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NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs InText Questions

CBSE Class 8 /

These NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs InText Questions

NCERT Intext Question Page No. 243

Question 1.
A machinery worth 10,500 depreciated by 5%. Find its value after one year
Answer:
(a) Here, P = 10, 500, R = 5% p.a.
₹ = 1 year, n = 1
A = p[1 + \(\frac{5}{100}\)]

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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

CBSE Class 8 /

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

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

NCERT Intext Question Page No. 139
Question 1.
Write two terms which are like
(i) 7xy
(ii) 4mn2
(iii) 21
Answer:
(i) Two terms like 7xy are: -3xy and 8xy.
(ii) Two terms like 4mn2 are: -6mn2 and 2n2m.
(iii) Two terms like 21 are: -51 and -7b.

NCERT Intext Question Page No. 143
Question 1.
Find 4x × 5y × 7z. First find 4x × 5y and multiply it by 7z; or first find 5y × 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Answer:
We have:
4x × 5y × 7z = (4x × 5y) × 7z
= 20xy × 7z = 140xyz
Also 4x × 5y × 7z = 4x × (5y × 7z)
= 4x × 35yz = 140xyz
We observe that
(4x × 5y) × 7x = 4 × (5y × 7z)
The product of monomials is associative, i.e. the order in which we multiply the monomials does not matter.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

NCERT Intext Question Page No. 144
Question 1.
Find the product
(i) 2x (3x + 5xy)
(ii) a2 (2ab – 5c)
Answer:
(i) 2x(3x + 5xy) = 2x × 3x + 2x × 5xy
= (2 × 3) × x × x + (2 × 5) × x × xy
= 6 × x2 + 10 × x2y = 6x2 + 10x2y

(ii) a2 (2ab – 5c) = a2 x 2ab + a2 2 – 5c
= (1 × 2) × a2 × ab + [1 × (-5)] × a2 × c
= 2 × a3b + (-5) × a2c = 2a3b – 5a2c

NCERT Intext Question Page No. 145
Question 1.
Find the product: (4p2 + 5p + 7) × 3p
Answer:
(4p2 + 5p + 7) × 3p
= (4p2 × 3p) + (5p + 3p) + (7 × 3p)
= [(4 × 3) × p2 × p2] + [(5 × 3) × p × p] + (7 × 3) × p
= 12 × p3 + 15 × p2 + 21 2 p
= 12p3 + 15p2 + 21p

NCERT Intext Question Page No. 149
Question 1.
Verify Identity (IV), for a = 2, b = 3, x = 5.
Answer:
We have
(x + a)(x + b) = x2 + (a + b)x + ab
Putting a = 2, b = 3 and x = 5 in the identity:
LHS= (x + a) (x + b)
= (5 + 2) (5 + 3)
= 7 × 8 = 56
RHS= x2 + (a + b)x + ab
= (5)2 + (2 + 3) x (2 x 3)
= 25 × (5) × 5 + 6
= 25 × (25) × 6
∴ LHS = RHS
∴ The given identity is true for the given values.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

Question 2.
Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?
Answer:
When a = b (each = y)
(x + a)(x + b) = x2 + (a + b)x + ab becomes
(x + y)(x + y) = x2 + (y + y)x + (y + y)
= x2 + (2y)x + y2
= x2 + 2xy + y2
= Yes, it is the same as Identity I

Question 3.
Consider, the special case of Identity (IV) with a = -c and b -c. What do you get? Is it related to Identity (II)?
Answer:
Identity IV is given by
(x + a)(x + b) = x2 + (a + b)x + ab
Replacing ‘a by (-c) and ‘b’ by (-c), we have (x – c)(x – c)
= x2 + [(-c) + (-c)]x + [(-c) × (-c)]
= x2 + [-2c]x + (c2)] = x2 – 2cx + c2
which is same as Identity II.

Question 4.
Consider the special case of Identity (IV) with b = -a. What do you get? Is it related to Identity (III)?
Answer:
The identity IV is given by
(x + a)(x + b) = x2 + (a + b)x + ab Replacing ‘b’ by (-a), we have:
(x + a)(x – a) = x2 + [a + (-a)]x + [a × (-a)]
= x2+ [0]x + [a2]
= x2+ 0 + (-a2)
= x2 -a2
which is same as the identity III.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions Read More »

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

CBSE Class 8 /

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]

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 1

(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

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

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

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

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

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

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

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

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]
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 3

= \(\frac{16}{9}\)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.

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

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

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

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 15 Introduction to Graphs Ex 15.3

Uncategorized /

These NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Exercise 15.3

Question 1.
Draw the graphs for the following tables of values, with suitable scales on the axes
(a) Cost of apples
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 1

(b) Distance travelled by a car
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 2
(i) How much distance did the car cover during the period 7.30 am to 8a.m.?
(ii) What was the time when the car had covered a distance of 100 km since it’s start?
(c) Interest on deposits for a year.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 3
(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on ? 2500 for a year.
(iii) To get an interest of ₹ 280 per year, how much money should be deposited?
Answer:
(a) (i) Draw x-axis and y-axis mutually perpendicular to each other.
(ii) Take a suitable scale (x-axis; 1cm = 1 unit) (y-axis; 1cm = 5 units)
(iii) Take the number of apples along the x-axis and mark the cost in ₹ along the y-axis.
(iv) Plot the points (1,5), (2,10); (3,15): (4, 20) and (5,25)
(v) On joining the points, we obtain the graph as a straight line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 4

NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1

(b) (i) Draw x-axis and y-axis mutually perpendicular to each other.
(ii) Choose suitable scale along x-axis and y-axis (x-axis; lcm = 1 unit; y-axis; lcm = 20 units.
(iii) Mark time (in hours) along x-axis and distance (in km) along y-axis
(iv) Plot the points (6, 40), (7, 80) (8, 120) and (9, 160)
(v) By joining the points, we get the required graph.
1. In the graph, draw a perpendic-ular at the point indicating 7:30 a.m. on the x-axis such that it meets the graph at A. From A draw a line parallel to x-axis to meet the y-axis at 100 km.
Distance travelled between 7.30 a.m and 8:00 am.
= (120 km – 80 km) 20 km
2. The time when the car had covered a distance of 100 km since its start was 7.30 a.m.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 5

(c) (i) Draw x-axis and y-axis mutually perpendicular to each other.
(ii) Take a suitable scale [x-axis 1cm = ₹ 1000 and y-axis lcm ? 40] mark the deposits along the x-axis and mark the interest along the y-axis.
(iii) Plot the points and join the points to get a straight line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 6
Answer:
(i) Yes, it passes through the origin.
(ii) From the graph, the interest on ₹ 2500 is ₹ 200.
(iii) From the graph, ₹ 3500 should be deposited to get an interest of ₹ 280.

Question 2.
Draw a graph for the following.
(i)
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 7
Is it a linear graph?
(ii)
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 8
Is it a linear graph?
Answer:
(i) Taking the side of the square along the x-axis and the perimeter along the y-axis (x-axis lcm = 1 unit, y-axis lcm = 4 units)
On plotting the points (2,8) (3, 12) (3.5,14) (5,20) and (6,24), we get the required graph as a straight line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 9
The graph is a linear graph.

NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1

(ii) Taking the side of the square along the x-axis and area (in cm2) along the y-axis (x-axis 1cm = 1 unit andy-axis 1 cm = 10 units)
Plot the points (2, 4) (3, 9) (4, 16) (5, 25) and (6, 36) in the required graph.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.3 10
The graph is not a straight line. It is not a linear graph.

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NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.2

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These NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Exercise 15.2

Question 1.
Plot the following points on a graph sheet. Verify if they lie on a line
(a) A (4,0), B (4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1),Q(2,2),R(3,3),S(4,4)
(c) K(2, 3) L(5, 3), M(5, 5), N(2, 5)
Answer:
In each case we draw the x-axis and the y-axis and plot the given points.
(a) On plotting the points A (4,0) B (4,2) C(4, 6) and D(4, 2.5) and then on joining them we find that they lie on the same line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.2 1
(b) On plotting the points P(l, 1), Q(2, 2), R(3, 3) and S(4, 4) and then on joining them we find that they all lie on the same line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.2 2
(c) Plotting the points K(2, 3), L(5, 3), M(5, 5) and N(2, 5) and on joining them we find that all of them do not lie on the same line.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.2 3

Question 2.
Draw the line passing through (2,3) and (3,2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Answer:
On plotting the points A (2, 3) and B(3, 2), we draw line that joins them. On extending, the line meets the x-axis at C(5,0) and the y-axis at D(0, 5).
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.2 4

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.2

Question 3.
Write the coordinates of the vertices of each of these adjoining figures.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.2 5
Answer:
(i) The coordinates of‘O’ are (0,0), A are (2,0), B are (2, 3) and C are (0, 3)
(ii) The coordinate of P are (4, 3), Q are (6, 1), R are (6, 5), S are (4, 7)
(iii) The coordinate of K are (10, 5), L are (7, 7) and M are (10, 8)

Question 4.
State whether True or False. Correct if they are false.
(i) A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y-coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0,0).
Answer:
(i) True
(ii) False: A point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis.
(iii) True

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NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1

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These NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Exercise 15.1

Question 1.
The following graph shows the temperature of a patient in a hospital recorded every hour.
(a) What was the patient’s temperature at 1 p.m?
(b) When was the patients temperature 38.5°C?
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 1
(c) The patients temperature was the same two times during the period given. What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patients’ temperature showed an upward trend?
Answer:
(a) The patient’s temperature at 1 p.m. was 36.5°C.
(c) The patient’s temperature was same at 1 p.m and 2 p.m.
(d) The patient temperature at 1.30 was 36.5°C because the temperature of the patient was constant, (ie 36.5°C) from 1 p.m to 2 p.m)
(e) The temperature of patient showed an upward trend during 9 a.m to 10 a.m to 11 a.m and 2 p.m. to 3 p.m.

NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1

Question 2.
The following line graphs shows the yearly sales figures for a manufacturing company.
(a) What were the sale in (i) 2002
(ii) 2006?
(b) What were the sales in (i) 2003
(ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 2
Answer:
(a) (i) Company’s sale in 2002 was ₹ 4 crores
(ii) Company’s sale in 2006 was ₹ 8 crores
(b) (i) Company’s sale in 2003 was ₹ 7 crores
(ii) Company’s sale in 2005 was ₹ 10 crores
(c) Difference in sales between 2002 and 2006
= ₹ 8 crores – ₹ 4 crores = ₹ 4 crores
(d) In 2005, there was the greatest difference between the sales as compared to its previous year 2004.

Question 3.
For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 3
(a) How high was Plant A after
(i) 2 weeks (ii) 3 weeks?
(b) How high was Plant B after (i) 2 weeks
(ii) 3 weeks?
(c) How much did Plant A grow during the 3rd week?
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did Plant A grow most?
(f) During which week did Plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.
Answer:
(a) (i) After two weeks, the plant A was 7 cm high.
(ii) After three weeks, the plant A was 9 cm high
(b) (i) After two weeks, the plant B was 7 cm high.
(ii) After three weeks, the plant B was 10 cm high.
(c) During the 3rd week, the plant A grew 2 cm (9 cm – 7cm)
(d) The plant B grew 3 cm (10 cm – 7cm) from the end of 2nd week to the end of 3rd week.
(e) The growth of the plant A
During the 1st week 2cm (2 cm – 0 cm).
During the 2nd week = 5cm (7 cm – 2 cm).
During the 3rd week = 2 cm (9 cm – 7 cm).
During the 2nd week, the plant A grew the most.
(f) The growth of the plant B
During the 1st week = 1 cm (1 cm – 0 cm) During the 2nd week = 6 cm (7 cm – 1 cm)
During the 3rd week = 3 cm (10 cm – 7 cm)
(g) Both the plants have shown the same height at the end of the 2nd week.

NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1

Question 4.
The following graph shows the temperature forecast and the actual temperature for each day of a week.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 4
(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?
Answer:
(a) The forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday.
(b) The maximum forecast temperature during the week was 35°C.
(c) The minimum actual temperature during the week was 15°C.
(d) Difference between the actual temperature and forecast temperature.
Monday = 17.5°C – 15°C = 2.5°C
Tuesday = 20°C – 20°C = 0°C
Wednesday = 30.0°C – 25°C = 5°C
Thursday = 22.5°C – 15°C = 7.5°C
Friday = 15°C – 15°C = 0°C
Saturday = 30°C – 25°C = 5°C
Sunday = 35°C – 35°C = 0°C
The maximum difference was on Thursday.

Question 5.
Use the tables below to draw linear graphs.
(a) The number of days a hill side city received snow in different years.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 5

(b) Population (in thousands) of men and women in a village in different years.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 6
Answer:
(a) Linear graph show snow fall received in different years.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 7

NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1

(b) Linear graph showing population of men and women in a village in different years:
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 8

Question 6.
A courier-person cycles from a town to neighbouring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 9
Answer:
(a) The time is taken along the x-axis The scale along x-axis is 4 units = 1 hour
(b) Total travel time
= 8 a.m to 10 a.m and further 10.30 a.m to 12 noon
= 3\(\frac { 1 }{ 2 }\) hours.
(c) Distance of the merchant from the town = 22km
(d) Yes, the stop age time = 10.00 am to 10.30am
(e) His fastest ride is between 8.00 am and 9.00 am

Question 7.
Can there be a time-temperature graph as follows? Justify your answer.
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 10
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 11
NCERT Solutions for Class 7 Maths Chapter 15 Introduction to Graphs Ex 15.1 12
Answer:
(i) It shows a time-temperature graph. It shows increase in temperature with increase in time.
(ii) It shows a time-temperature graph. It shows decrease in temperature with increase in time.
(iii) It cannot be time-temperature graph because it shows many-many different temperature uses at one particular time.
(iv) It shows a time-temperature graph. It shows a fixed temperature at difference times.

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