Joseph bought two varieties of rice, costing 5 cents per

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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?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7

The OA is B.

From 20%, we can obtain the cost price (x) of the mixture before profit,
120/100*x = 7 --> x = 35/6.

From the answers, we need to find an option that satisfies the above cost price of 35/6.

Option A (5+6*10)/11=65/11
Option B (5+6*5)/(6)=35/6.

Hence option B is the correct answer.

Has anyone another strategic approach to solve this PS question? Regards!
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by Shahrukh@mbabreakspace » Fri Jul 06, 2018 6:17 am
Let he buys x fraction of total at 5, remaining (1-x) at 6
Then CP= 5x+(1-x)6= 6-x
SP=7
(7-(6-x))/(6-x)=20/100

On solving x= 1/6 and 1-x= 5/6
So ratio in mixture= 1/6:5/6= 1:5

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by Scott@TargetTestPrep » Sat Jul 07, 2018 5:30 pm
AAPL wrote: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?

A. 1:10
B. 1:5
C. 2:7
D. 3:8
E. 5:7
Let x = the number of ounces of 5 cents per ounce rice and y = the number of ounces of 6 cents per ounce rice. We can create the following equation:

1.2(5x + 6y) = 7(x + y)

6x + 7.2y = 7x + 7y

0.2y = x

y = 5x

1/5 = x/y

Answer: B
.

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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