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The sum of all the multiples of 6 between –126 and 342, incl
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15 Sep 2017, 08:11
15 Sep 2017, 08:11
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Quantity A |
Quantity B |
The sum of all the multiples of 6 between –126 and 342, inclusive |
8,502 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The sum of all the multiples of 6 between –126 and 342, incl
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06 Dec 2017, 11:09
06 Dec 2017, 11:09
3
If we look closely, we only have to sum Integers from 127 to 342, incl. as the negative 126 Integers will cancel out the 126 positive Integers.
=> We have an evenly spaced sequence (Multiples of 6). Hence we can take the first value of the sequence + the last value of the sequence and divide the sum by 2 to obtain the mean of the sequence. We multiply the mean times the number of Integers.
(132+342)/(2) * ((342-132/6) +1)) = 357 * 36 = 8532 Hence A is the answer
=> We have an evenly spaced sequence (Multiples of 6). Hence we can take the first value of the sequence + the last value of the sequence and divide the sum by 2 to obtain the mean of the sequence. We multiply the mean times the number of Integers.
(132+342)/(2) * ((342-132/6) +1)) = 357 * 36 = 8532 Hence A is the answer
Re: The sum of all the multiples of 6 between –126 and 342, incl
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12 Dec 2017, 22:05
12 Dec 2017, 22:05
4
Carcass wrote:
Quantity A |
Quantity B |
The sum of all the multiples of 6 between –126 and 342, inclusive |
8,502 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Let me explain in a broader way-
Now we need the sum of multiples of 6 from -126 to 342.
SO to add these multiples we can find that number from -126 to +126 will cancel out, so we only need to find the sum of multiples of 6 from number 127 to 342
NOw there is a formula to find the number of multiples;
Multiple of x in the range = \(\frac{(last multiple of "x" in the range - first multiple of "x" in the range)}{x}\) +1
now we need to find the first multple of 6 in the range from 127 to 342.
The first multiple is 6 *22 = 132
and the last multiple is 6 * 57 =342
Now multple of 6 in the range from 127 to 342 =\(\frac{(342 - 132)}{6} + 1\)
= 36 numbers which are the multiple of 6 in the range from 127 to 342
Now
here we can also write as 6*22 + 6*23 + 6*24 + ...+6*54
or 6 (22 + 23 + 24 + ...)
SInce there are total of 36 numbers so the mean of the number * total number will give us the sum i.e
Mean of the numbers will be the 17th and 18th number i.e = 39 and 40 (in the range from 22 to 57)
So the sum (22 + 23 + 24 + ....+ 57)= \(\frac{(39 + 40)}{2} * 36\) = 1422
Therefore the total sum of multiple of 6 from the range 127 to 342 = 1422 * 6 =8532
* Note
There is a formula to find the sum of 'X' consecutive number
if it is odd then then sum will be the = middle number * X
if it is even then the sum will be = mean of the numbers in the range * X
