EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 191 200)

191.       Let the total number of staff members be X.

        Then, the number who can type or take shorthand = ( X – 3X / 4 ) = X/3.

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

         Let A and B represent the sets of persons who can type and take shorthand   respectively.

          Then, n (A ᴗ B ) = X/4, n (A) = X/5 and n (B0 =X/3.

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

192.       Hire charges =Rs.(60 x4 + 60 x 5 + 8/5 x 200) = Rs. 860 .

      Suppose Rohit had Rs .X with him initially. Then, X – 860 = ¼ x 860    X = 1075

193.       Let the total number of shots be X. then,

     Shots fired by A = 5/8 X; shots fired by B = 3/8 x.

     Killing shots by A = 1/3 of 5/8 X = 5X / 24; shots missed by B = 1/2 of 3/8X = 3/16 X.

     3x/16 = 27 or X = (27 x 16 / 3) = 144.Birds killed by A = 5X / 24 =(5 / 24 x 144 ) = 30.

194.       Number of alterations required in 1 shirt = (2/3 + 3/4 + 4/5) = 133/60.

      Number of alterations required in 60 shirts = (133 / 60 x 60 ) = 133.

195.       Let the largest fraction be X and the smallest be Y. the, x/y = 7/6 or y =6/7 X.

      Let the middle one be Z. Then, X + 6/7 X + z = 59 / 24 or Z = (59 /24 -13 X / 7)

      59/24 – 13x/7 + 1/3 = 7/6    13X/7 = 59/24 + 1/3 – 7/6 = 39/24   X =(39/24 x 7 / 13) =         7/8.

    So, X = 7/8, Y = 6/7 x 7/8 = 3/4 and z = 59/24 – 13/7 x 7/8 = 20/24 = 5/6.

    Hence, the fractions are 7/8, 5/6 and 3/4.      

      

196.       Suppose each tube contains x grams initially. Then,

     4[1/3 (X + 20)] =X + 2/3 (X + 20)  2/3 (X + 20)                          = X   x/3 = 40/3   Z = 40.

X = 40.

197.         Let the total number of apples be X. Then,

    Apples sold to 1^st customer = (X/2 + 1).remaining apples = X \- (X/2 + 1) = (X/2 -1).

    Apples sold to 2^nd customer = 1/3 (x/2 -1) + 1 = X/6 - 1/3  +  1 = (X/6 + 2/3)

    Remaining apples = (X/2 -1) – (X/6 + 2/3) – (X/2 –X/6) – (1 + 2/3) = ( X/3 – 5/3).

    Apples sold to 3^rd customer = 1/6 ( x/3 – 5/3) + 1 = (X/15 + 2/3).

    Remaining apples = (X/3 – 5/3) – (X/15 + 2/3 ) = ( X/3 – X/ 15) – (5/3 + 2/3 ) = (4X/ 15 –   7/3)

    4X / 15 – 7/3 = 3   4X / 15 = 163   X = (16 /3 x 15/ 4 ) = 20.

198.       Given exp. = (a + b)² +(a – B)² / a² + b², where a = 856, b = 167

= 2 (a + b)²/   a² + b² = 2.

199.       Given exp. = (a + b)² +(a – B)²/ a b =4ab / a b = 4(where a = 469,b = 174).

200.   2ab = (a² + b²) – (a – b)² = 29 – 9 = 20   a b = 10