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04 Nov 2019, 04:19
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D
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Difficulty:
95%
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Question Stats:
30% (03:26) correct
70%
(03:24)
wrong
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A software company named Hardsoft has three divisions, each staffed by at least 10 people. The ratio of men to women is 7:5 in Finance and Administration, 3:4 in Marketing and Sales and 8:7 in Research and Development. If there are 100 men working in Hardsoft and every Hardsoft employee works for one of the three divisions, then the difference between the maximum and minimum possible number of women working in Hardsoft is
(A) 39
(B) 40
(C) 41
(D) 42
(E) more than 42
Are You Up For the Challenge: 700 Level Questions
(A) 39
(B) 40
(C) 41
(D) 42
(E) more than 42
Are You Up For the Challenge: 700 Level Questions
04 Nov 2019, 06:20
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Tricky one imo
Number of men in Finance and Administration=7a
Number of men in Finance and Administration=5a
Number of men in Marketing and Sales=3b
Number of men in Marketing and Sales=4b
Number of men in Research and Development=8c
Number of men in Research and Development=7c
Given- 7a+3b+8c=100, where a>0, b>1 and c>0
If we gotta maximize number of women, we have to maximize b
As, a>0 and c>0; hence b<29
There are no possible solution for b= 27 or 28
At b=26, a=2 and c=1
maximum number of women =5a+4b+7c = 121
To minimize the number of women we have to minimize 'b' and maximize 'a'.
7a+3b+8c=100
minimum possible value of b=2
7a+8c=94
a(max)=10 and c=3
Minimum possible number of women= 5a+4b+7c=50+8+21=79
Difference= 121-79=42
Number of men in Finance and Administration=7a
Number of men in Finance and Administration=5a
Number of men in Marketing and Sales=3b
Number of men in Marketing and Sales=4b
Number of men in Research and Development=8c
Number of men in Research and Development=7c
Given- 7a+3b+8c=100, where a>0, b>1 and c>0
If we gotta maximize number of women, we have to maximize b
As, a>0 and c>0; hence b<29
There are no possible solution for b= 27 or 28
At b=26, a=2 and c=1
maximum number of women =5a+4b+7c = 121
To minimize the number of women we have to minimize 'b' and maximize 'a'.
7a+3b+8c=100
minimum possible value of b=2
7a+8c=94
a(max)=10 and c=3
Minimum possible number of women= 5a+4b+7c=50+8+21=79
Difference= 121-79=42
Bunuel
General Discussion
29 Feb 2020, 12:49
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Bunuel
so the number of man is 7x + 3y + 8z
The number of woman is 5x + 4y + 7z
Number of woman - Number of men = Number of woman - 100 = y - 2x -z
Since the number of man is fixed (100), we want to maximize y and minimize (2x+z) to get that max value. Similarly, we want to minimize y and maximize (2x+z) to get minimum number of woman.
Case 1: maximize y and minimize 2x+z
7x + 3y + 8z = 100. Since each division has at least 10 employees, so 12x >10 (x>0), 15z >20 (x>0) and 7y >10 (y>1)
The minimum value of 7x + 8z = 8 + 7 = 15. Next values are 22, 23, 30, 37, 38, 45...
Now, since 7x have x as an integer, so 100 = (7x+8z) must be divisible by 7. 100 has 1 as a remainder when divided by 3, so (7x+8z) should be expressed by the form 3k+1
The minimum "3k+1" number we have here is 22. So x = 100-22 = 78, y = 78/3 = 26. The number 22 reveals x = 2 and z = 1.
The number of woman is 10 + 104 + 7 = 121
Case 2: minimize y and maximize 2x +z
y > 1 and y is an integer, so y(min) = 2
7x + 8z = 100-6 = 94
Now, we want to maximize 2x+z, so it makes sense to go for x(max). That being said, there should be a minimum value of z so (94-8z) is divisible by 7
94 = 7*23 + 3, so 8z must have a remainder of 3 when divided by 7
8z = {8, 16, 24, 32...}
The smallest number which satisfy the aforementioned condition is 24. Which means z = 3 and x = 10.
The number of woman is 50 + 8 + 21 = 79
The answer is D: 121-79 = 42
