# Substitution ICSE Class-6th Concise Maths Selina Solutions Ch-20

Substitution ICSE Class-6th Concise Mathematics Selina Solutions Chapter-20 (Including Use of Brackets as Grouping Symbols). We provide step by step Solutions of Exercise / lesson-20 Substitution (Including Use of Brackets as Grouping Symbols) for ICSE Class-6 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-20 A, Exe-20 B, Exe-20 C and Revision Exercise to develop skill and confidence. Visit official Website  CISCE for detail information about ICSE Board Class-6 .

## Substitution ICSE Class-6th Concise Mathematics Selina Solutions Chapter-20 (Including Use of Brackets as Grouping Symbols)

–: Select Topics :–

Exe-20 A,

Exe-20 B,

Exe-20 C,

Revision Exercise

### Exercise – 20 ASubstitution (Including Use of Brackets as Grouping Symbols) for ICSE Class-6th Concise Selina Maths

#### Question- 1.

Fill in the following blanks, when :
x = 3,y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.

(i ) x + y = 3 + 6 = 9

(ii) y − x = 6 − 3 = 3

(iii) y/ x= 6/3 = 2

(iv) c ÷ b = c/b =32 / 8 = 4

(v) z ÷ x = z/x = 18/ 3 = 6

(vi) y × d = 6 × 0 = 0

(vii) d ÷ x = d/x = 0/3 = 0

(viii) ab + y = 2 × 8 + 6 = 16 + 6 = 22

(ix) a + b + x = 2 + 8 + 3 = 13

(x) b + z − d = 8 + 18 − 0 = 26

(xi) a − b + y = 2 − 8 + 6 = 8 − 8 = 0

(xii) z − a − b = 18 − 2 − 8 = 18 − 10 = 8

(xiii) d − a + x = 0 − 2 + 3 = 1

(xiv) xy − bd = 3 × 6 − 8 × 0 = 18 − 0 = 18

(xv) xz + cd = 3 × 18 + 32 × 0 = 54 + 0 = 54

#### Question -2.

Find the value of :

(i) p + 2q + 3r

= 1 + 2 × 5 + 3 × 2

= 1 + 10 + 6

= 17

(ii)

2a + 4b + 5c

= 2 × 5 + 4 × 10 + 5 × 20

= 10 + 40 + 100

= 150

(iii)

3a − 2b

= 3 × 8 − 2 × 10

= 24 − 20

= 4

(iv)

5x + 3y − 6z

= 5 × 3 + 3 × 5 − 6 × 4

= 15 + 15 − 24

= 30 − 24

= 6

(v)

2p − 3q + 4r −8s

= 2 × 10 − 3 × 8 + 4 × 6 − 8 × 2

= 20 − 24 + 24 − 16

= 20 − 16

= 4

(vi)

6m − 2n − 5p − 3q

= 6 × 20 − 2 × 10 − 5 × 2 − 3 × 9

= 120 − 20 − 10 − 27

= 120 − 57

= 63

#### Question -3.

Find the value of :

#### Question -4.

If a = 3, b = 0, c = 2 and d = 1, find the value of :

(i)

3a + 2b − 6c + 4d

= 3 × 3 + 2 × 0 − 6 × 2 + 4 × 1

= 9 + 0 − 12 + 4

= 13 − 12

= 1

(ii)

6a − 3b − 4c − 2d

= 6 × 3 − 3 × 0 − 4 × 2 − 2 × 1

= 18 − 0 − 8 − 2

= 18 − 10

= 8

(iii)

ab − bc + cd − da

= 3 × 0 − 0 × 2 + 2 × 1 − 1 × 3

= 0 + 0 + 2 − 3

= − 1

(iv)

abc − bcd + cda

= 3 × 0 × 2 − 0 × 2 × 1 + 2 × 1 × 3

= 0 − 0 + 6

= 6

(v)

a2 + 2b2 − 3c2

= 32 + 2 × 02 − 3 × 22

= 9 + 0 − 3 × 4

= 9 − 12

= − 3

(vi)

a2 + b2 − c2 + d2

= (3)2 + (0)2 − (2)2 + (1)2

= 9 + 0 − 4 + 1

= 6

(vii)

2a2 − 3b2 + 4c2 − 5d2

= 2 (3)− 3 (0)2 + 4 (2)2 − 5 (1)2

= 2 × 9 − 0 + 4 × 4 − 5 × 1

= 18 − 0 + 16 − 5

= 34 − 5

= 29

#### Question- 5.

Find the value of 5x2 – 3x + 2, when x = 2.

5x2 – 3x + 2

= 5(2)2 − 3(2) + 2

= 5(4) − 6 + 2

= 20 + 2 − 6

= 16

#### Question -6.

Find the value of 3x3 – 4x2 + 5x – 6, when x = -1.

3x3 – 4x2 + 5x – 6

= 3(− 1)− 4(− 1)2 + 5(− 1) − 6

= 3 × − 1 − 4 × 1 − 5 − 6

= − 3 − 4 − 5 − 6

= − 18

#### Question -7.

Show that the value of x3 – 8x2 + 12x – 5 is zero, when x = 1.

x3 – 8x2 + 12x – 5

= (1)3 − 8(1)+ 12(1) − 5

= 1 − 8 + 12 − 5

= 13 − 13

= 0

Hence proved.

#### Question- 8.

State true and false :
(i) The value of x + 5 = 6, when x = 1
(ii) The value of 2x – 3 = 1, when x = 0
(iii) $\frac { 2x-4 }{ x+1 }$ = -1,when x = 1

(i) True.

Verification x + 5 = 6,

When x = 1, 1 + 5 = 6

6 = 6

Hence proved

(ii) False.

Verification 2x − 3 = 1,

When x = 0

2 × 0 − 3 = 1

=  0 − 3

= − 3

−3 = 1

(iii) True.

Verification (2x-4 )/(x+1) = − 1,

When x = 1 = (2×1-4)/( 1+1 ) = − 1

= (2-4) / 2 =- 1= -2/2=-1

= − 1

#### Question- 9.

If x = 2, y = 5 and z = 4, find the value of each of the following :

(i)

(ii)

(iii)
zx = 42 = 4 × 4 = 16
(iv)
yx = 52 = 5 × 5 = 25
(v)

= (2)2−1 × (5)2 × (4)2−1

= 2 × 5 × 5 × 4

= 200

(vi)

= 5 × 2 × 25 × 16

= 4,000

(vii)

(viii)

= x2−1yx

= xyx

= (2) (5)2

= 2 × 25

= 50

#### Question- 10.

If a = 3, find the values of a2 and 2a.

a2 = (3)2 = 3 × 3 = 9

2a = (2)3 = 2 × 2 × 2 = 8

#### Question -11.

If m = 2, find the difference between the values of 4m3 and 3m4.

4m3 = 4 (2)3 = 4 × 2 × 2 × 2 = 32

3m4 = 3 (2)4 = 3 × 2 × 2 × 2 × 2 = 48

Now, a difference 3m4 – 4m3 = 48 – 32 = 16

### Substitution (Including Use of Brackets as Grouping Symbols) ICSE Class-6th Concise Mathematics Selina Solutions  Exercise – 20 B

Question-1

Evaluate :

(i) (23 – 15) + 4
(ii) 5x + (3x + 7x)
(iii) 6m – (4m – m)
(iv) (9a – 3a) + 4a
(v) 35b – (16b + 9b)
(vi) (3y + 8y) – 5y

(i) (23 – 15) + 4 = 8 + 4 = 12
(ii) 5x + (3x + 7x) = 5x + 10x = 15x
(iii) 6m – (4m – m) = 6m – 3m = 3m
(iv) (9a – 3a) + 4a = 6a + 4a = 10a
(v) 35b – (16b + 9b) = 35b – 25b = 10b
(vi) (3y + 8y) – 5y = 11y – 5y = 6y

#### Question- 2.

Simplify :

(i) 12x − (5x + 2x)

= 12x − 7x = 5x

(ii) 10m + (4n − 3n) − 5n

= 10m + n − 5n

= 10m − 4n

(iii) (15b − 6b) − (8b + 4b)

= 9b − 12b

= − 3b

(iv) − (− 4a − 8a)

= − (− 12a)

= 12a

(v) x − (x − y) − (− x + y)

= x − x + x + y − y

= x

(vi) p + (− q − r − s) − (p − q − r)

= p − q − r − s − p + q + r

= p − p − q + q − r + r − s

= − s

(vii) (a + b) − (c + d) − (e − f)

= a + b − c − d − e + f

(viii) 3x + (8x − 5x) − (7x − x)

= 3x + 3x − 6x

= 6x − 6x

= 0

(ix) a − (a − b − c)

= a − a + b + c

= b + c

(x) 6a2 + (2a2 − a2) − (a2 − b2)

= 6a+ a− a2 + b2

= 6a2 + b2

(xi) 2m − (3m + 2n − 6n)

= 2m − 3m − 2n + 6n

= − m + 4n

= 4n − m

(xii) − m − n − (− m) − m

= − m − n + m − m

= − m − n

(xiii) x + y − (x + y-x¯)

= x + y − (x + y − x)

= x + y − x − y + x

= x − x + x + y − y

= x

(xiv) 25y − (5x − 10y + 6x − 3y)

= 25y − 5x + 10y − 6x + 3y

= 25y + 10y + 3y − 5x − 6x

= 38y − 11x

(xv)

= 3x + (2x − x − 2)

= 3x + 2x − x − 2

= 4x − 2

(xvi)

= a − (2a − 4a − 3a)

= a − 2a + 4a + 3a

= 8a − 2a

= 6a

(xvii)

= 5x2 − (3x − x2 + 4)

= 5x2 − 3x + x− 4

= 5x2 + x2 − 3x − 4

= 6x2 − 3x − 4

(xviii)

= − (y − x) − (x + y − 2x − y)

= − y + x − x − y + 2x + y

= x − x + 2x − y − y + y

= 2x − y

Question -3.

Simplify :

(i) x − (y − z) + x + (y − z) + y − (z + x)

= x − y + z + x + y − z + y − z − x

= x + x − x − y + y + y + z − z − z

= x + y − z

(ii) x − [y + {x − (y + x)}]

= x − [y + {x − y − x}]

= x − [y + x − y − x]

= x − y − x + y + x

= x − x + x − y + y

= x

(iii) 4x + 3 (2x − 5y)

= 4x + 6x − 15y

= 10x − 15y

(iv) 2 (3a − b) − 5 (a − 3b)

= 6a − 2b − 5b + 15b

= 6a − 5a + 15b − 2b

= a + 13b

(v)

= p + 2 (q − r − p)

= p + 2q − 2r − 2p

= 2q − 2r − p

(vi)

= a − [− {− (a − b + c)}]

= a − [− {− a + b − c}]

= a − [+ a − b + c]

= a − a + b − c

= b − c

(vii) 3x − [5y − {6y + 2 (10y − x)}]

= 3x − [5y − {6y + 20y − 2x}]

= 3x − [5y − 6y − 20y + 2x]

= 3x − 5y + 6y + 20y − 2x

= 3x − 2x + 6y + 20y − 5y

= x + 21y

(viii)

= 5 {a2 − a (a − a + 2)}

= 5 {a2 − a2 + a2 − 2a}

= 5a2 − 5a+ 5a2 − 10a

= 5a2 − 10a

### Substitution (Including Use of Brackets as Grouping Symbols) Exercise – 20 C

ICSE Class-6th Concise Mathematics Selina Solutions

#### Question-1

Fill in the blanks :

(i) 2a + b − c = 2a + (b − c)

(ii) 3x − z + y = 3x − (z − y)

(iii) 6p − 5x + q = 6p − (5x − q)

(iv) a + b − c + d = a + (b − c + d)

(v) 5a + 4b + 4x − 2c = 4x − (2c − 5a − 4b)

(vi) 7x + 2z + 4y − 3 = − 3 + 4y + (7x + 2z)

(vii) 3m − 2n + 6 = 6 − (2n − 3m)

(viii) 2t + r − p − q + s = 2t + r − (p + q − s)

#### Question- 2.

Insert the bracket as indicated :

(i) x − 2y = − (2y − x)

(ii) m + n − p = − (p − m − n)

(iii) a + 4b − 4c = a + (4b − 4c)

(iv) a − 3b + 5c = a − (3b − 5c)

(v) x2 − y2 + z2 = x2 − (y2 − z2)

(vi) m2 + x2 − p2 = − (p2 − m2 − x2)

(vii) 2x − y + 2z = 2z − (y − 2x)

(viii) ab + 2bc − 3ac = 2bc − (3ac − ab)

### Revision Exercise of Substitution

(Including Use of Brackets as Grouping Symbols)

ICSE Class-6th Concise Selina Solutions Chapter-20

#### Question-1

Find the value of 3ab + 10bc – 2abc when a = 2, b = 5 and c = 8.

a = 2, b = 5, c = 8

∴ 3ab + 10bc − 2abc

= 3 × 2 × 5 + 10 × 5 × 8 − 2 × 2 × 5 × 8

= 30 + 400 − 160

= 430 − 160

= 270

#### Question -2.

If x = 2, y = 3 and z = 4, find the value of 3x2 – 4y2 + 2z2.

x = 2, y = 3, z = 4

∴ 3x2 − 4y2 + 2z2

= 3 (2)− 4 (3)2 + 2 (4)2

= 3 × 4 − 4 × 9 + 2 × 16

= 12 − 36 + 32

= 12 + 23 − 36

= 44 − 36

= 8

#### Question- 3. Substitution ICSE Class-6th Concise

If x = 3, y = 2 and z = 1; find the value of:
(i) xy
(ii) yx
(iii) 3x2 – 5y2
(iv) 2x – 3y + 4z + 5
(v) y2 – x2 + 6z2
(vi) xy + y2z – 4zx

(i) x = 3, y = 2, z = 1

xy = 32 = 3 × 3 = 9

(ii) x = 3, y = 2, z = 1

yx = 23 = 2 × 2 × 2 = 8

(iii) x = 3, y = 2, z = 1

3x2 − 5y2 = 3(3)2 − 5(2)

= 3 × 9 − 5 × 4

= 27 − 20

= 7

(iv) x = 3, y = 2, z = 1

2x – 3y + 4z + 5

= 2 × 3 − 3 × 2 + 4 × 1 + 5

= 6 − 6 + 4 + 5

= 15 − 6

= 9

(v) x = 3, y = 2, z = 1

y2 – x2 + 6z2

= (2)2 − (3)2 + 6(1)2

= 4 – 9 + 6 × 1

= 4 – 9 + 6

= 10 – 9

= 1

(vi) x = 3, y = 2, z = 1

xy + y2z – 4zx

= 3 × 2 + (2)× 1 − 4 × 1 × 3

= 6 + 4 − 12

= 10 − 12

= − 2

#### Question- 4.

If P = -12x2 – 10xy + 5y2, Q = 7x2 + 6xy + 2y2, and R = 5x2 + 2xy + 4y2 ; find :
(i) P – Q
(ii) Q + P
(iii) P – Q + R
(iv) P + Q + R

(i) P − Q

= (− 12x2 − 10xy + 5y2) − (7x2 + 6xy + 2y2)

= − 12x2 − 10xy + 5y2 − 7x2 − 6xy − 2y2

= − 12x2 − 7x2 − 10xy − 6xy + 5y2 − 2y2

= − 19x2 − 16xy + 3y2

(ii)  Q + P

= (7x2 + 6xy + 2y2) + (– 12x2 – 10xy + 5y2)

= 7x+ 6xy + 2y2 – 12x2 – 10xy + 5y2

= 7x – 12x2 + 6xy – 10xy + 2y2 + 5y2

= – 5x2 – 4xy + 7y2

(iii) P – Q + R

= (– 12x2 – 10xy + 5y2) – (7x2 + 6xy + 2y2) + (5x2 + 2xy + 4y2)

= – 12x2 – 10xy + 5y2 – 7x– 6xy – 2y2 + 5x2 + 2xy + 4y2

= – 12x2 – 7x2 + 5x2 – 10xy – 6xy + 2xy + 5y2 – 2y2 + 4y2

= – 14x2 – 14xy + 7y2

(iv) P + Q + R

= (– 12x2 – 10xy + 5y2) + (7x2 + 6xy + 2y2) + (5x2 + 2xy + 4y2)

= – 12x+ 7x2 + 5x2 − 10xy + 6xy + 2xy + 5y2 + 2y+ 4y2

= 0 − 2xy + 11y2

= − 2xy + 11y2

#### Question -5.

If x = a2 – bc, y = b2 – ca and z = c2 – ab ; find the value of :
(i) ax + by + cz
(ii) ay – bx + cz

(i) ax +by + cz

= a (a2 − bc) + b (b2 − ca) + c (c2 − ab)

= a3− abc + b3 − abc + c3 − abc

= a3 + b3 + c3 − 3abc

(ii) ay – bx + cz

= a (b2 − ca) − b (a2 − bc) + c (c2 − ab)

= ab2 − ca2 − a2b + b2c + c3 − abc

#### Question -6.

Multiply and then evaluate :
(i) (4x + y) and (x – 2y); when x = 2 and y = 1.
(ii) (x2 – y) and (xy – y2); when x = 1 and y = 2.
(iii) (x – 2y + z) and (x – 3z); when x = -2, y = -1 and z = 1.

(i) (4x + y) × (x − 2y)

= 4x (x − 2y) + y (x − 2y)

= 4x− 8xy + xy − 2y2

= 4x2 − 7xy − 2y2

Verification:

When x = 2, y = 1

L.H.S. = (4x + y) (x − 2y)

= (4 × 2 + 1) (2 − 2 × 1)

= (8 + 1) (2 − 2)

= 9 × 0

= 0

R.H.S. = 4x2 − 7xy − 2y2

= 4 (2)− 7 × 2 × 1 − 2 (1)2

= 4 × 4 − 14 − 2

= 16 − 16

= 0

∴ L.H.S. = R.H.S.

(ii) (x2 − y) × (xy − y2)

= x2 (xy − y2) − y (xy − y2)

= x3y − x2y2 − xy2 + y3

Verification:

When x = 1, y = 2

∴ L.H.S. = (x2 − y) (xy − y2)

= [(1)2 − 2] [1 × 2 − (2)2]

= (1 − 2) (2 − 4)

= − 1 × − 2

= 2

R.H.S. = x3y − x2y2 − xy2 + y3

= (1)× 2 − (1)2 (2)2 − 1(2)2 + (2)3

= 1 × 2 − 1 × 4 − 1 × 4 + 8

= 2 − 4 − 4 + 8

= 10 − 8

= 2

∴ L.H.S. = R.H.S.

(iii) (x − 2y + z) × (x − 3z)

= x (x − 3z) − 2y (x − 3z) + z (x − 3z)

= x2 − 2zx − 2xy + 6yz − 3z2

Verification:

When x = − 2, y = − 1, z = 1

L.H.S. = (x − 2y + z) × (x − 3z)

= [− 2 − 2 × (− 1) × 1] × [− 2 − 3 × 1]

= (− 2 + 2 + 1) × (− 2 − 3)

= 1 × (− 5)

= − 5

R.H.S. = x− 2zx − 2xy + 6yz − 3z2

= (− 2)2 − 2 (1) (− 2) − 2 (− 2) (− 1) + 6 (− 1) (1) − 3(1)2

= 4 + 4 − 4 − 6 − 3

= 8 − 13

= − 5

∴ L.H.S. = R.H.S.

#### Question -7. Substitution ICSE Class-6th Concise

Simplify :
(i) 5 (x + 3y) – 2 (3x – 4y)
(ii) 3x – 8 (5x – 10)
(iii) 6 {3x – 8 (5x – 10)}
(iv) 3x – 6 {3x – 8 (5x – 10)}
(v) 2 (3x2 – 4x – 8) – (3 – 5x – 2x2)
(vi)
(vii)

(i)  5 (x + 3y) – 2 (3x – 4y)

= 5x + 15y – 6x + 8y

= 5x – 6x + 15y + 8y

= – x + 23y

(ii) 3x – 8 (5x – 10)

= 3x – 40x + 80

= – 37x + 80

(iii) 6 {3x – 8 (5x – 10)}

= 6 {3x – 40x + 80}

= 18x – 240x + 480

= – 222x + 480

(iv) 3x – 6 {3x – 8 (5x – 10)}

= 3x – 6 {3x – 40x + 80}

= 3x – 18x + 240x – 480

= 243x – 18x – 480

= 225x – 480

(v)  2 (3x2 – 4x – 8) – (3 – 5x – 2x2)

= (6x2 – 8x – 16) – (3 – 5x – 2x2)

= 6x2 – 8x – 16 – 3 + 5x + 2x2

= 6x2 + 2x2 – 8x + 5x – 16 – 3

= 8x2 – 3x – 19

(vi)

= 8x − (3x − 2x + 3)

= 8x − 3x + 2x − 3

= 10x − 3x − 3

= 7x − 3

(vii)

= 12x− (7x − 3x− 15)

= 12x2 − 7x + 3x+ 15

= 12x2 + 3x2 − 7x + 15

= 15x− 7x + 15

#### Question- 8.

If x = -3, find the value of : 2x3 + 8x2 – 15.

∴ 2x3 + 8x2 – 15

putting x = -3

= 2 (– 3)3 + 8 (– 3)2 – 15

= 2 (– 27) + 8 (9) – 15

= – 54 + 72 – 15

= – 54 – 15 + 72

= – 69 + 72

= 3

End of Substitution ICSE Class-6th Concise Solutions :–

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