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Aptitude Day 33


Problems On Numbers

 1. The sum of the two number is 160. If one-third of the one exceeds one-seventh of the other by 10. Find the smaller number.

Solution:
                Let one number be x and other be (160-x).
                (x/3) - (160-x)/7 = 10        [Given condition]
                 7x-3(160-x) = 10
                  7x-480+3x = 10
                         10x= 490
                             x=49
               The two numbers are 49 and 160-x=(160-49) = 111.
               The smaller number is 49. 

 2. The difference of two numbers is 8 and one-sixth of their sum is 12. Find the numbers.

Solution:
                Let the numbers be x and y respectively.
                x-y = 8
                (x+y)/6 = 12
                x+y = 72
                By solving the highlighted equations, we get x = 40 and y = 32.
                These are the required 2 numbers.
                
             

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