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anushree01
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Interesting way,

you should also think about solving this way

Maximum production in 14 hours using only the faster machine:
14 × 4 = 56
Shortfall:
77 - 56 = 21
H makes 3 dozen/hr
21 ÷ 3 = 7
Minimum overlap = 7 hours.
Aayushh
I applied two concepts here - WorkRatetime and Averages

- 77 units produced in 14 hours means avg pace is 5.5

- To minimise the time both machines work, need to maximise the output from the along working one. Therefore the one that works at the speed of 4 (K).

One machine and both machines
- K (4) and H+K (3+4=7)

- Now 5.5 is the middle value. That means the working ratio is 1:1 ( we can use allegation method, Even if it were not the middle value we would have found the ratio. (5.5-4 : 5.5-7, 1.5:1.5, 1:1)

- 14 hours into 1:1 ratio ((14/2) *1) = 7 hours

7 hours


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Machine H worked alone for x hrs out of 14,
Machine K worked alone for y hrs out of 14,
Then, Machine H & K together worked for 14 - (x + y) hrs. We need to minimise this. For that, we need to maximise x+y.

3x+4y+7(14-x-y)=77.
Solve to get 4x + 3y = 21.
Possible solutions: x=3,y=3; x=0, y=7
x+y is maximum at x=0, y=7. Hence, 14 - (x+y) = 14 - 7 = 7.

Bunuel please could you confirm if this is also the right approach? thanks!
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Hi muskang2000,

Yes, your approach is completely valid, and it lands on the same answer the rest of the thread reaches (A, 7 hr). Let me confirm each piece so you can trust it.

Your setup is exactly right:
- H alone for x hours, K alone for y hours, both together for 14 - (x + y) hours.
- Total output: 3x + 4y + 7(14 - x - y) = 77, which cleanly simplifies to 4x + 3y = 21.
- Together-time = 14 - (x + y), so to minimize it you maximize x + y. All correct.

One small thing worth making airtight: you found two sample solutions (x=3, y=3 and x=0, y=7) and picked the larger x+y by inspection. That works here, but it's even safer to show why x=0, y=7 is the true maximum rather than just one of several options.

From 4x + 3y = 21, solve for x: x = (21 - 3y)/4. Then:

x + y = (21 - 3y)/4 + y = (21 + y)/4

This increases as y increases, so you want y as large as possible. Since x can't go below 0, set x = 0, which forces y = 7. That gives x + y = 7, and together-time = 14 - 7 = 7 hours.

Notice this matches the intuition everyone else used: push the faster machine (K) to run alone as much as possible, and bring H in only for the leftover. Your algebra is just the formal version of that same idea - so it's a perfectly good method to use on test day.

Answer: A

muskang2000
Machine H worked alone for x hrs out of 14,
Machine K worked alone for y hrs out of 14,
Then, Machine H & K together worked for 14 - (x + y) hrs. We need to minimise this. For that, we need to maximise x+y.

3x+4y+7(14-x-y)=77.
Solve to get 4x + 3y = 21.
Possible solutions: x=3,y=3; x=0, y=7
x+y is maximum at x=0, y=7. Hence, 14 - (x+y) = 14 - 7 = 7.

Bunuel please could you confirm if this is also the right approach? thanks!
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Suppose H works for x hr and K works for y and 14-x-y hr they work together:
Than:
3x + 4y + (14-x-y)7 = 77
3x + 4y -7x - 7y + 14.7 = 77
14.7 - 11.7 = 4x +3y
(14-11).7 = 4x + 3y
3.7 = 4x+3y
21= 4x + 3y (for least time let x = 0)
that gives y=7, machine K works for 7hrs
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