EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 231 241)

231.       Let the length of the piece be X meters. Then, cost of 1 m of piece =Rs. (35 / X)

( X +4) (35 / X – 1) = 35       35 – X + 140 / X – 4 = 35   140 / X – X = 4

X² + 4X – 140 = 0     ( Z + 14) (X – 10) = 0   X = 10.

232.       Let the cost of a chair and that of table be Rs. X and Rs. Y respectively.

   Then, 10x = 4y or y = 5/2x.

    15X + 2y = 4000      15X + 2 x 5/2 X = 4000    20X = 4000   X = 200.

   So, y = (5/2 x 200) = 500.

233.       Cost of 4 mangoes = cost of 9 lemons = Rs. (4.80 / 2 x 9) = Rs.  14. 40

       Cost of 1 mangoes = Rs/ (14.40 / 4) = Rs. 3.60.

       Cost of 5 apples = cost of 3 mangoes = Rs. (3.60 x 3) = Rs. 10.80.

        Cost of 9 oranges = cost of 5 apples = Rs. 10.80

       Cost of 1 orange = Rs. (10.80 / 9) = Rs. 1.20.

234.       Let the price of a saree and a shirt be Rs. Z and Rs. Y respectively.

      Then, 2X + 4u = 1600    ...(I)   and x + 6y = 1600..(II)

      Solving (I) and (II), we et: X = 400, Y = 200.

        Cost of 12 shirts =Rs. (12 x 200) = Rs. 2400.

235.       Let the cost of a table and that of a chair be Rs X and Rs. Y respectively.

     Then, 2X _+ cy = 3500 …(I)    and 3X + 2y = 40000

      Solving (I) and (II), we get: x = 1000 and y = 500.

236.       Let the fixed charge be Rs. X and variable charge be Rs. Y per km.

    Then, x + 16y = 156 ..(I)  and X + 24y = 204

    Solving (I) an (II0, we get: X = 60, y = 6.

     Cost of travelling 30 km = Rs. (60 + 30 x 6) = Rs. 240.

237.       Let the number of beeches n the class be. Then, 6 (X + 1) = 6 x 12 = 72.

               Hence, number of students in the class =  6 (X + 1) = 6 ( X + 1) = 6 x 12 = 72.

238.       Let the number of students in rooms A and B he X and y respectively. Then,

      X – 10 = y + 10    X –y = 20….(I)   and X = 20 = 2 (y – 20)   X – 2y = -60…(II)

      Solving (I) and (II), we get: X = 100, y = 80.

239.       Let the number of buffaloes be X and the number of ducks by Y.

     Then, 4X + 2y =2 (x + y) + 24     2x = 24     X = 12.

240.   Let the number of hens be X and the number of cows be y. then,

     X +y = 48..(I) and 2X + 4y = 140   X + 2y = 70.

     Solving (I) and (II), we get: X = 26, Y = 22.

241.   Suppose, Sanya and vidushi donate money to X and (x + 5 ) people respectively.

     Then, 100/X – 100 /X + 5 = 1 100 (+ 5) x²+ 5x – 500 = 0.

    ( X – 20) ( X + 25) = 0     X = 20.

   Total number of recipients of charity = X + ( X + 5) = 2X + 5 = 45.