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21 Oct 2012, 21:29
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Responding to a pm:
How do we apply the strategy discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... -of-thumb/
in the question given below? It seems to be time consuming.
How many positive integral solutions are there for the equation 7 x + 3 y=441?
a) 21
b) 20
c) 22
d) 19
Actually, it isn't. It is very simple.
The first thing we need to do is try to find one solution by brute force. What would be the easiest case? It would be very easy if 441 were divisible by 7 or 3. In fact, it is! Since 441 is divisible by 7, I already have the first solution: x = 63 and y = 0 because 7*63 + 0 = 441
What will be the next solution? Decrease x by 3 and increase y by 7 so the next solution is (60, 7)
It will go on and x will keep decreasing till it reaches 0 since 441 is divisible by 3 as well. The last non negative solution will be 7*0 + 3*147 = 441
x will take these values 63, 60, 57, ... 3, 0
Since we need POSITIVE integral solutions, we cannot accept either x or y equal to 0 so acceptable solutions will be x = 3, 6, 9, ... 60
How many values are these? n = (60 - 3)/3 + 1 = 20 (This is an AP. No of terms = (Last term - First term)/Common diff + 1)
Check this post if AP is not clear: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/03 ... gressions/
Answer (B)
How do we apply the strategy discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... -of-thumb/
in the question given below? It seems to be time consuming.
How many positive integral solutions are there for the equation 7 x + 3 y=441?
a) 21
b) 20
c) 22
d) 19
Actually, it isn't. It is very simple.
The first thing we need to do is try to find one solution by brute force. What would be the easiest case? It would be very easy if 441 were divisible by 7 or 3. In fact, it is! Since 441 is divisible by 7, I already have the first solution: x = 63 and y = 0 because 7*63 + 0 = 441
What will be the next solution? Decrease x by 3 and increase y by 7 so the next solution is (60, 7)
It will go on and x will keep decreasing till it reaches 0 since 441 is divisible by 3 as well. The last non negative solution will be 7*0 + 3*147 = 441
x will take these values 63, 60, 57, ... 3, 0
Since we need POSITIVE integral solutions, we cannot accept either x or y equal to 0 so acceptable solutions will be x = 3, 6, 9, ... 60
How many values are these? n = (60 - 3)/3 + 1 = 20 (This is an AP. No of terms = (Last term - First term)/Common diff + 1)
Check this post if AP is not clear: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/03 ... gressions/
Answer (B)
22 Oct 2012, 21:43
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Another query on the same topic:
A citrus fruit grower receives $15 for each crate of
oranges shipped and $18 for each crate of grapefruit
shipped. How many crates of oranges did the grower
ship last week?
(2) Last week the grower received a total of
$38,700 from the crates of oranges and
grapefruit shipped.
My question is,the number of crates will be integer values so why can we not test values for oranges and grape crates and arrive at unique solution? Also,how does one know that testing values on this particular equation will not yield unique values?
Say, no of crates of oranges is r and of grapefruit is g
15r + 18g = 38700
Here, you see that 15 and 18 are very small numbers compared with 38700. So if you have one solution for the values of (r, g), you will find many more (explained below). Obviously, since he ships the crates and makes this amount of money, there has to be an integral solution for the values of (r, g). This means that you will have many integral solutions. Let me show you with numbers:
38700 is divisible by 15 so one solution for (r, g) is (2580, 0). Now there will be many more solutions since you will keep decreasing 2580 by 18 and increasing 0 by 15 e.g. (2562, 15), (2544, 30) etc ... (you don't really need to calculate them... you know there will be many such values) So there is no unique value until and unless some other constraints are given.
If say the equation were something like 15r + 18g = 48, you are likely to find only one set of solution (2, 1). You can't reduce 2 by 18 since it will become negative. In this case, you have a unique solution.
A citrus fruit grower receives $15 for each crate of
oranges shipped and $18 for each crate of grapefruit
shipped. How many crates of oranges did the grower
ship last week?
(2) Last week the grower received a total of
$38,700 from the crates of oranges and
grapefruit shipped.
My question is,the number of crates will be integer values so why can we not test values for oranges and grape crates and arrive at unique solution? Also,how does one know that testing values on this particular equation will not yield unique values?
Say, no of crates of oranges is r and of grapefruit is g
15r + 18g = 38700
Here, you see that 15 and 18 are very small numbers compared with 38700. So if you have one solution for the values of (r, g), you will find many more (explained below). Obviously, since he ships the crates and makes this amount of money, there has to be an integral solution for the values of (r, g). This means that you will have many integral solutions. Let me show you with numbers:
38700 is divisible by 15 so one solution for (r, g) is (2580, 0). Now there will be many more solutions since you will keep decreasing 2580 by 18 and increasing 0 by 15 e.g. (2562, 15), (2544, 30) etc ... (you don't really need to calculate them... you know there will be many such values) So there is no unique value until and unless some other constraints are given.
If say the equation were something like 15r + 18g = 48, you are likely to find only one set of solution (2, 1). You can't reduce 2 by 18 since it will become negative. In this case, you have a unique solution.
24 Oct 2012, 04:01
Kudos
Bookmarks
Thanks a lot karishma for patiently answering a query
+1 KUDOS
all positive integral doubts cleared
now i am confident to solve such kind of questions.
+1 KUDOS
all positive integral doubts cleared
now i am confident to solve such kind of questions.
