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15 Sep 2020, 05:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
82% (01:12) correct
18%
(01:45)
wrong
based on 558
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A can manufacturing company has 5 identical machines, each of which produces cans at the same constant rate. How many cans will all 5 machines produce running simultaneously for z hours?
(1) Running simultaneously, 3 of the machines produce 72,000 cans in 2z hours.
(2) Running simultaneously, 2 of the machines produce 24,000 cans in z hours.
(1) Running simultaneously, 3 of the machines produce 72,000 cans in 2z hours.
(2) Running simultaneously, 2 of the machines produce 24,000 cans in z hours.
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15 Sep 2020, 05:54
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A can manufacturing company has 5 identical machines, each of which produces cans at the same constant rate. How many cans will all 5 machines produce running simultaneously for z hours ?
(1) Running simultaneously, 3 of the machines produce 72,000 cans in 2z hours
(2) Running simultaneously, 2 of the machines produce 24,000 cans in zz hours
Explanation : let the rate be x cans/hr.
S1:3 of the machines produce 72,000 cans in 2z hours i.e
3*2z*x=72000
x=12000/z
Now no. Of cans 5 machines running simultaneously for z hours will produce= 5*x*z
=5*(12000/z)*z
=60000
Hence S1 is sufficient.
S2:
Running simultaneously, 2 of the machines produce 24,000 cans in z hours
2*x*z=24000
x=12000/z
Now no. Of cans 5 machines running simultaneously for z hours will produce= 5*x*z
=5*(12000/z)*z
=60000
Hence S2 is also sufficient.
Thus D is the correct answer.
Regards
Atul Pandey
Posted from my mobile device
(1) Running simultaneously, 3 of the machines produce 72,000 cans in 2z hours
(2) Running simultaneously, 2 of the machines produce 24,000 cans in zz hours
Explanation : let the rate be x cans/hr.
S1:3 of the machines produce 72,000 cans in 2z hours i.e
3*2z*x=72000
x=12000/z
Now no. Of cans 5 machines running simultaneously for z hours will produce= 5*x*z
=5*(12000/z)*z
=60000
Hence S1 is sufficient.
S2:
Running simultaneously, 2 of the machines produce 24,000 cans in z hours
2*x*z=24000
x=12000/z
Now no. Of cans 5 machines running simultaneously for z hours will produce= 5*x*z
=5*(12000/z)*z
=60000
Hence S2 is also sufficient.
Thus D is the correct answer.
Regards
Atul Pandey
Posted from my mobile device
08 Oct 2020, 03:05
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Quote:
Step 1: Understanding the question
Let rate of each machine be x, rate of 5 machines = 5x
Using Rate * Time = Work
5x * z = w
w = 5xz
w is to be calculated
Step 2: Understanding statement 2 alone
(1) Running simultaneously, 3 of the machines produce 72,000 cans in 2z hours
3x * 2z = 72000
xz = 12000
Hence, w = 5*12000 = 60000
Sufficient
Step 3: Understanding statement 2 alone
(2) Running simultaneously, 2 of the machines produce 24,000 cans in z hours
2x * z = 24000
xz = 12000
Hence, w = 5*12000 = 60000
Sufficient
D is correct
