Data Interpretation Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 2
DIRECTIONS: Refer to the following information and answer the following questions.
Shyam Kumar multiplier is a smart mathematician. While negotiating has salary with his prospective employer he worked out and proposed a novel salary plan, which his employer accepted as he was one of the smartest salesman around. According to the plan, his salary would increase every day of his employment such that on any day his income would be two rupees more than the square of the number of the days he has been employed for with the company. Back home his wife Renu congratulated him on working out an excellent deal and then together they planned their expenses in tune with the new salary plan. As they expected their income to be increasing on a daily basis — they planned their expenses in a way that the expenses of any day would be met by the day’s income itself, moreover they planned to make a net saving (which would be the difference of incomeand the expenses of the day) out of the day’s income. Thus they expected their expenses to be one rupee more than twice the number of days Shyam Kumar would have been employed for on that day, with the company.
(1) On which day of his employment did Shyam Kumar Multiplier’s savings of a day exceed Rs. 100?
(2) Shyam Kumar Multiplier wanted to get into Limca Book of Records. He eyed the category which listed the person holding the record for the maximum salary increase in a single day. The person currently holding the record is Salim Khan who was given a raise of Rs. 12,000 in a single day. If Shyam Kumar were to beat his record he will have to remain in this job for a minimum of
(a) 5999 days
(b) 6000 days
(c) 6001 days
(d) 6002 days
(3) Shyam and Renu planned to put each day’s savings into their savings bank account. They also decided that they would regularly purchase Fixed Deposits in multiples of Rs. 300 from their savings in the account. They would purchase their first Fixed Deposit on
(a) 11th day
(b) 12th day
(c) 13th day
(d) 14th day
Salary = n^2 + 2 where n > 1 ………….. (1)
Expenses = 2n + 1 where n > 1 ……….. (2)
Saving = n^2 – 2n + 1 where n > 1 ………….. (3)
(1) (c) Putting the value of savings in equation (3), 100 = n^2 – 2n + 1 ⇒
n^2 – 2n – 99 = 0, we get n = 11.
Hence the savings would exceed Rs. 100 on 12th day.
(2) (c) Putting the value of n in eq.n (1)
Salary = n^2 + 2
When n = 6000
Salary = Rs. 36000002
When n = 6001
Salary = Rs. 36012003
∴ Increase in salary = 36012003 – 36000002 = Rs. 12001. Thus he will have to remain in his job for minimum 6001 days to break the record.
(3) (a) Putting different values of n starting from 1 in eqn 3 and adding all the savings up to 11th day,
Total saving = Rs. 385, so on 11th day they will buy a first fixed deposit.
DIRECTIONS: Refer to the following information and answer following questions.
Alord received a large order for stitching school uniforms from Mayflower school and Little Flower 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 10 hours a day. Each of the May-flower uniforms requires 20 min. for cutting the fabric, one hour for stitching, and 15 min. for stitching buttons and button holes, whereas the Little Flower uniform requires 30 min., 1 hour, and 30 min. respectively for these activities.
(1) What is the maximum number of little Flower uniforms that A lord can complete in a day?
(2) On particular day, Alord decided to complete 20 Little Flower uniforms. How many Mayflower uniforms can he complete on that day?
(3) If Alord decides to complete 30 Little Flower uniforms only and no other on a particular day, how many total man-hours will be idle?
(4) If he hires one more assistant, what is the maximum number of Mayflower uniforms that he can complete in a day?
(5) A lord has the option to hire one more employee of any category. Which category should he hire to get maximum increase in production capacity, assuming that he needs to stitch only Mayflower uniforms on that day?
(D) Cannot be determined
Refer to the following table
The time given in the table is for per uniform to be stitched.
(1) (C) No. of working hours for each person = 10 hour
∴ Total time available for cutters in a day = 20 hours = 1200 mins.
Max. No. of uniform that 2 cutters can cut for Little Flower in a day = 1200/30 = 40
Time required for stitching 40 uniform = 40 x 60 = 2400 min and Alord has 5 tailors for stitching
that means the Total available time is (5 X 10 x 60) = 3000 mins.
Hence maximum number of Little Flower uniforms can be completed in a day = 40.
(2) (A) To complete 20 Little Flower uniforms,
Time consumed by Cutters = 20 x 30 = 600 mins
Time consumed by Tailors = 20 x 60 = 1200 mins
Time consumed by Assistants = 20 x 30 = 600 mins
Time available for Cutters = (1200 – 600) = 600 mins
Time available for Tailors = (3000 – 1200) = 1800 mins
Time available for Assistant = (1200 – 600) = 600 mins
Max. No. of uniform that 2 cutters can cut for May Flowers uniform in a day = 600/20 = 30
Time required to stitch 30 uniforms = (30 x 60) = 1800 min and is exactly equal to the time available for tailors. Hence No. of Mayflower uniform that can be completed on that day is 30.
(3) (B) (1) To complete 30 Little Flower uniforms,
Time consumed by Cutters = (30 x 30) = 900 mins
Time consumed by tailors = (30 x 60) = 1800 mins
Time consumed by assistants = (30 x 30) = 900 mins
Time available for cutters = (1200 – 900) = 300 mins
Time available for tailors = (3000 – 1800) = 1200 mins
Time available for Assistant = (1200 – 900) = 300 mins
As they are not working in this available time it is the idle time.
Hence total man hours that are idle = (300 + 1200 + 300) = 1800 mins = 30 hours.
(4) (B) Hiring one more assistance increases the time available for stitching buttons and button holes.
Max. No. of uniform those cutter can cut for Mayflower in a day = 1200/20 = 60
Time required by Tailors to stitch 60 uniforms = (60 x 60) = 3600 mins, whereas time available is 3000 mins only.
Hence maximum No. of uniform that can be stitched in a day = 3000/60 = 50
Time consumed by assistant to stitch buttons to 50 uniform = (50 x 50) = 750 mins
Time available for assistant = (3 x 10 x 60) = 1800 mins.
Hence 50 uniforms can be completed in a day.
(5) (A) Max. number of uniforms that can be cut by those 2 cutters in a day for MayFlower is 60. And the buttons and buttonholes can be easily stitched by the 2 assistant for these 60 uniforms in a day in the time available for them. While 5 tailors can stitch only 50 uniforms in the available time.
Hence, Alord will hire 1 tailor to get maximum increase in production capacity in a day.