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Set P consists of the first n positive multiples of 3 and set Q
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23 Mar 2019, 09:00
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95% (hard)
Question Stats:
17% (04:06) correct 83% (02:41) wrong based on 24 sessions
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Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If 'm 'and 'n' are positive integers, is the difference between R and S odd?
I. m is odd and n is even II. m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer
Set P consists of the first n positive multiples of 3 and set Q
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Updated on: 24 Mar 2019, 09:54
kiran120680 wrote:
Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If 'm 'and 'n' are positive integers, is the difference between R and S odd?
I. m is odd and n is even II. m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer
As it can be hard to understand what we're dealing with just by reading, we'll first break things down into bite-size chunks. This is a Precise approach.
We know that R = 3 + 6 + 9 + ... for n multiples so R = 3(1 + 2 + 3... n), i.e. R is 3 * the sum of the first n numbers. Similarly, S = 5 * the sum of the first m numbers. Since R - S is odd only if both R,S are even or both are odd, we need to know if the sums 1+2+...n and 1+2+...m are even or odd. Using the standard arithmetic sum, 1+2+...n = n(n+1)/2 which is even if n or n+1 is a multiple of 4 and odd otherwise. Now that we know exactly what to look for, let's look at our statements.
(1) does not tell us if (m or m+1) and (n or n+1) are multiples of 4. (2) tells us that m+1 is a multiple of 4 but does not tell us if n or n+1 are a multiple of 4. Combining gives us no information on whether n or n+1 is a multiple of 4.
Re: Set P consists of the first n positive multiples of 3 and set Q
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24 Mar 2019, 09:40
DavidTutorexamPAL wrote:
kiran120680 wrote:
Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If 'm 'and 'n' are positive integers, is the difference between R and S odd?
I. m is odd and n is even II. m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer
As it can be hard to understand what we're dealing with just by reading, we'll first break things down into bite-size chunks. This is a Precise approach.
We know that P = 3 + 6 + 9 + ... for n multiples so P = 3(1 + 2 + 3... n), i.e. P is 3 * the sum of the first n numbers. Similarly, Q = 5 * the sum of the first m numbers. Since R - S is odd only if both R,S are even or both are odd, we need to know if the sums 1+2+...n and 1+2+...m are even or odd. Using the standard arithmetic sum, 1+2+...n = n(n+1)/2 which is even if n or n+1 is a multiple of 4 and odd otherwise. Now that we know exactly what to look for, let's look at our statements.
(1) does not tell us if (m or m+1) and (n or n+1) are multiples of 4. (2) tells us that m+1 is a multiple of 4 but does not tell us if n or n+1 are a multiple of 4. Combining gives us no information on whether n or n+1 is a multiple of 4.
(E) is our answer.
Little confused, is there any better way to solve this question.
The logic of this question is indeed hard. Try using numbers to make things concrete; this is a useful tool for approaching questions you don't really get. In this case:
(1) trying m = 1 and n = 2 gives R = 3+6 = 9 and Q = 5. Then R - S = 9 - 5 which is even. trying m = 3 and n = 2 gives R = 9 and S = 5 + 10 + 15 = 30. Then R - S = 30- 9 = 21 which is odd. Insufficient. (2) so we can still pick m = 3 and n = 2, giving R-S odd. We'll keep m = 3 and change to n = 4, giving S =30 like before and R = 3+6+9+12 = 30. Then R-S = 30 - 30 = 0 which is even. Insufficient.
Combined: Both examples given in (2), i.e. (m,n)=(3,2) and (m,n)=(3,4) work when combining the statements as well. So still insufficient.
Hence (E).
Also, I had a typo in the original explanation which I have now fixed. Hope it is clearer _________________
The logic of this question is indeed hard. Try using numbers to make things concrete; this is a useful tool for approaching questions you don't really get. In this case:
(1) trying m = 1 and n = 2 gives R = 3+6 = 9 and Q = 5. Then R - S = 9 - 5 which is even. trying m = 3 and n = 2 gives R = 9 and S = 5 + 10 + 15 = 30. Then R - S = 30- 9 = 21 which is odd. Insufficient. (2) so we can still pick m = 3 and n = 2, giving R-S odd. We'll keep m = 3 and change to n = 4, giving S =30 like before and R = 3+6+9+12 = 30. Then R-S = 30 - 30 = 0 which is even. Insufficient.
Combined: Both examples given in (2), i.e. (m,n)=(3,2) and (m,n)=(3,4) work when combining the statements as well. So still insufficient.
Hence (E).
Also, I had a typo in the original explanation which I have now fixed. Hope it is clearer
The logic of this question is indeed hard. Try using numbers to make things concrete; this is a useful tool for approaching questions you don't really get. In this case:
(1) trying m = 1 and n = 2 gives R = 3+6 = 9 and Q = 5. Then R - S = 9 - 5 which is even. trying m = 3 and n = 2 gives R = 9 and S = 5 + 10 + 15 = 30. Then R - S = 30- 9 = 21 which is odd. Insufficient. (2) so we can still pick m = 3 and n = 2, giving R-S odd. We'll keep m = 3 and change to n = 4, giving S =30 like before and R = 3+6+9+12 = 30. Then R-S = 30 - 30 = 0 which is even. Insufficient.
Combined: Both examples given in (2), i.e. (m,n)=(3,2) and (m,n)=(3,4) work when combining the statements as well. So still insufficient.
Hence (E).
Also, I had a typo in the original explanation which I have now fixed. Hope it is clearer
I too little confused with the solution. Now I understood. Thank you David
Set P consists of the first n positive multiples of 3 and set Q
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24 Mar 2019, 11:17
kiran120680 wrote:
Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If 'm 'and 'n' are positive integers, is the difference between R and S odd?
I. m is odd and n is even II. m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer
Here is how I solved. This is a Yes/No question and you need to find all the areas to see if you are getting consistent answers.
By looking at the statements, I got an idea that while testing conditions, n has to be odd and m has to be even! So I took two examples so it'll be easy to test/solve: Example 1: P={3,6,9} => n=3 and ΣP=18=R & Q= {5,10} =>m=2 and ΣQ=15=S Nowhere in the questions it is mentioned that the multiples should be consecutive. So I'm gonna take different numbers now. Example 2: P={3,9,15} => n=3 and ΣP=27=R & Q= {10,15} =>m=2 and ΣQ=25=S
Another hint: We gotta find if R-S=odd. As per number property rules, odd-even=odd or even-odd=odd(keep this in kind while taking examples. It'll help!)
Moving onto the statements
Statement (1): m is odd and n is even Now the role of examples come in. From example 1, R-S= 18-15=3 ODD......YES example 2, R-S= 27-25=2 EVEN...........NO Different answers.....INSUFFICIENT
Statement (2): m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x m=4x+3=odd (odd+even=odd) and n=2x= even Now this statement is as same as statement 1 Different answers.....INSUFFICIENT
Combining statement (1) + (2) would be same as above. Different answers. INSUFFICIENT
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17% (04:06) correct 
