Sunday, 15 March 2015

EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 121 130)

121.       3X – 5y = 5 ….(I) and x/  y  =5/7   7X= 5X = 5y  2X = 5y =0 .. (II)

Subtracting (II) from (I), we get: X =5.

Putting X =5 in (I), we get: y =2. So, x –y = 5 -2 =3.

122.       4 X + 3y = 18xy …(I) and 2 x -5y=  - 4xy

Dividing (I) and (II) by X y, we get: 3/X + 4/y = 18…..(III) and 5/X – 2/y = 4…(iv)

Multiplying (iv) by 2 and adding (III) to it, we get: 13/x = 26 or x = 1/2.

Putting X = ½ in (III), we get: y = 1/3.

123.       2X + y =17 …(I); Y + 2z =15 …(II) and x + y =9

Subtracting (III) from (I), we get: X =8.

Putting X =8 in (I), we get: y = 1. Putting =1 in (II-, we get: 2z =14 or z=7.

4  X + 3y X z = 4 x 8 + 3 x 1 + 7 = 42.

124.       3 X – 4y + z =7 .. (I) ;  2 X + 3y – z =  19…. (II) and x + 2y +2z = 24…..  (III)

Adding (I), (II), we get: 5x –y =26

Subtracting (I) from (II) and adding to (III), we get: 9y =36 or y =4.

Putting y =4 in (iv), we get: 5X – y =26.


Putting x =6, y =4 in (III), we get: 2z = 10 or z =5.

125.       2x = y = 15…(I); 2y + z = 25 ..(II) and 2 Z + x = 26.

              Adding (I), (II) and (III), we get: 3 (X + y + z) = 66 or x = Y + z = 22.

              From (II), we have; y = 25 –z/ 2. From (III), we have: x = 26 – 2z.

             (26 – 2z) + ( 25 –z/2) + z =22   77 – 3z = 33      z =11.


126.       2X = 3y = 31  …(I) ;  y – z =4  ..(II) and x + 2z = 11

            Multiplying (III) by 2 and subtracting from (I), we get: 3y – 4z =9…(IV)

            Solving (II) and (Iv), we get: y =7, z =3. Putting y =7 in (I), we get: X =5.

               X + y + z = (5 + 7 + 3)= 15.

127.       Given exp. = (3/4 x 4/3 x 5/3 x 3/5 /13/7 x 1/13) = 1/7.

128.       Given exp.= 1/2 x 2/3x ¾ x …..x (n -1) /n  = 1/n.

129.       Given exp. 3/2 x 4/3 x 5/4 x////x 121 /120 = 121/2 = 60.5.


130.   Given exp. = 5/3 7/5 x 9/7 x …..x 1003 / 1001 = 1003 / 4.

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