Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce, making a profit of 20 percent. What was the ratio of the mixture?
1:10
1:5
2:7
3:8
Explanation:
Let 1:k be the ratio in which Joseph mixed the two types of rice.
Then a sample of (1+k) ounces of the mixture should equal 1 ounce of rice of the first type, and k ounces of rice of the second type.
The rice of the first type costs 5 cents an ounce and that of the second type costs 6 cents an ounce. Hence, it cost him:
(1 ounce 5 cents per ounce) + (k ounces 6 cents per ounce) = 5+6k
Since he sold the mixture at 7 cents per ounce, he must have sold the net 1+k ounces of the mixture at 7(1+k).
Since he earned 20% profit doing this, 7(1+k) must be 20% more than 5+6k.
Hence, we have the equation
7(1+k)=$\left(1+\dfrac{20}{100}\right)$(5+6k)
7+7k=$\left(\dfrac{120}{100}\right)$(5+6k)
7+7k=$\dfrac{6}{5}$(5+6k)
7+7k=6+$\dfrac{36k}{5}$
1=$\dfrac{k}{5}$
k=5
Hence, the required ratio is 1:k= 1 : 5
Let 1:k be the ratio in which Joseph mixed the two types of rice.
Then a sample of (1+k) ounces of the mixture should equal 1 ounce of rice of the first type, and k ounces of rice of the second type.
The rice of the first type costs 5 cents an ounce and that of the second type costs 6 cents an ounce. Hence, it cost him:
(1 ounce 5 cents per ounce) + (k ounces 6 cents per ounce) = 5+6k
Since he sold the mixture at 7 cents per ounce, he must have sold the net 1+k ounces of the mixture at 7(1+k).
Since he earned 20% profit doing this, 7(1+k) must be 20% more than 5+6k.
Hence, we have the equation
7(1+k)=$\left(1+\dfrac{20}{100}\right)$(5+6k)
7+7k=$\left(\dfrac{120}{100}\right)$(5+6k)
7+7k=$\dfrac{6}{5}$(5+6k)
7+7k=6+$\dfrac{36k}{5}$
1=$\dfrac{k}{5}$
k=5
Hence, the required ratio is 1:k= 1 : 5