Sunday, 15 March 2015

EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 111 120)

111.       2 a + b/ a + 4b = 3    2a + b = 3 ( a + 4b)   a = -11b.

                                 A + b / a + 2b = -11b + b /-11b + 2b = -10b / -9b = 10 /9.

112.       ( 2 a + 3b) (2 c – 3 d ) = (2 a – 3b) (2c + 3 d)

= (2 a + 3b) /(2 a – 3b) = (2 c + 3d ) /( 2 c -3 d)  = 2 (a / b)+ 1/ (2 ( a/b ) -1= ( 2 (c/d) +1   / 2 (c/d) – 1   a / b = c/d.

113.       ( a + b + 2c + 3d) ( a – b – 2c + 3d ) = ( a – b+ 2 c – 3 d ) a + b – 2 c – 3 d)

              [(a + b) + (2c + 3d)] [(a – b ) –(2 c – 3 d)]

                      = [( a – b ) + (2c – 3d )] [( a +b) –(2 c + 3 d)]

              (a + b ) ( a – b ) – (a + b ) (2 c – 3d ) + ( a – b ) ( 2c + 3 d ) –(2c + 3d ) (2c – 3d)

                  =(a – b ) (a + b) – (a –b) *2c + 3d ) +( a + b) (2c =3d) –(2 c + 3 d) (2 c – 3d)

             ( a =) (2 c – 3d ) =3bd = 2ac + 3 ad – 2bc – 3bd

              4bc = 6ad  2 b c =3ad.

114.       X = y   1 – q = 2q + 1 3 q = 0    q -0.

115.       X/5 – X /6 = 4.  6 X – 5X /30  = 4   X =120.

116.       2 X – 2y = 21 / 22 X/y = (21 / 22 x 2/3) =7/11  X = 7/11 y.

4 X + 5y = 83   4 x 7/11  y + 5y = 83    28 /11 y + 5y = 83  83 y = 83 x 11    y = 11.


X = 7/11 y =( 7/11 x 11 )=7.

So, y – x = 11-7 =4.

117.       3x = y =5 19  …(I)                and X –y =9……(II)

Adding (I) and (II), we get: 4 x = 28 or X=7. Putting X = 7 in (I), we get:  y =2.

118.       A + b =5 ….(I)                and 3 a + 2b = 20 (II)

Multiplying (I) by 2 an subtracting from (II), we get: a – 10.

Putting a = 10 in (I), we get: b = - 5.

( 3 a + b) = 3 x 10 + (-5) = 30 -5 = 25.

119.       ( 2p + 3q) + ( 2p – q ) = 18 + 2  4p + 2q = 20 2 ( 2p + q) =20

2p + q =10.

120.   2X + y = 5..(I)               and 3X – 4y =2.

Multiplying (I) by 4 and adding (II) to it, we get : 11X = 22 or x =2.


Putting X =2 in (I), we get: y =1. So, 2xy = 2 x 2 x 2 =4. 

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