Ncert Solutions For Class 7 Maths Ex 12.3

Ncert Solutions For Class 7 Maths Chapter 12 Ex 12.3

 

Q1: If a=2, find the values of:

(i) a-2

(ii) 3a-5

(iii) 9-5a

(iv) 3a22a7

(v) 5m24

 

Sol:

(i) a-2 =2-2  (Putting a=2)

=0

 

(ii)  3a-5= 3×25   (Putting a=2)

=1

 

(iii)   9-5a=95×2 (Putting a=2)

= -1

 

(iv)  3a22a7=3×222×27  (Putting a=2)

=12-4-7

=1

 

(v)  5m24=5×224=54   (Putting a=2)

=1

 

Q2: If x=-2, find

(i)  4x+7

(ii)  3x2+4x+7

(iii) 2x33x2+4x+7

 

Sol:

(i)  4x+7=4(-2)+7   (Putting x= -2)

= -8+7=-1

 

(ii)  3x2+4x+7=3(2)2+4(2)+7

= -3(4)-8+7=-12-8+7

= -13

 

(iii) 2x33x2+4x+7=2(2)33(2)2+4(2)+7  (Putting x= -2)

= -2(-8)-3(4)+4(-2)+7

=  16-12-8+7

=3

 

Q3: Find the value of the following expressions, when x= -1:

(i) 5x-35

(ii) -2x+4

(iii) 3x2+6x+3

(iv) 6x23x6  

Sol:

(i) 5x-35 = 5(-1)-35 =-5-35              [Putting x= -1 ]

= -40

(ii)  -2x+4  = -2(-1)+4            [Putting x= -1 ]

= 2 + 4 = 6

(iii) 3x2+6x+3 = 3(1)2+6(1)+3      [Putting x= -1 ]

= 3-6+3 =0

(iv) 6x23x6  = 6(1)23(1)6        [Putting x= -1 ]

= 6+1-6 =1

 

Q 4: If x=2, y= -2, find the value of:

(i) x2+y2  

(ii) x2+xy+y2  

(iii) x2y2

 

Sol:

(i) x2+y2 = 22+(2)2                        [Putting a=2,  b= -2 ]

= 4 + 4 = 8

(ii) x2+xy+y2  = 22+2(2)+(2)2        [Putting a=2,b= -2 ]

= 4 – 4 + 4 = 4

(iii) x2y2 = (2)2(2)2                        [Putting a=2, b= -2]

= 4 – 4 = 0

 

Q5: When x=0,y= -1, find the value of the given expressions:

(i) 2x+2y

(ii) 2x2+y2+1  

(iii) 2x2y+2xy2+xy

(iv) x2+xy+2

Sol:

(i) 2x+2y = 2(0)+2(-1)     [Putting x=0,y= -1 ]

= 0 – 2 = -2

(ii) 2x2+y2+1 = 2(0)2+(1)2+1   [Putting x=0, y=-1 ]

= 0 + 1 + 1 = 2

(iii) 2x2y+2xy2+xy =  2(0)2(1)+2(0)(1)2+0(1)      [Putting x=0, y= -1]

= 0 + 0 + 0 = 0

(iv) x2+xy+2 = (0)2+(0)(1)+2    [Putting x=0, y= -1 ]

= 0 + 0 + 2 = 2

 

Q6: Simplify the following expressions and find the value at a= 2:

(i) a+7+4(a-5)

 (ii) 3(a+2)+5a-7  

(iii) 10a+4(a-2)

(iv) 5(3a-2)+4a+8    

 

Sol:

(i) a+7+4(a-5) = a+7+4a-20

=4a+a+7-20 =5a-13

= 5(2)-13 =10-13                                                                                   [Putting a=2 ]

= -3

 

(ii) 3(a+2)+5a-7 = 3a+6+5a-7

= 3a+5a+6-7  = 8a-1

= 8( 2) – 1                                                                                               [Putting a=2 ]

= 16 – 1 = 15

 

(iii) 10a+4(a-2) = 10a+4a-8

= 14a-8

= 14( 2) – 8                                                                                            [Putting a= 2 ]

= 28 – 8 = 20

 

(iv) 5(3a-2)+4a+8 = 15a-10+4a+8

=15a+4a-10+8 = 19a-2                                                                           [Putting =2  ]

= 19(2)-2 = 38-2

= 36

 

Q7: Simplify the expression given below and find the value at x=3, y= -1, z= -2  :

(i) 8x-10-3x+5

(ii) 10-5x+3x+6

(iii) 5y+3-2y+6

(iv) 5-8z-12-4z

(v) 3y-5z-6x+15

Sol:

(i) 8x-10-3x+5 = 8x-3x-10+5

=5x-5 = 5(3)-5                                                                                                [Putting x=3 ]

= 15-5 = 0

(ii) 10-5x+3x+6 = 10+6-5x+3x

= 16-2x = 16-2(3)                                                                                           [Putting x= 3 ]

=  16-6 =10

(iii) 5y+3-2y+6 = 5y-2y+3+6

= 3y+9 = 3(-1)+9                                                                                             [Putting y= -1 ]

= -3 + 9 = 6

(iv) 5-8z-12-4z = 5-12-8z-4z

= -7-12 z                                                                                                         [Putting z= -2 ]

= -7 -12(-2) = -7+24

= 17

(v) 3y-5z-6x+15

= 3(-1)-5(-2)-6(3)+15                                                                       [Putting x=3, y=-1, z=-2]

= -3+10-18+15

= 25-21

= 4

 

Q8:

(i) If x= 10, find the value of x33x25x+15 .

(ii) If y= -10, find the value of 2y23y+50

 

Sol:

(i) x33x25x+15

= 1033(10)25(10)+15                                                   [Putting x=10 ]

=1000-300-50+15

= 665

 

(ii) 2y23y+50

=2(10)23(10)+50                                                            [Putting y= -10 ]

=200+30+50

=280

 

Q9: What should be the value of p if the value of 2a2+ap=5 equals to 5, when a=0 ?

Sol:

2a2+ap=5

2(0)2+0p=5                                                                         [ Putting x= 0 ]

p=5

Hence, the value of p is -5.

 

Q10: Simplify the expression and find its value when x= 5 and y= -3:   

2(x2+xy)+3xy.

Sol:

 

Given:

2(x2+xy)+3xy

 

2x2+2xy+3xy

 

2x2+2xyxy+3

 

2x2+xy+3

 

2(5)2+(5)(3)+3             [Putting x=5, y= -3 ]

 

2(25)+(15)+3

 

5015+3 38