Question

Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3:5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is

Answer 1

1. Let the Variables Represent

Let:

• Number of persons standing ahead of Pinky = \( 3x \)
• Number of persons standing behind Pinky = \( 5x \)
• Here, \( x \) is a positive integer

The ratio is:

\[ \frac{3x}{5x} = \frac{3}{5} \]

2. Total Persons in the Queue

Total persons = Persons ahead + Pinky herself + Persons behind

\[ 3x + 1 + 5x = 8x + 1 \]

Given: Total is less than 300

\[ 8x + 1 < 300 \]

Subtract 1:

\[ 8x < 299 \]

Divide by 8:

\[ x < \frac{299}{8} \Rightarrow x < 37.375 \]

Maximum possible integer value of \( x \) is 37

3. Calculate Maximum Number Ahead of Pinky

\[ \text{Maximum } 3x = 3 \times 37 = 111 \]

Therefore, the maximum number of persons ahead of Pinky is: 111

Answer 2

1. Let the Surrounding People be Represented as

Pinky is surrounded by:

• \( 3a \) people in front
• \( 5a \) people behind

Where \( a \) is a positive integer.

 

2. Total Number of People

Total people in the queue = \( 3a + 1 + 5a = 8a + 1 \)

Since the total number of people is less than 300: \[ 8a + 1 < 300 \Rightarrow 8a < 299 \] \[ \Rightarrow a < \frac{299}{8} = 37.375 \]

So the maximum possible integer value of \( a \) is: \[ a = 37 \]

3. Maximum People in Front of Pinky

\[ 3a = 3 \times 37 = 111 \]

Hence, the greatest possible number of people in front of Pinky is: 111