Mr A.

asked • 07/06/19

Roger bought some $0.34 stamps and some $0.20 stamps, and spent $2.90. How many of each type of stamp did Roger buy?

Roger bought some $0.34 stamps and some $0.20 stamps, and spent $2.90. How many of each type of stamp did Roger buy?


For the most part I figured: 0.34x + 0.20y = $2.90


Since it doesn't signify the total number of stamps bought, I have made a few possible combinations.


If x = 0, y = 14.5.

If x = 8.5, y = 0.


Thus 'x' can't exceed 8.5 and 'y' can't exceed 14.5.


When 'x' is between 1-8, 'y' has the following possibilities with: x = 1, y = 12.8; x = 2, y = 11.1; ...etc.

x: {1, 2, 3, 4, 5, 6, 7, 8}

y: {12.8, 11.1, 9.4, 7.7, 6, 4.3, 2.6, 0.9}


Please help! What is the correct method and correct answer?

2 Answers By Expert Tutors

By:

Mr A.

Non-negative integer? I think I understand that it can't exceed parameters where (x = 8.5, y = 0) and (x = 0, y = 14.5). Thus, the numbers can't be negative -- where 'x' or 'y' are less than '0' -- nor can't exceed 8.5 and 14.5. I am not quite sure whether I had to list all these possibilities in order to find the correct answer. Is there another way, perhaps a shorter way, that doesn't require me to make a list or chart to guess from, is what I am wondering? Looking at the possibilities x = 5, y = 6 seems pertinent. However, it took me too long to figure out this problem.
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07/06/19

Edward C.

tutor
If you consider the amount he spends in cents, then obviously he must spend a multiple of 20 cents on the 20 cent stamps. If you multiply 34 cents by any integer that does not end in a 5 or a 0 and then subtract that amount from 290 you will not get a multiple of 20. Therefore the number of 34 cent stamps that he bought must be a multiple of 5, and the only multiples of 5 that lies between 0 and 8.5 are 0 and 5. 0 doesn’t work because 290 is not a multiple of 20, so he must have bought 5 34 cent stamps.
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07/06/19

Mr A.

Is there a way for me to answer this question without utilizing the method of trial and error -- listing all the values between these parameters -- to get a simple solution. If there is time limit, I wouldn't have the chance to answer this question without having to list all the variables of 'y' with respect to 'x'. Is there a way for me to answer this question by using "systems of equations," or something else, without utilizing the method of "trial and error." Or is my way the only way,
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07/06/19

Edward C.

tutor
If you take the time to read and understand my comment above, you will see that there is no trial and error involved. He cannot have bought zero 34 cent stamps because 290 is not a multiple of 20. He cannot have bought 1 34 cent stamp because 290 minus 34 ends in a 6 so it cannot be a multiple of 20. Similarly, he cannot have bought 2 34 cent stamps because 2*34 ends in an 8, and 290 minus something that ends in an 8 will end in a 2, so it cannot be a multiple of 20. Same idea for 3, 4, 6, 7, 8 and 9 34 cent stamps. He cannot have bought 10 34 cent stamps because 340 is greater than 290. So logically, the only number of 34 cent stamps he could have bought is 5, for 170 cents. The remaining 120 cents must have been spent on 6 20 cent stamps.
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07/06/19

Mr A.

Sir, I understand but If you think about what you are saying, you are expecting someone "to know" multiples of "20" in conjunction with "34" to see if it adds up to 290. What you are doing is calculating it in your head, which -- for the most part -- most people cannot do. Thus, without "trial and error," or plugging and chugging, you cannot figure out the most viable answer. And the answer has to lie in between the parameters I mentioned above: 'x' can't be < 0 & 'y' can't be < 0; 'x' can't be > 8.5 & 'y' can't be > 14.5 Thus, the answer has to be a "whole number," not "integer," since it has to be positive -- "0" or greater -- in between ranges I listed above. Therefore, rational answers cannot be accepted either. I was asking if there may be an another way that I may not be thinking of that doesn't require me the tedious task of "plugging and chugging" multiples of 20 or 34, or both, in conjunction to add up to 290. Or perhaps a more systematic way. How about using a graphing calculator TI-83 Plus? I hope I am making sense.
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07/06/19

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