ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-1.5 Questions for Rational and Irrational Number as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions
Board | ICSE |
Subject | Maths |
Class | 9th |
Chapter-1 | Rational and Irrational |
Topics | Solution of Exe-1.5 Questions |
Edition | 2024-2025 |
Rational and Irrational Number Exe-1.5
ML Aggarwal Class 9 ICSE Maths Solutions
Page 34
Question 1. Rationalize the denominator of the following:
(i) 3/4√5
(ii) 5√7 / √3
(iii) 3/(4 – √7)
(iv) 17/(3√2 + 1)
(v) 16/ (√41 – 5)
(vi) 1/ (√7 – √6)
(vii) 1/ (√5 + √2)
(viii) (√2 + √3) / (√2 – √3)
Answer:
(i) 3/4√5
Let us rationalize,
3/4√5 = (3×√5) /(4√5×√5)
= (3√5) / (4×5)
= (3√5) / 20
(ii) 5√7 / √3
Let us rationalize,
5√7 / √3 = (5√7×√3) / (√3×√3)
= 5√21/3
(iii) 3/(4 – √7)
Let us rationalize,
3/(4 – √7) = [3×(4 + √7)] / [(4 – √7) × (4 + √7)]
= 3(4 + √7) / [42 – (√7)2]
= 3(4 + √7) / [16 – 7]
= 3(4 + √7) / 9
= (4 + √7) / 3
(iv) 17/(3√2 + 1)
Let us rationalize,
17/(3√2 + 1) = 17(3√2 – 1) / [(3√2 + 1) (3√2 – 1)]
= 17(3√2 – 1) / [(3√2)2 – 12]
= 17(3√2 – 1) / [9.2 – 1]
= 17(3√2 – 1) / [18 – 1]
= 17(3√2 – 1) / 17
= (3√2 – 1)
(v) 16/ (√41 – 5)
Let us rationalize,
16/ (√41 – 5) = 16(√41 + 5) / [(√41 – 5) (√41 + 5)]
= 16(√41 + 5) / [(√41)2 – 52]
= 16(√41 + 5) / [41 – 25]
= 16(√41 + 5) / [16]
= (√41 + 5)
(vi) 1/ (√7 – √6)
Let us rationalize,
1/ (√7 – √6) = 1(√7 + √6) / [(√7 – √6) (√7 + √6)]
= (√7 + √6) / [(√7)2 – (√6)2]
= (√7 + √6) / [7 – 6]
= (√7 + √6) / 1
= (√7 + √6)
(vii) 1/ (√5 + √2)
Let us rationalize,
1/ (√5 + √2) = 1(√5 – √2) / [(√5 + √2) (√5 – √2)]
= (√5 – √2) / [(√5)2 – (√2)2]
= (√5 – √2) / [5 – 2]
= (√5 – √2) / [3]
= (√5 – √2) /3
(viii) (√2 + √3) / (√2 – √3)
Let us rationalize,
(√2 + √3) / (√2 – √3) = [(√2 + √3) (√2 + √3)] / [(√2 – √3) (√2 + √3)]
= [(√2 + √3)2] / [(√2)2 – (√3)2]
= [2 + 3 + 2√2√3] / [2 – 3]
= [5 + 2√6] / -1
= – (5 + 2√6)
Question 2. Simplify each of the following by rationalizing the denominator:
(i) (7 + 3√5) / (7 – 3√5)
(ii) (3 – 2√2) / (3 + 2√2)
(iii) (5 – 3√14) / (7 + 2√14)
Answer:
(i) (7 + 3√5) / (7 – 3√5)
Let us rationalize the denominator, we get
(7 + 3√5) / (7 – 3√5) = [(7 + 3√5) (7 + 3√5)] / [(7 – 3√5) (7 + 3√5)]
= [(7 + 3√5)2] / [72 – (3√5)2]
= [72 + (3√5)2 + 2.7. 3√5] / [49 – 9.5]
= [49 + 9.5 + 42√5] / [49 – 45]
= [49 + 45 + 42√5] / [4]
= [94 + 42√5] / 4
= 2[47 + 21√5]/4
= [47 + 21√5]/2
(ii) (3 – 2√2) / (3 + 2√2)
Let us rationalize the denominator, we get
(3 – 2√2) / (3 + 2√2) = [(3 – 2√2) (3 – 2√2)] / [(3 + 2√2) (3 – 2√2)]
= [(3 – 2√2)2] / [32 – (2√2)2]
= [32 + (2√2)2 – 2.3.2√2] / [9 – 4.2]
= [9 + 4.2 – 12√2] / [9 – 8]
= [9 + 8 – 12√2] / 1
= 17 – 12√2
(iii) (5 – 3√14) / (7 + 2√14)
Let us rationalize the denominator, we get
(5 – 3√14) / (7 + 2√14) = [(5 – 3√14) (7 – 2√14)] / [(7 + 2√14) (7 – 2√14)]
= [5(7 – 2√14) – 3√14 (7 – 2√14)] / [72 – (2√14)2]
= [35 – 10√14 – 21√14 + 6.14] / [49 – 4.14]
= [35 – 31√14 + 84] / [49 – 56]
= [119 – 31√14] / [-7]
= -[119 – 31√14] / 7
= [31√14 – 119] / 7
Question 3. Simplify:
[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]
Answer:
Let us simplify individually,
[7√3 / (√10 + √3)]
Let us rationalize the denominator,
7√3 / (√10 + √3) = [7√3(√10 – √3)] / [(√10 + √3) (√10 – √3)]
= [7√3.√10 – 7√3.√3] / [(√10)2 – (√3)2]
= [7√30 – 7.3] / [10 – 3]
= 7[√30 – 3] / 7
= √30 – 3
Now,
[2√5 / (√6 + √5)]
Let us rationalize the denominator, we get
2√5 / (√6 + √5) = [2√5 (√6 – √5)] / [(√6 + √5) (√6 – √5)]
= [2√5.√6 – 2√5.√5] / [(√6)2 – (√5)2]
= [2√30 – 2.5] / [6 – 5]
= [2√30 – 10] / 1
= 2√30 – 10
Now,
[3√2 / (√15 + 3√2)]
Let us rationalize the denominator, we get
3√2 / (√15 + 3√2) = [3√2 (√15 – 3√2)] / [(√15 + 3√2) (√15 – 3√2)]
= [3√2.√15 – 3√2.3√2] / [(√15)2 – (3√2)2]
= [3√30 – 9.2] / [15 – 9.2]
= [3√30 – 18] / [15 – 18]
= 3[√30 – 6] / [-3]
= [√30 – 6] / -1
= 6 – √30
So, according to the question let us substitute the obtained values,
[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]
= (√30 – 3) – (2√30 – 10) – (6 – √30)
= √30 – 3 – 2√30 + 10 – 6 + √30
= 2√30 – 2√30 – 3 + 10 – 6
= 1
Question 4. Simplify:
[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]
Answer:
Let us simplify individually,
[1/(√4 + √5)]
Rationalize the denominator, we get
[1/(√4 + √5)] = [1(√4 – √5)] / [(√4 + √5) (√4 – √5)]
= [(√4 – √5)] / [(√4)2 – (√5)2]
= [(√4 – √5)] / [4 – 5]
= [(√4 – √5)] / -1
= -(√4 – √5)
Now,
[1/(√5 + √6)]
Rationalize the denominator, we get
[1/(√5 + √6)] = [1(√5 – √6)] / [(√5 + √6) (√5 – √6)]
= [(√5 – √6)] / [(√5)2 – (√6)2]
= [(√5 – √6)] / [5 – 6]
= [(√5 – √6)] / -1
= -(√5 – √6)
Now,
[1/(√6 + √7)]
Rationalize the denominator, we get
[1/(√6 + √7)] = [1(√6 – √7)] / [(√6 + √7) (√6 – √7)]
= [(√6 – √7)] / [(√6)2 – (√7)2]
= [(√6 – √7)] / [6 – 7]
= [(√6 – √7)] / -1
= -(√6 – √7)
Now,
[1/(√7 + √8)]
Rationalize the denominator, we get
[1/(√7 + √8)] = [1(√7 – √8)] / [(√7 + √8) (√7 – √8)]
= [(√7 – √8)] / [(√7)2 – (√8)2]
= [(√7 – √8)] / [7 – 8]
= [(√7 – √8)] / -1
= -(√7 – √8)
Now,
[1/(√8 + √9)]
Rationalize the denominator, we get
[1/(√8 + √9)] = [1(√8 – √9)] / [(√8 + √9) (√8 – √9)]
= [(√8 – √9)] / [(√8)2 – (√9)2]
= [(√8 – √9)] / [8 – 9]
= [(√8 – √9)] / -1
= -(√8 – √9)
So, according to the question let us substitute the obtained values,
[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]
= -(√4 – √5) + -(√5 – √6) + -(√6 – √7) + -(√7 – √8) + -(√8 – √9)
= -√4 + √5 – √5 + √6 – √6 + √7 – √7 + √8 – √8 + √9
= -√4 + √9
= -2 + 3
= 1
Question 5. Give a and b are rational numbers. Find a and b if:
(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5
Answer:
(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
Let us consider LHS
[3 – √5] / [3 + 2√5]
Rationalize the denominator,
[3 – √5] / [3 + 2√5] = [(3 – √5) (3 – 2√5)] / [(3 + 2√5) (3 – 2√5)]
= [3(3 – 2√5) – √5(3 – 2√5)] / [32 – (2√5)2]
= [9 – 6√5 – 3√5 + 2.5] / [9 – 4.5]
= [9 – 6√5 – 3√5 + 10] / [9 – 20]
= [19 – 9√5] / -11
= -19/11 + 9√5/11
So when comparing with RHS
-19/11 + 9√5/11 = -19/11 + a√5
Hence, value of a = 9/11
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
Let us consider LHS
[√2 + √3] / [3√2 – 2√3]
Rationalize the denominator,
[√2 + √3] / [3√2 – 2√3] = [(√2 + √3) (3√2 + 2√3)] / [(3√2 – 2√3) (3√2 + 2√3)]
= [√2(3√2 + 2√3) + √3(3√2 + 2√3)] / [(3√2)2 – (2√3)2]
= [3.2 + 2√2√3 + 3√2√3 + 2.3] / [9.2 – 4.3]
= [6 + 2√6 + 3√6 + 6] / [18 – 12]
= [12 + 5√6] / 6
= 12/6 + 5√6/6
= 2 + 5√6/6
= 2 – (-5√6/6)
So when comparing with RHS
2 – (-5√6/6) = a – b√6
Hence, value of a = 2 and b = -5/6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5
Let us consider LHS
Since there are two terms, let us solve individually
{[7 + √5]/[7 – √5]}
Rationalize the denominator,
[7 + √5]/[7 – √5] = [(7 + √5) (7 + √5)] / [(7 – √5) (7 + √5)]
= [(7 + √5)2] / [72 – (√5)2]
= [72 + (√5)2 + 2.7.√5] / [49 – 5]
= [49 + 5 + 14√5] / [44]
= [54 + 14√5] / 44
Now,
{[7 – √5]/[7 + √5]}
Rationalize the denominator,
[7 – √5]/[7 + √5] = (7 – √5) (7 – √5)] / [(7 + √5) (7 – √5)]
= [(7 – √5)2] / [72 – (√5)2]
= [72 + (√5)2 – 2.7.√5] / [49 – 5]
= [49 + 5 – 14√5] / [44]
= [54 – 14√5] / 44
So, according to the question
{[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]}
By substituting the obtained values,
= {[54 + 14√5] / 44} – {[54 – 14√5] / 44}
= [54 + 14√5 – 54 + 14√5]/44
= 28√5/44
= 7√5/11
So when comparing with RHS
7√5/11 = a + 7/11 b√5
Hence, value of a = 0 and b = 1
Question 6. If {[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]} = p + q√5, find the value of p and q where p and q are rational numbers.
Answer:
Let us consider LHS
Since there are two terms, let us solve individually
{[7 + 3√5] / [3 + √5]}
Rationalize the denominator,
[7 + 3√5] / [3 + √5] = [(7 + 3√5) (3 – √5)] / [(3 + √5) (3 – √5)]
= [7(3 – √5) + 3√5(3 – √5)] / [32 – (√5)2]
= [21 – 7√5 + 9√5 – 3.5] / [9 – 5]
= [21 + 2√5 – 15] / [4]
= [6 + 2√5] / 4
= 2[3 + √5]/4
= [3 + √5] /2
Now,
{[7 – 3√5] / [3 – √5]}
Rationalize the denominator,
[7 – 3√5] / [3 – √5] = [(7 – 3√5) (3 + √5)] / [(3 – √5) (3 + √5)]
= [7(3 + √5) – 3√5(3 + √5)] / [32 – (√5)2]
= [21 + 7√5 – 9√5 – 3.5] / [9 – 5]
= [21 – 2√5 – 15] / 4
= [6 – 2√5]/4
= 2[3 – √5]/4
= [3 – √5]/2
So, according to the question
{[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]}
By substituting the obtained values,
= {[3 + √5] /2} – {[3 – √5] /2}
= [3 + √5 – 3 + √5]/2
= [2√5]/2
= √5
So when comparing with RHS
√5 = p + q√5
Hence, value of p = 0 and q = 1
Question 7. Rationalise the denominator of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732, upto three places of decimal:
(i) √2/(2 + √2)
(ii) 1/(√3 + √2)
Answer:
(i) √2/(2 + √2)
By rationalizing the denominator,
√2/(2 + √2) = [√2(2 – √2)] / [(2 + √2) (2 – √2)]
= [2√2 – 2] / [22 – (√2)2]
= [2√2 – 2] / [4 – 2]
= 2[√2 – 1] / 2
= √2 – 1
= 1.414 – 1
= 0.414
(ii) 1/(√3 + √2)
By rationalizing the denominator,
1/(√3 + √2) = [1(√3 – √2)] / [(√3 + √2) (√3 – √2)]
= [(√3 – √2)] / [(√3)2 – (√2)2]
= [(√3 – √2)] / [3 – 2]
= [(√3 – √2)] / 1
= (√3 – √2)
= 1.732 – 1.414
= 0.318
Question 8. If a = 2 + √3, find 1/a, (a – 1/a)
Answer:
Given:
a = 2 + √3
So,
1/a = 1/ (2 + √3)
By rationalizing the denominator,
1/ (2 + √3) = [1(2 – √3)] / [(2 + √3) (2 – √3)]
= [(2 – √3)] / [22 – (√3)2]
= [(2 – √3)] / [4 – 3]
= (2 – √3)
Then,
a – 1/a = 2 + √3 – (2 – √3)
= 2 + √3 – 2 + √3
= 2√3
Question 9. Solve:
If x = 1 – √2, find 1/x, (x – 1/x)4
Answer:
Given:
x = 1 – √2
so,
1/x = 1/(1 – √2)
By rationalizing the denominator,
1/ (1 – √2) = [1(1 + √2)] / [(1 – √2) (1 + √2)]
= [(1 + √2)] / [12 – (√2)2]
= [(1 + √2)] / [1 – 2]
= (1 + √2) / -1
= -(1 + √2 )
Then,
(x – 1/x)4 = [1 – √2 – (-1 – √2)]4
= [1 – √2 + 1 + √2]4
= 24
= 16
ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions
Page 35
Question 10. Solve:
If x = 5 – 2√6, find 1/x, (x2 – 1/x2)
Answer:
Given:
x = 5 – 2√6
so,
1/x = 1/(5 – 2√6)
By rationalizing the denominator,
1/(5 – 2√6) = [1(5 + 2√6)] / [(5 – 2√6) (5 + 2√6)]
= [(5 + 2√6)] / [52 – (2√6)2]
= [(5 + 2√6)] / [25 – 4.6]
= [(5 + 2√6)] / [25 – 24]
= (5 + 2√6)
Then,
x + 1/x = 5 – 2√6 + (5 + 2√6)
= 10
Square on both sides we get
(x + 1/x)2 = 102
x2 + 1/x2 + 2x.1/x = 100
x2 + 1/x2 + 2 = 100
x2 + 1/x2 = 100 – 2
= 98
Question 11. If p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5), find the values of
(i) p + q
(ii) p – q
(iii) p2 + q2
(iv) p2 – q2
Answer:
Given:
p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5)
(i) p + q
[(2-√5)/(2+√5)] + [(2+√5)/(2-√5)]
So by rationalizing the denominator, we get
= [(2 – √5)2 + (2 + √5)2] / [22 – (√5)2]
= [4 + 5 – 4√5 + 4 + 5 + 4√5] / [4 – 5]
= [18]/-1
= -18
(ii) p – q
[(2-√5)/(2+√5)] – [(2+√5)/(2-√5)]
So by rationalizing the denominator, we get
= [(2 – √5)2 – (2 + √5)2] / [22 – (√5)2]
= [4 + 5 – 4√5 – (4 + 5 + 4√5)] / [4 – 5]
= [9 – 4√5 – 9 – 4√5] / -1
= [-8√5]/-1
= 8√5
(iii) p2 + q2
We know that (p + q)2 = p2 + q2 + 2pq
So,
p2 + q2 = (p + q)2 – 2pq
pq = [(2-√5)/(2+√5)] × [(2+√5)/(2-√5)]
= 1
p + q = -18
so,
p2 + q2 = (p + q)2 – 2pq
= (-18)2 – 2(1)
= 324 – 2
= 322
(iv) p2 – q2
We know that, p2 – q2 = (p + q) (p – q)
So, by substituting the values
p2 – q2 = (p + q) (p – q)
= (-18) (8√5)
= -144√5
Question 12. If x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1), find the value of x2 + 5xy + y2.
Answer:
Given:
x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1)
x + y = [(√2 – 1)/( √2 + 1)] + [(√2 + 1)/( √2 – 1)]
By rationalizing the denominator,
= [(√2 – 1)2 + (√2 + 1)2] / [(√2)2 – 12]
= [2 + 1 – 2√2 + 2 + 1 + 2√2] / [2 – 1]
= [6] / 1
= 6
xy = [(√2 – 1)/( √2 + 1)] × [(√2 + 1)/( √2 – 1)]
= 1
We know that
x2 + 5xy + y2 = x2 + y2 + 2xy + 3xy
It can be written as
= (x + y)2 + 3xy
Substituting the values
= 62 + 3 × 1
So we get
= 36 + 3
= 39
— : End of ML Aggarwal Rational and Irrational Number Exe-1.5 Class 9 ICSE Maths Solutions :–
Return to :- ML Aggarawal Maths Solutions for ICSE Class-9
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