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# Ex 1.3 NCERT Solutions - Number System Class 9 Notes | EduRev

## Class 9 : Ex 1.3 NCERT Solutions - Number System Class 9 Notes | EduRev

The document Ex 1.3 NCERT Solutions - Number System Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.
All you need of Class 9 at this link: Class 9

Exercise 1.3

Q.1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36/100
Solution: = 0.36 (Terminating)

(ii)1/11

Solution:   Solution:  = 4.125 (Terminating)

(iv) 3/13
Solution:  (v) 2/11
Solution:  (vi) 329/400
Solution: = 0.8225 (Terminating)

Q.2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]
Solution: Q.3. Express the following in the form p/q, where p and q are integers and q 0.
(i) Solution: Assume that  x = 0.666…
Then,10x = 6.666…
10x = 6 + x
9x = 6
x = 2/3

(ii) $0.4\overline{7}$
Solution:
$0.4\overline{7} = 0.4777..$
= (4/10)+(0.777/10)
Assume that x = 0.777…
Then, 10x = 7.777…
10x = 7 + x
x = 7/9
(4/10) + (0.777../10) = (4/10) + (7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9 × 10) = 7/90)
= (36/90) + (7/90) = 43/90 Solution: Assume that  x = 0.001001…
Then, 1000x = 1.001001…
1000x = 1 + x
999x = 1
x = 1/999

Q.4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Assume that x = 0.9999…..Eq (a)
Multiplying both sides by 10,
10x = 9.9999…. Eq. (b)
Eq.(b) – Eq.(a), we get
(10x = 9.9999)-(x = 0.9999…)
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible.
Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

Q.5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.
Solution:

1/17
Dividing 1 by 17:  There are 16 digits in the repeating block of the decimal expansion of 1/17.

Q.6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example:
1/2 = 0. 5, denominator q = 21
7/8 = 0. 875, denominator q =23
4/5 = 0. 8, denominator q = 51
We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

Q.7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

• √3 = 1.732050807568
• √26 =5.099019513592
• √101 = 10.04987562112

Q.8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Solution: Three different irrational numbers are:

• 0.73073007300073000073…
• 0.75075007300075000075…
• 0.76076007600076000076…

Q.9. Classify the following numbers as rational or irrational according to their type:
(i)√23
Solution:
√23 = 4.79583152331…
Since the number is non-terminating non-recurring therefore, it is an irrational number.
(ii)√225

Solution:
√225 = 15 = 15/1
Since the number can be represented in p/q form, it is a rational number.
(iii) 0.3796
Solution:
Since the number,0.3796, is terminating, it is a rational number.
(iv) 7.478478
Solution:
The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…
Solution:
Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

Exercise 1.4

Q.1. Visualise 3.765 on the number line, using successive magnification.
Solution: Q.2. Visualise on the number line, up to 4 decimal places.
Solution:  Exercise 1.5

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

Exercise 1.6

Q.1. Find:

(i) 641/2
Solution:
641/2 = (8×8)1/2
= (82)½
= 81 [⸪2×1/2 = 2/2 =1]
= 8

(ii) 321/5
Solution:
321/5 = (25)1/5
= (25)
= 21 [⸪5×1/5 = 1]
= 2

(iii) 1251/3
Solution:
(125)1/3 = (5×5×5)1/3
= (53)
= 51 (3×1/3 = 3/3 = 1)
= 5

Q.2. Find:
(i) 93/2
Solution:
93/2 = (3×3)3/2
= (32)3/2
= 33 [⸪2×3/2 = 3]
=27

(ii) 322/5
Solution:
322/5 = (2×2×2×2×2)2/5
= (25)2⁄5
= 22 [⸪5×2/5= 2]
= 4

(iii)163/4
Solution:
163/4 = (2×2×2×2)3/4
= (24)3⁄4
= 23 [⸪4×3/4 = 3]
= 8

(iv) 125-1/3
Solution:
125-1/3 = (5×5×5)-1/3
= (53)-1⁄3
= 5-1 [⸪3×-1/3 = -1]
= 1/5

Q.3. Simplify:
(i) 22/3×21/5
Solution:
22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]
= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/33)7
Solution:
(1/33)= (3-3)7 [⸪Since,(am)= am x n____ Laws of exponents]
= 3-21

(iii) 111/2/111/4
Solution:
111/2/111/4 = 11(1/2)-(1/4)
= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 71/2×81/2
Solution:
71/2×81/2 = (7×8)1/2 [⸪Since, (am×b= (a×b)m ____ Laws of exponents]
= 561/2

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