testing

1. Let the sides of the triangle (in cm) be a, b and c.

                a2(b + c) + b2 (a + c) + c2 (a + b) = 6abc

                a2b + a2c + b2a + b2c + c2a + c2b – 6abc = 0

                a(b2 + c2)  + b(a2 + c2) + c(a2 + b2) – 6abc = 0

                a(b2 + c2 – 2bc) + b(a2 + c2 – 2ac) + c(a2 + b2 – 2ab) = 0

                a(b – c)2 + b(a – c)2 + c(a – b)2 = 0

                Since, a > 0, and b > 0 and c > 0

                only possibility is (b – c)2 = (a – c)2 = (a – b)2 = 0

                \b – c = a – c       = a – b = 0 i.e., a = b = c

                \The triangle is equilateral.                                                                                                                                                                                                                                                                                                                                                               Choice (1)

 

2.             k =

                If , each of them equals .

                k =

                If p + q + r ¹ 0, then k = 1.

                If p + q + r = 0, p + q = –r.

                \k == –.  \ k = 1 or – 1/2                                                                                                                                                                                                                                                                  Choice (3)

 

3.             Let the rates of X, Y and Z be x units/day, y units/day and z units/day

x = 3(y) + 3(z)                …… (1)

                y = ……. (2)

From (1) and (2),

3z = x – 3y = 4y – x

                2x = 7y Þ x =

                (1) Þ z = y/6

 

\ x : y : z = = 21 : 6 : 1

               

                \Required ratio = = 2 : 7 : 42

 

4.             Let the usual speed of Rahul be s kmph.

                Let his usual time be t hours.

                Distance = st km = (s – 3)(t + 1/2) km

                = (s + 3)km

               

                st = (s – 3) and st = (s + 3)

               

                st = st – 3t + and

               

                st = st + 3t –

 

                3t = and 3t =

 

                3t =

 

               

                \ s = 27