GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The sum of the positive integers from 1 through n can be cal
[#permalink]
14 Apr 2018, 02:31
14 Apr 2018, 02:31
Expert Reply
1
Bookmarks
00:00
Question Stats:
85% (01:49) correct
14% (01:07) wrong based on 87 sessions
Hide Show timer Statistics
The sum of the positive integers from 1 through n can be calculated by the formula \(\frac{n(n + 1)}{2}\)
Quantity A |
Quantity B |
The sum of the multiples of 6 between 0 and 100 |
The sum of the multiples of 8 between 0 and 100 |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: The sum of the positive integers from 1 through n can be cal
[#permalink]
14 Apr 2018, 22:48
14 Apr 2018, 22:48
3
Find the value of \(n\) for each quantity
For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))
sum \(= 17 * \frac{96}{2}= 816\)
For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))
sum = \(13 * 96/2 = 624\)
clearly \(A\) is bigger
For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))
sum \(= 17 * \frac{96}{2}= 816\)
For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))
sum = \(13 * 96/2 = 624\)
clearly \(A\) is bigger
Re: The sum of the positive integers from 1 through n can be cal
[#permalink]
15 Apr 2018, 12:20
15 Apr 2018, 12:20
4
amorphous wrote:
Find the value of \(n\) for each quantity
For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))
From formula, sum \(= 17 * \frac{18}{2} = 17*9\)
For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))
From formula, sum = \(13 * \frac{14}{2} = 13*7\)
clearly \(A\) is bigger
For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))
From formula, sum \(= 17 * \frac{18}{2} = 17*9\)
For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))
From formula, sum = \(13 * \frac{14}{2} = 13*7\)
clearly \(A\) is bigger
Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13
So A is bigger
