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Rasheed bought two kinds of candy bars, chocolate and toffee, that came in packages of 2 bars each. He handed out 2/3 of

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Rasheed bought two kinds of candy bars, chocolate and toffee, that came in packages of 2 bars each. He handed out 2/3 of

by BTGmoderatorDC » Tue May 12, 2020 8:57 pm

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Rasheed bought two kinds of candy bars, chocolate and toffee, that came in packages of 2 bars each. He handed out 2/3 of the chocolate bars and 3/5 of the toffee bars. How many packages of chocolates bar did Rasheed buy?

(1) Rasheed bought 1 fewer package of chocolate bars than toffee bars.

(2) Rasheed handed out the same number of each kind of candy bar.


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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that came in packages of 2 bars each. He handed out 2/

by deloitte247 » Sun May 17, 2020 2:46 pm

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Let the chocolate candy bar = c
Let the toffee candy bar = t
Rasheed handed out 2/3 of 2c - chocolate candy bars
He also handled out 3/5 of 2T - toffee candy bars
Target question => how many packages of chocolate bars did Rasheed buy?

Statement 1: Rasheed bought 1 fewer package of chocolate bars than toffee bars.
c = t - 1
The value of c and t are both unknown. So, we cannot evaluate c, therefore, statement 1 is NOT SUFFICIENT.

Statement 2: Rasheed handed out the same number of each kind of candy bar.
(2/3) * 2c = (3/5) * 2t
The value of c and t are both unknown. So, we cannot evaluate c, therefore, statement 2 is NOT SUFFICIENT.

Combining both statements together:
c = t - 1 or t = c + 1
$$\frac{2}{3}\cdot\frac{2c}{1}=\frac{3}{5}\cdot\frac{2t}{1}$$
$$\frac{4c}{3}=\frac{6t}{5}$$
Substitute t = c + 1
$$\frac{4c}{3}=\frac{6\left(c+1\right)}{5}$$
$$\frac{4c}{3}=\frac{6c+6}{5}$$
$$20c=18c+18$$
$$20c-18c=18$$
$$c=\frac{18}{2}=9$$
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