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To fill a larger concert hall, a madrigal singing group consisting of
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Bunuel
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To fill a larger concert hall, a madrigal singing group consisting of [#permalink] New post  08 Feb 2019, 02:59
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To fill a larger concert hall, a madrigal singing group consisting of sopranos, altos, and basses, in a 5:7:3 ratio, needs 40 singers. What is the least number of basses the group will need?

A. 8
B. 9
C. 10
D. 11
E. 12
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Re: To fill a larger concert hall, a madrigal singing group consisting of [#permalink] New post  08 Feb 2019, 03:19
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Bunuel wrote:
To fill a larger concert hall, a madrigal singing group consisting of sopranos, altos, and basses, in a 5:7:3 ratio, needs 40 singers. What is the least number of basses the group will need?

A. 8
B. 9
C. 10
D. 11
E. 12


sopranos: altos : basses = 5:7:3

Requirement = 40 singers

i.e. Total Number of Basses = 3*40/(5+7+3) = 8

Answer: Option A
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Re: To fill a larger concert hall, a madrigal singing group consisting of [#permalink] New post  08 Feb 2019, 05:26
Bunuel wrote:
To fill a larger concert hall, a madrigal singing group consisting of sopranos, altos, and basses, in a 5:7:3 ratio, needs 40 singers. What is the least number of basses the group will need?

A. 8
B. 9
C. 10
D. 11
E. 12



basses can be determined
3* 40/15 = 8
IMO A
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Re: To fill a larger concert hall, a madrigal singing group consisting of [#permalink] New post  28 Nov 2020, 12:16
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Bunuel wrote:
To fill a larger concert hall, a madrigal singing group consisting of sopranos, altos, and basses, in a 5:7:3 ratio, needs 40 singers. What is the least number of basses the group will need?

A. 8
B. 9
C. 10
D. 11
E. 12

Solution:

If we let the number of sopranos, altos, and basses be 5x, 7x and 3x, respectively, we can create the inequality:

5x + 7x + 3x ≥ 40

15x ≥ 40

x ≥ 40/15 = 8/3 = 2 ⅔

Since x must be a whole number, x is at least 3. Therefore, the number of basses is at least 3(3) = 9.

Answer: B
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Re: To fill a larger concert hall, a madrigal singing group consisting of [#permalink]
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