750 = 6000 ଟଙ୍କା । ଏବଂ Bର ଆୟ = 7x = 7 × 750 = 5250 ଟଙ୍କା ।

Question 17.
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ପୁତ୍ରର ବୟସର ତିନିଗୁଣ ହେବ ଓ 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ପୁତ୍ର ବୟସର ସାତଗୁଣ ଥିଲା । ତେବେ ସେମାନଙ୍କର ବର୍ତ୍ତମାନ ବୟସ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପିତାର ବର୍ତ୍ତମାନ ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ y ବର୍ଷ ।
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ହେବ = (x + 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ହେବ = (y + 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, x + 5 = 3(y + 5)
⇒ x + 5 = 3y + 15 ⇒ x – 3y = 10 …(i)
ପୁନଶ୍ଚ, 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ଥିଲା = (x – 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ଥିଲା = (y – 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) = 7(y – 5)
⇒ x -5 = 7y – 35 ⇒ x – 7y = -30 …(ii)
ସମୀକରଣ (i)ରୁ ସମୀକରଣ (ii)କୁ ବିୟୋଗ କଲେ, (x – 3y) – (x – 7y) = 10 – (-30)
⇒ x – 3y – x + 7y = 40 ⇒ 4y = 40 ⇒ y = 10
ସମୀକରଣ (i) ରେ y = 10 ସ୍ଥାପନ କଲେ, x – 3 × 10 = 10 ⇒ x = 40
∴ ପିତାର ବର୍ତ୍ତମାନ ବୟସ 40 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 10 ବର୍ଷ ।

Question 18.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧ୍ଵ ଓ ପ୍ରସ୍ଥ 2ମି. କମ୍ ହେଲେ, କ୍ଷେତ୍ରଫଳ 28 ବ.ମି. କମିଯାଏ; ମାତ୍ର ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ବକ ହେଲେ କ୍ଷେତ୍ରଫଳ 33 ବ. ମି. ବଢ଼ିଯାଏ । ମୂଳ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ x ମି. ଓ Y ମି. ।
∴ କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = xy ବର୍ଗ ମି. ।
ପ୍ରଶ୍ନନୁସାରେ, ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧୂକ ଏବଂ ପ୍ରସ୍ଥ 2 ମି. କମ୍ ହେଲେ କ୍ଷେତ୍ରଫଳ 28 ବର୍ଗ ମି. କମିଯାଏ ।
ତେଣୁ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x + 2) ମି. ଏବଂ (y – 2) ମି. ହେବ ।
(x + 2) (y – 2) = xy – 28 ⇒ xy + 2y – 2x – 4 = xy – 28
⇒ 2y – 2x = -28 + 4 ⇒ y – x = \(\frac{-24}{2}\) ⇒ y – x = -12
⇒ x – y = 12 … (i)
ସେହିପରି ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ଯକ ହେଲେ, କ୍ଷେତ୍ରଫଳ 33 ବର୍ଗ ମି. ବୃଦ୍ଧିପାଏ ।
∴ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x – 1) ମି, ଏବଂ (y + 2) ମି. ହେବ ।
∴ (x – 1) (y + 2) = xy + 33 ⇒ xy + 2x – y – 2 = xy + 33
⇒ 2x – y = 33 +2 ⇒ 2x – y = 35 ……(ii)

‘x’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ 23 – y = 12 = y = 11
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ମି. = (23 × 11) ବର୍ଗ ମି. = 253 ବର୍ଗ ମି. ।

Question 19.
50କୁ ଏପରି ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି ରୂପେ ପ୍ରକାଶ କର ଯେପରିକି ସଂଖ୍ୟା ଦ୍ଵୟର ବ୍ୟକ୍ରମର ସମଷ୍ଟି \(\frac{1}{12}\) ହେବ ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟା ଦୁଇଟି x ଓ y ।
ସଂଖ୍ୟା ଦୁଇଟିର ସମଷ୍ଟି 50 । ⇒ x + y = 50 ….. (i)
x ଓ y ର ବ୍ୟକ୍ରମ ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଓ \(\frac{1}{y}\) ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……..(ii)
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ⇒ \(\frac{x+y}{xy}=\frac{1}{12}\) ⇒ \(\frac{50}{xy}=\frac{1}{12}\) (∵ x + y = 50)
⇒ xy = 600 ……..(iii)
ଆମେ ଜାଣିଛେ, x – y = \(\sqrt{(x+y)^2-4xy}\) = \(\sqrt{(50)^2-4×600}\) = \(\sqrt{100}\)
⇒ x – y = 10 ……(iv)
ସମୀକରଣ (i) ଓ (iv)କୁ ଯୋଗକଲେ, 2x = 60 ⇒ x = 30
ସମୀକରଣ (i)ରେ x = 30 ସ୍ଥାପନ କଲେ, 30 + y = 50 ⇒ y = 20
∴ ସଂଖ୍ୟାଦ୍ବୟ 30 ଓ 20 ।

Question 20.
ଗୋଟିଏ ଭଗ୍ନ ସଂଖ୍ୟାର ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକ-ତୃତୀୟାଂଶ ନେଲେ, ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ଓ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନ ସଂଖ୍ୟାଟିକୁ ଲଘିଷ୍ଠ ଆକାରରେ ଲେଖୁଲେ ତାହା \(\frac{1}{4}\) ହୁଏ । ଭଗ୍ନ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\), ଯାହାର ଲବ x ଏବଂ ହର y ।
ପ୍ରଶ୍ନନୁସାରେ, ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକତୃତୀୟାଂଶ ନେଲେ ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ।
\(\frac{1}{3}\)(x+y) = y – 4 ⇒ x + y = 3y – 12 ⇒ x – 2y= – 12 …….. (i)
ପୁନଶ୍ଚ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନାଂଶଟିକୁ ଲଘିଷ୍ଠ ଆକାରକୁ ଆଣିଲେ ତାହା \(\frac{1}{4}\) ହୁଏ ।
\(\frac{x}{y+1}=\frac{1}{4}\) ⇒ 4x = y + 1 ⇒ 4x – y = 1 ……….(ii)

‘y’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, x – 2 × 7 = -12
⇒ x – 14 = -12 ⇒ x = 2
∴ ଭଗ୍ନାଂଶଟି \(\frac{x}{y}=\frac{2}{7}\)