At a school fundraiser, a snack table sold four types of items: fruit

Question

At a school fundraiser, a snack table sold four types of items: fruit cups, granola bars, sandwich boxes, and salad bowls. A fruit cup cost $3, a granola bar cost $5, a sandwich box cost $9, and a salad bowl cost $14. If the snack table sold 21 items in total, at least one of each type, and collected $180, how many granola bars were sold?

(1) The number of granola bars sold was equal to the number of sandwich boxes sold.

(2) The snack table sold 4 fruit cups and 5 sandwich boxes.

ankushsambare · Answer

Let F, G, S, and B be the numbers of fruit cups, granola bars, sandwich boxes, and salad bowls.
Given:
F + G + S + B = 21
3F + 5G + 9S + 14B = 180
(1) G = S
Substitute S = G:
F + 2G + B = 21
3F + 14G + 14B = 180
Eliminating F gives:
8G + 11B = 117
This does not yield a unique value of G (and in fact has no valid integer solution), so Statement (1) is not sufficient.
(2) F = 4 and S = 5
Then:
G + B = 12
Also,
3(4) + 5G + 9(5) + 14B = 180
57 + 5G + 14B = 180
5G + 14B = 123
Substitute G = 12 − B:
5(12 − B) + 14B = 123
60 + 9B = 123
B = 7
Therefore,
G = 12 − 7 = 5
A unique value of G is obtained.
 Answer: B — Statement (2) alone is sufficient, but Statement (1) alone is not sufficient.

SS990 · Answer

Note: this is a B trap DS question. Given: $/no f g sb sal total $ 3 5 9 14 no. ? ? ? ? 21 spend 3f 5g 9sb 14s 180 Deduction: f+g+sb+s=21; 3f+5g+9sb+14s=180 3f+3g+3sb+3s=63.. so 2g+6sb+11s=117 to find: value of g 1) g=s so 2g+6sb+11s= 8g+11s=117 g=15 - (11s+3)/8 s has to be odd for 11s+3 to be even since g cant be fraction. Also 11s

BongBideshini · Answer

Let:
 
 F = fruit cups 
 G = granola bars 
 S = sandwich boxes 
 B = salad bowls  
Given:
F + G + S + B = 21
3F + 5G + 9S + 14B = 180
At least one of each type.
We need G.

Statement (1) 
G = S
Then:
F + 2G + B = 21
3F + 14G + 14B = 180
Substitute F = 21 − 2G − B:
3(21 − 2G − B) + 14G + 14B = 180
63 − 6G − 3B + 14G + 14B = 180
 8G + 11B = 117 
G = (117 − 11B)/8 => will give an integer number, i.e., remainder will be zero
Modulo 8:
117 − 11B ≡ 5 − 3B ≡ 0 (mod 8)
3B ≡ 5 (mod 8)
B ≡ 7 (mod 8)
Possible positive value:
B = 7
Then:
8G + 77 = 117
8G = 40
G = 5

Statement (1) is sufficient.

Statement (2) 
F = 4 and S = 5.
Then:
 G + B = 12-------(i) 
and
3(4) + 5G + 9(5) + 14B = 180
12 + 45 + 5G + 14B = 180
 5G + 14B = 123 -----(ii) 
From (i) and (ii), the value of G can easily be found.

Statement (2) is sufficient.

Answer: D.

At a school fundraiser, a snack table sold four types of items: fruit

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GMAT Data Sufficiency (DS) Questions

At a school fundraiser, a snack table sold four types of items: fruit

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