EXERCISE 4 (OBJECTIVE TYPE QUESTIONS SOLUTION IN 101 110)

101.       6 a + 4 b /6 a – 5 b  = 6 ( a/ b) + 4 / 6 ( a / b – 5 = 6 x 4/3 + 4/ 6 x 4/3 – 5 = 8 + 4 8-5 = 12/3 = 4.

102.       X/2y = 6/7  X/y = (2 x 6/7 ) = 12/7.

X – y / X + y +X / y – 1 / X/y + 1 + 14 /19 = 12/7 -1 / 12 / 7 + 1 + 14 / 19 = (5/7) / 

(19.7) + 14/19

= (5/7 x 7/19 ) + 14/19 = 5/19 + 14/19  = 19/19 = 1.

103.       1/b = 4/5 and b/c = 15/16 = ( a/b x b/c ) = (4 / 5 x 15/16)  a /c = 3/4

 c² - a²/c² + a² = 1 – (a/c)²/ 1+ (a²/ c²) = 1 – (a/c)²/1 + (a/c)² = 1 – 9/16 / 1 + 9 /16 = (7 

/16) / (25 /16) = 7 / 25

= 7/25.

104.       (a – c ) – (b + d ) = 6 and (c –d ) ( a + b ) =3

( a – c) – (b + d) = 6 and (c –a) –( b + d) =3

(b + d) = (a –c) -6 and (b + d) = (c – a) -3

( a –c ) = -6 = (c –a) -3

2 (a –c) =3

( a-c) = 3/2 =1.5

105.       X = a / a – 1 = 1 + 1 / a -1 = 1 + y.                     X >y

106.       A is positive and a < 1     1/a > 1.                      ( a + 1/a ) > 2

107.       A /X + y/b = 1  a /X = 1 –y/b = b – y /b    X / a =b / b- y

B /y + z/c = 1    z /c = 1 – b/y = y – b / y c/z = y / y-b = -y / (b – y)

X/a + c/z = b/( b – y ) = (b – y / (b – y ) =1.

108.       a² + b² = 45 ….(I)    and  b² + c² = 40

Subtracting, we get: a² - c² =5  ( a + c) (a – c) =5.

(a + c) =5 and (a – c) =1.

Solving, we get: a =3, c =2. Putting c =2 in (II), we get b =6.

109.       a/3 = b/4 = c/7 = k (say). Then, a = 3k, b = 4k, c = 7k. 

a + b + c / c = 3 k + 4 k + 7k / 7k = 14k / 7k = 2.

110.   3X = 7 =7 X +  5    7 X – 3 X =2      4 X = 2     X =1/2.

Now, 3X + 7 = X² + p  3/2 + 7 =1/4 + P

P = 17/2 – ¼ = 33/4 = 8  1/4.

= 8 1/4