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Swagatalaxmi
Joined: 05 Sep 2023
Last visit: 17 Nov 2025
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Schools: Wharton '27 (D) Columbia '27 (D) Booth '27 (WL) MIT '27 (D) Kellog '27 (D) Stern '27 (WL) Yale '27 (D) Fuqua '27 (D) Ross '27 (A$) Stanford '27 (S) Darden '27 (II)
GMAT Focus 1: 695 Q85 V88 DI80
Posts: 133
09 May 2024, 10:52
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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38%
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In a photo contest, prize money of $30, $20, and $10 was awarded to the first-, second-, and third-place photos, respectively, in several categories. No other prize money was awarded. Photographers could submit an unlimited number of photos to the contest but could enter each photo in only 1 category. Julia received a total of $110 in prize money from the contest. How many of her photos were awarded second place in their categories?
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.
(2) A total of 6 of Julia's photos received awards.
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.
(2) A total of 6 of Julia's photos received awards.
09 May 2024, 11:52
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In a photo contest, prize money of $30, $20, and $10 was awarded to the first-, second-, and third-place photos, respectively, in several categories. No other prize money was awarded. Photographers could submit an unlimited number of photos to the contest but could enter each photo in only 1 category. Julia received a total of $110 in prize money from the contest. How many of her photos were awarded second place in their categories?
Let the numbers of first, second, and third place photos be x, y, and z.
30x + 20y + 10z = 110
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.
30x + 20y + 10(2x) = 110
Notice that since 10 is multiplied by 2x, the only way to get 110, which is not a multiple of 20, is to have an odd number of 30s. So, x must be odd.
If x = 1,, the addition can work.
If x = 3, 2x = 6. So, the total will be greater than 110.
Thus, x = 1, and we can figure out the rest.
Sufficient.
(2) A total of 6 of Julia's photos received awards.
Try different combinations.
x = 0, y = 5, and z = 1 works.
0(30) + 5(20) + 1(10) = 110
x = 1, y = 3, and z = 2 works.
1(30) + 3(20) + 2(10) = 110
Insufficient.
Correct answer: A
Let the numbers of first, second, and third place photos be x, y, and z.
30x + 20y + 10z = 110
(1) Twice as many of Julia's photos were awarded third place as were awarded first place in their categories.
30x + 20y + 10(2x) = 110
Notice that since 10 is multiplied by 2x, the only way to get 110, which is not a multiple of 20, is to have an odd number of 30s. So, x must be odd.
If x = 1,, the addition can work.
If x = 3, 2x = 6. So, the total will be greater than 110.
Thus, x = 1, and we can figure out the rest.
Sufficient.
(2) A total of 6 of Julia's photos received awards.
Try different combinations.
x = 0, y = 5, and z = 1 works.
0(30) + 5(20) + 1(10) = 110
x = 1, y = 3, and z = 2 works.
1(30) + 3(20) + 2(10) = 110
Insufficient.
Correct answer: A
General Discussion
22 May 2025, 10:36
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MartyMurray
Twice as many of Julia's photos were awarded third place as were awarded first place in their categories - doesnt it mean x=2z
