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If a and b are both integers, how many possible solutions are there to
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Updated on: 19 Jan 2015, 04:57
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50 = 3a + 2b 7 > –a If a and b are both integers, how many possible solutions are there to the system above? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8
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Originally posted by u0422811 on 22 May 2011, 20:32.
Last edited by Bunuel on 19 Jan 2015, 04:57, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: If a and b are both integers, how many possible solutions are there to
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23 May 2011, 18:35
u0422811 wrote: 50 = 3a + 2b
7 > –a
If a and b are both integers, how many possible solutions are there to the system above?
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8 Remember that –a = a e.g. –3 = 3 = 3 If a < 7, this implies that the absolute value of a is less than 7. So a could range from 6 to 6 (since a can only be an integer). Those are 13 values. Now look at this: 50 = 3a + 2b 50  3a = 2b (50  3a)/2 = b Since b has to be integer too, 50  3a must be divisible by 2. Since 50 is even, a should be even too (Even  Even = Even). From 6 to 6, there are 7 even numbers and 6 odd numbers. So there are 7 possible solutions (a = 6, b = 34; a = 4, b = 31 etc)
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Re: If a and b are both integers, how many possible solutions are there to
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22 May 2011, 22:10
50 = 3a + 2b Since the sum of 3a and 2b is to be even (50) either both are odd or both are even. since 2b cant be odd, 3a and 2b should both be even. for 3a to be even, a can be : 2,4,6,8, and 0. however, 7 > –a
This means that 7<a<7 If a is positive/zero: it can have values 0,2,4 and 6. cant be 8. 4 solutions.
But we have to consider negative values for a too. Similar to above negative value of 3a will mean subtracting a number from 2a. for subtracting from an even number (2a) to result in 50, the product 3a has to end with an even digit. so a can be 6,4,2.
e.g. if a = 4, 3a= 12 and b= 31
we have 7 solutions uptil now, considering all possible values that 2 can take within this bracket of 7<a<7 and a and b being integers.
Note: We dont have to think of values of b, since a is the one with constraints. if b was ve, a will have to be positive and a much higher valuse than 7, so these are not to be considerd. Answer is 7 possible solutions.



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Re: If a and b are both integers, how many possible solutions are there to
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25 May 2011, 22:12
possible values 7<a<7 for all even 'a's there is an integer value for b. thus the solution points are 0, + 2,4,6.
hence 7. D



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Re: If a and b are both integers, how many possible solutions are there to
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19 Jan 2015, 13:19
Hi All, In this question, you can use Number properties to your advantage (as some of the others posters have pointed out). If you got "stuck" on this question, then you should note that the answer choices are relatively small and the math is basic arithmetic. This means that you can use "brute force" to get to the answer. You can easily determine the solutions by taking some notes and doing some basic math. We're told: 1) A and B are both INTEGERS 2) 3A + 2B = 50 3) 7 > A We're asked for the total number of POSSIBLE solutions to the equation. From the answers, we know that there are at least 4 and at most 8. Let's find them.... Keep things simple at first...and take notes so that you can make deductions... IF.... A = 0, then B = 25 IF.... A = 1, then B = 23.5 This is NOT a possible solution (since B is NOT an integer). This tells us that A CANNOT be ODD, since that causes B to become a noninteger. From here on, we WON'T WASTE TIME testing ODD numbers for A.... IF.... A = 2, then B = 22 IF.... A = 4, then B = 19 Notice the pattern emerging: when A increases by 2, B decreases by 3..... IF.... A = 6, then B = 16 We're not allowed to go any higher (the inequality limits us here). So far, we have 4 answers. The question NEVER stated that A and B couldn't be negative though....and that 'absolute value' IS there for a REASON..... IF.... A = 2, then B = 28 IF... A = 4, then B = 31 IF... A = 6, then B = 34 Now, we're not allowed to go any lower (again, the inequality limits us here). We now have 3 additional answers. 4 + 3 = 7 total answers. Final Answer: GMAT assassins aren't born, they're made, Rich
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If a and b are both integers, how many possible solutions are there to
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20 Jul 2015, 00:58
i wonder what the stats were if one of the answers was "13". i must admit that i would mark that one...



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Re: If a and b are both integers, how many possible solutions are there to
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18 Oct 2015, 09:54
u0422811 wrote: 50 = 3a + 2b 7 > –a
If a and b are both integers, how many possible solutions are there to the system above?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8 My take is Option D. Used a simple approach. Going by 2nd equation, possible values of a are : 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6 1st equation: 50 = 3a + 2b => b = (503a) / 2 As per the question stem, both a and b are integers. Hence 50  3a should be even. Hence, a should be even  This gives us 6 values. Now remember 50  3a = even, if a is 0. So we have 7 values for a (6, 4, 2, 0, 2, 4, 6). Total 7 solutions. Thanks, Chanakya Hit kudos if you like the explanation!!



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If a and b are both integers, how many possible solutions are there to
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Updated on: 18 Oct 2015, 12:31
u0422811 wrote: 50 = 3a + 2b 7 > –a
If a and b are both integers, how many possible solutions are there to the system above?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8 Its very simple.Before solve this solution lets discus 2 things: 1) when we write x this means the same as x. 2) Mod is very easy concept if you solve mod question by considering as a distance. when a mod is written as x(a) = b, this means the distance from point 'a' (both side left and right of 'a' on number line) is b. x(a) < b means the distance is between the two extreme distance(left and right side of 'a' on number line, considering the max distance is 'b' from 'a'  as per this scenario.....hence the value of 'a' must be between these two extremes. x(a) > b means the distance is greater than the distance of 'b'..i.e the value of a could be anywhere more than 'b'. Now come to the question. First its givena < 7 ==> a < 7 ===> a0 < 7==> the distance from zero is less than 7. So the point will be 7 and 7, as distance from 0 to (7) is 7 and distance from 0 to 7 is 7. Thus, the value of a will lie in between 7 to 7....i.e the value of a (integer given in ques) would be 6, 5, 4,3,2,1,0,1,2,3,4,5,6. Now, lets move to equation 3a + 2b = 50 ==> b = 25 (3/2)a. According to question, b is an integer, hence to make b integer a must be divisible by 2. Now remove the value which can not be divisible by 2 from the possible values of a. It will remain with 6,4,2,0,2,4,6...i.e total = 7... hence ans is D+1 Kudos if it helped you.
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Originally posted by VikashAlex on 18 Oct 2015, 12:20.
Last edited by VikashAlex on 18 Oct 2015, 12:31, edited 1 time in total.



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Re: If a and b are both integers, how many possible solutions are there to
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18 Oct 2015, 12:26
damn..I did an error an wrote 7<a and this messed up everything



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Re: If a and b are both integers, how many possible solutions are there to
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18 Oct 2015, 18:19
Hi mvictor, It's perfectly fine to make mistakes during practice, but there's a bigger 'takeaway' from all of this. Silly/little mistakes tend to 'kill' Test Takers, and regardless of how far you might be from your score goal, they're likely hurting your performance. You'd be amazed how many of the questions you face on Test Day are 'gettable'... as long as you take the necessary notes, stay organized and do the work ON THE PAD. GMAT assassins aren't born, they're made, Rich
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Re: If a and b are both integers, how many possible solutions are there to
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27 Feb 2019, 15:08
u0422811 wrote: 50 = 3a + 2b 7 > –a
If a and b are both integers, how many possible solutions are there to the system above?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8 Given,
a<7
7<a<7
7>a>7
7<a<7.
a = 6,  5, 4, 3, 2, 1, 0, 1 ,2, 3, 4, 5, 6,
3a + 2b = 50
2b = 50  3a
b = 50  3a / 2.
Now , both a and b are integers thus 3a has to be even. In order to do so, a has to be even.
There are 7 even values of a. For each even value of a , b will have an integer value.
Thus , correct answer is D.




Re: If a and b are both integers, how many possible solutions are there to
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