Monday, 16 March 2015

EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 171 180)

171.       Let the fraction be a/b = 512 /27              a / b x a/b x a/b =512 /27   

      (a/b) ³ = (8/3)³       a/b = 8/3 = 2  2/c.

172.       Maximum internal assessment score = (47 / 50 x 10) = 9. 4.

      Minimum internal assessment score = (14/50 x 10) =2.8.

       Required difference = (9.4 – 2.8) = 6.6.

173.       Let savings in N.S.C and P.P.F. . be Rs. x and Rs. (150000 – x ) respectively. Then, 1/3 X   = 1/2 (15000000 –X) x/3 + X/2 = 75000    5x/6 = 75000      X = (75000 x 6/5) = 90000.

                 Savings in public provident fund = Rs. (150000 – 90000) = Rs. 60000.

174.       Let there be (X + 1) members. The,

     Father’s share =1/4, share of each other members = 3/4X.

     3 (3/4X) = ¼     4X = 36       X = 9.

175.       Let salary =Rs. X. then, tips = Rs. (5/4 X).

             Total income = Rs. (x + 5/4 X) = Rs. (9X / 4).

               Required fraction = (5X  / 4 x 4 / 9X ) = 5/9.

176.       Let C’s share = Rs. X. then, B’s share = Rs. (x/4), A’s share = Rs. (2/3 x x/4) =Rs. X/6.


      X/6 + X/4 + X = 1360 17X / 12   = 1360     X = (1360 x 12 / 17) =Rs. 960.

      Hence, B’s share = Rs. (960/4) = Rs. 240.

177.       Let Tanya’s share = Rs. X. then, Veena’s share = Rs. (x/2)

      Amita’s share = Rs. (2/3 x X/2) = Rs. (X/3). Total bill = Rs. (X+ X/2 + X/3) = 
Rs. (11X/6).

     Required fraction = (X / 2) x 6 /11x) = 3/11.

178.       Let the capacity of the tank be X litres. Then, ¼ X = 135     X = 135 x 4 = 540.

     Required fraction = (180 / 540) = 1/3.

179.       Let the capacity of the tank be X litres.

      Then, 6/7X – 2/5X = 16     30X – 14X = 16 x 35     16X = 560     X = 35.

180.   Let the capacity of the bucket be X litres. Then,

      Capacity of 1 large bottle = X/4; capacity of 1 small bottle = X/7.

      Fluid left in large bottle = ( x/4 – X/7) = 3X/ 28


                                    Required fraction = (3X / 28 / X 4) = (3X / 28 x 4/X) = 3/7. 

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