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As a bicycle salesperson, Norman earns a fixed salary of $20 [#permalink]

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15 Jan 2008, 10:20

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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

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

Right off the bat it's obvious that II works. Unfortunately that only eliminates option A. Let's try statement 1.

1. y > 2x. This doesn't necessarily have to be true. It would be true if each bicycle made him the same amount of commission, but that's not true. There are different tiers of commission that make it possible to sell fewer than 2x bikes and still make more than twice as much money.

Example:

Selling 6 bikes: $20 + 6(6) = $56 Selling 11 bikes: $20+6(6)+5(12) = $116 which is more than twice as much as $56, but you don't have to sell 2x bikes.

2. y > x

no way you can sell fewer bikes than last week and earn twice the money

3. y>3

This is true because even if he sold 0 bikes last week. He still made $20 base rate. If y = 3 then he made $20+3(6) = $38 which doesn't even double his base rate of $20. y must be > 3 for this to work.

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

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

II is obvs. III is also pretty obvs. If we think of the least scenario: 26 (1 bike sold) if the salarly is twice as large, then it def has to be more than 3 bikes.

I: we can have 26 and thus we need to have 52 for the 2nd week salary. This is obvs > than 2x

but we can have $128 we have 6,6. Then to get double we would need 256. well x=12, that means y would be greater than 24 for I, but we can obviously see that 12*18 +128 is not going to be 256 so y does not have to be greater than 2x.

mbafall - let me try an equation based approach here:

The salary equation is a function of number of bicycles sold. Depending on the number of cycles sold the salary equation (for B bicycles) would be:

S = 20+6*B for B = (0,6) S = 20+36+12*(B-6) or 12*B-16 for B = (7,12) and S= 20 + 36 + 72 +18*(B-12) or 18*B-88 for B > 12

We need to check for three scenarios, y>3, Y>x and Y>2x.

Now the value for Salary is linearly related to the number of bicycles under all conditions, so y would be always greater than x if salary from y bicycles is to be greater than salary from x bicycles.

To test whether y needs to be greater than 3, lets see how equations behave when y is in the region (0,6)

If y is within (0,6) then salary for week 2 is 20+6*y and the equation is 20+6*y > 2*(20+6*x) or 6*y > 12*x + 20 or y > 2*x + 3.66 If x = 0 then minimum value for y is 3.66 so y > 3

To test for y > 2x, lets just check at border points, lets say if x=12 then salary is 128 in week 1 and for salary to be 256 in week 2, the value of y would be governed by 18*y-88 (as we already know that y>x) so 18*y-88>256 or Y >344/18 or y > 19.xx or Y=20 would suffice so y need not be greater than 2x. Hence only 2 and 3 must be true.

While the equation based approach is more sound - in the current context where equation is a step function, it makes more sense to plug numbers along various ranges and see. Plugging in approach is always likely to be faster.

mbafall - let me try an equation based approach here:

The salary equation is a function of number of bicycles sold. Depending on the number of cycles sold the salary equation (for B bicycles) would be:

S = 20+6*B for B = (0,6) S = 20+36+12*(B-6) or 12*B-16 for B = (7,12) and S= 20 + 36 + 72 +18*(B-12) or 18*B-88 for B > 12

We need to check for three scenarios, y>3, Y>x and Y>2x.

Now the value for Salary is linearly related to the number of bicycles under all conditions, so y would be always greater than x if salary from y bicycles is to be greater than salary from x bicycles.

To test whether y needs to be greater than 3, lets see how equations behave when y is in the region (0,6)

If y is within (0,6) then salary for week 2 is 20+6*y and the equation is 20+6*y > 2*(20+6*x) or 6*y > 12*x + 20 or y > 2*x + 3.66 If x = 0 then minimum value for y is 3.66 so y > 3

To test for y > 2x, lets just check at border points, lets say if x=12 then salary is 128 in week 1 and for salary to be 256 in week 2, the value of y would be governed by 18*y-88 (as we already know that y>x) so 18*y-88>256 or Y >344/18 or y > 19.xx or Y=20 would suffice so y need not be greater than 2x. Hence only 2 and 3 must be true.

While the equation based approach is more sound - in the current context where equation is a step function, it makes more sense to plug numbers along various ranges and see. Plugging in approach is always likely to be faster.

Well done! i was trying to form one equation for all three cases. Clearly three different equations are required based on the range!

Thanks

gmatclubot

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26 Feb 2011, 11:24

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