A total fixed amount of $N$ thousand rupees is given to three persons $A,B,C$ every year, each being given an amount proportional to her age. In the first year, $A$ got half the total amount. When the sixth payment was made, $A$ got six-seventh of the amount that she had in the first year; $B$ got Rs. 1000 less than that she had in the first year; and $C$ got twice of that she had in the first year. Find $N$.

Step 1: Let the total amount given each year be $T$ (in thousands). Let the initial ages of $A,B,C$ be $a,b,c$ respectively.

Step 2: In the first year, the amounts are proportional to $a:b:c$. Since $A$ got half of the total amount:

Step 3: The amount $A$ got in the first year is $A_1=T/2$.

Step 4: When the sixth payment was made, their ages increased by 5. The new ages are $a+5,b+5,c+5$. The new total age sum is:

Step 5: In the sixth year, $A$ got $6/7$ of $A_1$, which is:

Using the age ratio in the sixth year:

Step 6: Since $a=10$, the initial age sum is $2a=20$, and $b+c=10$.

Step 7: Let's write the sixth-year equations for $B$ and $C$ (in thousands):

• $C_1=\frac{c}{20}T$

• $C_6=\frac{c+5}{2(10)+15}T=\frac{c+5}{35}T$

Since $C$ got twice of $C_1$:

Step 8: Since $b+c=10$, we have $b=8$.

Step 9: Now use the equation for $B$:

• $B_1=\frac{b}{20}T=\frac{8}{20}T=0.4T$

• $B_6=\frac{b+5}{35}T=\frac{13}{35}T$

Since $B$ got Rs. 1000 less (which is 1 in thousands):

Step 10: Thus, $N=35$ thousand rupees.