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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Question 1.
Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\))
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a² + b²) (-a² + b²)
(vii) (6x – 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix) \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{5}\right)\)
(x) (7a – 9b) (7a – 9b)
Answer:
(i) (x + 3) (x + 3) = (x + 3)²
= x² + 2 × x × 3 + 3²
= x² + 6x +9
[Using the identity (a + b)² = a²+ 2ab + b²]

(ii) (2y + 5) (2y + 5) = (2y + 5)²
= (2y)² + 2 × 2y × 5 + 5²
= 4y² + 20y + 25
[Using the identity (a + b)² = a² + 2ab + b² ]

(iii) (2a – 7) (2a – 7) = (2a – 7)²
= (2a)² – 2 × 2a × 7 + (7)²
[using the identity (a – b)² = a² -2ab + b²]
= 4a² – 28a + 49

(iv) (3a – \(\frac { 1 }{ 2 }\)) (3a – \(\frac { 1 }{ 2 }\)) = (3a – \(\frac { 1 }{ 2 }\))²
= (3a)² – 2 × 3a × \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)²
= 9a² – 3a + \(\frac { 1 }{ 4 }\)
[using the identity (a – b)² = a² – 2ab + b²]

(v) (1.1m – 0.4) (1.1m + 0.4)
= (1.1m)² – (0.4)² = 1.21m² – 0.16
[using the identity (a × b)(a – b) = (a² – b²)]

(vi) (a² + b²) (-a² + b²)
= (b² + a²) (b² – a²) = (b²)² – (a²)²
= b⁴ – a⁴
[using identity (a + b) (a – b) = a² – b² ]

(vii) (6x – 7) (6x + 7)
= (6x)² – 7² = 36x² – 49
[using the identity (a + b)(a – b) = a² – b²]

(viii) (- a + c) (- a + c) = (-a + c)²
= (-a)² -2 × a × c + c² = a² – 2ac + c²
[Using identity (a – b)² = a² – 2ab + b²]

(x) (7a – 9b) (7a – 9b) = (7a – 9b)²
= (7a)² – 2 × 7a × 9b × (-9b)²
[using identity (a – b)² = a² – 2ab + b²]
= 49a² – 126ab + 81b²

Question 2.
Use the identity (x + a) (x + b) = x² (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a² + 9) (2a² + 5)
(vii) (xyz – 4) (xyz – 2)
Answer:
(i) (x + 3)(x + 7)
= x² + (3 + 7) x + 3 × 7 = x² + 10x + 21

(ii) (4x + 5)(4x + 1)
= (4x)² + (5 + 1) 4x + 5 × 1
= 16x² + 6 × 4x + 5
= 16x² + 24x + 5

(iii) (4x – 5) (4x – 1)
= (4x)² + (-5 -1) 4x + (-5) (-1)
= 16x² + (-6) 4x + 5
= 16x² – 24x + 5

(iv) (4x + 5) (4x – 1)
= (4x)² + (5 – 1) 4x + 5 x (-1)
= 16x² + (4) 4x – 5
= 16x² + 16x – 5

(v) (2x + 5y) (2x + 3y)
= (2x)² + (5y + 3y) 2x + 5y × 3y
= 4x² + (8y) 2x + 15y²
= 4x² + 16xy + 15y²

(vi) (2a² + 9) (2a² + 5)
= (2a²)² + (9 + 5) 2a² + 9 × 5
= 4a⁴ + (14) 2a² + 45
= 4a⁴ + 28a² + 45

(vii) (xyz – 4) (xyz – 2)
= (xyz)² + (-4 -2) xyz + (-4) (-2)
= x²y²z² + (- 6) xyz + 8
= x²y²z² – 6xyz + 8

Question 3.
Find the following squares by using the identities.
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y)²
(iv)(\(\frac { 2 }{ 3 }\)m + \(\frac { 3 }{ 2 }\)n)²
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Answer:
(i) (b – 7)² = b² – 2 b 7 + (-7)²
= b² – 14b + 49
[Using the identity (a – b)² = a² – 2ab + b² ]

(ii) (xy + 3z)² = (xy)² + 2 (xy) 3z + (3z)²
= x²y² + 6xyz + 9z²
[using the identity (a x b)² = a² + 2ab + b² ]

(iii) (6x² – 5y)²
= (6x²)² – 2 × 6x² × 5y + (5y)²
= 36x² – 60x²y + 25y²
[Using the identity (a – b)² = a² – 2ab + b²

(iv) ( \(\frac { 2 }{ 3 }\)m + \(\frac { 3 }{ 2 }\)n)²

[Using the identity (a + b)² = a² + 2ab + b² ]

(v) (0.4p – 0.5q)²
= (0.4p)² – 2 × 0.4p × 0.5q + (0.5q)²
= 0.16p² – 0.4pq + 0.25q²
[Using the identity (a – b)² = a² – 2ab + b²]

(vi) (2xy + 5y)²
= (2xy)² + 2 × 2xy × 5y + (5y)²
= 4x²y² + 20xy² + 25y²

Question 4.
Simplify:
(i) (a² – b²)²
(ii) (2x + 5)² – (2x – 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (4m + 5n)² + (5m + 4n)²
(v) (2.5p – 1.5q)² – (1.5p -2.5q)²
(vi) (ab + bc)² – 2ab²c
(vii) (m² – n²m²)² + 2m³n²
Answer:
(i) (a² – b² )² = (a²)² – 2 × a² × b² + (b²)²
= a⁴ – 2a² b² + b⁴
[using (a – b)² = a² – 2ab + b²]

(ii) (2x + 5)² – (2x – 5)²
= (2x)² + 2 × 2x × 5 + (5)² – [(2x)² – 2 × 2x × 5 + 5²]
[using (a + b)² = a² + 2ab + b²]
(a – b)² = a² – 2ab + b²
= 4x² + 20x + 25 – (4x² – 20x + 25)
= 4x² + 20x + 25 – 4x² + 20x – 25
= 4x² – 4x² + 20x + 20x + 25 – 25
= 0 + 40x + 0
= 40x

(iii) (7m – 8n)² + (7m + 8n)²
= (7m)² – 2 × 7m × 8n + (8n)² + (7m)² + 2 × 7m × 8n + (8n)²
[Using (a – b)² = a² – 2ab + b² and (a + b)² = a² + 2ab + b² ]
= 49m² – 112mn + 64n² + 49m² + 112mn + 64n²
= 49m² + 49m² – 112mn + 112 mn + 64n² + 64n²
= 98m² + 0 + 128n²
= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²
= (4m)² + 2 × 4m × 5n + (5n)² + (5m)² + 2 × 5m × 4n + (4n)2
[using (a + b)² = a² + 2ab + b² ]
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 16m² + 25m² + 40mn + 40mn + 25n² + 16n²
= 41m² + 80mn + 41n²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= (2.5p)² – 2 × 2.5p × 1.5q + (1.5q)² – [(1.5p)² – 2 × 1.5p × 2.5q + (2.5q)²]
[Using (a – b)² = a² – 2ab + b²]
= 6.25p² – 7.5pq + 2.25q² – (2.25p² – 7.5pq + 6.25q²)
= 6.25p² – 7.5pq + 2.25q² – 2.25p² + 7.5pq – 6.25q²
= 6.25p² – 2.25p² + 2.25q² – 6.25q² – 7.5pq + 7.5pq
= 4p² – 4q² + 0 = 4p² – 4q²

(vi) (ab + bc)² – 2ab²c
= (ab)² + 2 × ab × bc + (bc)² – 2ab²c
[using (a + b)² = a² + 2ab + b²]
= a²b² + 2ab²c + b²c² – 2ab²c
= a²b² + b²c² + 2ab²c – 2ab²c
= a²b² +b²c² + 2ab²c – 2ab²c
= a²b² + b²c² + 0
= a²b² + b²c²

(vii) (m² – n²m)² + 2m³n²
= (m²)² – 2 × m² × n²m + (n²m)² + 2m3n2
= m⁴ – 2m³n² + n⁴m² + 2m³n²
= m⁴ + n⁴m² – 2m³n² + 2m³n²
= m⁴ + n⁴m² + 0
= m⁴ + n⁴m²

Question 5.
Show that:
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) (\(\frac { 4 }{ 3 }\) m – \(\frac { 3 }{ 4 }\) n) + 2mn = \(\frac { 16 }{ 9 }\)m² + \(\frac { 16 }{ 9 }\)n²
(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0
Answer:
(i) L.H.S. = (3x + 7)² – 84x
= (3x)² + 2 × 3x × 7 + 7² – 84x
[using (a + b)² = a² + 2ab + b² ]
= 9x² + 42x + 49 – 84x
= 9x² + 42x – 84x + 49
= 9x² – 42x + 49
R.H.S. = (3x – 7)²
= (3x)² – 2 × 3x × 7 + 7² = 9x² – 42x + 49
R.H.S = L.H.S. Hence, proved

(ii) L.H.S. = (9p – 5q)² + 180pq
= (9p)² – 2 x 9p x 5q + (5q)² + 180pq
[Using the formula (a – b)² = a² – 2ab + b²]
= 81p² – 90pq + 25q² + 180pq
= 81p² – 90pq + 180pq + 25q²
= 81p² + 90pq + 25q²
R.H.S. = (9p + 5q)²
[Using (a + b)² = a² + 2ab + b² ]
= (9p)² + 2 x 9p x 5q + (5q)²
= 81p² + 90pq + 25q²
R.H.S. = L.H.S.
∴ Hence, proved.

(iii) L.H.S = ( \(\frac{4}{3}\)m – \(\frac{3}{4}\)n)² + 2mn
[Using (a – b)² = a² – 2ab + b²]

= \(\frac{16}{9}\)m² + \(\frac{9}{16}\)n² = RHS
L.H.S. = R.H.S.
Hence, proved.

(iv) L.H.S. = (4pq + 3q)² – (4pq – 3q)²
= (4pq)² + 2 × 4pq × 3q + (3q)² – [(4pq)² – 2 × 4pq × 3q + (3q)²]
[Using (a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b²]
= 16p²q² + 24pq² + 9q² – [ 16p²q² – 24pq² + 9q²]
= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²
= (16p²q² – 16p²q²) + 24pq²+24pq² + 9q² – 9q²
= 0 + 48pq² + 0s
= 48pq²
L.H.S. = R.H.S.
Hence, proved.

(v) L.H.S.
=(a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)
= a² – b² + b² – c² + c² – a²
[Using the identity (a + b) (a – b) = a2 – b2]
= a² – a² + b² – b² + c² – c²
= 0 + 0 + 0 = 0
L.H.S. = R.H.S.
Hence, proved.

Question 6.
Using identities, evaluate.
(i) 71²
(ii) 99²
(iii) 102²
(iv) 998²
(v) 5.2²
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.9²
(ix) 1.05 × 9.5
Answer:
(i) 71² = (70 + 1)2 = 70² + 2 × 70 × 1 + 1² [Usingtheidentity(a + b)² = a² + 2ab + b²]
= 4900 + 140 + 1 = 5041

(ii) 99² = (100 – 1)² = 100² – 2 × 100 × 1 + 1²
[Usingtheidentity(a – b)² = a² – 2ab + b²]
= 10000 – 200 + 1 = 9801

(iii) (102)² = (100 + 2)² = 100² + 2 × 100 × 2 + 2²
[Using (a + b)2 = a² + 2ab + b²]
= 10000 + 400 + 4 = 10404

(iv) (998)² = (1000 – 2)²
= 1000² – 2 × 1000 × 2 + 2²
[Using (a – b)² = a² – 2ab + b²]
= 1000000 – 4000 + 4 = 996004

(v) 5.2² = (5 + 0.2)² = 5² + 2 × 5 × 0.2 + (0.2)²
[Using (a + b)² = a² + 2ab + b² ]
= 25 + 2 + 0.04 = 27.04

(vi) 297 × 303 = (300 – 3) (300 + 3) = 300² × 3²
[Using (a + b) (a – b) = a² – b² ]
= 90000 – 9 = 89991

(vii) 78 × 82 = (80 – 2) (80 + 2) = 80² – 2²
[Using (a + b) (a – b) = a² – b²]
= 6400 – 4 = 6396

(viii)(8.9)² = (9 – 0.1)² = 9² – 2 × 9 × 0.1 + (0.1)²
[Using (a – b)² = a² – 2ab + b²]
= 81 – 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5 = \(\frac{1}{10}\) × 10.5 × 9.5
= \(\frac{1}{10}\)[(10 + 0.5)(10 – 0.5)]
= \(\frac{1}{10}\) [10² – 0.5²]
[Using (a + b) (a – b) = a² – b²]
= \(\frac{1}{10}\)[100 – 0.25] = \(\frac{1}{10}\) × 99.75
= 9.975

Question 7.
Using a² – b² = (a + b) (a – b) find
(i) 51² – 49²
(ii) (1.02)² – (0.98)²
(iii) 153² – 147²
(iv) 12.1² – 7.9²
Answer:
(i) 51² – 49² = (51 + 49)(51 – 49)
= (100) × 2 = 200

(ii) (1.02)² – (0.98)²
= (1.02 + 0.98) (1.02 – 0.98)
= 2 × 0.04 = 0.08

(iii) 153² – 147² = (153 + 147) (153 – 147)
= 300 × 6 = 1800

(iv) (12.1)² – (7.9)²
= (12.1 + 7.9) (12.1 – 7.9)
= 20 × 4.2 = 84

Question 8.
Using (x + a) (x + b) = x² + (a + b) x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
Answer:
(i) 103 × 104 = (100+ 3) (100 + 4)
= 100² + (3 + 4) 100 + 3 × 4
= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2= (5 + 0.1) (5 + 0.2)
= 5² + (0.1 +0.2) 5+ 0.1 × 0.2
= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 – 2)
= 100² + (3 – 2) 100 + (3) (- 2)
= 10000 + 100 – 6 = 10094

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)
= 10² + (- 0.3 – 0.2) 10 + (- 0.3) (- 0.2)
= 100 + (- 0.5) × 10 + 0.06
= 100 – 5 + 0.06 = 95.06.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Read More »

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions

CBSE Class 8 / Prasanna

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions

NCERT Intext Question Page No. 250

Question 1.
Write the following numbers in generalised form.;
(i) 25
(ii) 73
(iii) 129
(iv) 302
Answer:
(i) 25 = 20 + 5 = 2 × 10 + 5 × 1 = 10 × 2 + 5
(ii) 73 = 70 + 3 = 7 × 10 + 3 × 1 = 10 × 7 + 3
(iii) 129 = 100 + 20 + 9
= 1 × 100 + 2 × 10 + 9 × 1
= 100 × 1 + 10 × 2 + 9
(iv) 302 = 3 × 100 + 0 × 10 + 2 × 1
= 100 × 3 + 10 × 0 + 2

Question 2.
Write the following in the usual form.
(i) 10 × 5 + 6
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b
Answer:
(i) 10 × 5 + 6 = 50 + 6 = 56
(ii) 100 × 7 + 10 × 1 + 8
= 700 + 10 + 8 = 718
(iii) 100 × a + 10 × c + b
= 100a + 10c + b = a c b

NCERT Intext Question Page No. 257

Question 1.
If the division N + 5 leave a remainder of 3, what might be the one’s digit of N?
Answer:
The unit digit when divided by 5 must leave a remainder of 3.50, the one’s digit must be either 3 or 8.

Question 2.
If the division N + 5 leaves a remainder of 1, what might be the ones digit of N?
Answer:
The one’s digit when divided by 5 must have a remainder of 1. So the one’s digit must be either 1 or 6.

Question 3.
If the division N + 5 leaves a remainder of 4. What might be the one’s digit of N?
Answer:
The one’s digit, when divided by 5 must leave a remainder of 4. So the one’s digit must be either 4 or 9.

NCERT Intext Question Page No. 259

Question 1.
Check the divisiblity of the following number by 9.
(i) 108
(ii) 616
(iii) 294
(iv) 432
(v) 927
Answer:
(i) 108
Sum of the digits = 1 + 0 + 8 = 9 which is divisible by 9.
∴ 108 is divisible by 9.

(ii) 616
Sum of the digits = 6 + 1 + 6
= 13 which is not divisible by 9.
∴ 616 is not divisible by 9.

(iii) 294
Sum of the digits = 2 + 9 + 4 = 15
which is not divisible by 9.
∴ 294 is not divisible by 9.

(iv) 432
Sum of the digit = 4 + 3 + 2 = 9
which is divisible by 9.
∴ 432 is divisible by 9.

(v) 927
Sum of the digits = 9 + 2 + 7 = 18
which is divisible by 9.
∴ 927 is divisible by 9.

Question 9.
Check the divisibility of the following number by 3.
(i) 108
(ii) 616
(iii) 294
(iv) 432
(v) 927
Answer:
(i) 108
Sum of the digits =1 + 0 + 8 = 9
which is divisible by 3.
∴ 108 is divisible by 3.

(ii) 616
Sum of the digits = 6 + 1 + 6 = 13
which is not divisible by 3.
∴ 613 is also not divisible by 3.

(iii) 294
Sum of the digits = 2 + 9 + 4 = 15
which is divisible by 3.
∴ 294 is also divisible by 3.

(iv) 432
Sum of the digits = 4 + 3 + 2 = 9
which is divisible by 3.
∴ 432 is also divisible by 3.

(v) 927
Sum of the digits = 9 + 2 + 7 = 18 which is divisible by 3.
∴ Thus 927 is also divisible by 3.

NCERT Intext Question Page No. 251

Question 1.
Check what the result would have been if Sundaram had chosen the numbers shown below.
1. 27
2. 39
3. 64
4. 17
Answer:
1. Chosen number = 27
Number with reversed digits = 72
Sum of the two numbers = 27 + 72 = 99
Now, 99 = 11 [9] = 11 [2 + 7]
= 11 [Sum of the digits of the chosen number]

2. Chose number = 39
Number with reversed digits = 93
Sum of the two numbers = 39 + 93 = 132
Now, 132 ÷ 11 = 12
i. e„ 132 = 11 [12] = 11 [3 + 9]
= 11 [Sum of the digits of the chose number]

3. Chosen number = 17
Number with reversed digits = 71
Sum = 17 + 71 = 88
Now, 88 = 11 [8] = 11 [1 + 7]
= 11 [Sum of the digits of the chosen number]

Question 2.
Check what the result would have been if Sundaram had chosen the numbers shown.
1. 17
2. 21
3. 96
4. 37
Answer:
1. Chosen number =17
Number with reversed digits = 71
Difference = 71 – 17 = 54 = 9 × [6]
= 9 x [Difference of the digits of the chosen number (7 – 1 = 6)]

2. Chosen number = 21
Number with reversed digits = 12
Difference = 21 – 12 = 9 = 9 × [1]
= 9 x [Difference of the digits of the chosen number (12 -1 = 1)]

3. Chosen number = 96
Number with reversed digits = 69
Difference = 96 – 69 = 27 = 9 × [3]
=9 x [Difference of the digits of the chosen number (9 – 1 = 3)]

4. Chosen number = 37
Number with reversed digits = 73
Difference = 73 – 37 = 36 = 9 × [4]
=9 x [Difference of the digits of the chosen number (7-3 = 4)]

NCERT Intext Question Page No. 252

Question 1.
Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.
1. 132
2. 469
3. 737
4. 901
Answer:
1. 132 Chosen number = 132
Reversed number = 231
Difference = 231 – 132 = 99
We have 99 ÷ 99 = 1, remainder = 0

2. 469 Chosen number = 469
Reversed number = 964
Difference = 964 – 469 = 495
We have 99 ÷ 99 = 5, remainder = 0

3. Chosen number = 737
Reversed number = 737
We have Difference = 737 – 737 = 0
0 ÷ 99 =0, remainder = 0

4. Chosen number = 901
Reversed number = 109
Difference = 901 – 109 = 792
We have 792 ÷ 99 = 8, remainder = 0
Forming three-digit number with given three digits

NCERT Intext Question Page No. 253

Question 1.
Check what the result would have been if Sundaram had chosen the numbers shown below.
1. 417
2. 632
3. 117
4. 937
Answer:
1. Chosen number = 417
Two other numbers with the same digits are 741 and 174
Sum of the three numbers

We have 1332 ÷ 37 = 36, remainder = 0

2. Chosen number = 632
Two other numbers are 263 and 326
Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

3. Chosen number =117
Other numbers are 711 and 171
Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

4. Chosen number = 937
Other two numbers are 793 and 379
Sum of the three numbers

We have 2109 + 37 = 57, remainder = 0

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions Read More »

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

CBSE Class 8 / Prasanna

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1

Question 1.
Find the values of the letters in each of the following and give reasons for the steps involved.

Answer:
1. Let us see the sum in units column. It is A + 5 and we get 2 from this
(A+ 5 = 7 + 5= 12). So, A has to be 7
For sum in ten’s column, we have
1 + 3 + 2 = B
∴ B = 6

Question 2.
Here, there are three letters A, B and C whose values are to be found out.
Answer:
Let us see the sums in units column.
It is A + 8 and we get 3 from that so A has to be 5 (A + 8 = 5 + 8 = 13)
Now, for the sum in tens column, we have
I + 4 + 9= 14
∴ B = 4, C = 1
∴ The given puzzle is solved,
i. e. A = 5, B = 4 and C = 1

Question 3.
Units digit of A x A is A
Answer:
∴ A = 1 or 5 or 6
When, A = 1
II ≠ 9A
∴ A ≠ 1
When, A = 5
15 x 5 ≠ 9A;
∴ A ≠ 5
When, A = 6
16 x 6 = 96
∴ A = 6

Question 4.
Here, there are two letters A and B whose values are to be found out.
Answer:
B + 7 gives A
and A + 3 gives 6
The possible values are 0 + 7 = 7
A = 7 but 7 + 3 ≠ 6 so it is not acceptable.
1 + 7 = 8
A = 8 but 8 + 3 ≠ 6 so not acceptable
2 + 7 = 9
A = 9 but 9 + 3 ≠ 6 so not acceptable
3 + 7= 10
A = 0 but 1+0 + 3 ≠ 6 so, not acceptable.
4 + 7= 11
A = 1 but 1 + 1 + 3 ≠ 6 so not acceptable.
5 + 7= 12
A = 2 Also 1 + 2 + 3 = 6
So B = 5 works and then we get A as 2
∴ the puzzle is solved as shown below

i. e. A = 2 and B = 5.

5. We need B x 3 = □ B
Since 5 x 3 = 15
∴ Possible value of B can be 5.
Also 0 x 3 = 0, i.e., B = 0 can be another possible value
∵ A x 3 = A + 0 = A
∴ Possible value of A = 5 or A = 0
∴ Since C ≠ 0
∴ Possible value of A = 5
30, B must be
Thus, B = 0.

6. B can be either 5 or 0.
∵ A x 5 = A
∴ B must be 0
Again A can either 5 or 0
∴ C ≠ 0
∴ A ≠ 0
∴ A must be equal to 5
Thus, we have
Therefore, A = 5, B = 0, C = 2

7. B can be 2, 4, 6 or 8
We need product 111 or 222, or 333 or 444 or 888 out of them 111 and 333 are rejected Possible products are 222, 444 or 888 To obtain
The possible value is B = 4

∴ A can be either 2 or 7

∴ A x 6 = 7 x 6 is the accepted value

Thus, A = 7 and B = 4

Thus, A = 7 and B = 9

9. ∵ 8 – 1 = 7
∴ B = 7
7 + 4= 11 ∴ A = 4

Now, we have
Thus A = 4 and B = 7

10. 10 – 2 = 8
∴ A = 8
Also 9 – 8 = 1

∴ B = 1
Now, we have
Thus, A = 8 and B = 1

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Read More »

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

CBSE Class 8 / Prasanna

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2

Question 1.
It 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer:
21Y5 is a multiple of 9
∴ 2 + 1 + y + 5 = 8 + y must be divisible by 9
(8 + y) should be 9, 18, 27,… etc.
Since y is a digit.
8 + y = 9
y = 9 – 8 = 1
∴ The value of y = 1.

Question 2.
If 31z5 is a multiple of 9, where Z is a digit. What is the value of z? You will find that there are two answers for the last problem. Why is this so?
Answer.
31z5 = 3 + l + z + 5 = 9 + z
31z5 is divisible by 9
∴ 9 + z must be equal to 9, 18, 27,… etc.
Z is a single digit number.
9 + z is one of these numbers.
9 + z = 9 then z = 0
9 + z=18 thenz= 18-9 = 9
∴ The value of z = 0 or 9.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer:
Since 24x is a multiple of 3, so sum of digits i.e. 6 + x is a multiple of 3.
6 + x is one of these numbers: 0, 3,6, 9 12, 15, 18,… But since x is a digit,
∴ 6 + x = 6 or 9 or 12 or 15
∴ x = 0, or 3 or 6 or 9
The value of x = 0 or 3 or 6 or 9

Question 4.
31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer:
Sum of the digits =3+l+z+5=9+z 9 + z is a multiple of 3
So, 9 + z is one of the numbers: 0, 3, 6, 9, 12, 15,…
But since z is a digit, 9 + z = 9 or 12 or 15 or 18 or 21…
∴ The Value of z = 0 or 3 or 6 or 9 or 15 or 18.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Read More »

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

CBSE Class 8 / Prasanna

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a+ 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq -2q²)
(vi) \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\) and \(\left(a^{2}-\frac{2}{3} b^{2}\right)\)
Answer:
(2x + 5) (4x – 3) = 2x (4x – 3) + 5 (4x – 3)
= 2x × 4x + 2x (-3) + 5 × 4x + 5 (- 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

(ii) (y – 8) (3y – 4) = y (3y – 4) – 8 (3y – 4)
= y × 3y + y (-4) – [8 × 3y + 8 (-4)]
= 3y² – 4y – (24y – 32)
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

(iii) (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – [0.5m × 2.5l + 0.5m × 0.5m]
= 6.25l² + 1.25lm – 1.25lm – 0.25m²
= 6.25l² – 0.25m²

(iv) (a + 3b) (x + 5) = a × (x + 5) + 3b × (x + 5)
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq – 2q²)
= 2pq × (3pq – 2q²) + 3q² × (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴

= 3a⁴ -2a²b² +12a²b² – 8b⁴
= 3a⁴ + 10a²b² – 8b⁴

Question 2.
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q)
Answer:
(i) (5 – 2x) (3 + x) = 5 × (3 + x) – 2x (3 + x)
= 5 × 3 + 5 × x – (2x) × 3 – 2x (x)
= 15 + 5x – 6x – 2x² = 15 – x – 2x²

(ii) (x + 7y)(7x – y) = x × (7x – y) + 7y × (7x – y)
= x × 7x – x × y + 7y × 7x – 7y × y
= 7x² – xy + 49xy – 7y²
= 7x² + 48xy – 7y²

(iii) (a² + b)(a + b²) = a² × (a + b²) + b × (a + b²)
= a² × a + a² × b² + b × a + b × b²
= a^{3</sup + a²b² + ab + b³}

(iv) (p² – q²) (2p + q)
= p² × (2p + q) – q² (2p + q)
= p² × 2p + p² × q – q² × 2p – q² × q
= 2p³ + p²q – 2pq² – q³

Question 3.
Simplify
(i) (x² – 5) (x + 5) + 25
(ii) (a² + 5) (b³ + 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y) (x² – xy + y²)
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 45x + 12y
(viii) (a + b + c) (a + b – c)
Answer:
(i) (x² – 5) (x + 5) + 25
= x² × (x + 5) – 5 (x + 5) + 25
= x²× x + x² × 5 – 5 × x – 5 × 5 + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

(ii) (a² + 5) (b³ + 3) + 5
= a² × (b³ + 3) + 5 × (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5³ + 20

(iii) (t + s²) (t² – s) = t × (t² – s) + s² × (t² – s)
= t × t² – t × s + s² × t² – S² × s
= t³ – ts + s²t² – s³

(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
= a × (c – d) + b × (c – d) + a × (c + d) – b × (c + d ) + 2 × ac + 2 × bd
= a × c – a × d+ b × c – b × d + a × c + a × d – b × c – b × d + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2bd
= 4ac – 2bd + 2bd = 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y (2x + y) + x × (x – y) + 2y × (x – y)
= x × 2x + x × y + y × 2x + y × y + x × x + x × (-y) + 2y × x + 2y × (-y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y² = 2x² + x² + xy + 2xy + 2xy – xy + y² – 2y²
= 3x² + 5xy – xy – y²
= 3x² + 4xy – y²

(vi) (x + y) (x² – xy + y²)
= x × (x² – xy + y²) + y × (x² – xy + y²)
= x × x² + x × (-xy) + x × y² + y × x² + y × (-xy) + y × y²
= x³ – x²y + xy² + x²y – xy² + y³
= x³ – x²y + x²y + xy² – xy² + y³
= x³ + y³

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) – 4y × (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × 1.5x + 1.5x × 4y + 1.5x × 3 – 4y × 1.5x – 4y × 4y – 4y × 3 – 4.5x + 12y
= 2.25xz + 6xy + 4.5x – 6xy – 16y2
– 12y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x – 16y² – 12y + 12y
= 2.25x² + 0 + 0 – 16y² + 0
= 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)
= a × (a + b – c) + b × (a + b – c) + c × (a + b – c)
= a × a + a × b + a × -c + b × a + b × b – b × c + c × a + b × c – c × c
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + ab + ab – ac + ac – bc + bc + b² – c²
= a² + 2ab + o + o + b² – c²
= a² + b² – c² + 2ab

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Read More »

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

CBSE Class 8 / Prasanna

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Question 1.
Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) (4p) × (q + r)
= (4p × q) + (4p × r)
= 4pq + 4pr

(ii) (ab) × (a – b)
= (ab × a) – (ab × b)
= a²b – ab²

(iii) (a + b) (7a²b²)
= (a × 7a²b²) + (b × 7a²b²)
= 7a³b² + 7a²b³

(iv) (a² – 9) × 4a
= (a² × 4a) – (9 × 4a)
= 4a³ – 36a

(v) (pq + qr + rp) × 0
= (pq × 0) + (qr × 0) + (rp × 0)
= 0 + 0 + 0
= 0

Question 2.
Complete the table.

Solution:
(i) a × (b + c + d)
= (a × b) + (a × c) + (a × d)
= ab + ac + ad

(ii) (x + y – 5) (5xy)
= (x × 5xy) + (y × 5xy) – (5 × 5xy)
= 5x²y + 5xy² – 25xy

(iii) p × (6p² – 7p + 5)
= (p × 6p²) + p × (-7p) + p × 5
= 6p³ + (-7p²) + 5p
= 6p³ – 7p² + 5p

(iv) 4p²q² × (p² – q²)
= (4p²q² × p²) + (4p²q² × -q²)
= 4p⁴q² + (-4p²q⁴)
= 4p⁴q² – 4p²q⁴

(v) (a + b + c) × abc
= (a × abc) + (b × abc) + (c × abc)
= a²bc + ab²c + abc²

Question 3.
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)\)
(iii) \(\left(-\frac{10}{3} \mathrm{pq}^{3}\right) \times\left(\frac{6}{5} \mathrm{p}^{3} \mathrm{q}\right)\)
(iv) x × x² × x³ × x⁴
Solution:
(i) (a²) × (2a²²) × (4a²⁶)
= (1 × 2 × 4) × (a² × a²² × a²⁶)
= 8 × (a^2+22+26)
= 8a⁵⁰

(iv) x × x² × x³ × x⁴
= x^1+2+3+4
= x¹⁰

Question 4.
(a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3
(ii) x = \(\frac{1}{2}\)
(b) Simplify a (a² + a + 1) + 5 and find its value for
(i) a = 0
(ii) a = 1
(iii) a = -1.
Solution:
(a) 3x (4x – 5) + 3
= (3x × 4x) + (3x × -5) + 3
= 12x² – 15x + 3
(i) when x = 3
12x² – 15x + 3
= 12 × (3)² – 15(3) + 3
= 12 × 9 – 45 + 3
= 108 – 45 + 3
= 111 – 45
= 66

(ii) When x = \(\frac{1}{2}\)
12x² – 15x + 3

(b) a(a² + a + 1) + 5
= (a × a²) + (a × a) + (a × 1) + 5
= a³ + a² + a + 5

(i) When a = 0
a³ + a² + a + 5
= (0)³ + (0)² + 0 + 5
= 0 + 0 + 0 + 5
= 5

(ii) When a = 1
a³ + a² + a + 5
= (1³) + (1²) + 1 + 5
= 1 + 1 + 1 + 5
= 8

(iii) When a = -1
a³ + a² + a + 5
= (-1)³ + (-1)² + (-1) + 5
= (-1) + 1 – 1 + 5
= -1 + 1 – 1 + 5
= 4

Question 5.
(a) Add: p(p – q), q(q – r) and r (r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c (-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= p × p – p × q + q × q + q(-r) + r × r + r × (-p)
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – (pq + qr + rp)

(b) 2x (z – x – y) + 2y(z – y – x)
= 2x × z + 2x (-x) + 2x (-y) + 2yz + 2y (-y) + 2y(-x)
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= -2x² – 2y² – 4xy + 2yz + 2xz

(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) + (4l × -3m) + (4l × 2l) + (-3l × 1) + (-3l × -4m) + (-3l × 5n)
= 40ln + (-12lm) + 8l² + (-3l²) + 12lm + (-15ln)
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 40ln – 15ln – 12lm + 12lm + 8l² – 3l²
= 25ln + 5l²
= 5l² + 25ln

(d) Simplify the I part we get 3a (a + b + c) – 2b(a – b + c)
= (3a × a) + (3a × b) + (3a × c) – [(2b × a) + 2b(-b) + (2b) × c]
= 3a² + 3ab + 3ac – (2ab – 2b² + 2bc)
= 3a² + 3ab + 3ac – 2ab + 2b² – 2bc
= 3a² + 2b² + 3ab – 2ab + 3ac – 2bc
= 3a² + 2b² + ab + 3ac – 2bc
Simplify the 2nd part 4c × (-a + b + c) = (4c × – a) + (4c × b) + (4c × c)
According to the given question = -4ac + 4bc + 4c²
2nd part – 1st part
= -4ac + 4bc + 4c² – (3a² + 2b² + ab + 3ac – 2bc)
= -4ac + 4bc + 4c² – 3a² – 2b² – ab – 3ac + 2bc
= -4ac – 3ac + 4bc + 2bc + 4c² – 3a² – 2b² – ab
= -7ac + 6bc + 4c² – 3a² – 2b² – ab
= -3a² – 2b² + 4c² – ab + 6bc – 7ac

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Read More »