Question

I+II+III+IV=140
I=IV+15
II,III,IV=nr consecutive
I,II,III,IV=?

Answer 1

I+II+III+IV=140

I=IV+15

II,III,IV=nr consecutive

=> IV=I-15; III=I-16; II=I-17

I+I-17+I-16+I-15=140

4I-48=140

I=(140+48):4=188:4

I=47

II=47-17=30

III=47-16=31; sau III=II+1=31

IV=47-15=32;  sau IV=III+1=32

R: I=47; II=30;  III=31;  IV=32

   47+30+31+32=140 (A)