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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:**

Solution:

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:**

Solution:

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**

Solution:

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.

QUESTION: 4

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

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**

Solution:

QUESTION: 5

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

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**

Solution:

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?

Solution:

► 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

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.

Solution:

► 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

QUESTION: 8

Sumit constructs a wall working in a special way and takes 12 days to complete it. If S_{n} is the length of the wall (in m) that he constructs on the nth day, then

S_{n} = 2n, 0 ≤ n ≤ 4

S_{n} = 8, for n = 5

S_{n} = 3n - 7, 6 ≤ n ≤ 12

Find the total length of the wall he constructs in the first 10 days.

Solution:

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

► Total length =113 m

QUESTION: 9

Alok, Mithilesh, and Bimlesh started a work and after completing 1 / 5^{th} 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?

Solution:

► 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.

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?

Solution:

► 40% work in 40h → 60% work in 60h.

► Hence, working hours = 60 / 4 = 15 h

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. Find the maximum number of T.M. High School uniforms that Hi-Choice Dressers can complete in a day.**

Solution:

► For T.M. high school uniform, the 1^{st} set of uniform can be completed at the end of 2 hours. (1^{st} 30 min are required for cutting of 2 uniforms, next 1 hour for next part and last 30 min for 3rd part which will be after 2 hours).

► Then in every 30 min 2 more uniforms will be completed as there are 2 assistants to perform the 3rd step.

► So, there will be total 16 sets (1 set = 2 uniforms) by 2 assistants from 2 hour to 10 hour and adding the set received at the end of 2 hours.

► There must be total 17 sets of uniforms which is equal to 34 uniforms.

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. 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?**

Solution:

► 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.

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 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?**

Solution:

► 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

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. 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?**

Solution:

► 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.

QUESTION: 15

**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?**

Solution:

► 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.

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