Step 1: Set up.
Let a, b, c, d be the numbers of Re. 1, Rs. 2, Rs. 5 and Rs. 10 notes, each at least 1. The given condition is \(c+d=a+b+1\) and the price is \(P=a+2b+5c+10d\).
Step 2: Statement A alone.
Total notes = 13 gives \((a+b)+(c+d)=13\), so \(2(a+b)+1=13\), i.e. \(a+b=6,\ c+d=7\). Then \(P=(6+b)+(35+5d)=41+b+5d\) with \(1\le b\le 5,\ 1\le d\le 6\). For example \(b=1,d=1\) gives \(P=47\) and \(b=2,d=1\) gives \(P=48\). Not unique, so A alone is insufficient.
Step 3: Statement B alone.
Take \((a,b,c,d)=(1,2,1,3)\): \(c+d=4=a+b+1\), and \(P=1+4+5+30=40\). Take \((a,b,c,d)=(2,4,6,1)\): \(c+d=7=a+b+1\), and \(P=2+8+30+10=50\). Both prices are multiples of Rs. 10, so B alone does not fix the price.
Step 4: Both together.
We need \(41+b+5d\equiv 0 \pmod{10}\), i.e. \(b+5d\equiv 9 \pmod{10}\). Since \(5d\) ends in 0 or 5 and \(1\le b\le5\), we need \(b=4\) and \(d\) odd, so \(d=1,3,5\). That gives \(P=50,\,60,\,70\) (for \((a,b,c,d)=(2,4,6,1),(2,4,4,3),(2,4,2,5)\)). The price is still not unique.
Final Answer:
The correct option is (3): The question cannot be answered on the basis of the two statements.