Aptitude Day 62

 Answers of Day 61:
        1. 25/48.
        2. 15, 4.

Problems on Numbers:

1. The sum of the numerator and denominator of a fraction is 7. If 1 is added to the numerator and 3 is subtracted from the denominator, it becomes 3/2. Find the fraction.

Solution:
                Let the numrerator be x and the denominator be y.
                x+y = 7.
                (x+1)/(y-3) = 3/2
                (7-y+1)/(y-3)=3/2
                2(8-y) = 3(y-3)
                16-2y = 3y-9
                5y = 25
                y = 5.
                x = 2.
               The fraction is 2/5.

2. The difference between the numerator and the denominator is 3. If 3 is added denominator, the fraction is decreased by 1(1/2). Find the numerator. 

Solution:
                Let the numrerator be x and the denominator be y.
                x-y = 3
                x/y - x/(y+3) = 1(1/2)
                (3+y)/y - (3+y)/(y+3) = 3/2
                (3+y)/y - 1 = 3/2
                2(3+y-y) = 3y
                6=3y
                y=2.
                x= 5.
               The numerator is 5.