Track / Stage 5 / 209 of 400 #809 of 2000

Problem 809

AIME late

Algebra Difficulty 5.5 Find the answer

2. An old problem. A landlord, having calculated that a cow is four times more expensive than a dog, and a horse is four times more expensive than a cow, took 200 rubles with him to the city and spent all the money to buy a dog, two cows, and a horse. How much does each of the purchased animals cost?

Your answer

Official solution

2. Let's write the conditions of the problem as equalities:

$K=4 C, L=4 K$

Substituting $K$ with $4C$ in the second equality, we get:

$\text { L }=16 \mathrm{C} .$

Since a dog, two cows, and a horse cost 200 rubles, we can write the following equation:

$C+2 K+L=200$

Using the previously established relationships for $L$ and $K$ , we get the equation

$C+8 C+16 C=200$

Combining like terms $(25 \mathrm{C}=200)$ and solving for the unknown factor in the product $\mathrm{C}=8$ . Multiplying this number by 4 and 16, we determine the price of the cow and the horse ( $8 \times 4=32,8 \times 16=128$ ).

Answer: the dog costs 8 rubles, the cow 32 rubles, and the horse 128 rubles.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.