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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In a certain supermarket, a triangular display of cans is ar
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gmat_guy666
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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
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
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Hmmm,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]
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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GMATinsight
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It's a typical Questions that GMAT asks so you should take them very intelligently.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
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OA [spoiler]C
[/spoiler]
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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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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