The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 and 7 are subtracted from the 3rd and 4th, all the four numbers will be equal. The numbers are

Let the numbers be $$a,b,c,d$$

=> $$a+b+c+d$$ = 48 --------Eqn(1)

When 5 & 1 are added to first two, => $$(a+5) and (b+1)$$

and when 3 & 7 are subtracted from last two, => $$(c-3) and (d-7)$$

According to question :

=> $$a+5 = b+1 = c-3 = d-7 = k$$ (let)

Now, in eqn(1)

$$(a+5) + (b+1) + (c-3) + (d-7)$$ = 48 + (5+1-3-7)

=> $$k+k+k+k$$ = 48-4

=> $$k$$ = 11

=> Numbers are : $$a = k-5 = 11-5 = 6$$

Similarly, $$b$$ = 10

$$c$$ = 14

$$d$$ = 18