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Class 8 R S Aggarwal Maths solutions

Contents

13.      Time and Work Ex 13A,

14.      POLYGONS Ex 14A

15.      QUADRILATERALS Ex 15

16.      PARALLELOGRAMSEx16A

18.      Area of trapezium and a polygonEx18A,B

 

 

Q3.      Time taken by (A+B) together to finish the work = 6 days.

            Time taken by A to finish the work = 9 days.

            Work done by (A+B) in 1 day =

                Work done by A in 1 day =   

            B’s 1 day work = { (A+B)’s 1 day work – A’s 1 day work }.

                                    =   -  =     =  .     www.rsmaths99.com

Hence, B alone can finish the work in 18 days.

 

Q4.      Time taken by Raju & Siraj to finish the work = 6 hrs.

            Time taken by Raju to finish the work  15 hrs.

            Work done by Raju + Siraj in 1 hr= .

                Work done by Raju in 1 hr = .      www.rsmaths99.com

            Work done by Siraj in 1 hr = {( Raju & Siraj 1 hr works) – Raju’s 1 hr work}

                                                      =    -  =  =   .

Hence, Siraj alone can finish the work in 10 hrs.

 

Q5.      Time taken by A to finish the work  = 10 days.

            Time taken by B to finish the work  = 12 “

            Time taken by C to finish the work  = 15 “

            Work done by (A+B+C) in 1 day = ( ) days.

=  =    =  .   Hence, A+B+C) can finish the work in 4 days.

 

Q6.      Time taken by A to finish the work  = 24 hrs

            Time taken by B to finish the work  = 16 hrs

            Time taken by A+B+ C to finish the work  = 8 hrs

            Work done by A in 1 hr =  hrs

               Work done byB in 1 hr =  hrs   www.rsmaths99.com

            Work done by A+B+C in 1 hr =   hrs

            Work done by (A+B) in 1 hrs = ( ) hrs =   =

            Work done by C alone =  -  =  =

            Hence, C can finish the work in 48 hrs.

 

Q7.      Time taken by A+B+ C to finish the work  = 8hrs       

               Time taken by A to finish the work  = 20 hrs

            Time taken by B to finish the work  = 24 hrs   
            Work done by (A+B) in 1 hr =

Work done by A in 1 hr =       www.rsmaths99.com

            Work done by B in 1 hr =

            Work done by A+B  together in 1 hr = + =  =

            Work done by C alone in 1 hr =  -  =  =  =  = 30 hrs.

 

 

Q8.      Time taken by A to finish the work =16 days.

            Time taken by B to finish the work = 12 days

 Work done by A in 1 day =

            Work done by B in 1 day =

Work done by (A+B) in 1 day =   +   =  =

            Work done by A in 2 day =  x 2 =

            Work left 1 -     = 

 If    work done in 1 day   www.rsmaths99.com

“  1      “          “   

“           “           “        x  = 6 days

            Hence , 6+2 days =8 days.

 

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Q9.      Time taken by A to finish the work =14 days.

            Time taken by B to finish the work = 21 days

            Work done by A in 1 day =

            Work done by B in 1 day =

            Work done by (A+B) in 1 day =   +   =  =

            Work done by (A+B) in 6 day =  x 6 =  .

               Remaining work (1 -   )  =

Now, B does 1 work in 21 days

Hence, B will do  work in  x 21= 6 days. 

 Hence, total number  of days taken = 6 + 6 = 12 days.

 

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Q10.    A can do   part of work in 16 days

               A   “          1 work               “  16 x    = 24 days

            B can do   part of work in 3 days

               B    “         1  work   in             3 x   = 12 days

            Work done by (A+B) in 1 day =  +  =    =  = .

               Hence,   work done in 1 day

               Then,    1 “          “       “         1 x 8 = 8 days.

 

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Q11.    Time taken by A to finish the work =15 days.

            Time taken by B to finish the work = 12 days

            Time taken by C to finish the work =20 days

            Work done by A in 1 day =

            Work done by B in 1 day =

            Work done by C in 1 day =

               Work done by (A+B+C) in 1 day =  =  =  

               Work done by (A+B+C) in  2 day =  =  

               Remaining work = 1 -  =

               Work done by (A+B) in 1 day =  +  =    =  

            If  work done in 1 day

               Then, 1 “          “       “    “

             Hence,    “     “     “           =  4 days.

 

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Q12.       Work done by A+B in 1 day =

            Work done by B+C in 1 day =

            Work done by C+A in 1 day =

            2(A+B+C)’s  1 day work =  +  =   =

               Hence, (A+B+C)’s  1 day work =  x  =    = 16 days.

 

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Q13.    Work done by A+B in 1 day =

            Work done by B+C in 1 day =

            Work done by C+A in 1 day =

            2(A+B+C)’s  1 day work =  +  =   =

  Hence, (A+B+C)’s  1 day work =  x  =   days.

A’s 1 day’s work =  =    =   .  Hence,  A alone finish the work in 30 days.

 

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Q14.    Time taken by A to fill the tank = 10 hrs

            Time taken by B to fill the tank = 15 hrs

            Work done by A in 1 hr  =

                Work done by b in 1 hr  = .

               Work done by (A+B) in 1 hr =   =   =   .  

               Hence, time taken by (A+B) to fill the tank = 6 hrs.

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Q15.    Time taken by A to fill the tank = 5 hrs

            Time taken by B to empty the tank = 6 hrs

            Work done by A in 1 hr  =

                Work done by B in 1 hr  =   - 

               Work done by (A+B) in 1 hr =   =  

               Hence, time taken  to fill the tank completely = 30 hrs.

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Q16.    Hints: Work done by (A+B +C) in 1 hr =  +  =   =  .

               Hence, it will take =  hrs  =  2hrs and x 60 = 2 hrs 40 min.

 

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Q17.    Work done by A in 1 min =

            Work done by B in 1 min  =

                Work done by C in 1 min =   - 

                Work done by (A+B +C) in 1 min =  -   =   =  

                Hence, time taken  to fill the tank completely = 20 minutes.

 

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Q18.    Time taken by A to fill the cistern = 9 hrs

                Work done by A in 1 hr  =

               Work done by the leak in 1 hr = -   =  =90 hrs.

 

Q19.    Work done by A in 1 hr  =

                Work done by B in 1 hr  =   

                Work done by (A+B) in 1 hr =   =  

           

 Work done by (A+B) in 2 hr =  x 2  =  

Remaining part =  1 -  =  

Now,   part is filled by B in 1 hr   www.rsmaths99.com

            Then,  1 “       “          B  ‘   8hrs

            Hence,    “        “             B   “    8 x   =  =  2  hrs  =  2 hrs   x 60 = 3 hrs 20 min.

           

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Curve- A figure traced out on a plane surface with the help of a sharp pencil without lifting it, is called curve. It may be open curve and closed curve.

 

Simple closed curve:- A closed curve which does not intersect itself is called a simple closed curve.

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Polygons:- A simple closed curve made up of only line segments is called a polygon.

Each straight line in a polygon is called its side.

 

Regular polygon:- A polygon having all sides equal and all angles equal is called a regular polygon. Eg. Equilateral triangle and square.

 

Irregular polygon:- A polygon which are not regular are called irregular polygons. Eg. A rectangle and rhombus.

 

For a regular polygon of n sides, we have

[i] each exterior angle =  

[ii] each interior angle = 1800 – (each exterior angle).

 

In a convex polygon of n sides, we have

[i] sum of all exterior angles = 4 right Ls

[ii]   sum of all interior angles = (2n - 4) right Ls

 

Number of diagonals in a polygon of n sides =    

 

 

     

 

Q3.      Prove that the sum of all the angles a quadrilateral is 360o.

            Let ABCD be a quadrilateral.

            Join A and C.

            Now, we know that sum of the angles of a triangle is 1800.

            In ABC L 2+ L4 + LB = 1800  -----(1)

            In ADC  L1 + L3 + LD = 1800 ----(2)

           

            Adding (1) and (2)     www.rsmaths99.com

            L 2+ L4 + LB + L1 + L3 + LD = 3600

            Or, ( L1 + L2 + L3 + L4 ) + LB + LD = 3600

           

            Or,  LA + LC + LB + LD = 3600

            Hence, the sum of all the angles of a quadrilateral is 360o.

           

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Q4.      Let the measure of the forth angle be xo .

            We know that the sum of the angle of the quadrilateral = 360o.

             76 + 54+ 108+ xo = 360o.        x = 122o.

 

Q5.      Let the angle of the quadrilateral be (3x)o,(5x)o, (7x)o, (9x)o.

            We know that the sum of the angle of the quadrilateral = 360o.

 

             3x + 5x+ 7x + 9x = 360o.       x = .

            Now, 3 x 15 = 45o, 5 x 15 = 75o, 7 x 15 = 105o, 15 x 9 = 135o.

 

Q6.      75o x 3 + y = 360o.       => y = 360 – 225 = 135o.

 

Q7.      Let the three equal angle be x = (3x)o.  www.rsmaths99.com

             (3x)o + 120o = 360o.      360 – 120 = 240     x = 80o.

Q8.      Let the two equal angle be x = (2x)o.

             2x + 85 + 75 = 360      => 2x = 360 – 160 = 200   x = 100o .

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Given ABCD is a parallelogram and LA = 110o. since the sum of any two adjacent angle of a parallelogram is 180o. we have

           

            LA +  LB  = 1800.       =>    110 + LB  = 1800   => LB = 70o

            Also, LB +  LC = 180O                          (Since, LB +  LC are adjacent angle ).

 

π  70 + LC = 180        => LC = 110o.

 

Also, LC + LD = 180O                           (Since, LC + LD are adjacent angle ).

 

π  110 + LD = 180      => LD = 70O.

‘OR’

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LA +  LB  = 1800.       =>    110 + LB  = 1800   => LB = 70o

If LA = 110o   and LB = 70o   then in parallelogram opposite angles are equal .

 LC = 110o    and    LD = 70O.

 

   Q2.   Let the adjacent angle be = LA  & LB  . then,  LA +  LB  = 1800.

 

            Since, both are equal     2 LA = 180o            LA = 90o .

 

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Q3.      Let two adjacent angle are LA  & LB  and LA = 4x &   LB = 5x.

 

            Then,    4x + 5x = 180             => 9x = 180  => x = 20

            Since, LA = 4x                 LA = 4 x 20 = 80o.

            Since, LB = 5x                LB = 5 x 20 = 100o.

            Then, in parallelogram opposite angles are equal .

                          LA = LC         LC = 80o .

            And,      LB = LD         LD = 100o .

 

Q4.      Let LA = 3x – 4, and   LB = 3x + 16.

We know, the sum of two adjacent angles of a parallelogram is 180o.

 

             3x – 4 + 3x + 16 = 180        =>    6x +12 = 180      =>  x = 28.

 

            Then, 1st adjacent angle LA = 3x – 4                            = 3 x 28 + 4 = 80o.

            And,   2nd      “           “       LB =  3x +16            = 3 x 28 + 16 = 100o.

            Now, in parallelogram opposite angles are equal .

                          LA = LC         LC = 80o .

            And,      LB = LD         LD = 100o .  www.rsmaths99.com

 

Q5.      Let ABCD is a ||grams. If  LA + LC = 130o .

 

            Since, in parallelogram opposite angles are equal .

 LA = LC          2 LA = 130                 LA = 65o .      LC = 65o .  

Now, the sum of two adjacent angles of a parallelogram is 180o.

  LA + LB = 180o        => 65 + LB = 180o     => LB = 115o .

            Now, in parallelogram opposite angles are equal .

             LC = LD = 650 .

 

Q6.      Let ABCD is a ||gm. Given, AB = DC = 5x & BC = AD = 3x.

            We know,    perimeter = sum of all sides

                                    64        =  5x + 3x +5x + 3x   = 16x.     x = 4 cm.

            Side AB = 5 x 4 = 20 cm.                    BC = 3 x 4 = 12 cm.

 

Q7.      Let ABCD be a ||gm. and AB = x.     ATQ, if AB = x then, BC = 10 +x.

            Now, AB = x         CD = x          (since, in ||gm opp. Sides are equal).

            And, BC = 10 + x       AD = 10 + x . 

            Now, we know,

                                    Perimeter = sum of all sides

                                     x + 10 + x + x + 10 +x = 140

                                      x = 30 cm. Hence, AB = 30 cm & BC = 40 cm.

 

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Q8.      In BMC and   DNA

 

            L DNA =  LBMC = 900

                        LBMC = LDAN (alternate angles)

            BC = DA (Opposite sides)

            By AAS congruency criteria

             BMC  DNA    

            Yes, it is true that BM is equal to DN

            (by corresponding parts of congruent triangles BMC and DNA)

 

 

         

 

 

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