Aptitude Day 38

Problems on Numbers

1. 35 is divided into two parts such that the sum of their reciprocals is 7/60. Find the two parts.

Solution:

                 Let the 2 parts be x and (35-x).

                 (1/x)+1/(35-x) =  7/60

                 (35-x+x)/[x(35-x)] =7/60

                 60(35)= 7(35x-x²)

                 2100 = 245x-7x²

                        7x²-245x+2100=0

                        x²-35x+300=0

                        x²-20x-15x+300=0

                        x(x-20)-15(x-20)=0

                        (x-20)(x-15)=0

                             x=20 and 15.

                The 2 numbers are 20 and 15.

2. If 3 numbers are added in pairs, we get 17, 13 and 14. Find the numbers.

Solution:

                  Let the3 numbers be x, y and z.

                        x + y = 17 ---------- Equation 1

                        y + z = 13 ---------- Equation 2

                        z + x = 14 ---------- Equation 3

                  The above 3 conditions are given.

                  Now adding the 3 equations, we get

                        2(x +y +z) = 44

                           x + y + z = 22

                           x + 13 = 22

                                  x = 9.

                   Substitute x value in Equation 3, we get

                         z+9= 14

                             z= 5

                  Substitute z value in Equation 2, we get

                         y+5=13

                             y=8.

The 3 numbers are 9, 8 and 5.