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# In a certain supermarket, a triangular display of cans is arranged in

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Math Expert
Joined: 02 Sep 2009
Posts: 59039
In a certain supermarket, a triangular display of cans is arranged in  [#permalink]

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16 Oct 2019, 00:14
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:03) correct 33% (02:41) wrong based on 30 sessions

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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?

A. 10
B. 11
C. 13
D. 14
E. 16

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Re: In a certain supermarket, a triangular display of cans is arranged in  [#permalink]

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16 Oct 2019, 00:22
Bunuel 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?

A. 10
B. 11
C. 13
D. 14
E. 16

Let the number of cans in the first row = x
Each row below has 3 more than the row above it
--> Total number of cans in 8 rows = x + (x+3) + (x+6) + . . . . . . (x+21) = 8x + 84

Total is less than 95
--> 8x + 84 < 95
--> 8x < 11
--> Only possible value of x = 1

--> Number of cans in 5th row = x + 12 = 1 + 12 = 13

IMO Option C
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Posts: 3138
Re: In a certain supermarket, a triangular display of cans is arranged in  [#permalink]

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20 Oct 2019, 12:36

Solution

Given
• In a certain supermarket, a triangular display of cans is arranged.
• The display has 8 rows, numbered 1 through 8 from top to bottom.
• Each successively numbered row contains 3 more cans than the row immediately above it.

To find
• The number of cans in the fifth row if there are fewer than 95 cans in the entire display.

Approach and Working out

Let row 1 has ‘a’ cans.
• Hence, row 2 has ‘a + 3’ cans.
• Row 3 = ‘a + 6’ cans
• Similarly, each successive row has 3 more cans.
o Hence, cans are in an arithmetic progression with:
o a = being the number of cans in the first row and,
o 3= being the common difference,
o 8 = number of rows

• Therefore, total cans =$$\frac{8}{2}$$(2a + (8-1)3) = 4(2a + 21)
o 4(2a + 21) < 95
o 2a + 21 < $$\frac{95}{4}$$
o 2a < $$\frac{95}{4}$$ – 21
o 2a < $$\frac{11}{4}$$
o a < $$\frac{11}{8}$$
o a < 1 + $$\frac{3}{8}$$

• Since the number of balls has to be a positive integer, a must be 1.
• Thus, number of balls in the 5th row = 1 + 4*3 = 13

Thus, option C is the correct answer.

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Re: In a certain supermarket, a triangular display of cans is arranged in  [#permalink]

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23 Oct 2019, 19:10
1
Bunuel 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?

A. 10
B. 11
C. 13
D. 14
E. 16

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.

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Re: In a certain supermarket, a triangular display of cans is arranged in   [#permalink] 23 Oct 2019, 19:10
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