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Each of the 1,800 households that participated in a survey owned 02 Dec 2022, 02:30
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89% (02:15) correct 11% (02:38) wrong based on 103 sessions

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Each of the 1,800 households that participated in a survey owned either one car, two cars, or no cars. If 740 of the households owned only one car and at least 1/3 of the households owned two cars, what is the greatest possible value of the ratio of the number of households that owned no cars to the number of households that owned two cars?

A. 17/30
B. 23/30
C. 9/10
D. 1
E. 30/23


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B
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Each of the 1,800 households that participated in a survey owned 02 Dec 2022, 02:56
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File comment: Answer should be B
IMG_20221202_152452__01.jpg
IMG_20221202_152452__01.jpg [ 1.81 MiB | Viewed 2984 times ]

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Each of the 1,800 households that participated in a survey owned 02 Dec 2022, 03:08
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Bunuel
Each of the 1,800 households that participated in a survey owned either one car, two cars, or no cars. If 740 of the households owned only one car and at least 1/3 of the households owned two cars, what is the greatest possible value of the ratio of the number of households that owned no cars to the number of households that owned two cars?

A. 17/30
B. 23/30
C. 9/10
D. 1
E. 30/23

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Required Ratio = No Car / Two Car

Households with one car = 740

Households with two cars >= 600

To maximize ratio minimize the denominator and maximize the numerator.

Min. households with two cars = 600

Max. household with two cards = 1800 - (600 + 740) = 460

Required Ratio = \(\frac{460 }{ 600} = \frac{23}{30}\)

Option B
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Each of the 1,800 households that participated in a survey owned 02 Dec 2022, 23:24
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Hi gmatophobia,

Could you please explain how the max household with 2 cars is 460? Thank you.
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Each of the 1,800 households that participated in a survey owned 03 Dec 2022, 01:45
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meowmeowmeow22
Hi gmatophobia,

Could you please explain how the max household with 2 cars is 460? Thank you.
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File comment: If a is constant and we want to maximize c and minimize b, we have to take the least value of b, thus maximizing c.
So take b = 600
And thus c=1800-(740+600)= 460
IMG_20221202_152452__01.jpg
IMG_20221202_152452__01.jpg [ 1.81 MiB | Viewed 2690 times ]

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Each of the 1,800 households that participated in a survey owned 04 Dec 2022, 10:57
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Given: Each of the 1,800 households that participated in a survey owned either one car, two cars, or no cars.

Asked: If 740 of the households owned only one car and at least 1/3 of the households owned two cars, what is the greatest possible value of the ratio of the number of households that owned no cars to the number of households that owned two cars?

Total number of households = 1800
Number of households owning only 1 car = 740
Number of households owning 2 cars = y >= 1800/3 = 600
Let the number of households owning 0 car = x

740 + y + x = 1800
y >=600

Ratio of number of households that owned no cars to the number of households that owned two cars = x/y
To maximize the ratio of x/y, we have to maximise x and minimise y.

y = 600
x = 1800 - 600 - 740 = 460
x/y max = 460/600 = 23/30

IMO B
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