ML Aggarwal Solutions Class 9 Math 1st Chapter Rational and Irrational Numbers Exercise 1.5
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions First Chapter Rational and Irrational Numbers Exercise 1.5. APC Solution Class 9 Exercise 1.5.
(1) Rationalise the denominator of the following:
(i) 3/4√5
(ii) 5√7/√3
(iii) 3/4-√7
(iv) 17/3√2+1
(v) 16/√4 – 5
(vi) 1/√7 – √6
(vii) 1/√5 + √2
(viii) √2+√3/√2-√3
Solution:
(i) 3/4√5 = 3×√5/4√5×√5 = 3√5/4×5 = 3√5/20
(ii) 5√7/√3 = 5√7×√3/√3×√3 = 5√21/3
(iii) 3/4-√7 = 3/4-√7 = 4+√7/4+√7 = 12+3√7/42– (√7)2
= 12+3√7/16-7
= 3 (4+√7)/9
= 4+√7/3
(iv) 17/3√2+1 = 17/3√2+1 = 3√2-1/3√2-1
= 17(3√2-1)/(3√2)2 – (1)2
= 17(3√2 – 1)/9×2-1
= 17 (3√2 – 1)/17
= 3√2 – 1
(v) 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 = 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 = 1/√5+√2 × √5-√2/√5-√2
= √5-√2/(√5)2 – (√2)2 = √5-√2/5-2
= √5-√2/3
(viii) √2+√3/√2-√3 = √2+√3/√2-√3 = √2+√3/√2+√3
= (√2+√3)2/(√2)2 – (√3)2
= (√2)2 + 2√2× √3 + (√3)2/2-3
= 2+2√6+3/-1
= -5 – 2√6
(2) Simplify each of the following by rationalising the denominator:
(i) 7+3√5/7-3√5
(ii) 3-2√2/3+2√2
(iii) 5-3√14/7+2√14
Solution:
(i) 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 + 2×7×3√5 + (3√5)2/49 – 9×5
= 49+42√5+45/49-45
= 94+42√5/4
= 2 (47+21√5)/4
= 47+21√5/2
(ii) 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
= 3 – 2√2)2/32 – (2√2)2
= 32 – 2×3×2√2 + (2√2)2/9 – 4×2
= 9 – 12√2 + 8/1
= 17 – 12√2
(iii) 5-3√14/7+2√14 = 5-3√14/7+2√14 × 7-2√14/7-2√14
= 35-10√14–21√14+6×14/72-(2√14)2
= 35-31√14+84/49-56
= 119-31√14/-7
= -119+31√14/7
(3) Simplify: 7√3/√10+√3 – 2√5/√6+√5 3√2/√15+3√2
Solution:
= 7√3/√10+√3 – 2√5/√2+√5 – 3√2/√15+3√2
= 7√3/√10+√3 × √10-√3/√10-√3 – 2√5/√6+√5 × √6-√5/√6-√5 – 3√2/√15+3√2 × √15-3√2/√15-3√2
= 7√3 (√10+√3)/(√10)2 – (√3)2 – 2√5 (√6 – √5)/(√6)2 – (√5)- 3√2 (√15 – 3√2/(√15)2 – (3√2)2
= 7√3 (√10 – √3)/7 – 2√5(√6-√5)/1 – 3√2 (√15 – 3√2)/-3
= √30 – 3 – 2√30 + 2× 5 + √30 – 3 × 2
= 2√30 – 2√30 – 3 + 10 – 6
= 10 – 9
= 1
(4) Simplify: 1/√4+√5 + 1/√5+√6 + 1/√6+√7 + 1/√7+√8 + 1/√8+√9
Solution:
1/√4+√5 + 1/√5+√6 + 1/√6+√7 + 1/√7+√ + 1/√8+√9
= 1/√4+√5 × √4-√5/√4-√5 + 1/√5+√6 × √5-√6/√5-√6 + 1/√6+√7 × √6-√7/√6-√7 + 1/√7+√8 × √7-√8/√7-√8 + 1/√8+√9 × √8-√9/√8-√9
= √4-√5/(√4)2-(√5)2 + √5-√6/(√5)2-(√6)2 +
√6-√7/(√6)2 – (√7)2 + √7-√8/(√7)2 – (√8)2 + √8-√9/(√8)2 – (√9)2
= √4-√5/1 + √5-√6/1 √6-√7/-1 + √7-√8/-1 + √8-√9/-1
= – √4 + √5 – √5 + √6 – √6 + √7 – √7 + √8 – √8 + √9
= – √4 + √9
= 3-2
= 1
(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
Solution:
Given, a, b are rational numbers.
(i) 3-√5/3+2√5 = -19/11 + a√5
Or, 3-√5/3+2√5 × 3-2√5/3-2√5 = -19+11a√5 /11
Or, 3×3 – 3×2√5 – 3√5 + 2 × 5/32 – (2√5)2
= 11a√5 – 19/11
Or, 9 – 6√5 – 3√5 + 10/9-20 = 11a√5-19/11
Or, 19-9√5/-11 = 11a √5-19/11
Or, -19 + 9√5 = 11a√5 – 19
Or, 11a √5 = 9√5
Or, a = 9√5/11√5
Or, a = 9/11
(ii) √2+√3/3√2 – 2√3 = a – b√6
Or, √2+√3/3√2-2√3 × 3√2+2√3/3√2+2√3 = a – 3√6
= a – b√6
Or, 12+2√6+3√6+6/18-12 = a – b√6
Or, 18 + 5√6/6 = a – b√6
Or, 18/6 + 5/6 √6 = a – b√6
Or, 3 + 5/6 √6 = a – b√6
On comparing both sides we get,
a = 3, – b√6 = 5/6 √6
Or, b = – 5/6 √6/√6
= -5/6
(iii) 7+√5/7-√5 – 7-√5/7+√5 = a + 7/11 b√5
Or, (7 + √5) (7 + √5) – (7-√3) (7-√5)/(7-√5)(7+√5) = a + 7/11 b√5
Or, (7+√3)2-(7-√5)2/72 – (√5)2 = a + 7/11 b√5
Or, 49+2×7×√5 + (√5)2-49+2×7×√5 – (√5)2/49-5
= a + 7/11 b√5
Or, 28√5/44 = a + 7/11 b√5
Or, 0 + 28√5/44 = a + 7/11 b√5
On comparing both sides we get,
a = 0, 7/11b√5 = 26√5/44
Or, 7/11 b√5 = 7√5/11
Or, b = 1
(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.
Solution:
7+3√5/3+√5 – 7-3√5/3-√5 = p + q√5
Or, (7+3√5) (3-√5) – (7-3√) (3+√5)/(3+√5)(3-√5) = p + q√5
Or, 18√5 – 14√5/9-5 = P + q√5
Or, 4√5/4 = P + q√5
Or, √5 = p + q√5
Or, 0 + √5 = p + q√5
On comparing both sides we get,
P = 0, q√5 = √5
Or, 9 = 1
(7) Rationalise the denominator of the following and hence evaluate by taking √2 = 1.414 and √3 = 1.732, up to three places of decimal:
(i) √2/2+√2
(ii) 1/√3+√2
Solution:
Given, √2 = 1.414, √5 = 1.732
(i) √2/2+√2
= √2/√2×√2+√2
= √2/√2 (√2 + 1)
= 1/√2+1
= 1/√2+1 × √2-1/√2-1
= √2-1/(√2)2-12
= √2-1/2-1
= √2-1
= 1.414-1
= 0.414
[∵√2 = 1.414]
= 0.414
(ii) 1/√3 + √2
= 1/√3+√2 × √3-√2/√3-√2
= √3-√2/(√3)2-(√2)2
= √3-√2/3-2
= √3-√2/1
= 1.732 – 1.414 [∵√4 = 1.414 & √3 = 1.732]
= 0.318
(8) If a = 2 + √3, then find the value of a – 1/a,
Solution:
a = 2 + √3
∴ a – 1/a
= a2 – 1/a
= (2 + √3)2-1/(2+√3)
= 22 + 2 × 2 × √3 + (√3)2-1/2+√3 × 2-√3/2-√3
= (4 + 4√3 + 3 – 1) (2 – √3)/22 + (√3)2
= (6+ 4√3) (2 – √3)/4-3
= 12 – 6√3 + 8√3 – 4 × 3/1
= 12 + 2√3 – 12
= 2√3
(9) If x = 1 – √2, find the value of (x – 1/x)4
Solution:
Given, x = 1 – √2
∴ (x – 1/x)
= x2-1/x
= (2-1√2) (1+√2)/1-2
= 2 (1-√2) (1+√2)/-1
= -2 {12 – (√2)2}
= -2 {12 – (√2)2}
= -2 (1 – 2)
= – 2 × -1
= 2
∴ x – 1/x = 2
Or, (x – 1/x)4 = 24
= 16 (Ans)
(10) If x = 5 – 2√6, find the value of x2 + 1/x2
Solution:
Given, x = 5 – 2√6 —- (i)
∴ 1/x = 1/5-2√6
= 1/5-2√6 × 5+2√6/5+2√6
= 5+2√6/52-(2√6)2
= 5+2√6/25-24
= 5+2√6
∴ 1/x = 5 + 2√6 —– (ii)
Adding equation (i) & (ii) we get,
x + 1/x = (5 – 2√6) + (5 + 2√6)
Or, (x + 1/x) = 10
Squaring both sides we get,
(x + 1/x)2 = 102
Or, x2 + 2 × x × 1/x × 1/x2 = 100
Or, x2 + 1/x2 = 100 – 2
Or, x2 + 1/x2 = 98
(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
Solution:
(i) p + q
= 2-√5/2+√5 + 2+√5/2-√5
= 2-√5/2+√5 × 2-√5/2-√5 + 2+√5/2-√5 × 2+√5/2+√5
= (2-√5)2/22-(√5)2 + (2+√5)2/22 – (√5)2
= (2-√5)2/-1 + (2+√5)2/-1
= -1 {22 – 2 × 2 ×√5 + (√5)2} – {22 + 2 × 2 × √5 + (√5)2}
= -4 + 4√5 – 5 – 4 – 4√5 – 5
= -18
(ii) p – q
= 2-√5/2+√5 – 2+√5/2-√5
= 2-√5/2+√5 × 2-√5/2+√5 – 2+√5/2-√5 × 2+√5/2+√5
= (2-√5)2/22-(√5)2 – (2+√5)2/22-(√5)2
= (2-√5)2/4-5 – (2+√5)2/4-5
= – (2 – √5)2 + (2 + √5)2
= (2 + √5)2 – (2 – √5)2
= (2 + √5 + 2 – √5) (2 + √5 – 2 + √5)
= 4 × 2√5
= 8√5
(iii) P2 + q2
Previously we found p + q = -18
Squaring both sides.
(p + q)2 = (-18)2
Or, p2 + 2p + q2 = 324
Or, p2 + q2 = 324 – 2pq
Or, p2 + q2 = 2 (162 – pq)
Putting the value of pq on right hand side we get,
p2 + q2 = 2 (162 – 2-√5/2+√5 × 2+√5/2-√5)
Or, p2 + q2 = 2 (162 – 1)
Or, p2 + q2 = 2 × 161
Or, p2 + q2 = 322
(iv) p2 – q2
Previously we found that,
p+q = -18 — (i) & p – q = 8√5 —- (ii)
∴ p2 – q2
= (p + q) (p – q)
Putting the values from equation (ii) we get,
= 18 × 8√5
= -144√5
(12) If x = √2-1/√2+1 and y = √2+1/√2-1, then find the value of x2 + 5xy + y2
Hint, x2 + 5xy + y2 = (x + y)2 + 3xy
Solution:
Given, x = √2-1/√2+1, y = √2+1/√2-1
We know, (x + y)2 = x2 + 2xy + y2
∴ (x + y)2 + 3xy = x2 + 2xy + y2 + 3xy
[Adding 3xy on both sides)
∴ (x2 + y)2 + 3xy = x2 + y2 + 5xy —– (i)
Now, x2 + y = √2-1/√2+1 + √2+1/√2-1
= √2-1/√2+1 × √2-1/√2-1 + √2+1/√2-1 × √2+1/√2+1
= (√2-1)2/(√2)-(1)2 + (√2+1)2/(√2)2-12
= (√2 – 1)2/2-1 + (√2+1)2/2-1
= 2 – 2 √2 + 1 + 2 + 2√2 + 1
= 6
∴ (x + y) = 6 —- (ii)
Putting the value of x + y in equation (i) we get,
(x + y)2 + 3xy = x2 + y2 + 5xy
Or, x2 + y2 + 5xy = 6 + 3xy
Putting the value of xy in night-hand side we get,
Or, x2 + y2 + 5xy = 6 + 3 × √2-1/√2-1 × √2+1/√2-1
Or, x2 + y2 + 5xy = 6+3
Or, x2 + y2 + 5xy = 9
Multiple Choice questions:
(1) Choose the correct statement:
(a) Reciprocal of every rational number is a rational number.
(b) The square roots of all positive integers are irrational numbers.
(c) The product of a rational and an irrational number is an irrational number
(d) The difference of a rational number and an irrational number is an irrational number.
Solution:
(d) The difference of a rational number and an irrational number is an irrational number.
(2) Every rational number is
(a) A natural number
(b) An integer
(c) A real number
(d) A whole number
Solution:
(c) A real number
(3) Between two rational numbers
(a) There is no rational number
(b) There is exactly one rational number
(c) There are infinitely many rational numbers
(d) There are only rational numbers and no irrational numbers.
Solution:
(c) There are infinitely many rational numbers
(4) Decimal representation of a rational number cannot be
(a) Terminating
(b) Non-terminating
(c) Non-terminating repeating
(d) Non-terminating non-repeating
Solution:
(d) Non-terminating non-repeating
(5) The product of any two irrational numbers is
(a) Always an irrational number
(b) Always a rational number
(c) Always an integer
(d) Sometimes rational, sometimes irrational
Solution:
(a) Always an irrational number
(6) The division of two irrational numbers is
(a) A rational number
(b) An irrational number
(c) Either a rational number or an irrational number
(d) Neither rational number nor irrational number
Solution:
(c) Either a rational number or an irrational number
(7) Which of the following is an irrational number?
(a) √4/9
(b) √12/√3
(c) √7
(d) √81
Solution:
(c) √7
(8) Which of the following numbers has terminating decimal representation?
(a) 3/7
(b) 3/5
(c) 1/3
(d) 3/11
Solution:
(b) 3/5 = 3×2/5×2 = 6/10 = 0.6
(9) Which of the following is an irrational number?
(a) 0.14
(b) 0.1416
(c) 0.1416
(d) 0.4014001400014…
Solution:
(d) 0.4014001400014..
(10) Which of the following numbers has non-terminating repeating decimal expansion?
(a) 11/80
(b) 17/160
(c) 63/240
(d) 93/420
Solution:
(c) 63/240
(11) A rational numbers between √2 and √3 is
(a) √2+√3/2
(b) √2×√3/2
(c) 1.5
(d) 1.8
Solution:
(c) 1.5, √2 = 1.44, √5 = 1.732
(12) The decimal expansion of 2 – √3 is
(a) Terminating and non-repeating
(b) Terminating and repeating
(c) Non-terminating and non-repeating
(d) Non-terminating and repeating
Solution:
(c) Non-terminating and non-repeating
(13) The decimal expansion of the rational number 33/22×5 will terminate after
(a) One decimal place
(b) Two decimal places
(c) Three decimal places
(d) Four decimal places
Solution:
33/22×5 = 33×5/22×52 = 33×5/102 – 33×5/100
(b) Two decimal places. Since no. of zeros is 2.
(14) √10×√15 is equal to
(a) 6√5
(b) 5√6
(c) √25
(d) 10√5
Solution:
√10 × √15
= √2 × √5 × √3 × √5
= 5 × √6
= 5√6 (c)
(15) 2√3 + √3 is equal to
(a) 2√6
(b) 6
(c) 3√3
(d) 4√6
Solution:
2√3 + √3
= 3√3 (c)
(16) The value of √8 + √18 is
(a) √26
(b) 2 (√2 + √3)
(c) 5√2
(d) 6√2
Solution:
√8 + √18
= √2 ×√4 + √2 × √9
= √2 (2+3)
= 5√2 (c)
(17) The number (2 – √3)2 is
(a) A natural number
(b) An integer
(c) A rational number
(d) An irrational number
Solution:
(2 – √3)2 = 22 – 2 × 2 ×√3 + (√3)2
= 4 – 4√3 + 3
= 7 – 4√3 is an irrational number
Ans – (d)
(18) If x a positive rational number which is not a perfect square, then -5√x is
(a) A negative integer
(b) An integer
(c) A rational number
(d) An irrational number
Solution:
(d) Irrational number
(19) If x, y are both positive rational numbers, then (√x + √y) (√x – √y) is
(a) A rational number
(b) An irrational number
(c) Neither rational nor irrational number
(d) Both rational as well as irrational number
Solution:
(√x + √y) (√x – √y) = (√x)2– (√y)2. x – y is a rational number (Ans (c))
(20) After rationalising the denominator of 7/3√3-2√2, we get the denominator as
(a) 13
(b) 19
(c) 5
(d) 35
Solution:
7/3√3-2√2 = 7/3√3-2√2 × 3√3+2√2/3√3+2√2 = 21√3+14√2/(3√3)2-(2√2)2
= 7(3√3+2√2)/17-8 = 7(3√3+2√2)/19 (Ans – (b))
(21) The number obtained on rationalising the denominator of 1/√7-2 is
(a)√7+2/3
(b) √7-2/3
(c) √7+2/5
(d) √7+2/45
Solution:
1/√7-2 = 1/√7-2 ×√7+2/√7+2
= √7+2/(√7)2-22
= √7+2/7-4
= √7+2/3
Ans – (a) √7+2/3