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As a bicycle salesperson, Norman earns a fixed salary of ...

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As a bicycle salesperson, Norman earns a fixed salary of ...

by hengirl03 » Wed Sep 03, 2008 9:52 am
When I did this problem originally, use the picking numbers technique to find the answer. However, picking numbers didn't really work for I, since some numbers works and other didn't. My question is, how do you know that you are picking all of the numbers that you need to find the answer?

As a bicycle salesperson, Norman earns a fixed salary of $20 per week plus $6 per bicycle for the first six bicycles he sells, $12 per bicycle for the next six bicycles he sells, and $18 per bicycle for every bicycle sold after the first 12. This week, Norman earned more than twice as much as he did last week. If he sold x bicycles last week and y bicycles this week, which of the following statements must be true?

I. y > 2x

II. y > x

III. y > 3

Answer: II and III
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by mayur00 » Wed Sep 03, 2008 11:20 am
Plug in x= 1 to 5 and find closest value of y, you'll find that
1. None satisfy the given condition that his earning in one month was 2x the other.
2. Shows that x>3, since x=1 to 5 don't satisfy given condition
3. y>x (obvious)
4. As x increases from 1 to 5, y increases from 5 to something less than 10. Which shows that y < 2x
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by hengirl03 » Wed Sep 03, 2008 12:01 pm
For I: if x = 4 and y = 9. Then Norman earned ($20+24 = 44) this week, and ($20+36+36 = 92). So y> 2x.

However, if x = 6 and y = 11. Then Norman earned $56 in the first week, and 116 the next. In this case, $116 > 2 × $56, yet y < 2x.



I still don't know how I can make sure that I choose all of the right answers when solving this problem. Then I first did this problem, I chose x = 4 and y = 9. But I didn't consider the situation where x=6 and y=11.
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by pepeprepa » Wed Sep 03, 2008 12:19 pm
It is clear that statement one is the most complicated.

y>2x must be true?
Let's try to find a counter-example, if we cannot so it must be true.
In your example, you want Norman to earn more than twice as much as he did last week.
If we chose x=10 and y=18 we have y<2x
Revenue with 10: 20 + 6*6 + 4*12 = 104
Revenue with 18: 20 + 6*6 + 6*12 + 6*18 = 236

We respect the conditions given by the question and we find that it is possible that Norman earns more than twice as much as he did last week and at the same time with y<2x
That's a counter-example to the statement "y>2x must be true".
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by pepeprepa » Wed Sep 03, 2008 12:22 pm
hengirl03 wrote:For I: if x = 4 and y = 9. Then Norman earned ($20+24 = 44) this week, and ($20+36+36 = 92). So y> 2x.

However, if x = 6 and y = 11. Then Norman earned $56 in the first week, and 116 the next. In this case, $116 > 2 × $56, yet y < 2x.



I still don't know how I can make sure that I choose all of the right answers when solving this problem. Then I first did this problem, I chose x = 4 and y = 9. But I didn't consider the situation where x=6 and y=11.
I did not precisely read your post.
Indeed, when a question asks you " X must be true", you only need to find a counter-example which respects all the conditions given by the question but wich contradicts X. That's what you did in your post, so do not chose I/
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by alanforde800Maximus » Sat Aug 04, 2018 1:46 am
Can we deploy some smart numbers to solve this problem without delving into algebra much?
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by [email protected] » Tue Aug 07, 2018 6:27 pm
Hi All,

To start, we will need to 5 answer choices for reference (as the ability to 'prove' or 'disprove' an answer can help us to eliminate answer choices):

A) I only
B) II only
C) I and II
D) II and III
E) I, II, and III

This Roman Numeral question can be solved with "brute force"; let's map out the possibilities and look for patterns. Based on the given information, here's a table of how much money Norman would make in 1 week (based on the number of bikes sold):

Bikes = Money
0 = $20

1 = $26
2 = $32
3 = $38
4 = $44
5 = $50
6 = $56

7 = $68
8 = $80
9 = $92
10 = $104
11 = $116
12 = $128

13 = $144
14 = $162
Etc. ($18 per additional bike)

We're told that Norman sold X bicycles last week and Y bicycles this week. We also know that he earned MORE THAN TWICE the money he earned in the prior week, so we have to use THAT fact to evaluate what the possibilities could be (within the table above).

II. Y > X

Roman Numeral II is easiest, so we'll start there. Since Norman earned MORE MONEY, he had to have sold MORE bicycles. Thus Y MUST be greater than X (so no heavy work is required here).
Roman Numeral II is TRUE.

III. Y > 3

Here, we can look at the "top" of the table and talk through the possibilities.

If last week, Norman sold ___ bikes last week, then how many would he need to have sold this week, at the MINIMUM, to make MORE than twice the money?
0 bikes....4 or more bikes were sold
1 bike.....6 or more bikes were sold

The second number will just get bigger and bigger. This proves that Y MUST be greater than 3.
Roman Numeral III is TRUE.

I. Y > 2X

For this Roman Numeral, we can continue the work that we did in Roman Numeral II; we have to look to see whether Y is ALWAYS greater than 2X or not...

2 bikes....7 or more bikes were sold
3 bikes....8 or more bikes were sold
4 bikes....9 or more bikes were sold
At this point, notice the ratio of Y to X is getting smaller....?
5 bikes...10 or more bikes were sold.

This last example PROVES that Y isn't always greater than 2X.
Roman Numeral I is NOT ALWAYS TRUE.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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