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Re: A child bus ticket costs [#permalink]
GreenlightTestPrep wrote:
Sonalika42 wrote:
A child bus ticket costs $0.40 and an adult’s ticket costs $0.70. If a group of 11 pays $5.90 ?


Quantity A
Quantity B
Number of children
Number of Adults




IMPORTANT CONCEPT: If the number of children equaled the number of adults, then the AVERAGE ticket price would equal the AVERAGE of $0.40 and $0.70 (which is $0.55)
In this question, the average ticket price = $5.90/11 ≈ $0.54
Since $0.54 is closer to $0.40 (price of child ticket) than it is to $0.70 (price of adult ticket), we can conclude that there must be more children than adults.

Answer:
Show: ::
A


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I love this! it is very helpful
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Re: A child bus ticket costs [#permalink]
1
Based on the condition,
Let's say number of Children = x
number of Adults = y

So,

0.4x + 0.7y = 5.9
Also,
x + y = 11

By solving the two equations simultaneously, we'll get:

x = 6, and y = 5

So

Choice A correct.
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Re: A child bus ticket costs [#permalink]
0.4x + 0.7y = 5.9
x + y = 11


x = 6, and y = 5
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A child bus ticket costs [#permalink]
1
Why doesn't the following logic work?

x - number of children
(11 - x) - number of adults

[0.4x + 0.7(11 - x)] / 11 = 0.59

4x + 77 - 7x = 5.9*11

-3x = -12.1

x = 4.1

The answer is wrong, but why?



-----------------------------------------------

Another way, which now after thinking is more correct, but still doesn't work, why?

Average 0.59/11 = 0.0536

[0.4x + 0.7(11 - x)] / 11 = 0.0536

[0.4x + 0.7(11 - x)]= 0.59

[4x + 7(11 - x)]= 5.9

-3x = 5.9 - 77

x = 23.7
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