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n200080 [17]
3 years ago
10

John divided his souvenir hat pins into two piles. The two piles had an equal number of pins. He gave his brother one-half of on

e-third of one pile. John had 66 pins left. How many pins did John originally have?
Mathematics
1 answer:
Serggg [28]3 years ago
5 0

Answer: John had 72 pins originally.

Step-by-step explanation:

Let x represent the total number of pins that John had originally.

John divided his souvenir hat pins into two piles. The two piles had an equal number of pins. This means that each pile has x/2 pins.

He gave his brother one-half of one-third of one pile. This means that the number of pins that he gave his brother is

1/2 × 1/3 × x/2 = x/12 pins

The total amount left is

x/2 + (x/2 - x/12) = x/2 + 5x/12 = 11x/12

John had 66 pins left. Therefore

11x/12 = 66

11x = 66 × 12 = 792

x = 792/11 = 72

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Answer:

\frac{5}{12}

Step-by-step explanation:

The given problem is a subtraction problem between fractions. The first step in a problem like this is to convert both fractions to a common denominator. In other words, multiply both the numerator and denominator of a fraction by the same value such that the denominator of one of the fractions is the same as the other fraction. The numerator of a fraction is the value over the fraction bar, whereas the denominator is the value under it. Multiplying any fraction by a number over itself is the same as multiplying it by one, thus, the equation will remain true, and be equivalent to the original expression.

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\frac{5}{12}

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