Joseph bought two varieties of rice, costing 6 cents per ounce and 8 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 10 cents per ounce, making a profit of 50 percent. What was the ratio of the mixture?
Joseph bought two varieties of rice, costing 6 cents per ounce and 8 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 10 cents per ounce, making a profit of 50 percent. What was the ratio of the mixture?
The correct option is D 2:1
Let1: 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 price of the first type costs 6 cents an ounce and that of the second type costs 8 cents an ounce.
Hence, it cost him: (1 ounce at 6 cents per ounce ) + (k ounces at 8 cents per ounce) = 6 + 8k
Since he sold the mixture at 10 cents per ounce, he must have sold the net 1+k ounces of the mixture at 10(1+k).
He earned 50% profit doing this, 10(1+k) must be 50% more than 6+8k.
Hence, we have the equation 10+10k = 9 + 12 k
⇒ k = 12
Therefore, the required ratio is 1: k = 2: 1
2:1
Let1: 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 price of the first type costs 6 cents an ounce and that of the second type costs 8 cents an ounce.
Hence, it cost him: (1 ounce at 6 cents per ounce ) + (k ounces at 8 cents per ounce) = 6 + 8k
Since he sold the mixture at 10 cents per ounce, he must have sold the net 1+k ounces of the mixture at 10(1+k).
He earned 50% profit doing this, 10(1+k) must be 50% more than 6+8k.
Hence, we have the equation 10+10k = 9 + 12 k
⇒ k = 12
Therefore, the required ratio is 1: k = 2: 1
