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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 28 Jan 2024, 04:22
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 bottles of soda per minute. If both machines start operating at the same time and at their respective constant rates, in how many seconds will machine Y have capped 12 more bottles of soda than machine X?

A. 15
B. 18
C. 20
D. 21
E. 24

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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 28 Jan 2024, 04:25
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guddo
Machine X caps 20 bottles of soda per minute, and machine Y caps 60 bottles of soda per minute. If both machines start operating at the same time and at their respective constant rates, in how many seconds will machine Y have capped 12 more bottles of soda than machine X?

A. 15
B. 18
C. 20
D. 21
E. 24

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2024-01-28_15-19-07.png
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Machine Y caps 40 more bottles per minute than machine X, which is 40/60 = 2/3 more bottles per second. Therefore, to cap 12 more bottles, machine Y will require 12/(2/3) = 18 seconds.

Answer: B.
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 28 Jan 2024, 05:20
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Question asks how many seconds...

So I converted the rates into seconds.
X-1/3 and Y-1 bottle/second..
Then I looked at the options and found that 20 secnds will not give clear answer.So i left and moved on to option B which gives 6 bottles for X and 18 bottles for Y.

Posted from my mobile device
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 28 Jan 2024, 06:28
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guddo
Machine X caps 20 bottles of soda per minute, and machine Y caps 60 bottles of soda per minute. If both machines start operating at the same time and at their respective constant rates, in how many seconds will machine Y have capped 12 more bottles of soda than machine X?
A. 15
B. 18
C. 20
D. 21
E. 24
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Attachment:
2024-01-28_15-19-07.png
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Sol:

Option-B

Rate for X (bottle/sec) = Ry= 20/60 = 1/3 bottle/sec
Rate for Y (bottle/sec) = Rx=60/60 = 1 bottle/sec

No.of bottles capped by Y = No. of bottle capped by X + 12
Use RTW equation
Say, t be the time;

Ry*t = Rx + 12
(1)*t = (1/3)*t + 12
(1-1/3)t = 12
2/3*t -= 12
t = 36/2 = 18 seconds
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 28 Jan 2024, 06:36
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guddo
Machine X caps 20 bottles of soda per minute, and machine Y caps 60 bottles of soda per minute. If both machines start operating at the same time and at their respective constant rates, in how many seconds will machine Y have capped 12 more bottles of soda than machine X?

A. 15
B. 18
C. 20
D. 21
E. 24

Show Spoiler
Attachment:
2024-01-28_15-19-07.png
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Machine X caps 20 bottles of soda per minute ⇒ Time taken by Machine X to cap one bottle of soda = \(\frac{60}{20} = 3\) seconds

Machine Y caps 60 bottles of soda per minute ⇒ Time taken by Machine Y to cap one bottle of soda = \(\frac{60}{60} = 1\) second

Hence, in every 3-second interval, Machine Y caps 2 extra bottles of soda than Machine X does.

To cap 12 more bottles of soda Machine Y will take \(\frac{2}{3} * 12 = 18\) seconds

Option B
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 02 Feb 2025, 07:39
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guddo
Machine X caps 20 bottles of soda per minute, and machine Y caps 60 bottles of soda per minute. If both machines start operating at the same time and at their respective constant rates, in how many seconds will machine Y have capped 12 more bottles of soda than machine X?

A. 15
B. 18
C. 20
D. 21
E. 24

Show Spoiler
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2024-01-28_15-19-07.png
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Y caps 60 and X caps 20 bottles in a min (60 seconds)

i.e. Y caps 40 bottles more than X in 60 seconds

use unitary methods (divide both time and bottle by 10 to get time 4 bottles)

i.e. Y caps 4 bottles more than X in 6 seconds

Multiply by 3 to convert 4 into 12

i.e. Y caps 4*3=12 bottles more than X in 6*3=18 seconds

Answer: Option B
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Machine X caps 20 bottles of soda per minute, and machine Y caps 60 02 Mar 2026, 02:58
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