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What is the sum of the multiples of 7 from 84 to 140?

This topic has 5 expert replies and 1 member reply

What is the sum of the multiples of 7 from 84 to 140?

Post Mon Sep 25, 2017 5:59 pm
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?

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Post Fri Dec 15, 2017 10:23 am
Vincen wrote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016
We can use the following equation to determine the quantity:

(largest multiple in the set - smallest multiple in the set)/7 + 1 = number of multiples of 7

(140 - 84)/7 + 1

56/7 + 1 = 9

Next, let’s determine the average of this evenly spaced set.

average = (first number + last number)/2

average = (84 + 140)/2 = 224/2 = 112

Thus, the sum is 9 x 112 = 1,008.

Answer: C

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Post Fri Dec 15, 2017 1:26 pm
Vincen wrote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?
In other words, 84 + 91 + 98 + . . . 140 = ?

Since the values are EQUALLY SPACED, we can use the rule: SUM = [(FIRST + LAST)/2][# of values]

NUMBER of values
Here's a nice rule: If x and y are multiples of k, then the number of multiples of k from x to y inclusive = [(y-x)/k] + 1
So, for example, the NUMBER multiples of 3 from 6 to 21 inclusive = [(21 - 6)/3] + 1 = [15/3] + 1 = 6

So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/7] + 1
= [56/7] + 1
= 9

------------------------------------

Now apply the formula:
SUM = [(FIRST + LAST)/2][# of values]
= [(84 + 140)/2][9]
= [224/2][9]
= [112][9]
= 1008
= C

Cheers,
Brent

_________________
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Post Fri Dec 15, 2017 10:23 am
Vincen wrote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016
We can use the following equation to determine the quantity:

(largest multiple in the set - smallest multiple in the set)/7 + 1 = number of multiples of 7

(140 - 84)/7 + 1

56/7 + 1 = 9

Next, let’s determine the average of this evenly spaced set.

average = (first number + last number)/2

average = (84 + 140)/2 = 224/2 = 112

Thus, the sum is 9 x 112 = 1,008.

Answer: C

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Post Fri Dec 15, 2017 1:26 pm
Vincen wrote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?
In other words, 84 + 91 + 98 + . . . 140 = ?

Since the values are EQUALLY SPACED, we can use the rule: SUM = [(FIRST + LAST)/2][# of values]

NUMBER of values
Here's a nice rule: If x and y are multiples of k, then the number of multiples of k from x to y inclusive = [(y-x)/k] + 1
So, for example, the NUMBER multiples of 3 from 6 to 21 inclusive = [(21 - 6)/3] + 1 = [15/3] + 1 = 6

So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/7] + 1
= [56/7] + 1
= 9

------------------------------------

Now apply the formula:
SUM = [(FIRST + LAST)/2][# of values]
= [(84 + 140)/2][9]
= [224/2][9]
= [112][9]
= 1008
= C

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
Come see all of our free resources

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Post Fri Oct 13, 2017 10:28 am
Quote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?
Hi Vincen,
Let's take a look at your question.

First of all we need to find the number of terms between 84 to 140 inclusive.
Since we are talking about multiples of 7, lets find out the term numbers of 84 and 140 in the sequene of multiples of 7.
$$\frac{84}{7}=12$$
84 is 12th term.

$$\frac{140}{7}=20$$
140 is 20th term.

Total number of terms = n = 20 - 12 + 1 = 9

Sum of n terms in geometric series$$ =\frac{n}{2}\left(a_1+a_n\right)$$
$$=\frac{9}{2}\left(84+140\right)$$
$$=\frac{9}{2}\left(224\right)$$
$$=\frac{9}{2}\left(84+140\right)$$
$$=9\left(112\right)$$
$$=1008$$

Therefore, Option C is correct.
Hope this helps.

I am available, if you'd like any follow up.

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Post Tue Sep 26, 2017 9:48 pm
Vincen wrote:
What is the sum of the multiples of 7 from 84 to 140, inclusive?

A) 896
B) 963
C) 1008
D) 1792
E) 2016

The OA is C.

Is there any way to solve it without listing the numbers?
Sum of the multiples of 7 from 84 to 140, inclusive = 84 + 91 + ... + 140.

Number of multiples of 7 from 84 to 140 = 140/7 - 84/7 + 1 = 20 - 12 + 1 = 9.

Now, write these numbers twice, once in ascending order and then in descending order:

S = 84 + 91 + .... + 133 + 140
S = 140 + 133 + .... + 91 + 84

The sum of each pair is 224. As the total number of pairs are 9.

2S = 224*9 or S = 112*9 = 1008.

Thus, the correct answer is C.

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