Question

Half of the sum of two is 42 one third of their differences is 4 find two numbers

Answer 1

Question;

Half of the sum of two numbers is 42. If one third of their difference is 4 ,then find the two numbers?

Solution;

Let the two required numbers be ;

x and y.

Now,

According to question,

Half of the sum of the numbers is 42.

=> (x+y)/2 = 42

=> x+y = 42×2

=> x+y = 84 --------(1)

Also,

One third of their difference is 4.

=> (1/3)(|x-y|) = 4

=> |x-y| = 4×3

=> |x-y| = 12

=> x-y = ±12 --------(2)

Now , adding eq-(1) and (2),

we get;

=> x+y+x-y = 84±12

=> 2x = (84+12) or (84-12)

=> 2x = 96 or 72

=> x = 96/2 or 72/2

=> x = 48 or 36

now,

using eq-(1), we have;

=> x+y=84

=> y=84-x

# case(1):

when, x = 48

then , y = 84 - 48 = 36

#case(2):

when, x = 36

then, y = 84 - 36 = 48

However, we got the same pair of numbers in both the cases ;

ie, 36 and 48.

Thus, the required numbers are ;

36 and 48.