Exercise 1.5 NCERT Solutions: Number System

# NCERT Solutions for Class 9 Maths - Exercise 1.5 Number System

Q.1. Classify the following numbers as rational or irrational:
(i) 2 –√5
Solution:
We know that, √5 = 2.2360679…
Here, 2.2360679…is non-terminating and non-recurring.
Now, substituting the value of √5 in 2 –√5, we get,
2-√5 = 2-2.2360679… = -0.2360679
Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23
Solution:
(3 +23) –√23 = 3+23–√23
= 3
= 3/1
Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7
Solution:
2√7/7√7 = ( 2/7)× (√7/√7)
We know that (√7/√7) = 1
Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7
Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2
Solution:
Multiplying and dividing numerator and denominator by √2 we get,
(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)
We know that, √2 = 1.4142…
Then, √2/2 = 1.4142/2 = 0.7071..
Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2
Solution:
We know that, the value of = 3.1415
Hence, 2 = 2×3.1415.. = 6.2830…
Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

Q.2. Simplify each of the following expressions:
(i) (3+√3)(2+√2)
Solution:
(3+√3)(2+√2 )
Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )
Solution:
(3+√3)(3-√3 ) = 32-(√3)2 = 9-3
= 6

(iii) (√5+√2)2
Solution:
(√5+√2)= √52+(2×√5×√2)+ √22
= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)
Solution:
(√5-√2)(√5+√2) = (√52-√22) = 5-2 = 3

Q.3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution:
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

Q.4. Represent (√9.3) on the number line.
Solution:
Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC = 1 unit.
Step 2: Now, AC = 10.3 units. Let the centre of AC be O.
Step 3: Draw a semi-circle of radius OC with centre O.
Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.
Step 5: OBD, obtained, is a right angled triangle.
Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1
OB = OC – BC
⟹ (10.3/2)-1 = 8.3/2
Using Pythagoras theorem,
We get,
OD= BD+ OB2
⟹ (10.3/2)2 = BD2+(8.3/2)2
⟹ BD2 = (10.3/2)2-(8.3/2)2
⟹ (BD)= (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)
⟹ BD2 = 9.3
⟹ BD =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

Q.5. Rationalize the denominators of the following:
(i) 1/√7
Solution:
Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)
Solution:
Multiply and divide 1/(√7-√6) by (√7+√6)
[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)
= (√7+√6)/√72-√6[denominator is obtained by the property, (a+b)(a-b) = a2-b2]
= (√7+√6)/(7-6)
= (√7+√6)/1
= √7+√6

(iii) 1/(√5+√2)
Solution:
Multiply and divide 1/(√5+√2) by (√5-√2)
[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)
= (√5-√2)/(√52-√22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2]
= (√5-√2)/(5-2)
= (√5-√2)/3

(iv) 1/(√7-2)
Solution:
Multiply and divide 1/(√7-2) by (√7+2)
1/(√7 - 2)×(√7 + 2)/(√7 + 2) = (√7 + 2)/(√7 - 2)(√7 + 2)
= (√7 + 2)/(√7- 22) [denominator is obtained by the property, (a + b)(a - b) = a2-b2]
= (√7 + 2)/(7 - 4)
= (√7 + 2)/3

The document NCERT Solutions for Class 9 Maths - Exercise 1.5 Number System is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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## FAQs on NCERT Solutions for Class 9 Maths - Exercise 1.5 Number System

 1. What are the NCERT Solutions for Exercise 1.5 in the Number System Class 9?
Ans. The NCERT Solutions for Exercise 1.5 in the Number System Class 9 are detailed explanations and solutions for the questions given in Exercise 1.5 of the Number System textbook prescribed by NCERT for Class 9 students. These solutions help students understand and solve the problems related to number systems effectively.
 2. Where can I find the NCERT Solutions for Exercise 1.5 in the Number System Class 9?
Ans. The NCERT Solutions for Exercise 1.5 in the Number System Class 9 can be found on various online platforms, educational websites, and e-learning portals. Students can also refer to the official NCERT website or download dedicated mobile applications that provide these solutions. Additionally, many coaching institutes and teachers also upload these solutions on their websites or YouTube channels.
 3. How can NCERT Solutions for Exercise 1.5 in the Number System Class 9 help me in my exam preparation?
Ans. NCERT Solutions for Exercise 1.5 in the Number System Class 9 can be immensely helpful in exam preparation. These solutions provide step-by-step explanations and solutions to the questions, helping students understand the concepts and techniques required to solve them. By practicing these solutions, students can improve their problem-solving skills, gain clarity on various topics, and enhance their overall understanding of the number system.
 4. Are the NCERT Solutions for Exercise 1.5 in the Number System Class 9 available in multiple languages?
Ans. Yes, the NCERT Solutions for Exercise 1.5 in the Number System Class 9 are available in multiple languages. Apart from English, these solutions can be found in Hindi and other regional languages as well. This ensures that students from different linguistic backgrounds can access and understand the solutions effectively, thus promoting inclusivity in education.
 5. Can I rely solely on the NCERT Solutions for Exercise 1.5 in the Number System Class 9 for my exam preparation?
Ans. While the NCERT Solutions for Exercise 1.5 in the Number System Class 9 are highly beneficial for exam preparation, it is recommended to use them as a reference and not solely rely on them. It is advisable to refer to other study materials, textbooks, and practice sample papers to get a comprehensive understanding of the subject. Additionally, seeking guidance from teachers and solving additional problems will further strengthen your preparation.

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