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Class 7  RS Aggarwal Maths Solutions

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5.        Exponents 5A

6.        ALGEBRIC EXPRESSIONS 6a, 6B, 6C, 6D

 

 

EXERCISE – 5A.    EXPONENTS

Q1.         Write in power notation:

                [ii]  x x x x  =                                      

Q2.         [i]                [iv]   

Q3.         Express as a rational number:

                [i]            [iii]   

Q4.         Express as a rational number:

                [i] (4)-1  =    .      [ii]   (-6)-1 = .         [iv] 

Q5.         Find the reciprocal:

                [i]            [iv]   (-4)3  =                                                                                                                                     

Q6.         40 + 50 = 1 + 1 = 2.       Iv]    60  x 70 = 1x 1 = 1.

Q7.         [i]   x  x x x         

[iii]    

[v]  .

      x 16  = x 16 =

 

Q8.         [i]  x  =   =  =  .

[ii]    x   =

                               

Q9.         Express as a rational no.:

                [i]  5-3 =     ii] (-2)-5  =    

[Vii]              

[ix] =   

 

Q10.       [i]                

                [iii]    =      

                [iv]     

 

Q11.       (-5)-1 x y = (8)-1  =>  

 

Q12.       3-3 x y = 4  =>

 

Q13.       (-30)-1

 

Q14.      

                    =>X = -1.

 

Q15.      

 

Q16.      

 

Q17.       [i]       52n x 53 = 59 => 52n +3 =59  =>2n + 3 = 9  => 2n = 9 – 3 = 6.  => n =

                [ii]      8 x 22n +2 => 23 x 2n+2 = 25 => 2n+5 = 25  => n + 5= 5  => n = 1.

                [iii]   62n +1 x  =>  ]   62n +1x 6 -2 = 63  => 62n – 1 = 63   => 2n -1  = 3   => n =  = 2.

 

Q18.       2 n – 7 x 5n – 4 = 1250.  => 2n – 4 x 2-3  x 5n – 4 = 1250.   =>

π  (2 x 5)n – 4 = (10)4 =>  n- 4 = 4   =>  n = 8.

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EXERCISE – 6A   ALGEBRIC EXPRESSIONS                                       

Q1.         ADD:

                [i]           5X + 7X + 6X = 18X

                [ii]         

                [iii]         4a2 b.

                [iv]        

                [v]          4x – 4y – 7z.

                [vi]         2x2 -3y2 + 5x2 +6y2 – 3x2 – 4y2  => 4x2 – y2.

                [ix]          + +                               

                               

                [x]       

                           +  +      

π                                                        

                        .

Q2.         Subtract:

                [i]           7xy +8xy = 15xy                                [ii] – 3x2 -x2 = -4x2   

                [iv]         a 2+ b2 + 2ab –a2 – b2 + 2ab = 4ab.

Q3.         (a+3b - 4c)+(4a – b + 9c)+(-2b +3c -a)= a+3b - 4c+ 4a – b + 9c -2b +3c –a

= a + 3b - 4c + 4a –b +9c – 2b + 3c –a = 4a + 8c – 2a + 3b - 4c = 2a + 3b + 4c.

Q4.         8m  –  7n  +  6p2

                -3m - 4n  -    p2

-----------------------------------------

               5m  - 11n + 5p2                                                   2m + 4n – 3p2

                                                                                                -m  - n    -   p2

                                                                                                -----------------------

                                                                                                M +  3n – 4p2

-5m  -11n +5p2

(-)      (+)    (-)

----------------------

-4m +14n – 9p2

 ----------------------

Q5.         8a     – 6a2 + 9

                -10a + 8a2  - 8

                ----------------------

                -2a   + 2a2  + 1                                                 = - 3 –(-2a   + 2a2  + 1) = -3 + 2a -2a2 – 1.

                                                                                            = - 2a2  +  2a  -4.

                                                                                                                               

Q6.         [ii]          x2 –x -  

                [iv]                      

 

                                =   - +            

                                =   -  +    = 

-------------------------X--------------------------

 

EXERCISE – 6B     

                FIND THE PRODUCTS:

Q1.         3a2 x 8a4 = 24a2+4 = 24 a6.         Q3.  (-4ab) x (-3a2 bc)  = 12 a3 b2 c .

 

Q5.         .           

 

Q12.       2a2 b x -5 ab2 c x – 6 b c2                  =  - 10 a3 b3c – 6 b c2 = 60 a3 b4 c3.

 

Q14.      

Q18.      

Verification:  a = 2, b = 3.  

LHS =

RHS = .  Thus, LHS = RHS.

Q20.       2.3 a5 b2 x 1.2 a2 b2  = 2.76 a7 b4.      Verification:  a= 1, b = 0.5.

                LHS = 2.3 x 1 x 0.5 x 0.5 x 1.2 x 1 x 0.5 x 0.5 = 0.1725.

                RHS = 2.76 x 1 x 0.5 x 0.5 x 0.5 x 0.5 = 0.1725.                        Thus, LHS = RHS.

Q24.      

                Verification:  a= 1, b = 2, c = 3.

                LHS = 3 =

               

                RHS = Thus, LHS = RHS.

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EXERCISE – 6C   MULTIPLICATION OF A MONOMIAL AND BINOMIAL   

DISTRIBUTIVE LAW OF MULTIPLICATION   P X (Q + R) = (P XQ)+(PXR).

Q1.         4a (3a + 7b)  =  12a2 + 28ab .

Q11.        x       

 = .

Q16.       24x2 – 48x3.     Evaluate: x = 2,  24 x 22(1 – 2 x 2) =96 – 384= -288.

                24x2 – 48x3 = 24 x 2 x 2 – 48 x 2x2x2 = 96 -384 = -288.

                Simplify:

Q20.       ab – ac + bc – ba + ca - cb   =  ab – ab – ac + ca + bc – bc = 0.

Q21.       Ab – ac – bc + ab –ac –ac + bc = ab + ab – ac – ac = 2ab – 2ac.

Q22.       3x2 + 2x +4 – 6x2 -3x = -3x2 –x + 4.

Q23.       X2 + 4x + 3x4 -3x + 4x2 + 4 =   6x2 + 5x2 + x + 4.

Q24.       2x2 + 3x – 6x4 + x2 + x   = 3x2 - 6x4 + 4x.

Q25.       A3b – a2 b3 + 4a2 b3 – 2a2 b2 – a3 b + 2a3 b2  = 3a2 b3 .

Q26.       4s2t – 4st2 – 6s2t + 6s2 t2 – 6t2 s2 + 3t2 s + 2s2t – 2st2 

= 4s2t– 6s2t+ 2s2t– 4st2 +3st2– 2st2 =  - 3st2.

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EXERCISE -6D  MULTIPLICATION OF TWO BINOMIALS

Q1.         (5X+7)(3X+4)

                = 5X(3X+4)+7(3X+4)

                =15X2=20X+21X+28

                =15X2+41X+28

Q7          (3m-4n)(2m-3n)

                =6m2-9mn-8nm+12n2

                =6m2-17mn+12n2

Q8          (4a+3b)

               =(4a+3b)+(4a+3b)

               = x 4a +  x 3b +  x 4a+ x 3b

               =3a2 +++ 2b2

                =3a2+ + 2ab

                =3a2+ +2b2

Q17         =   x4  +

Q25.       (x2 – xy + y2) (x + y) = x2(x + y) – xy(x + y ) + y2(x + y) = x3 + x2y – x2y – xy2 + y2x + y3.

                X3 + y3.

Q27.       (3x + 4)(2x - 3)+ (5x - 4)(x + 2)  = 3x(2x - 3)+ 4 (2x - 3)+ 5x(x + 2) – 4(x + 2).

                6x2 – 9x + 8x – 12 + 5x2 + 10x – 4x – 8 . = 11x2 – 13x + 18 – 20 = 11x2 + 5x -20.

 

Q31.       (3x2 + 5x - 7)(x - 1) - (x2 -2x + 3)(x + 4).

                = [3x2(x - 1) + 5x (x - 1) - 7(x - 1)] – [x2 (x + 4) – 2x (x + 4) +3 (x + 4)].         

                =[ 3x3 – 3x2 + 5x2 – 5x -7x +7] –[ x3 + 4x2 -2x2 - 8x + 3x + 12].

                =  3x3 – 3x2 + 5x2 – 5x -7x +7 – x3 - 4x2 +2x2 + 8x - 3x - 12.

                = 3x3 + 2x2  - 12x + 7 – x3 – 2x2 + 5x – 12  =   2x3 -7x – 5 .

                OR

                3x2 + 5x – 7

                x – 1

                ------------------------

                3X3 + 5x2   -   7x

                      -3x2       -  5x   +  7

                -------------------------------

                3x3 + 2x2  - 12x  +  7

                --------------------------------          

               

x2   -  2x  +  3

                x + 4

-----------------------------------

                X3   –  2x2 + 3x

                      +  4x2  - 8x + 12

                -------------------------------------

                x3   + 2x2 -  5x +12 

                -------------------------------------

               

3x3 + 2x2  - 12x + 7 –( x3   + 2x2 -  5x +12  )

= 3x3 + 2x2  - 12x + 7 – x3 – 2x2 + 5x – 12  = 2x3 -7x – 5.

-----------------------------------------------------------------X--------------------------------------------------------

EXERCISE – 7A LINEAR EQUATION IN ONE VARIABLE

Solve the equation:

Q1.         3x – 5 = 0 => 3x = 5  => x =

Q2.         8x – 3 = 9 – 2x  => 8x + 2x = 9 + 3  => x = .

Q3.         7 – 5x = 5 – 7x => 5x + 7x = 5 – 7 => 12x = -12  => x =

Q4.         3 == 2x = 1 – x   => 2x + x = 1 – 3  => 3x = -2  => x = .

Q5.         2(x -2) + 3(4x - 1)= 0   => 2x -4 + 12x -3 = 0  => 2x + 12x = 4 +3  => x = .

Q6.         5(2x - 3) -3(3x - 7)= 5  => 10 x -15 – 9x +21 = 5 => x + 6 = 5 => x = 5-6 = 1.

Q7.         2x -

Q8.          =>

Q9.          =   =

Q10.       3x + 2x +4 = 20 – 2x +5  =>7x = 21   => x = 3.

Q11.       13y – 52 – 3y +27 -5y -20 = 0   => 5y = -45   =>y = -9.

Q12.       2m + 5 = 3(9m -10)= 9m – 30  => 2m – 9m = -30 -5 => -7m = -35 => m = 5.

Q13.       18x + 12 -30x + 5 = 3x -24 -35x + 30 + 9x => 17 – 12x = -23x + 6  => 11 = -11x => x = -1.

Q14.       T – 2t – 5 -5 + 10t = 6 + 8t – 3t + 12 => 9t -10 = 18 + 5t  => 4t = 28  => t = 7.

Q15.      

Q16.       = 3  =>  21x – 7 -5x = 3 x 35  => 16x = 105 + 7 = 122  => x =                                                                                         

----------------------X-----------------------

EXERCISE – 7B

Q1.         2x – 7 = 45  => 2x = 45 +7   => x =

Q2.        3x + 5 = 44  => 3x = 44 – 5  => x =

Q3.         4 + 2x = 2x =

Q4.         X +

Q5.         X +  = 55   => = 55   => 5x = 55 x 3   => x =

Q6.         X + 4 – x = 45   => 4x – x = 45  => 3x = 45  => x =

Q7.         (x - 21)=(71 - x)  => x + x = 71 + 21   =>2x = 92   => x =

Q8.         X -

Q9.         X +  and

Q10.      

Q11.      

Q12.       X + (x + 1)=63  => x + x +1 =63   => 2x = 62   => x = 31    and   x + 1 = 31.

Q13.       X + (x + 2)=76  => 2x + 2 = 76  =>  2x = 76 – 2 = 74  => x =

Q14.       x + (x+ 2)+(x +4)= 90   => 3x + 6 = 90    =>  x =

Q15.      

π  10x = 168 + 552 = 720   => x = 72. Next part  (184 - 72)= 112.

Q16.       5x + 900 – 10x = 500  => -5x = - 400  =>  x = 80. Then, 90 – 80 = 10.

Q17.       50x + 2(25x) 34 x 100P  => 50x + 50x = 3400P  => x =. => x = 34.

π  No. of 50P coin = 34. And no. of 25P coin = 34 x 2 = 68.

Q18.       Let cousin age = y . then Raju’s age =  y – 19.

Now , after 5 yrs their age will be y +5 & y -19 +5. Now, ATQ,

Hence, cousin’s age = 52, & Raju’s age = 52 – 19 = 33.

Q19.       Let son’s age = y, then   Father’s age = y + 30.

After 12yrs,  Son’s age= y + 12 & father’s age = (y + 30)+ 12.

 y + 42 = 3(y + 12)  => y + 42 = 3y +36  => 42 – 36 = 3y – y => 2y = 6 => y = 3.

 son’s age = 3yrs. & father’s age = 3 + 30 = 33yrs.           

Q20.       Let the sonal’s age = 7y & Manoj’s age = 5y. then ratio = . after 10 yrs,

               

                 Present age of sonal = 7y = 7 x5 = 35yrs. & Manoj’s age = 5 x 5 = 25yrs.

Q21.       Let present age of son be y yrs.   5 yrs ago.   Age of son = (y - 5)yrs.

                ATQ, Age of Father = 7(y - 5) = 7y – 35.

                Present age of Father = 7y – 35 +5 = 7y – 30 ----------------(1)

                After 5 yrs            Age of son = (y + 5)yrs ,  & age of Father = 3(y + 5) =3y + 15.

                  Present age of Father = 3y + 15 – 5 = 3y + 10 -------------------(2)

                 7y – 30 = 3y + 10  => 4y = 40 => y = 10.

                 present age  of son = y = 10 yrs.  & present age of Father = 3y + 10 = 30 + 10 = 40 yrs.

Q22.       Let the present age of Manoj = x yrs.   After 12 yrs, Age of Manoj = (x + 12 yrs)

4 yrs ago, Age of Manoj = (x - 4) yrs.  ATQ,  (x + 12) = 3 (x - 4)  => x + 12 = 3x – 12.

π  X – 3x = -24    => - 2x = -24  => x = 12.   present age og Manoj = 12 yrs.

Q23.       Let total marks = y. then ATQ, 40% of y = 185 + 15 .  =>

                Y = = 500.   Total marks = 500.

Q24.       Let ones digit = y . Then tens digit = 8 – y.    no. y + 10(8 - y) = y + 80 – 10y = 80 – 9y.

                Also ATQ, 80 – 9y + 18 = 10y + 8 – y  => 98 – 9y = 9y + 8  => 18y = 90. => y = 5.

                 tens digit = 8 – 5 = 3.     no. = 10 x 3 + = 35.

                OR  (alternate method)

                Let the no. be 10x + y ATQ, x + y = 8  ------------- (1)

                Also no. + 18 = reversed no.  => 10 x + y + 18 = 10y + x  => 9x – 9y = -18.

π  X – y = -2. => x = y -2.    From (1)=> y – 2 +y = 8. => 2y = 10 => y = 5.

 (1) => x = 8 – 5 = 3.    no. = 10 x + y = 10 x 3 x 5 = 35.

Q25.       Let price of 1 chair = Rs. x . Then, price of ! table = Rs(x + 75).

                Now, price of 3 tables + 2 chair = Rs. 1850.  => 3(x + 75) + 2x = 1850.

π  3x + 2x = 1850 – 225.  =. 5x = 1625.   => x =

 price of 1 chair = x = Rs. 325. & price of 1 table = x + 5 = 325 + 75 = Rs. 400.

Q26.       We know, CP =                

                OR,     let CP = Rs.y. Then,   y + y x 105 = 495.  Y +

                110y = 495 x 100    => y =  = Rs. 450.

------------------------------X---------------------------------

EXERCISE – 8A     RATIO AND PROPORTION

Q1 – Q6

Q7.         Sum of ratio terms = (7 + 8)= 15.

                Kunal’s share = Rs.

Q8.         Sum of ratio terms = () =  .

                Rajan’s share = Rs.

                Kamal’s share =

Q9.         Sum of ratio terms = (1 + 3 + 4)= 8.

                A’s share =    B’s share =

                C,s share =

Q10.      

                3x – 2x = 32 – 27     x = 5.

Q11.        =

π  -15x + 7x = 231 – 255   => -8x = 624  => x =     

Q12.      

               

Q13.        No. =3 x 5 = 15 & 3 x 9 = 27.

Q14.       .   LCM of 3y & 4y = 12y.    => 12y = 180  => y = 15.

               

Q15.       A’s age = 8y & B’s age = 3y    => 

π   5y = 30   => y = 6.      A’s age = 6 x 8 = 48 & B’s age = 6 x 3 = 18.

Q16.       Let the alloy’s weight be y gm.  now, sum of the ratio = 9 + 5 = 14.

                ATQ,   

                  weight of Zinc in the alloy = 75.6 – 48.6 = 27gm.

Q17.       Let he no. of boys = 8y & no. of girls = 3y. ATQ, 3y = 375  => y = 125.

                no. of boys = 8 x 125 = 1000.

Q18.       Let monthly Income = 11y.  &  Saving =2y.  ATQ, 2y = 2500  => y = 1250.

                Monthly Income = 11 x 1250 = Rs.13750.

               

Q19.       Let the no. of Re.1, 50P and 25P coins be 5y : 8y : 4y respectively.

                Then,  =Rs. 750            

π  5y + 4y +y = 750  => y = 75.   5y = 5 x 75 = 375. 8y = 8 x 75 =600,  4y = 4 x 75 = 300.

Hence, the no. of 1Re, 50p, & 25P coins are 375, 600 & 300 respectively.

Q20.      

π  68x – 39x = 143 – 85   => 29x = 58   => x = 2.

Q21.       X : y = 3 : 4   =>

           =

Q22.     X : y = 6 : 11  =>

π 

Q23.     Let the two nos. = 5x, 7x.  ATQ,   5x + 7x = 720  => x = 60

                 1st no. = 5x = 5 x 60 = 300.

                And 2nd no. = 7x = 7 x 60 = 420.

Q24.     Which ratio is greater?

i)                   (5 : 6) or (7 : 9)  =  

  (5 : 6) > (7 : 9).   [LCM of 6,9 = 18]

 

ii)                              

     (2 : 3)>(4 : 7).  [LCM of 3 & 7 =21]

 

iii)                 (1:2)<(4:7)

 

iv)               Hints:  .

Q25.     (i)            [LCM of 6,9,18= 18]

(17 :21).

-----------------------X-----------------------------------

EXERCISE – 8B   

Q1.         We have: product of extremes = (30x 60 ) = 1800.

                Product of means = (40 x 45) = 1800.  product of extremes = product of means.

                Hence, 30, 40, 45, 60 are in proportion.

Q2.         product of extremes = (7x 36 )= 252.       Product of means = (6 x 49)= 294. 

                 product of extremes product of means. Hence, 36,49,6,7 are not in proportion.

Q3.         Given,   2: 9 :: x : 27.  But, prod of extremes = prod of means. 

                 2 x 27 = 9 x X    => X = 

Q4.         X x 16 = 8 x 35   => X =

Q5.         X x 60 = 35 x 48    =>  X =  = 28.

Q6.         Let the no. be y, then ATQ,  8 : 36 :: 6 x Y.    =>  8 x y = 36 x 6  => y =

Ii] Let the no. be y, then ATQ, 5 : 7 :: 30 : y   => 5 xy = 30 x 7  => y =

Iii]  hints: 2.8 x y = 14 x 3.5  => y =

Q7.         Since, 36, 54, X are in continued proportion, we have:

                36 : 54 :: 54 : X   => 36 x X = 54 x 54   [since, POE = POM]  =>  X =

Q8.         27 : 36 :: 36 : X  => 27 x X = 36 x 36 => X =

Q9.         Let the third proportional to 8 and 12 be X. Then, 8 : 12 :: 12 : X

π  8 x X = 12 x 12  [since POE = POM]  =>   X =

Ii] hints: 12 : 18 :: 18 : X => 12 x X = 18 x 18 => X = 27.

Iii] hints: 4.5 : 6 :: 6 : X => 4.5 x X = 6 x6 => X = 8.

Q10.       ATQ, 7:X :: X: 28 => 28 x 7 = X x X  => 7x 2 x 2 x 7 = X2  => X = 7 x 2 = 14.

Q11.       Let  the mean proportion be X. then, ATQ,

                I] 6 : X :: X : 24  => X x X = 6 x 24 => X2 = 144    => X = 12.

                Ii]   3: X :: X : 27   => X x X = 3 x 27   => X2 = 81   => X = 9.

                Iii]  0.4 : X::X : 0.9   => X x X = 0.4 x 0.9   => X2 = 0.36  =>  X2 = (0.6)2  => X = 0.6  .

Q12.       Let the no. to be added be y. Then, ATQ,

                (5 + y): (9 : y): (7 + y): (12 + y)  => (5 + y)x(12 + y)  = (7 + y)x (9 : y)   [since, POE = POM] 

π  60 + 5y + 12y + y2 = 63 + 9y + 7y+ y2  =>  60 + 17y = 63 +16y  => y = 3.

Q13.