Question

In a farm Raju had some goats and some ducks. He makes a head count

and a leg count and finds that there are 150 legs altogether. If there are 40

ducks and goats altogether,how many are ducks and how many are goats ?​

Answer 1

Treating this as an algebra word problem :)

Let d and g be the number of ducks and goats, respectively.

d + g = 14 (they have 1 head each), ergo 2d + 2g = 28

2d + 4g = 40 (each duck has 2 legs, each goat has 4 legs)

Subtracting to knock out the 2d term, we have 2g = 12, and g = 6. Then d = 8.

Check: 2(8) + 4(6) = 16 + 24 = 40 legs, as expected.

~~~~~~~~

Generally, this kind of word problem boils down to 2 equations for 2 unknowns (or n of each, for some n > 2). That’s a system of linear equations. We solve those in the usual way, by combining equations (equivalently, rows in a matrix) using arithmetic operations:

- An equality is preserved under scaling, so you can multiply through by any constant (including negative ones).

- Two equalities are preserved under addition or subtraction, so you can add two equations together to knock out an identical term in each.

Keep doing that until you have an equation with only one variable, such as the “2g = 12” equation above. Plug that result into the remaining N-1 equations, and now you have a smaller problem (with fewer unknowns), so this process is finite.

And yes, you should always apply this general method, because it’s general. For trivial problems, you can get away with a wordy argument based on special-case reasoning, but that approach breaks down for 5x5 puzzles.

Answer 2

Step-by-step explanation:

Treating this as an algebra word problem :)

Let d and g be the number of ducks and goats, respectively.

d + g = 14 (they have 1 head each), ergo 2d + 2g = 28

2d + 4g = 40 (each duck has 2 legs, each goat has 4 legs)

Subtracting to knock out the 2d term, we have 2g = 12, and g = 6. Then d = 8.

Check: 2(8) + 4(6) = 16 + 24 = 40 legs, as expected.

~~~~~~~~

Generally, this kind of word problem boils down to 2 equations for 2 unknowns (or n of each, for some n > 2). That’s a system of linear equations. We solve those in the usual way, by combining equations (equivalently, rows in a matrix) using arithmetic operations:

- An equality is preserved under scaling, so you can multiply through by any constant (including negative ones).

- Two equalities are preserved under addition or subtraction, so you can add two equations together to knock out an identical term in each.

Keep doing that until you have an equation with only one variable, such as the “2g = 12” equation above. Plug that result into the remaining N-1 equations, and now you have a smaller problem (with fewer unknowns), so this process is finite.

And yes, you should always apply this general method, because it’s general. For trivial problems, you can get away with a wordy argument based on special-case reasoning, but that approach breaks down for 5x5 puzzles.