In a certain supermarket, a triangular display of cans is ar

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gmat_guy666: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Tue Apr 23, 2013 12:51 am; Thanked: 6 times

by gmat_guy666 » Tue Oct 14, 2014 6:31 am

In a certain supermarket, a triangular display of cans is arranged in 8 rows, numbered 1 through 8 from top to bottom. Each successively numbered row contains 3 more cans than the row immediately above it. If there are fewer than 95 cans in the entire display, how many cans are in the fifth row?

10

11

13

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16

OA [spoiler]C
[/spoiler]

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Source: Beat The GMAT — Problem Solving | Tue Oct 14, 2014 6:31 am

gmat_winter: Senior | Next Rank: 100 Posts; Posts: 34; Joined: Tue Oct 14, 2014 8:23 am; Thanked: 1 times

by gmat_winter » Tue Oct 14, 2014 8:28 am

Test the answer:

1- x

2-x+3

3-x+6

4-x+9

5-x+12

.
.
.
.

You see x has to be more than 12 on the 5th Day . so eliminate A) & B)

Pick 13 i.e x=1 satisfies all the condition (fewer than 95 cans in the entire display)

Cheers

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Oct 14, 2014 8:33 am

gmat_guy666 wrote:In a certain supermarket, a triangular display of cans is arranged in 8 rows, numbered 1 through 8 from top to bottom. Each successively numbered row contains 3 more cans than the row immediately above it. If there are fewer than 95 cans in the entire display, how many cans are in the fifth row?

10
11
13
14
16

OA [spoiler]C
[/spoiler]

Hmmm,

Let's see what happens if there's 1 can in the top (1st) row.
Row #1: 1 can
Row #2: 4 cans
Row #3: 7 cans
Row #4: 10 cans
Row #5: 13 cans

Done.
Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Wed Oct 15, 2014 3:48 am

gmat_guy666 wrote:In a certain supermarket, a triangular display of cans is arranged in 8 rows, numbered 1 through 8 from top to bottom. Each successively numbered row contains 3 more cans than the row immediately above it. If there are fewer than 95 cans in the entire display, how many cans are in the fifth row?

10
11
13
14
16

OA [spoiler]C
[/spoiler]

It's a typical Questions that GMAT asks so you should take them very intelligently.

Another mathematical standard approach to this question sis using Arithmetic Progressions

First term of the sequence is a
Common difference = 3
No. of terms = 8
Sum of all the terms = (n/2) [2a+(n-1)d]
Given Sum of n terms < 95
(8/2) [2a+(8-1)3] < 95
4 [2a+21] < 95
8a + 84 < 95
8a < 11
a < 11/8
But since a has to be an integer therefore a = 1 [only possibility]

Now:
nth term of sequence = [a+(n-1)d]
5th term of sequence = [1+(5-1)3] = 13

Answer: Option C

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8139; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Oct 27, 2019 6:34 pm

gmat_guy666 wrote:In a certain supermarket, a triangular display of cans is arranged in 8 rows, numbered 1 through 8 from top to bottom. Each successively numbered row contains 3 more cans than the row immediately above it. If there are fewer than 95 cans in the entire display, how many cans are in the fifth row?

10

11

13

14

16

OA [spoiler]C
[/spoiler]

We can let x = the number of cans in the first row. Thus, x + 3, x + 2(3), ..., x + 7(3) are the number of cans in the second, third, ..., eighth rows, respectively. Since we have an evenly spaced set, we can use the formula sum = avg * quantity where:

Avg = (largest number + smallest number)/2
Quantity = 8

We can create the inequality:

sum < 95

avg * quantity < 95

(x + 7(3) + x)/2 * 8< 95

(2x + 21)/2 * 8 < 95

(2x + 21) * 8 < 190

16x + 168 < 190

16x < 22

x < 22/16

Since x must be a whole number, x = 1. Therefore, the number of cans in the fifth row is x + 4(3) = 1 + 12 = 13.

Answer: C

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Founder and CEO
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