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

[gmat_guy666]

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]

[gmat_winter]

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

[Brent@GMATPrepNow]

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

[GMATinsight]

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
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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

[Scott@TargetTestPrep]

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