kaplan question of Average

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pradeepkaushal9518: Legendary Member; Posts: 1309; Joined: Wed Mar 17, 2010 11:41 pmFollowed by: 5 members; Thanked: 33 times

by pradeepkaushal9518 » Tue May 04, 2010 7:19 am

if the average (arithmetic mean) of 18 consecutive odd integers is 534,then the least of these integers is

1. 517
2.518
3.519
4.521
5.525

plz explain yr answers

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Source: Beat The GMAT — Problem Solving | Tue May 04, 2010 7:19 am

rn4gmat: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Sun Apr 18, 2010 7:44 pm; Thanked: 3 times

by rn4gmat » Tue May 04, 2010 8:15 am

Let the first term be X
The 18th term will be : X+34

So :

X + X + 34 = 2*534

2X = 1068-34 = 1034

so X = 517.

So the least term is 517.

Hope this helps.

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Tue May 04, 2010 8:20 am

Least no = x
18th no = x +34
x+x+34 = (534)*2 = 1068
2x = 1034
x = 517

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pradeepkaushal9518: Legendary Member; Posts: 1309; Joined: Wed Mar 17, 2010 11:41 pmFollowed by: 5 members; Thanked: 33 times

by pradeepkaushal9518 » Tue May 04, 2010 8:23 am

thanks buddy

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akhpad: Legendary Member; Posts: 809; Joined: Wed Mar 24, 2010 10:10 pmFollowed by: 4 members; Thanked: 50 times

by akhpad » Tue May 04, 2010 8:23 am

Let first term be X
X, X+2, X+4, X+6
Sum = 18X + (2+4+6+ ... 17th terms) = 18X + 2(1+2+3 ... 17th terms) = 18X + 17*18
Average = X + 17 = 534
X = 517

Answer: A

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sk818020: Master | Next Rank: 500 Posts; Posts: 214; Joined: Mon Mar 29, 2010 1:46 pm; Location: Houston, TX; Thanked: 37 times; GMAT Score:700

by sk818020 » Tue May 04, 2010 8:32 am

If you don't want to do any math you can solve this one just by realizing that the average of any even set of odd integers will also be the median and that the the 9th and 10th members of that set will be 1 less and 1 more than the average. For this, the 9th = 533 and 10th = 535.

Counting down to the 1st element, 531=8th, 529=7th, 527=6th, 525=5th, 523=4th, 521=3rd, 519=2nd, 517=1st.

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pradeepkaushal9518: Legendary Member; Posts: 1309; Joined: Wed Mar 17, 2010 11:41 pmFollowed by: 5 members; Thanked: 33 times

by pradeepkaushal9518 » Tue May 04, 2010 8:35 am

f you don't want to do any math you can solve this on just by realizing that the average of any even set of odd integers will also be the median and that the the 9th and 10th members of that set will be 1 less and 1 more than the average. For this, the 9th = 533 and 10th = 535.

Counting down to the 1st element, 531=8th, 529=7th, 527=6th, 525=5th, 523=4th, 521=3rd, 519=2nd, 517=1st.

ya buddy for yr method

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