1. 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
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By: anil on 05 May 2019 01.51 am
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
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=> $$a+b+c+d$$ = 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