Q1.Complete the following:
(i) Every point on the number line corresponds to a _______ number which may be either _______ or _______.
(ii) The decimal form of an irrational number is neither _______nor _______ .
(iii) The decimal representation of the rational number8/27 is _______ .
(iv) 0 is _______number. [Hint: a rational / an irrational]
Sol:
(i) Every point on the number line corresponds to a real number which may be either rational or irrational.
(ii) The decimal form of an irrational number is neither recurring nor terminating.
(iii) The decimal representation of the rational number is 0.296.
(iv) 0 is a rational number.
Q2: Find the value of’(0.6)0 − (0.1)−1(3/8)−1 × (3/2)3 + (-1/3)−1
Sol:
(0.6)0 − (0.1)−1(3/8)−1 × (3/2)3 + (-1/3)−1
= 1 x 1/0.18/3 × 27/8 + (3)
= 1 − 109 − 3 = −96 = −32
Q3: Find the value of 4(216)-2/3− 1(256)-3/4
Sol:
4(216)-2/3− 1(256)-3/4
= 4 × (216)2/3 − (256)3/4
= 4 × (6 × 6 × 6)2/3 − (4 × 4 × 4 × 4)3/4
= 4 × 63 × 2/3 − 44 × 3/4
= 4 × 62 − 43
= 4 × 36 − 64
= 144 − 64 = 80
Q1: Give three rational numbers lying between and .
Sol:The rational number lying between 13 and 12 is calculated as follows:
12 × ( 13 + 12 ) = 1 × (2 + 3)3 × 2 = 512.
Therefore, 13 < 512 < 12.
Now, the rational number lying between 13 and 512 is calculated as:
12 × ( 13 + 512 ) = 1 × (4 + 5)12 × 2 = 38.
Therefore, 13 < 38 < 512.
The rational number lying between 512 and 12 is:
12 × ( 512 + 12 ) = 1 × (5 + 6)12 × 2 = 1124.
Therefore, 512 < 1124 < 12.
Hence, the three rational numbers lying between 13 and 12 are: 38, 512, and 1124.
Q2.Identify a rational number among the following numbers :
14 + √3, 2√2, 1 and π/4
Sol: 1 is a rational number.
Q3.Simplify: (√5 + √3)2.
Sol:
Here,using (a+b)2 = a2 + 2ab +b2
we get,
(√5 + √3)2= (√5)2+ 2(√5)(√3) + (√3)2
= 5 + 2√15 + 3
= 8 + 2√15
Q4.Simplify : √45 – 3√20 + 4√5
Sol: Simplify each square root.
√45 = √9 × 5 = 3√5
√20 = √4 × 5 = 2√5
Substitute these value in given equation : √45 – 3√20 + 4√5
we get ,
= 3√5 – 6√5 + 4√5
= √5.
Q5.Express 1.8181… in the form pq where p and q are integers and q ≠ 0.
Sol:Let x =1.8181… …(i)
[multiplying eqn. (i) by 100]
we get ,
100x = 181.8181… …(ii)
99x = 180
[subtracting (i) from (ii)]
we get,
x = 180/99
Hence, 1.8181… = 180/99 = 20/11
Q1.Simplify the expression: (11 + √11) (11 – √11)
Sol:Using Identity: (a – b) (a+b) = a2 – b2
(i) (11 + √11) (11 – √11)
= 112 – (√11)2
= 121 – 11
= 110
Q2.Rationalise the denominator :
3√2 + 12√5 − 3
Sol:Multiply and divide the given number by 2√5 + 3
3√2 + 12√5 − 3
= (3√2 + 1) × (2√5 + 3)(2√5 − 3)(2√5 + 3)
= 6√10 + 9√2 + 2√5 + 3(20 − 9)
= 6√10 + 9√2 + 2√5 + 311
Q3. Simplify and find the value of
(a) (729)1/6
(b) (64)2/3
(c) (243)6/5
(d) (21)3/2 x (21)5/2
(e) (81)1/3(81)1/12
Sol:
(a) (729)1/6
Prime factorize 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
(729)1/6 = (36)1/6 = (3)(6/6) = 3
( Used : Rule of exponents used: (ax)y = axy)Q4. Rationalize the denominator:
18 + 3√5
Sol: Multiply and divide by the (8 - 3√5):
= 18 + 3√5 × 8 − 3√58 − 3√5
= 8 − 3√582 − (3√5)2
= 8 − 3√564 − 45
= 8 − 3√519
Q5 : 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.
Sol:
Thus, 1/17 = 0.0588235294117647….
Therefore, 1/17 has 16 digits in the repeating block of digits in the decimal expansion.
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