1. **State the problem:** The principal has a sum of money to buy 50 computers. There are two types of computers, P and Q. 2. **Given information:** - Buying 50 computers of type P leaves 3000 units of money left. - Buying 50 computers of type Q spends three-fourths of the total money. - Price of Q is 740 less than price of P. 3. **Define variables:** Let the total sum of money be $S$. Let the price of computer P be $p$. Let the price of computer Q be $q$. 4. **Write equations from the problem:** From I: Buying 50 P computers and having 3000 left means $$S = 50p + 3000$$ From II: Buying 50 Q computers spends three-fourths of the money, so $$50q = \frac{3}{4}S$$ Also, price relation: $$q = p - 740$$ 5. **Substitute $q$ in the second equation:** $$50(p - 740) = \frac{3}{4}S$$ 6. **Substitute $S$ from the first equation into the above:** $$50(p - 740) = \frac{3}{4}(50p + 3000)$$ 7. **Expand both sides:** $$50p - 37000 = \frac{3}{4} \times 50p + \frac{3}{4} \times 3000$$ $$50p - 37000 = 37.5p + 2250$$ 8. **Bring all terms to one side:** $$50p - 37.5p = 2250 + 37000$$ $$12.5p = 39250$$ 9. **Solve for $p$:** $$p = \frac{39250}{12.5}$$ $$p = 3140$$ 10. **Find $q$:** $$q = p - 740 = 3140 - 740 = 2400$$ 11. **Find total sum $S$:** $$S = 50p + 3000 = 50 \times 3140 + 3000 = 157000 + 3000 = 160000$$ **Final answer:** The principal has a total sum of money equal to **160000**.