NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Thank you for reading this post, don't forget to subscribe!
Table of Contents
Exercise Set 2.1
Question 1:
Find the degrees of the following polynomials:
निम्नलिखित बहुपदों की घात ज्ञात कीजिए।
(i) 2x2 – 5x + 3
(ii) y3 + 2y – 1
(iii) – 9
(iv) 4z – 3
Solution
Degree = Highest power
(i) 2x2 – 5x + 3
\(\quad\quad\) Degree = 2
(ii) y3 + 2y – 1
\(\quad\quad\) Degree = 3
(iii) – 9
\(\quad\quad\) Degree = 0
(iv) 4z – 3
\(\quad\quad\) Degree = 1
Question 2:
Write polynomials of degrees 1, 2 and 3.
घात 1, 2 और 3 के बहुपद लिखिए।
Solution
(1) A polynomial of degree 1 = 2x + 5
(2) A polynomial of degree 2 = x2 + 2x + 5
(3) A polynomial of degree 3 = x3 – x2 + 2x + 5
Question 3:
What are the coefficients of x2 and x3 in the polynomial x4 – 3x3 + 6x2 – 2x + 7?
बहुपद x4 – 3x3 + 6x2 – 2x + 7 में x2 और x3 के गुणांक क्या हैं?
Solution
x4 – 3x3 + 6x2 – 2x + 7
coefficients of x2 = 6
coefficients of x3 = -3
Question 4:
What is the coefficient of z in the polynomial 4z3 + 5z2 – 11?
बहुपद 4z3 + 5z2 – 11 में z का गुणांक क्या है?
Solution
4z3 + 5z2 – 11
coefficient of z = 0
Question 5:
What is the constant term of the polynomial 9x3 + 5x2 – 8x –10?
बहुपद 9x3 + 5x2 – 8x –10 का अचल पद क्या है?
Solution
9x3 + 5x2 – 8x –10
constant term = – 10
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Exercise Set 2.2
Question 1:
Find the value of the linear polynomial 5x – 3 if:
रेखीय बहुपद 5x – 3 का मान ज्ञात कीजिए, यदि:
(i) x = 0 (ii) x = –1 (iii) x = 2
Solution
Let P(x) = 5x – 3
(i) x = 0,
\(\quad\) P(0) = 5(0) – 3
\(\quad\quad\quad\) = 0 – 3 = – 3
(ii) x = -1,
\(\quad\) P(-1) = 5(-1) – 3
\(\quad\quad\quad\) = -5 – 3 = – 8
(iii) x = 2,
\(\quad\) P(2) = 5(2) – 3
\(\quad\quad\quad\) = 10 – 3 = 7
Question 2:
Find the value of the quadratic polynomial 7s2 – 4s + 6 if:
द्विघात बहुपद 7s2 – 4s + 6 का मान ज्ञात कीजिए, यदि:
(i) s = 0 (ii) s = –3 (iii) s = 4
Soution
Let P (s) = 7s2 – 4s + 6
(i) s = 0,
\(\quad\) P (0) = 7(0)2 – 4(0) + 6
\(\quad\quad\quad\) = 0 – 0 + 6 = 6
(ii) s = -3,
\(\quad\) P (-3) = 7(-3)2 – 4(-3) + 6
\(\quad\quad\quad\) = 63 + 12 + 6 = 81
(ii) s = 4
\(\quad\) P (4) = 7(4)2 – 4(4) + 6
\(\quad\quad\quad\) = 112 – 16 + 6 = 102
Question 3:
The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.
सलील की माँ की वर्तमान आयु सलील की वर्तमान आयु की तीन गुना है। 5 वर्ष बाद, उनकी आयुओं का योग 70 वर्ष होगा। उनकी वर्तमान आयु ज्ञात कीजिए।
Solution
Let Salil’s present age be x years
and Salil’s mother’s present age be 3x years
After 5 years, Salil’s age = (x + 5) years
After 5 years, Mohter’s age = (3x + 5) years
ATQ,
Sum = 70 years
\(\;\) x + 5 + 3x + 5 = 70
\(\Rightarrow\) 4x + 10 = 70
\(\Rightarrow\) 4x = 70 – 10 = 60
\(\Rightarrow\) x = \(\frac{60}{4}\) = 15
Hence, Salil’s present age = 15 years
and Salil’s mother’s present age = 3(15) = 45 years
Question 4:
The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
दो धनात्मक पूर्णांकों के बीच का अंतर 63 है। उन दोनों पूर्णांकों का अनुपात 2:5 है। उन दोनों पूर्णांकों को ज्ञात कीजिए।
Solution
Let two integers be 2x and 5x.
ATQ,
Difference = 63
\(\;\) 5x – 2x = 63
\(\Rightarrow\) 3x = 63
\(\Rightarrow\) x = \(\frac{63}{3}\) = 21
Therefore, two integers are 2\(\times\)21 and 5\(\times\)21 = 42 and 105.
Question 5:
Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹ 88, how many coins does she have of each type?
रूबी के पास पाँच रुपये के सिक्कों की तुलना में दो रुपये के सिक्के तीन गुना अधिक हैं। यदि उसके पास कुल ₹88 हैं, तो उसके पास प्रत्येक प्रकार के कितने सिक्के हैं?
Solution
Let number of five rupee-coins be x.
and number of two-rupee coins = 3x
ATQ,
Total Value = ₹ 88
\(\Rightarrow\) 5x + 2(3x) = 88
\(\Rightarrow\) 5x + 6x = 88
\(\Rightarrow\) 11x = 88
\(\Rightarrow\) x = \(\frac{88}{11}\) = 8
Therefore, number of five rupee-coins = 8
and number of two-rupee coins = 3(8) = 24
Question 6:
A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
एक किसान 300 फीट लंबी बाड़ को दो अलग-अलग आकार के टुकड़ों में काटता है। बड़ा टुकड़ा छोटे टुकड़े से चार गुना लंबा है। दोनों टुकड़ों की लंबाई क्या है?
Solution
Let shorter piece be x feet
and longer piece = 4x feet
ATQ,
Total Length = 300 feet
\(\Rightarrow\) x + 4x = 300
\(\Rightarrow\) 5x = 300
\(\Rightarrow\) x = \(\frac{300}{5}\) = 60
Therefore, shorter piece = 60 feet
and longer piece = 4(60) = 240 feet
Question 7:
If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
यदि किसी आयत की लंबाई उसकी चौड़ाई के दुगुने से 3 अधिक है और उसका परिमाप 24 सेमी है, तो आयत के आयाम क्या हैं?
Solution
Let width be x cm
and length = (2x + 3) cm
ATQ,
Perimeter = 24 cm
\(\Rightarrow\) 2 (length + width) = 24
\(\Rightarrow\) 2 (2x + 3 + x) = 24
\(\Rightarrow\) 3x + 3 = \(\frac{24}{2}\)
\(\Rightarrow\) 3x + 3 = 12
\(\Rightarrow\) 3x = 12 – 3 = 9
\(\Rightarrow\) x = \(\frac{9}{3}\) = 3
Therefore, width = 3 cm
and length = 2(3) + 3 = 6 +3 = 9 cm
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Exercise Set 2.3
Solve the following:
Question 1:
A student has 500 in her savings bank account. She gets 150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month.
एक छात्रा के बचत बैंक खाते में ₹500 हैं। उसे हर महीने ₹150 जेब खर्च के रूप में मिलते हैं। दूसरे महीने से प्रत्येक महीने के अंत में उसके पास कितनी राशि होगी? nवें महीने में उसके पास कितनी राशि होगी, इसे दर्शाने के लिए एक रेखीय व्यंजक ज्ञात कीजिए।
Solution
Total amount = ₹ 500
Monthly pocket money = ₹ 150
At the end of :
2nd month = 500 + 2 \(\times\) 150 = 500 + 300 = ₹ 800
3rd month = 500 + 3\(\times\) 150 = 500 + 450 = ₹ 950
4th month = 500 + 4 \(\times\) 150 = 500 + 600 = ₹ 1,100, and so on…
Now,
Amount = nth month = 500 + 150n
\(\quad\quad\quad\quad\quad\quad\) = 150n + 500.
Question 2:
A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour.
एक रैली 120 सदस्यों के साथ शुरू होती है। प्रत्येक घंटे, 9 सदस्य समूह छोड़ देते हैं। 1, 2, 3, … घंटों के बाद कितने सदस्य शेष रहेंगे? nवें घंटे के अंत में सदस्यों की संख्या को दर्शाने के लिए एक रेखीय व्यंजक ज्ञात कीजिए।
Solution
Total members = 120
Members drop out each hour = 9
After 1 hour = 120 – 1(9) = 120 – 9 = 111
After 2 hour = 120 – 2(9) = 120 – 18 = 102
After 3 hour = 120 – 3(9) = 120 – 27 = 93
Now,
At the end of the nth hour = 120 – n (9) = 120 – 9n.
Question 3:
Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.
मान लीजिए किसी आयत की लंबाई 13 सेमी है। यदि उसकी चौड़ाई (i) 12 सेमी, (ii) 10 सेमी, (iii) 8 सेमी हो, तो उसका क्षेत्रफल ज्ञात कीजिए। आयत के क्षेत्रफल को दर्शाने वाला रैखिक पैटर्न ज्ञात कीजिए।
Solution
Given, Length of a rectangle = 13 cm
(i) breadh = 12 cm
\(\;\) Area of a rectangle = length \(\times\) breadth = 13 \(\times\) 12 = 156 cm2.
(ii) breadh = 10 cm
\(\;\) Area of a rectangle = length \(\times\) breadth = 13 \(\times\) 10 = 130 cm2.
(iii) breadh = 8 cm
\(\;\) Area of a rectangle = length \(\times\) breadth = 13 \(\times\) 8 = 104 cm2.
Now, let breadth be x cm
\(\;\) Area of a rectangle = length \(\times\) breadth = 13 \(\times\) x = 13x cm2. = Linear Pattern
Question 4:
Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.
मान लीजिए किसी आयताकार डिब्बे (बॉक्स) की लंबाई 7 सेमी और चौड़ाई 11 सेमी है। यदि उसकी ऊँचाई (i) 5 सेमी, (ii) 9 सेमी, (iii) 13 सेमी हो, तो उसका आयतन ज्ञात कीजिए। आयताकार डिब्बे के आयतन को दर्शाने वाला रैखिक पैटर्न ज्ञात कीजिए।
Solution
Given length = 7 cm and breadth = 11 cm
(i) height = 5 cm
\(\;\) Volume of a rectangular box = length \(\times\) breadth \(\times\) height
\(\quad\quad\quad\quad\quad\quad\quad\quad\) = 7 \(\times\) 11 \(\times\) 5 = 385 cm3.
(ii) height = 9 cm
\(\;\) Volume of a rectangular box = length \(\times\) breadth \(\times\) height
\(\quad\quad\quad\quad\quad\quad\quad\quad\) = 7 \(\times\) 11 \(\times\) 9 = 693 cm3.
(iii) height = 13 cm
\(\;\) Volume of a rectangular box = length \(\times\) breadth \(\times\) height
\(\quad\quad\quad\quad\quad\quad\quad\quad\) = 7 \(\times\) 11 \(\times\) 13 = 1001 cm3.
Now,
let height = h cm
\(\;\) Volume of a rectangular box = length \(\times\) breadth \(\times\) height
\(\quad\quad\quad\quad\quad\quad\quad\quad\) = 7 \(\times\) 11 \(\times\) h = 77h cm3.
Thus, linear pattern = 77 h.
Question 5:
Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
सरिता 500 पृष्ठों की एक पुस्तक पढ़ रही है। वह प्रतिदिन 20 पृष्ठ पढ़ती है। 15 दिनों के बाद कितने पृष्ठ शेष रह जाएंगे? इसे एक रैखिक पैटर्न के रूप में व्यक्त कीजिए।
Solution
Total pages = 500
Pages read per day = 20
Pages read after 15 days = 20 \(\times\) 15 = 300
Pages left after 15 days = 500 – 300 = 200
Now,
Let pages left after n days = 500 – 20 \(\times\) n = 500 – 20n
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Exercise Set 2.4
Question 1:
Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
मान लीजिए एक पौधे की ऊँचाई 1.75 फीट है और वह हर महीने 0.5 फीट बढ़ता है।
(i) 7 महीनों के बाद उसकी ऊँचाई ज्ञात कीजिए।
(ii) t = 0 से 10 महीनों तक के लिए मानों की एक सारणी बनाइए और दिखाइए कि ऊँचाई h हर महीने कैसे बढ़ती है।
(iii) h और t के बीच संबंध दर्शाने वाला एक व्यंजक ज्ञात कीजिए और समझाइए कि यह रैखिक वृद्धि क्यों दर्शाता है।
Solution
Given: Initial height = 1.75 feet and growth per month = 0.5 feet
(i) Height after 7 months = 1.75 + 7 \(\times\) 0.5 = 1.75 + 3.5 = 5.25 feet
(ii) Table of vlaues of t varying from 0 to 10 months
| t (month) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| h (feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
(iii) Expression :
Let height after t months = h
h = 1.75 + 0.5t
This represents linear growth becasue height increase by a constant amount (0.5 feet) every month.
Question 2:
A mobile phone is bought for ₹ 10,000. Its value decreases by ₹ 800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
एक मोबाइल फोन ₹10,000 में खरीदा गया। इसका मूल्य हर वर्ष ₹800 घटता है।
(i) 3 वर्षों के बाद फोन का मूल्य ज्ञात कीजिए।
(ii) t = 0 से 8 वर्षों तक के लिए मानों की एक सारणी बनाइए और दिखाइए कि फोन का मूल्य v समय के साथ कैसे घटता है।
(iii) v और t के बीच संबंध दर्शाने वाला एक व्यंजक ज्ञात कीजिए और समझाइए कि यह रैखिक ह्रास क्यों दर्शाता है।
Solution
Given: Initial value of a mobile phone = ₹ 10,000
and decrease per year = ₹ 800
(i) The value of the phone after 3 years = 10,000 – 3 \(\times\) 800
\(\quad\quad\quad\quad\quad\quad\quad\quad\quad\) = 10,000 – 2,400 = ₹ 7,600
(ii) Table of values for t varying from 0 to 8 years
| t (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Value of the phone | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) Expression:
Let value after t years = v
v = 10000 – 800t
This represents linear decay because the value decreases by a constant amount (₹ 800) every year.
Question 3:
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
एक गाँव की प्रारंभिक जनसंख्या 750 है। हर वर्ष पास के शहर से 50 लोग गाँव में आकर बसते हैं।
(i) 6 वर्षों के बाद गाँव की जनसंख्या ज्ञात कीजिए।
(ii) t = 0 से 10 वर्षों तक के लिए मानों की एक सारणी बनाइए और दिखाइए कि जनसंख्या P हर वर्ष कैसे बढ़ती है।
(iii) P और t के बीच संबंध दर्शाने वाला एक व्यंजक ज्ञात कीजिए और समझाइए कि यह रैखिक वृद्धि क्यों दर्शाता है।
Solution
Given: Initial population of a village = 750 and move to the village = 50
(i) Population after 6 years = 750 + 50 \(\times\) 6 = 750 + 300 = 1050
(ii) Table of values for t varying from 0 to 10 years
| t (month) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| h (feet) | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) Expression:
Let population after t years = P
P = 750 + 50t
This represents linear growth because the population increases by a constant number (50 people) every year.
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Question 4:
A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
एक दूरसंचार कंपनी एक रिचार्ज योजना के लिए ₹600 लेती है। रिचार्ज के बाद यह प्रीपेड बैलेंस प्रतिदिन ₹15 कम होता है।
(i) x दिनों तक योजना उपयोग करने के बाद शेष बैलेंस b(x) को दर्शाने वाला एक समीकरण लिखिए और समझाइए कि यह रैखिक ह्रास क्यों दर्शाता है।
(ii) कितने दिनों बाद बैलेंस समाप्त हो जाएगा?
(iii) x = 1 से 10 दिनों तक के लिए मानों की एक सारणी बनाइए और दिखाइए कि बैलेंस b(x) समय के साथ कैसे घटता है।
Solution
Let remaining balance after x days = b(x)
b(x) = 600 – 15x
This represent linear decay because the balances decreases by a constnat amount (₹ 15) every day.
(ii) When balalnce run out
\(\;\) b(x) = 0
\(\;\) 600 – 15x = 0
\(\;\) 15x = 600
\(\;\) x = \(\frac{600}{15}\) = 40
So, the balance will run out after 40 days.
(iii) Tabe of Values for x varyingg from 1 to 10 days
| x (days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| b(x) (₹) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Exercise Set 2.5
Question 1:
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹ 400. When she accessed 14 modules, her bill was ₹ 500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
एक लर्निंग प्लेटफ़ॉर्म एक निश्चित मासिक शुल्क और प्रत्येक डिजिटल लर्निंग मॉड्यूल के लिए अतिरिक्त शुल्क लेता है। एक छात्रा देखती है कि जब उसने 10 मॉड्यूल एक्सेस किए, तो उसका बिल ₹400 था। जब उसने 14 मॉड्यूल एक्सेस किए, तो उसका बिल ₹500 था। यदि मासिक बिल y, एक्सेस किए गए मॉड्यूल की संख्या x पर y=ax+b संबंध के अनुसार निर्भर करता है, तो a और b के मान ज्ञात कीजिए।
Solution
Number of modules accessed = x and bill amount = ₹ y
Given: when x = 10, y = ₹ 400 and x = 14, y = ₹ 500
Using, y = a x+ b
For x = 10, y = 400; 400 = 10a + b … (1)
For x = 14, y = 500; 500 = 14a + b … (2)
Subtracting (1) from (2);
4a = 100 \(\Rightarrow\) a = \(\frac{100}{4}\) = 25
Putting the value of a in (1); 400 = 10 (25) + b
\(\Rightarrow\) b = 400 – 250 = 150
So, a = 25 and b = 150
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Question 2:
A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹ 800. When she used it for 15 hours, her bill was ₹ 1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
एक जिम एक निश्चित मासिक शुल्क और बैडमिंटन कोर्ट के उपयोग के प्रति घंटे के हिसाब से अतिरिक्त शुल्क लेता है। जिम का उपयोग करने वाली एक छात्रा ने देखा कि जब उसने बैडमिंटन कोर्ट 10 घंटे इस्तेमाल किया, तो उसका बिल ₹800 था। जब उसने 15 घंटे उपयोग किया, तो उसका बिल ₹ 1100 था। यदि मासिक बिल y, बैडमिंटन कोर्ट के उपयोग के घंटों x पर y=ax+b संबंध के अनुसार निर्भर करता है, तो a और b के मान ज्ञात कीजिए।
Solution
Number of hours = x and monthy bill = ₹ y
Given: when x = 10, y = ₹ 800 and x = 15, y = ₹ 1100
Using, y = a x+ b
For x = 10, y = 800; 800 = 10a + b … (1)
For x = 15, y = 1100; 1100 = 15a + b … (2)
Subtracting (1) from (2);
5a = 300 \(\Rightarrow\) a = \(\frac{100}{4}\) = 60
Putting the value of a in (1); 800 = 10 (60) + b
\(\Rightarrow\) b = 800 – 600 = 200
So, a = 60 and b = 200
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Question 3:
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
डिग्री सेल्सियस (°C) और डिग्री फ़ारेनहाइट (°F) में मापे गए तापमान के बीच संबंध पर विचार कीजिए, जो °C=a°F+b द्वारा दिया गया है। a और b के मान ज्ञात कीजिए, यह जानते हुए कि बर्फ 0°C (32°F) पर पिघलती है और पानी 100°C (212°F) पर उबलता है।
(संकेत: जब °C=0, तब °F=32 और जब °C=100, तब °F=212। इस जानकारी का उपयोग करके a और b ज्ञात करें, और इस प्रकार °C और °F के बीच रैखिक संबंध प्राप्त करें।)
Solution
Given: when °C = 0, °F = 32 and °C = 100, °F = ₹ 212
Using, °C = a °F + b
For °C = 0, °F = 32; 0 = 32a + b … (1)
For °C = 100, °F = 212; 100 = 212a + b … (2)
Subtracting (1) from (2);
180a = 100 \(\Rightarrow\) a = \(\frac{100}{180}\) = \(\frac{5}{9}\)
Putting the value of a in (1); 0 = 32 (\(\frac{5}{9}\)) + b
\(\Rightarrow\) b = -\(\frac{160}{9}\)
So, a = \(\frac{5}{9}\) and b = -\(\frac{160}{9}\)
Here, the linear relationship is °C = \(\frac{5}{9}\) °F – \(\frac{160}{9}\)
\(\Rightarrow\) °C = \(\frac{5}{9}\) (°C – 32)
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Exercise Set 2.6
Question 1:
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
(i) y = 4x, y = 2x, y = x
(ii) y = – 6x, y = – 3x, y = – x
(iii) y = 5x, y = –5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = –2x – 3, y = –2x, y = 2x + 3
Solution
(i) y = 4x, y = 2x, y = x
(ii) y = – 6x, y = – 3x, y = – x
(iii) y = 5x, y = –5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = –2x – 3, y = –2x, y = 2x + 3
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
End of Chapter Exercise
Question 1:
Write a polynomial of degree 3 in the variable x, in which the coefficient of the x2 term is –7.
Solution
A polynomial of degree 3 in the variable x
ax3 + bx2 + cx + d
Given : the coefficient of x2 = – 7 so b = – 7
Polynomial = x3 – 7x2 + 2x + 1
Question 2:
Find the values of the following polynomials at the indicated values of the variables.
(i) 5x2 – 3x + 7 if x = 1
(ii) 4t3 – t2 + 6 if t = a
Solution
(i) 5x2 – 3x + 7 if x = 1
\(\quad\) Putting x = 1
\(\quad\) = 5(1)2 – 3(1) + 7
\(\quad\) = 5 – 3 + 7
\(\quad\)= 9
(ii) 4t3 – t2 + 6 if t = a
\(\quad\) Putting t = a
\(\quad\) = 4(a)3 – (a)2 + 6
\(\quad\) = 4a3 – a2 + 6
Question 3:
If we multiply a number by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, we get \(\frac{-7}{12}\). Find the number.
Solution
Let a number be x
ATQ,
\(\quad\) \(\frac{5}{2}\)x + \(\frac{2}{3}\) = \(\frac{-7}{12}\)
\(\Rightarrow \; \; \) \(\frac{5}{2}\)x = \(\frac{-7}{12}\) – \(\frac{2}{3}\)
\(\Rightarrow \; \; \) \(\frac{5}{2}\)x = \(\frac{-7 – 8}{12}\) = \(\frac{-15}{12}\)
\(\Rightarrow \; \; \) x = \(\frac{-15}{12} \times \frac{2}{5}\)
\(\Rightarrow \; \; \) x = \(\frac{-1}{2} \)
So, the number is \(\frac{-1}{2}\).
Question 4:
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution
Let the smaller number be x
and the larger number = 5x
ATQ,
\(\quad\) 2(x + 21 )= 5x + 21
\(\Rightarrow\) 2x + 42 = 5x + 21
\(\Rightarrow\) 5x – 2x = 42 – 21
\(\Rightarrow\) 3x = 21
\(\Rightarrow\) x = 7
So, the smaller number = 7
and the larger number = 5 (7) = 35
Question 5:
If you have ₹ 800 and you save ₹ 250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.
Solution
Initial amount = ₹ 800
saving every month = ₹ 250
(i) Amount after 6 months
\(\quad\) = 800 + 250 (6)
\(\quad\) = 800 + 1500
\(\quad\) = ₹ 2300
(ii) Amount after 2 years, i.e., 24 months
\(\quad\) = 800 + 250 (24)
\(\quad\) = 800 + 6000
\(\quad\) = ₹6,800
Linear patern
A = 800 + 250n, where n is the number of months.
Question *6.
The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Solution
Let ten’s digit be x
and unit’s digit be y.
Number = 10 \(\times\) ten’s digit + unit’s diigt
\(\quad\quad\quad\)= 10x + y
The interchanged number = 10y + x
ATQ,
\(\quad\) 10x + y + 10y + x = 143
\(\Rightarrow\) 11x + 11y = 143
\(\Rightarrow\) 11 (x + y) = 143
\(\Rightarrow\) x + y = \(\frac{143}{11}\)
\(\Rightarrow\) x + y = 13 … (i)
and, differ = 3
\(\Rightarrow\) x – y = 3 … (ii)
Adding (i) and (ii), 2x = 16
\(\Rightarrow\) x = 8
Putting the value of x in (i), 8 + y = 13 \(\Rightarrow\) y = 5
Hence the original numbers are 10(8) + 5 = 80 + 5 = 85
and interchanged number = 58
Question *7:
Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Question *8:
If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = \(\frac{9}{5}\)
(x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Question *9:
The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Question *10.
The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Question *11:
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.
Find the polynomials p(x) and q(x).
Question *12:
Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Question *13:
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, –1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
Solution
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
Question *14:
What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution
Given: f(x) = ax + a, a > 0
\(\Rightarrow\) f(x) = a (x + 1)
Since a > 0, al the lines have positive slope.
y-intercept: Putting x = 0;
f(0) = a (0 + 1) = a
So, the y-intercept is (0, a)
x-intercept: f(x) = 0;
\(\Rightarrow\) a (x + 1) = 0
Since, a > 0, a is not zero.
So, x + 1 = 0
or x = – 1
Thus, every line cuts the x-axis at the same point (-1, 0).
Therefore, all linear functions of the form f(x) = ax + a, a>0 have the following in common:
(i) all have positve slope.
(ii) all cut the y-axis above the origin.
(iii) all pass through the fixed point (-1 0).
NCERT Solutions for Class 9 Ganita Manjari Chapter 2 Introduction to Linear Polynomials
