CAT 1996 Question Paper

Percentage capacity of production for Lipton coffee = 64.8%
Let's say maximum production = $$x$$
So 1.64 = $$x \times \frac{64.8}{100}$$
or $$x = 2.53$$  

Maximum unutilized capacity => Minimum unitized capacity

MAC has the least utilized capacity.

Let's total coffee production is = x

So $$x \times \frac{61.3}{100} = 11.6$$

x = 18.9 (nearly)

The price per kilogram sales of each of the four brands is given below.

Brooke Bond - Rs. 122.16
Nestle - Rs. 131.77
Lipton - Rs. 121.03
MAC - Rs. 118.71

Hence, the correct answer is Nestle

Summation of sales value of all coffees (excluding others) = 31.15 + 26.75 + 15.25 + 17.45 = 90.6

Total value = 132.8

Others = 132.8 - 90.6 = 42.2

% share for others = $$ \frac{42.2}{132.8} \times 100 $$ = 31.77 $$\approx$$ 32

In given table, value of cost for 5th month = $$(N) \times (C)$$ (No.of employer for 5th month = N and cost per person for 5th month = C)
So value of cost for 5th month before overrun = $$4 \times 10 = 40$$
value of cost for 5th month after overrun = $$ 5 \times 20 = 100 $$
Change = 60
%change = $$\frac{60}{40} \times 100$$ = 150

As coding is done in 6-8 months,
So total person employed = 4+5+5 = 14
cost per person for 6-8 month = 10000 
Total cost will be = $$(14 \times 10000)$$ = 140,000

In new technique, increment in 5th month cost ('000)  = $$(5 \times 20 - 4 \times 10)$$ = 60 

Total incurred cost for testing = $$(4+1)\times 15 = 75$$
Incurred cost for specification = $$(2+3) \times 40 = 200$$
Incurred cost for coding = $$(4+5+5) \times 10 = 140$$
Incurred cost for design = $$(3+4+5) \times 20 = 240$$

Hence design is most expensive for the company.

Lowest cost per person is for coding and maintenance i.e. 10 rs.
But for consecutive five months avg. for coding and any other will be more than consecutive 5 months for maintenance.
So avg. cost for five months of maintenance is = $$\frac{(3+3+1+1+1) \times 10}{9} = 10$$
Which is lowest among all others