Puzzles worth an equation

The rest of the chapter is puzzles. Each one is solved the same way: name the unknown, turn each sentence into an equation, solve.

The farm. 55 heads, 150 legs, horses and hens. Let there be h horses, so 55 − h hens. Legs: 4h + 2(55 − h) = 150, so 4h + 110 − 2h = 150, so 2h = 40 and h = 20. Twenty horses, thirty-five hens.

You can also do it without letters. If all 55 were hens there would be 110 legs. We are 40 legs short, and swapping a hen for a horse adds 2 legs, so we need 20 swaps: 20 horses. Same answer; it is worth seeing both ways.

Mother and daughter. A mother is 5 times her daughter's age; in 6 years she will be 3 times it. Let the daughter be d, so the mother is 5d. In six years: 5d + 6 = 3(d + 6). Then 5d + 6 = 3d + 18, so 2d = 12 and d = 6. The daughter is 6 and the mother 30.

The cowherds. Let Gauri have g cows, so Naina has 2g. If Naina gives her three, they are equal: g + 3 = 2g − 3, so g = 6. Gauri has 6 and Naina 12.

The dosa cart. Rent is ₹5000 a day, and each dosa costs ₹10 to make.

• Selling 100 dosas at price p: income 100p, costs 5000 + 1000. Profit 100p − 6000 = 2000 gives p = ₹80.
• At ₹50 each, selling n dosas: 50n − (5000 + 10n) = 2000, so 40n = 7000 and n = 175 dosas.

The fractions. 1/3; (1 + 3)/(5 + 7) = 4/12 = 1/3; (1 + 3 + 5)/(7 + 9 + 11) = 9/27 = 1/3. Every one is 1/3. Why? The first n odd numbers add up to n², and the next n odd numbers add up to (2n)² − n² = 3n²: always three times as much.

Karim and the Genie. Each round round the tree doubles Karim's coins, and then he pays the genie 8. After the third doubling he has exactly 8 coins, just what he owes, so he ends with nothing.

Let him start with x coins. After round one he has 2x − 8. After round two, 2(2x − 8) − 8 = 4x − 24. In round three the coins double to 2(4x − 24) = 8x − 48, and that is the 8 he has to pay. So 8x = 56 and x = 7.

Counted out: 7 → 14 → 6 → 12 → 4 → 8 → 0. He could never win: doubling 7 gains him only 7, but the genie takes 8.