Consider the diagram with current distribution as shown below.
As voltage is same in parallel combination, so ‌60I=(15+5)I1 ‌60I=20I1‌ or ‌I1=3I . . . (i) Similarly, (15+5)I1=10(1−I−I1) ‌⇒‌‌2I1=1−I−I1 ‌⇒‌‌3I1=1−I ‌⇒‌‌3(3I)=1−I [Using Eq.(i)] ‌⇒‌‌10I=1 ‌I=‌