The fruit punch in a bowl was 12% orange juice. Several cups

Question

The fruit punch in a bowl was 12% orange juice. Several cups of pure orange juice were then added, and as a result, 20% of the punch is now orange juice. By what fraction did the quantity of orange juice increase?

A) \(\frac{1}{2}\)
B) \(\frac{2}{3}\)
C) \(\frac{3}{4}\)
D) \(\frac{5}{6}\)
E) \(\frac{5}{3}\)

A)  \frac{1}{2} 
B)  \frac{2}{3} 
C)  \frac{3}{4} 
D)  \frac{5}{6} 
E)  \frac{5}{3}

stne · Answer

Initially let there be  x  amt of fruit punch , later let  y  amt of pure orange juice be added.

The amt of non orange juice remains constant.

.88(x) = .8(x+y)

x=10y

So  .12 (10y ) = 1.2y  -> initial concentration of orange juice given .

.2 (11y ) = 2.2y  -> After adding extra pure orange juice, final concentration given ...

2.2y -1.2y = y  -> extra orange juice

Required fraction  \frac{y}{1.2y} = \frac{5}{6}

Ans D

Hope it helped.

adconsectetur · Answer

Weighted average

FP ------ Mixed FP ------- Pure Orange juice

12% ---- 20% ----- 100%

if FP = 10 then Pure Orange juice = 1 and Mixed FP = 11

OJ = 1.2 ----> OJ =2.2 Fraction increase = 5/6.

dzhou751 · Answer

You start with 12/100 for the fraction of orange juice cups and you end up with 20/100. Add x (number of cups) to both numerator and denominator for 12/100 and cross multiply to get 1200 + 100x = 2000 + 20x. 80x + 800 and x = 10. Therefore, you added 10 cups and it started at 12 cups: 10/12 = 5/6.

The fruit punch in a bowl was 12% orange juice. Several cups

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The fruit punch in a bowl was 12% orange juice. Several cups

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