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

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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?


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OA [spoiler]C
[/spoiler]
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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)

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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?

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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
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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?


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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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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?


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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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