1. Solving the equation 6a - 8t = 3 we get a > t

Ans : A

2. D PQK is right angled triangle


because m Ð PKQ = 90⁰ [Angle in a semi-circle]
h is an acute angle


\ h = Ð 900

Ans : B

3. x / z = 51


x = 5z x + y +z    =   28

5z + 4 + z   =   28

            6z   =   24


             z    =    4


       x    =   20             y + z   =    8


       20   >   8

  Ans : A

4. The side opposite the angle measuring 60⁰ is Ö 3/2 ´ hyp

p    =   60(r - 30⁰ )       \ Ð PQR   =   90⁰

p   >   r

Ans : A

5. t + t² < t³ for all values of t greater than 1.

Consider    t   =   2
           t + t²   =    2 + 2²
                      =    6
                t³   =    2³
                     =   8
       \ 6   <   8


Ans : B

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