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multiples of 7

This topic has 3 expert replies and 6 member replies
j_shreyans Legendary Member Default Avatar
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multiples of 7

Post Mon Sep 08, 2014 10:17 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the sum of the multiples of 7 from 84 to 140, inclusive?

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

    OAC

    How to approach this kind of question?

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    abhasjha Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Sep 08, 2014 10:27 pm
    This question tests your knoweldge of AP series

    84 = 12x7 .. so 84 is 12the term
    140= 20 X7 ..... 140 is 20th term ....

    so total number of terms between 84 and 140 = 9 . ( using the formula total numbers between b and a inclusisve is( b-a ) +1

    now the sum of an AP series is... n/2 ( first term + last Term ) = 9/2 (84 +140 ) = 1008

    Answer C .

    Post Mon Sep 08, 2014 10:30 pm
    j_shreyans 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

    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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    Last edited by Brent@GMATPrepNow on Fri Feb 12, 2016 8:13 pm; edited 1 time in total

    Thanked by: manik11
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    prada Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Feb 12, 2016 4:06 pm
    Brent@GMATPrepNow wrote:
    j_shreyans 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

    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)/3] + 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
    Hi Brent,

    I think here
    So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/3] + 1

    you mean 7 in lieu of 3?

    Thanked by: Brent@GMATPrepNow
    Post Fri Feb 12, 2016 8:14 pm
    prada wrote:
    you mean 7 in lieu of 3?
    You're absolutely right. Good catch.
    I edited my response accordingly.

    Cheers,
    Brent

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    waleedijaz Newbie | Next Rank: 10 Posts Default Avatar
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    Post Sat Apr 30, 2016 4:07 am
    You're absolutely right. Good catch???


    waleeed

    YTarhouni Newbie | Next Rank: 10 Posts Default Avatar
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    8 messages
    Post Tue Sep 05, 2017 6:45 pm
    When it comes to calculating the number of item in a sequence of multiples, i find it easy to divide first term and last term by the multiple then apply the formula of (last-First+1).

    In this case:
    84/7=12
    140/7=20
    #items=20-12+1=9
    from here you can follow any approach that makes you comfortable.

    Imperiex Junior | Next Rank: 30 Posts Default Avatar
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    26 messages
    Post Tue Sep 05, 2017 9:44 pm
    j_shreyans 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

    OAC

    How to approach this kind of question?
    Solution:

    Sum of the multiples of 7 from 84 to 140, inclusive,
    84 + 91 + 98 + 105 + ... +140

    This above series represents an arithmetic series with a common difference d = 7.
    We know that sum of a finite arithmetic series can be calculated using the formula,

    Sum = (First term + Last term)(n/2) where n is the number of terms in the series.
    Therefore, to find sum, we need to find the number of terms in the series above.

    We know that the terms in the arithmetic sequence are the multiples of 7. Therefore,
    First term = 84 = 7 x 12 = 12th multiple of 7
    Last term = 140 = 7 x 20 = 20th multiple of 7

    The number of terms in the series 84 + 91 + 98 + 105 + ... +140
    = 9 (84 and 140 inclusive)
    Therefore, n = 9

    Now,
    Sum = (First term + Last term)(n/2)
    Sum = (84 + 140)(9/2)
    Sum = (224)(9/2)
    Sum = (224 x 9) / 2
    Sum = 2016 / 2
    Sum = 1008

    Therefore, Option C is the correct answer.

    Imperiex Junior | Next Rank: 30 Posts Default Avatar
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    Posted:
    26 messages
    Post Tue Sep 05, 2017 9:58 pm
    A streamlined solution:
    84 / 7 = 12
    140 / 7 = 20
    So, it would be seven multiple of 12, 13, 14, 15, 16, 17, 18, 19 and 20.
    Thus the sum of the multiples of 7 from 84 to 140, inclusive, is:
    7(12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20) = 7 (144) = 1008.

    Answer: C

    Post Mon Sep 11, 2017 10:19 am
    j_shreyans 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

    OAC
    '

    We can use the following formula:

    sum = average x quantity

    Since we have an evenly spaced set of integers, we can calculate the average of the set by using this formula:

    average = (smallest integer in set + greatest integer in set)/2

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

    quantity = (140 - 84)/7 + 1 = 56/7 + 1 = 9

    Thus:

    sum = 112 x 9 = 1008

    Answer: C

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