Test: Arun Sharma Based Level 2: Time & Work - CUET Commerce MCQ

Calculation of Gas Discharge Rates:

- Burner A discharges gas at 1 cc/minute
- Burner B discharges gas at 2 cc/minute

Capacity of Gas Cylinder:

- Capacity of the gas cylinder is 1000 cc of gas

Heat Generated and Wastage:

- Heat generated is equal to 1 kcal per cc of gas
- Wastage of heat is present

Calculating Time to Empty Cylinder:

- Initially, the gas cylinder is half full, which is 500 cc
- Burner A is opened to 90% of its capacity, which is 0.9 * 1 cc/minute = 0.9 cc/minute
- Burner B is opened to 80% of its capacity, which is 0.8 * 2 cc/minute = 1.6 cc/minute
- Total discharge rate when both burners are opened simultaneously = 0.9 + 1.6 = 2.5 cc/minute
- Time taken to empty the cylinder = 500 cc / 2.5 cc/minute = 200 minutes

Therefore, the correct answer is C: 200 minutes.

To determine the best way to utilize the maximum amount of heat with the fastest speed of cooking, we need to consider both the discharge rates of gas from the two burners and the wastage of heat as described.

Key points to consider:

1. Burner A can discharge gas at a rate of 1 cc/min (maximum).
2. Burner B can discharge gas at a rate of 2 cc/min (maximum).
3. The capacity of the gas cylinder is 1000 cc of gas, and each cc of gas generates 1 kcal of heat.
4. The question is asking for the maximum amount of heat with the fastest speed of cooking, which means we need to balance speed (gas discharge rate) and heat utilization (minimizing wastage).

From the given options, we need to evaluate the discharge rates:

• Option 1: The first burner is opened up to 50% of its aperture.

  • Burner A’s maximum discharge is 1 cc/min, and at 50% aperture, it would discharge 0.5 cc/min.
  • While this would reduce the speed of gas discharge, it might minimize heat wastage, but it wouldn't maximize speed for fast cooking.

• Option 2: The second burner is opened up to 25% of its aperture.

  • Burner B’s maximum discharge is 2 cc/min, and at 25% aperture, it would discharge 0.5 cc/min (same as burner A in option 1).
  • Again, this would likely reduce heat wastage, but the speed of gas discharge is still limited to 0.5 cc/min.

• Option 3: Either (a) or (b).

  • Both options would result in a gas discharge rate of 0.5 cc/min, so they would have similar results in terms of cooking speed and heat utilization.

• Option 4: None of these.

  • This would imply that neither option (a) nor (b) is optimal.

Conclusion:

To achieve maximum heat utilization with the fastest cooking speed, we would ideally want to minimize wastage while maintaining a high gas discharge rate. However, based on the options given, either (a) or (b) provides the same discharge rate of 0.5 cc/min, balancing the need for reduced wastage and reasonable cooking speed.

Therefore, the correct answer is:

3. Either (a) or (b).

- Rate of Discharge
- Burner A: 1 cc/minute
- Burner B: 2 cc/minute

- Total Capacity of Gas Cylinder: 1000 cc

- Amount of Heat Generated: 1 kcal/cc

- Wastage of Heat:
- Burner A: 0%
- Burner B: 50%

- Calculation:
- Amount of gas used by Burner A in 1 minute = 1 cc
- Amount of gas used by Burner B in 1 minute = 2 cc * 50% = 1 cc
- Total gas used in 1 minute = 1 cc + 1 cc = 2 cc
- Heat utilized in 1 minute = 2 cc * 1 kcal/cc = 2 kcal
- Total heat utilized for cooking if a full gas cylinder is burnt = 2 kcal/minute * 1000 minutes = 2000 kcal
- Wastage of heat by Burner B = 50% of 2000 kcal = 1000 kcal
- Heat utilized for cooking = Total heat - Wastage = 2000 kcal - 1000 kcal = 1000 kcal

 

First, let's find the rate at which each burner discharges gas when opened at the given percentages.

For burner A, opened at 25%:
Rate of discharge = 1 cc/min * 25% = 0.25 cc/min

For burner B, opened at 50%:
Rate of discharge = 2 cc/min * 50% = 1 cc/min

Now, let's find the wastage of heat for each burner based on the given discharge rates.

For burner A, with a discharge rate of 0.25 cc/min:
Wastage = 10% (since it falls within the 0 - 0.5 cc/min range)

For burner B, with a discharge rate of 1 cc/min:
Wastage = 25% (since it falls within the 1 - 1.5 cc/min range)

Now, we can find the effective heat generated by each burner after accounting for wastage.

For burner A:
Effective heat = 1 kcal/cc * (1 - 10%) = 1 kcal/cc * 90% = 0.9 kcal/cc

For burner B:
Effective heat = 1 kcal/cc * (1 - 25%) = 1 kcal/cc * 75% = 0.75 kcal/cc

Next, we need to find the total amount of gas discharged by both burners until the gas cylinder is empty. Since the capacity of the cylinder is 1000 cc, we can find the time it takes for both burners to empty the cylinder.

Total rate of discharge = 0.25 cc/min (burner A) + 1 cc/min (burner B) = 1.25 cc/min

Time to empty the cylinder = 1000 cc / 1.25 cc/min = 800 minutes

Now, we can find the total heat generated by each burner during these 800 minutes.

Heat generated by burner A = 0.9 kcal/cc * 0.25 cc/min * 800 min = 180 kcal

Heat generated by burner B = 0.75 kcal/cc * 1 cc/min * 800 min = 600 kcal

Finally, we can find the total heat available for cooking by adding the heat generated by both burners.

Total heat available = 180 kcal (burner A) + 600 kcal (burner B) = 780 kcal

So, the amount of heat available for cooking would be 780 kcal.

► The original time requirement was 9 days and so now 18 days are needed.
► From 2^nd day onwards, on each odd day 1 worker is being added and on each even day, 2 workers are leaving. So, in 2 days overall 1 worker is leaving. So, from 3^rd to 18^th there are 8 such odd-even day pairs (3^rd and 4^th day to 17^th and 18^th day).
► So, at the beginning of 3^rd day there must been have 8 workers. On 2^nd day, 2 workers were withdrawn. So, originally there must been have 10 workers

► Let x = Number of days it rained in the morning and had clear afternoons.
y = Number of days it rained in the afternoon and had clear mornings.
z = Number of days it rained in the morning or afternoon.
► So, according to question,
⇒ x + y = 7
⇒ x + z = 5
⇒ y + z = 6
Adding all three equations, x + y + z = 9
So, d = 9 days

Let us make a table of the units of work everyday :

    ► Total length =113 m

► Bimlesh does 12/40 = (3/10)^th part of the work in 12 days
1/5 = 2/10 of the work is completed by all three working together.
► Remaining work = 1 - (3/10 + 2/10) = 5/10 = 1/2 is done by Alok and Mithilesh in 20 days.
► Both will complete whole work together in 40 days.
► Since efficiency of both is same, Hence each one will take 80 days to complete the work alone.

  ► 40% work in 40h → 60% work in 60h.
  ► Hence, working hours = 60 / 4 = 15 h

► Using solution form above, the total time needed to complete 20 uniforms for T.M. High school must be 6 hours and 30 minutes (1st set of 2 uniforms is received after 2 hours and then in each 30 min, a set of 2 uniforms is received).
► Now by 6 hours and 30 minutes, the cutters and tailors must have completed the 1^st 2 steps for the uniforms for A.R. Academy. The uniforms for A.R. academy can be produced as 1 set of 2 uniforms per 15 minutes.
► From 6 hours 30 minutes to 10 hours there are 14 such sets of uniforms which can be produced.
► So, total 28 A.R. Academy uniforms can be completed.

► To complete 30 T.M. high school uniforms, the time needed is 9 hours. So, all machines must remain idle for 1 hour. There are 9 people so 9 man-hours are idle.
► Further, during the time production was on, both the people involved in cutting remain idle for last 1.5 hours. The 2 people involved in finishing remain idle for 1^st 1.5 hours. So, there are total 6 man-hours for these people.
► For 1st 30 minutes and last 30 minutes, all 5 tailors were idle. So, 5 man-hours are there. After 30 minutes 3 tailors were idle for 30 minutes which means 1.5 man-hours. After 1 hour, 1 tailor was idle for 30 minutes. 30 minutes later, 1 idle tailor will get new work but still 1 from 2 tailors who just finished the work will remain idle.
► So, from 1 hour to 8 hours exactly 1 tailor will remain idle. So, 7 man hours are idle. For 8 hour to 8 hour 30 minutes time duration, 3 tailors were idle which means 1.5 man-hours.
► The total man-hours which will go idle is = (9 + 6 + 5 + 1.5 + 7 + 1.5) = 30  man-hours

► If they have 1 more assistant, then they have 2 cutters, 5 tailors and 3 assistants and they are producing uniforms only for A.R. Academy.
► In 1st 20 minutes, cloth for 2 uniforms is ready and it will be tailored in 1 hour 20 minutes and by 1 hour 35 min the finishing will be done. So, in 1 hour 35 minutes, we will have 1 set of 2 uniforms.
► In next 20 minutes (i.e., at 1 hour 55 min), we will get 1 more set of 2 uniforms.
► But for next 20 min (i.e., at 2 hour 15 min) we will get only 1 uniform due to the non-availability of 1 tailor at 1 hour when 2 clothes were available but only 1 tailor was available. So, this way we will get 5 uniforms per hour.
► So, 5 uniforms at 2 hour 15 min, 10 uniforms at 3 hours 15 min…40 uniforms at 9 hour 15 minutes. By 9 hours 55 min, we will receive 44 uniforms.
► So, we will be able to produce 44 uniforms of A.R. Academy.

► There are 2 cutters who can cut the cloth for uniform in 20 min. So, in 1 hour they can cut cloth for the 6 uniforms.
► There are 5 tailors who can prepare 5 uniforms in 1 hour. There are 2 assistants who can perform the required finishing in 15 min. So they can produce 8 uniforms in 1 hour.
► So, we can observe that to increase the rate of uniform production we must add 1 more tailor, so that the uniforms will be produced by tailors at rate of 6 per hour which is same as the rate by which the cutting is being done.

(A + B)'s one day's % work = 

100/15 = 6.66%

B's one day's % work = 

100/20 = 5%

A's one day's % work = 6.66 - 5 = 1.66%
Thus, A need = 

100/1.66 = 60 days to complete the work