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

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

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## 15 Questions MCQ Test General Test Preparation for CUET - Test: Arun Sharma Based Level 2: Time & Work

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Test: Arun Sharma Based Level 2: Time & Work - Question 1

### Directions for Question: Study the following and answer the questions that follow. A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas. The amount of heat generated is equal to 1 kcal per cc of gas. However, there is wastage of the heat as per follows : @ (Include higher extremes) Q. If both burners are opened simultaneously such that the first is opened to 90% of its capacity and the second is opened to 80% of its capacity, the amount of time in which the gas cylinder will be empty (if it was half full at the start) will be:

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 1

Test: Arun Sharma Based Level 2: Time & Work - Question 2

### Directions for Question: Study the following and answer the questions that follow. A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas. The amount of heat generated is equal to 1 kcal per cc of gas. However, there is wastage of the heat as per follows: @ (Include higher extremes) Q. The maximum amount of heat with fastest speed of cooking that can be utilised for cooking will be when:

Test: Arun Sharma Based Level 2: Time & Work - Question 3

### Directions for Question: Study the following and answer the questions that follow. A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas. The amount of heat generated is equal to 1 kcal per cc of gas. However, there is wastage of the heat as per follows : @ (Include higher extremes) Q. The amount of heat utilised for cooking if a full gas cylinder is burnt by opening the aperture of burner A 100% and that of burner B 50% is

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 3

At 1cc/minute, the loss of heat is 20%. Hence, when 1000 cc of the gas is used, out of the 1000 kcal of heat generated 200 kcal will be lost.

Test: Arun Sharma Based Level 2: Time & Work - Question 4

Directions for Question: Study the following and answer the questions that follow.

A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.
The amount of heat generated is equal to 1 kcal per cc of gas.

However, there is wastage of the heat as per follows:

@ (Include higher extremes)

Q. For Question 3, if burner A had been opened only 25% and burner B had been opened 50%, the amount of heat available for cooking would be

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 4

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.

Test: Arun Sharma Based Level 2: Time & Work - Question 5

Directions for Question: Study the following and answer the questions that follow.

A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.
The amount of heat generated is equal to 1 kcal per cc of gas.

However, there is wastage of the heat as per follows :

@ (Include higher extremes)

Q. For Question 4, the amount of time required to finish a full gas cylinder will be

Test: Arun Sharma Based Level 2: Time & Work - Question 6

A group of workers can complete a certain job in 9 days. But it so happens that every alternate day starting form the second day, two workers are withdrawn from the job and every alternate day starting from the third day, one worker is added to the group. In such a way, the job is finished by the time, there is no worker left. If it takes the double time to finish the job now, find the number of workers who started the job?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 6

► The original time requirement was 9 days and so now 18 days are needed.
► From 2nd 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 3rd to 18th there are 8 such odd-even day pairs (3rd and 4th day to 17th and 18th day).
► So, at the beginning of 3rd day there must been have 8 workers. On 2nd day, 2 workers were withdrawn. So, originally there must been have 10 workers

Test: Arun Sharma Based Level 2: Time & Work - Question 7

A student studying the weather for d days observed that
(i) It rained on 7 days, morning or afternoon,
(ii) When it rained in the afternoon, it was clear in the morning,
(iii) There were five clear afternoons, and
(iv) There were six clear mornings. Then, d equals.

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 7

► 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

Test: Arun Sharma Based Level 2: Time & Work - Question 8

Sumit constructs a wall working in a special way and takes 12 days to complete it. If Sn is the length of the wall (in m) that he constructs on the nth day, then
Sn = 2n, 0 ≤ n ≤ 4
Sn = 8, for n = 5
Sn = 3n - 7, 6 ≤ n ≤ 12
Find the total length of the wall he constructs in the first 10 days.

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 8

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

► Total length =113 m

Test: Arun Sharma Based Level 2: Time & Work - Question 9

Alok, Mithilesh, and Bimlesh started a work and after completing 1 / 5th of the work Bimlesh left. Alok and Mithilesh then worked for 20 days. Bimlesh then took over from Alok and Mithilesh and completed the remaining portion of the work in 12 days. If Bimlesh takes 40 days to complete the work, in how many days would Alok alone or Mithilesh alone complete the work if the efficiencies with which they work is the same?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 9

► 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.

Test: Arun Sharma Based Level 2: Time & Work - Question 10

Sixty-for men working 8 hrs a day plan to complete a piece of work in 9 days. However, 5 days later they found that they had completed only 40% of the work. They now wanted to finish the remaining portion of the work in 4 more days. How many hours per day should they need to work in order to achieve the target?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 10

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

Test: Arun Sharma Based Level 2: Time & Work - Question 11

Direction for Question : Read the passage below and solve the questions based on it.
Hi-Choice Dressers received a large order for stitching uniforms from A.R. Academy and T.M. High School. He has two cutters who will cut the fabric, five tailors who will do the stitching, and two assistants to stitch the buttons and button holes. Each of these nine persons will work for exactly 10hrs a day. Each of the A.R. Academy uniform requires 20 min for cutting the fabric, 1hr for doing the stitching and 15 min for stitching the buttons and the button holes. The T.M. High School uniform requires 30 min, 1hrs and 30 min, respectively, for the same activities.

Q. On a particular day, Hi-Choice Dressers decided to complete 20 T.M. High School uniforms. How many A.R. Academy uniforms can it complete on that day?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 11

► 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 1st 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.

Test: Arun Sharma Based Level 2: Time & Work - Question 12

Direction for Question : Read the passage below and solve the questions based on it.
Hi-Choice Dressers received a large order for stitching uniforms from A.R. Academy and T.M. High School. He has two cutters who will cut the fabric, five tailors who will do the stitching, and two assistants to stitch the buttons and button holes. Each of these nine persons will work for exactly 10hrs a day. Each of the A.R. Academy uniform requires 20 min for cutting the fabric, 1hr for doing the stitching and 15 min for stitching the buttons and the button holes. The T.M. High School uniform requires 30 min, 1hrs and 30 min, respectively, for the same activities.

Q. If Hi-Choice Dressers decided to complete 30 T.M. High School uniforms only and no other uniform on that particular day, how many total man-hours will go idle?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 12

► 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 1st 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

Test: Arun Sharma Based Level 2: Time & Work - Question 13

Direction for Question : Read the passage below and solve the questions based on it.
Hi-Choice Dressers received a large order for stitching uniforms from A.R. Academy and T.M. High School. He has two cutters who will cut the fabric, five tailors who will do the stitching, and two assistants to stitch the buttons and button holes. Each of these nine persons will work for exactly 10hrs a day. Each of the A.R. Academy uniform requires 20 min for cutting the fabric, 1hr for doing the stitching and 15 min for stitching the buttons and the button holes. The T.M. High School uniform requires 30 min, 1hrs and 30 min, respectively, for the same activities.

Q. If Hi-Choice Dressers hires one more assistant, what is the maximum number of A.R. Academy uniforms that can be completed in a day?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 13

► 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.

Test: Arun Sharma Based Level 2: Time & Work - Question 14

Direction for Question : Read the passage below and solve the questions based on it.
Hi-Choice Dressers received a large order for stitching uniforms from A.R. Academy and T.M. High School. He has two cutters who will cut the fabric, five tailors who will do the stitching, and two assistants to stitch the buttons and button holes. Each of these nine persons will work for exactly 10hrs a day. Each of the A.R. Academy uniform requires 20 min for cutting the fabric, 1hr for doing the stitching and 15 min for stitching the buttons and the button holes. The T.M. High School uniform requires 30 min, 1hrs and 30 min, respectively, for the same activities.

Q. Hi-Choice Dressers has the option to hire one more employee of any category. Whom should it hire to get the maximum increase in the production capacity, assuming that it needs to stitch only A.R. Academy uniforms on that day?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 14

► 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.

Test: Arun Sharma Based Level 2: Time & Work - Question 15

If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

Detailed Solution for Test: Arun Sharma Based Level 2: Time & Work - Question 15

(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

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