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by arora007 » Mon Oct 04, 2010 4:18 am
At a factory, each employee's daily pay ,S, is defined by S=75 + (ny/4) where n is the number of widgets buit by the employee that day and y is the number of years of employment. Abe, Bon and Cindy have 3,6 and 8 years of employment respectively. Each day, Abe works twice as long as Bob and twice as long as Cindy, but Bob builds widgets 50% faster than Abe and 40% slower than Cindy. If Abe, Bob and Cindy's combined daily pay is $675, how many widgets do Abe and Bob build together each day?
A)150
B)160
C)170
D)180
E)190
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by sumit.sinha » Mon Oct 04, 2010 4:46 am
arora007 wrote:At a factory, each employee's daily pay ,S, is defined by S=75 + (ny/4) where n is the number of widgets buit by the employee that day and y is the number of years of employment. Abe, Bon and Cindy have 3,6 and 8 years of employment respectively. Each day, Abe works twice as long as Bob and twice as long as Cindy, but Bob builds widgets 50% faster than Abe and 40% slower than Cindy. If Abe, Bob and Cindy's combined daily pay is $675, how many widgets do Abe and Bob build together each day?
A)150
B)160
C)170
D)180
E)190
IMO D.
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by Rahul@gurome » Mon Oct 04, 2010 8:40 pm
Solution:
If Bob and Cindy work for ‘t’ hours everyday, Abe works for ‘2t’ hours everyday.
Also if the rate of work of Abe is ‘r’, then rate of work of Bob is ‘1.5r’ and rate of work of Cindy is (100/60) * 1.5r = ‘2.5r’.
Or ratio of number of widgets built by Abe : no. built by Bob : no. built by Cindy is
2t*r : t*1.5r : t*2.5r = 2 : 1.5 : 2.5 = 4 : 3 : 5.

Let the number of widgets built by Abe, Bob and Cindy be 4x, 3x and 5x respectively.

So (75 + 4x*3/4) + (75 + 3x*6/4) + (75 + 5x*8/4) = 675.
Or 70x = 450*4.
Number of widgets built by Abe and Bob is 4x+3x = 7x = 45*4 = 180.
The correct answer is (D).
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by arora007 » Tue Oct 05, 2010 6:35 am
OA is D
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