EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 181 190)

181.       Let the capacity of 1 bucket =X. then, capacity of tank = 25X.

     New capacity of bucket = 2/5 X.

      Required number of buckets = 25X/ (2X \ 5) = (25 X x 5/2X) = 125/2 = 62 1/2.

182.       Suppose initially peter had Rs .X. then,

     Amount received by Michael = Rs. (x/4).

     Amount remaining with peter = Rs. ( x – x/4) = Rs. (3x/4).

     Amount received by Sam = Rs. (1/2 x x/4) =Rs. (x/8).

     3X/ 4 – X/8 = 500    %x = 4000    X = 800.

183.       A’s share = 1/3. Remainder = (1 -1/3) =2/3.

      B’s share = 2/5 of 2/3 = 4/15. Rest = (2/3 -4/15 ) = 6/15 = 2/5.

      C’s share =D’s share = ½ of 2/5 = 1/5.

184.       Part read on first day = 3/8.remaining part = (1 -3/8) = 5/8.

     Part read on second day = 4/5 of 5/8 = ½. Unread part = [1 – (3/8 + 1/2)] = 1/8.

     Let the number of pages be X. then, 1/8 x = 30 or x =30 x 8 = 240.

185.       Wife’s share = 1/2 .remaining part = (1 -1/2) =1/2.

     Share of 3 sons = (2 /3 of ½ ) =1/3. Remaining part = (1/2 – 1/3 ) = 1/6.

     Each daughter’s share = ¼ x 1/6 = 1/24.

     Let the total money be Rs. X. then, 1/24 X = 20000      X = 20000 x 24 = 480000.

     Each son’s share = Rs. [1/3 x 480000)] = Rs. 53, 333.33.

186.       Out of 5 girls, 1 took part in fete, out of 8 boys, 1 took part in fete.

      Out of 13 students, 2 took part in fete.

       Hence, 2/13 of the total number took part in fete.

187.       French men = 1/5; French women = (1/5 + 2/3 x 1/5 ) = 5/15 = 1/3.

      French people = (1/5 +1/3 ) =8/15.        Not –French = (1 – 8/15 ) = 7/15.

188.       Girls – 3/5; boys = (1 -3/5) = 2/5.

      Fraction of students absent = 2/9 of 3/5 + ¼ of 2/5 = 6/45 + 1/10 = 21/90 = 7/30.

       Fraction of students present = (1 – 7/30) = 23 / 30.

189.       Number of boys who participate = 100.

           1/3 of boys = 100 or total number of boys = 300.

          Number of girls who participate = 200.

         ½ of girls = 200 or total number of girls = 400.

          Hence, total number of students = (300 + 400 ) = 700.

190.   Let the number of votes cast be x. the, number of votes required = 3x/4.

             Counted votes = 2x/3. Uncounted votes = (X -2X/ 3 ) = x/3.

            Votes won by the candidate = 5/6 of  3X / 4 = 5x/8

            Remaining votes required = ( 3X / 4 – 5x/ 8 )  =x/8.

            Required fraction = (x/8) / (x/3) = (x/8 x 3/x) = 3/8.