Ncert Solutions For Class 9 Maths Ex 2.5

Ncert Solutions For Class 9 Maths Chapter 2 Ex 2.5

Q.1.Use suitable identities to find the following products:

 

(i)(x + 4) (x + 10) 

Answer: (x+4)(x+10)=x2+(4+10)x+(4×10)=x2+14x+40
(ii)(x + 8) (x – 10)     

Answer: (x+8)(x10)=x2+(8+(10))x+(8×(10))=x2+(810)x80=x22x80

 

(iii)(3x + 4) (3x – 5)

Answer: (3x+4)(3x5)=(3x)2+4+(5)3x+4×(5)=9x2+3x(45)20=9x23x20
(iv)(y2+32)(y232)

Answer: (y2+32)(y232)=(y2)2(32)2=y494

 

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   103×107=(100+3)×(100+7)=(100)2+(3+7)(100+(3×7))=10000+1000+21=11021

 

(ii) 95 × 96  

Answer: 95×96=(90+5)×(90+6)=(90)2+90(5+6)+(5×6)=8100+990+30=9120

 

(iii) 104 × 96
Answer: 104×96=(100+4)×(1004)=(100)2(4)2=1000016=9984

 

Q.3. Factorize the following using appropriate identities:

(i)9x2+6xy+y2

Answer: 9x2+6xy+y2=(3x)2+(2×3x×y)+y2=(3x+y)2=(3x+y)(3x+y)

 

(ii) 4y24y+1

Answer: 4y24y+1=(2y)2(2×2y×1)+12=(2y1)2=(2y1)(2y1)

 

 (iii) x2y2100

Answer: x2(y10)2=(xy10)(x+y10)

 

Q.4. Expand each of the following, using suitable identities:

(i) (x+2y+4z)2

(ii) (2xy+z)2

(iii) (2x+3y+2z)2

 (iv) [14a12b+1]2

Answer:

(i) (x+2y+4z)2

Using identity,

(x+2y+4z)2=x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)=x2+4y2+16z2+4xy+16yz+8xz
(ii)  (2xy+z)2
Using identity,

(2xy+z)2=(2x)2+(y)2+z2+(2×2x×y)+(2×y×z)+(2×z×2x)=4x2+y2+z24xy2yz+4xz

 

(iii) (2x+3y+2z)2

Using identity,

(2x+3y+2z)2=(2x)2+(3y)2+(2z)2+(2×2x×3y)+(2×3y×2z)+(2×2z×2x)=4x2+9y2+4z212xy+12yz8xz
(iv) [14a12b+1]2
 Using identity,
[14a12b+1]2=(14a)2+(12b)2+12+(2×14a×12b)+(2×12b×1)+(2×1×14a)=116a2+14b2+114abb+12a

 

Q.5. Factorize:
(i) 4x2+9y2+16z2+12xy24yz16xz

(ii) 2x2+y2+8z222xy+42yz8xz
Answer:

(i) 4x2+9y2+16z2+12xy24yz16xz

Answer:
4x2+9y2+16z2+12xy24yz16xz

=(2x)2+(3y)2+(4z)2+(2×2x×3y)+(2×3y×4z)+(2×4z×2x)=(2x+3y4z)2=(2x+3y4z)(2x+3y4z)
(ii) 2x2+y2+8z222xy+42yz8xz
Answer:

2x2+y2+8z222xy+42yz8xz
=(2x)2+(y)2+(22z)2+(2×2x×y)+(2×y×22z)+(2×22z×2x)=(2x+y+22z)2=(2x+y+22z)(2x+y+22z)

 

Q.6. Write the following cubes in expanded form:

(i)(2x+1)3

(ii) (2a3b)3

(iii) [32x+1]3

(iv) [x23y]3

Answer:

(i)(2x+1)3

(2x+1)3=(2x)3+13+(3×2x×1)(2x+1)=8x3+1+6x(2x+1)=8x3+12x3+6x+1

 

(ii) (2a3b)3

(2a3b)3=(2a)3(3b)3(3×2a×3b)(2a3b)=8a327b318ab(2a3b)=8a327b336a2b+54ab2

 

(iii) [32x+1]3

[32x+1]3=(32x)3+13+(3×32x×1)(32x+1)=278x3+1+92x(32x+1)=278x3+1+2742+92x=278x3+2742+92x+1
(iv) [x23y]3
[x23y]3=(x)3(23y)3(3×x×23y)(x23y)=(x)3827y32xy(x23y)=(x)3827y32x2y+43xy2

 

Q.7. Evaluate the following using suitable identities: 

(i)(99)3

(ii)(102)3

(iii) (998)3

Answer:

(i) (99)3=(1001)3

(1001)3=(100)313(3×100×1)(1001)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300 = 970299


(ii) (102)3=(100+2)3
(100+2)3=(100)3+23+(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200= 1061208

 

(iii) (998)3=(10002)3

=(1000)323(3×1000×2)(10002)

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000 = 994011992

 

Q.8. Factorise each of the following:
(i)
8a3+b3+12a2b+6ab2

(ii) 8a3b312a2b+6ab2

(iii)27 – 125a3 – 135a + 225a2   

(iv) 64a327b3144a2b+108ab2

(v) 27p3121692p2+14p

Answer:

(i) 8a3+b3+12a2b+6ab2

8a3+b3+12a2b+6ab2=(2a)3+b3+3(2a)2b+3(2a)(b)2=(2a+b)3=(2a+b)(2a+b)(2a+b)
(ii) 8a3b312a2b+6ab2

8a3b312a2b+6ab2=(2a)3b33(2a)2b+3(2a)(b)2=(2ab)3=(2ab)(2ab)(2ab)
(iii)27 – 125a3 – 135a + 225a2   

27125a3135a+225a2=33(5a)33(3)2(5a)+3(3)(5a)2=(35a)3=(35a)(35a)(35a)

 

 (iv) 64a327b3144a2b+108ab2

64a327b3144a2b+108ab2=(4a)3(3b)33(4a)2(3b)+3(4a)(3b)2=(4a3b)3=(4a3b)(4a3b)(4a3b)
(v) 27p3121692p2+14p 

27p3121692p2+14p

=(3p)3(16)33(3p)2(16)+3(3p)(16)2=(3p16)3=(3p16)(3p16)(3p16)

 

Q.9. Verify:

(i) x3+y3=(x+y)(x2xy+y2)

(ii) x3y3=(xy)(x2+xy+y2)

Answer:

(i) x3+y3=(x+y)(x2xy+y2)Weknowthat,(x+y)3=x3+y3+3xy(x+y)x3+y3=(x+y)33xy(x+y)x3+y3=(x+y)[(x+y)23xy]Taking(x+y)commonx3+y3=(x+y)[(x2+y2+2xy)3xy]x3+y3=(x+y)(x2+y2xy)
(ii) x3y3=(xy)(x2+xy+y2)
x3y3=(xy)(x2+xy+y2)Weknowthat,(xy)3=x3y33xy(xy)x3y3=(xy)3+3xy(xy)x3y3=(xy)[(xy)2+3xy]Taking(x+y)commonx3y3=(xy)[(x2+y22xy)+3xy]x3+y3=(xy)(x2+y2+xy)

 

Q.10. Factorize each of the following:

 (i) 27y3+125z3

 (ii) 64m3343n3

Answer:

(i) 27y3+125z3

27y3+125z3=(3y)3+(5z)3=(3y+5z)[(3y)2(3y)(5z)+(5z)2]=(3y+5z)(9y215yz+25z)2
(ii) 64m3343n3

64m3343n3=(4m)3(7n)3=(4m+7n)[(4m)2+(4m)(7n)+(7n)2]=(4m+7n)(16m2+28mn+49n)2

 

Q.11. Factorise : 27x3+y3+z39xyz

Answer:

27x3+y3+z39xyz=(3x)3+y3+z33(3x)(y)(z)=(3x+y+z)(3x)2+y2+z23xyyz3xz=(3x+y+z)(9x2+y2+z23xyyz3xz)
Q.12. Verify that:

x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

Answer:

We know that,
x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzxz)x3+y3+z33xyz=12×(x+y+z)[2(x2+y2+z2xyyzxz)]=12(x+y+z)(2x2+2y2+2z22xy2yz2xz)=12(x+y+z)[(x2+y22xy)+(y2+z22yz)+(x2+z22xz)]=12(x+y+z)[(xy)2+(yz)2+(zx)2]

 

Q.13. If x + y + z = 0, show that x3+y3+z3=3xyz.

Answer:

We know that,
x3+y3+z3=3xyz= (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Now put (x + y + z) = 0,  
x3+y3+z3=3xyz=(0)(x2+y2+z2xyyzxz)x3+y3+z33xyz=0

 

Q.14. Without actually calculating the cubes, find the value of each of the following:
(i) (12)3+(7)3+(5)3

(ii) (28)3+(15)3+(13)3

Answer:

(i) (12)3+(7)3+(5)3
(12)3+(7)3+(5)3=0+3(12)(7)(5)(Since.(12)+(7)+(5)=0)UsingIdentity=1260
(ii) (28)3+(15)3+(13)3
x + y + z = 28 – 15 – 13 = 0

(28)3+(15)3+(13)3=0+3(28)(15)(13)=16380

 

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : 25a235a+12

(ii) Area : 35y2+13y12

Answer:

(i) Area : 25a235a+12
25a235a+12=25a215a20a+12=5a(5a3)4(5a3)=(5a4)(5a3)

Possible expression for length = 5a – 4
Possible expression for breadth = 5a – 3

 

(ii) Area : 35y2+13y12

35y2+13y12=35y215y+28y12=5y(7y3)+4(7y3)=(5y+4)(7y3)

Possible expression for length = (5y + 4)
Possible expression for breadth = (7y – 3)

 

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 

(i) Volume : 3x212x
(ii) Volume : 12ky2+8ky20k

Answer:

(i) Volume : 3x212x
3x212x
= 3x(x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

 

(ii) Volume : 12ky2+8ky20k

12ky2+8ky20k=4k(3y2+2y5)=4k(3y2+5y3y5)=4k[y(3y+5)1(3y+5)]=4k(3y+5)(y1)

Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1) 

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