IIFT 2020

Let the labour employed by the company be 2$$x$$ and the capital be $$x$$.

Thus, the total cost of running production for an hour = $$500\times2x+1500\times x=2500x$$

Total cost for running the production for a month = $$5$$ Crores = $$5\times 10^7$$

Total number of hours the production is done = $$x=\frac{5\times 10^7}{2500}=20000$$

Since the production is cut in half, the total hours = $$0.5\times = 20000=10000$$

Labour costs decreased by 20% and the capital cost increased by 20%.

Thus, the total cost per hour = $$500\times 2x\times 0.8 + 1500\times x\times 1.2=2600x$$

Since $$x=10000$$, the total cost = $$2600\times 100000=2.6$$Crores.

Hence, the answer is option B.

The total number of possible ways from AB to EF = $$4\times 4\times 4\times 4=256$$

If a curved path is taken in a step, then the next step cannot be a curved path.

Let C denotes a curved path and S denotes a straight path.

Thus, the invalid paths are as follows: CCCC, CCCS, CCSC, CSCC, SCCC, CCSS, SCCS, SSCC. Thus, a total of 8 paths.

Total number of arrangements for these 8 paths = $$8\times 2\times 2\times 2\times 2=128$$

Thus, the total number of valid paths = $$256-128=128$$

Hence, the answer is option A.

Cubing the given equation, we have :

$$\left(a+b\right)^3=\ a^3+\ b^3+3\cdot a\cdot b\left(a+b\right)$$

$$11+2x+11-2x+3\cdot\left(11+2x\right)^{\frac{1}{3}}\left(11-2x\right)^{\frac{1}{3}}\left(a+b\right)$$

Given a+b = 2.

Hence :

22 + $$3\cdot\left(11+2x\right)^{\frac{1}{3}}\left(11-2x\right)^{\frac{1}{3}}\left(2\right)$$

= 22 + 6*($$\left(121-4x^2\right)^{\frac{1}{3}}$$) = 8

$$Cubing\ this\ we\ get\ :\ 121-4x^2=-\frac{343}{27}$$

$$4x^2=\ 121+\frac{343}{27}=\ \frac{3610}{27}$$

$$x^2\ =\ 33.42$$

One root would be $$\ +\ \sqrt{\ 33.42}\ and\ the\ other-\sqrt{\ 33.42}$$.

The product of the roots would be approximately -33.

Let Sumit's age at the time of graduating MBA [in 2013] be $$X$$ years. Thus, his wife's age that year will be $$X-3$$ years.

The average age of the family in 2013 = 41 years.

Thus, the total age of the family in 2013 = 7 x 41 = 287

Sumit got married in 2015. Thus, his wife's age in 2015 will added to the family's total.

The total age of family in 2015 = 287 + 14 + $$X$$ - 3 + 2 = $$X$$ + 300

In 2017, Sumit had a daughter.

The total age of the family in 2017 = $$X$$ + 300 + 8(2) = $$X$$ + 316

The total age of the family in 2019 = $$X$$ + 316 + 9(2) = $$X$$ + 334

We are given that the average age of the family in 2019 is the same as that in 2013.

Thus, $$X$$ + 334 = 41 x 9 = 369

$$X$$ = 369 - 334 = 35 years

Thus, the age of Sumit at the time of graduating MBA is 35 years.

Hence, the answer is option B.

Let the cost of fruit be Rs. 100 for 1000gm.

The fruit seller marks the price by 15% and then gives a discount of 8%. Thus, the selling price of fruit = Rs. 105.8

The fruit seller also cheats the customers by 10% of the weight. Thus, he sells 900gm for Rs. 105.8

Cost Price = $$\frac{100}{1}=Rs. 100/kg$$

Selling price = $$\frac{105.8}{0.9}=Rs. 117.56%$$

Percentage profit = $$\frac{117.56-100}{100}\times100=17.56\%$$

hence, the answer is option B.

From the first equation: $$16^x\cdot256^y=64$$ ==> $$4^{2x}\cdot4^{4y}=4^3$$ 
Comparing powers: 2x + 4y = 3 .....(A)

From second equation: $$3^{2x+y}=3^2$$
Comparing powers: 2x + y = 2 ......(B)

Solving A and B, we get option C as answer

Let the rate of interest of first borrowing be R% p.a. 
The principal of first borrowing is kept for an entire 12 month span.
Hence, simple interest payable on it at end of financial year = (12,00,000*1*R)/100 = 12,000*R rupees

Now, rate of second borrowing = 0.9R % p.a. 
The principal of second borrowing is kept for 8 months total till financial year end. 
Hence, simple interest payable on it at end of financial year = (9,00,000*R*8)/(12*100) = 5400R rupees

Now, 12000R + 5400R = 1,39,200
Hence, R = 8% p.a. 

It has been given that radii OE = OF = 2 cm. Consequently, AB = 4 cm {side of the square}. AC = $$4\sqrt{2}$$ cm

Further, let the radius of the smaller circles be $$r$$ cm. Then, AP = $$r\sqrt{2}$$ cm. Thus, AQ = $$r\sqrt{2}+r$$ or $$r\left(\sqrt{2}+1\right)$$ cm.

There are two such regions on line AC. Hence, we can rewrite AC = $$2\times\ r\left(\sqrt{2}+1\right)+4$$. This is equal to $$4\sqrt{2}$$ cm. On equating, we obtain $$r=\left(6-4\sqrt{2}\right)\ $$ cm. 

We have been asked to find the total area covered by the five circles = $$4\pi\ +4\times\ \pi\ \left(6-4\sqrt{2}\right)^2\ cm^2$$

Hence, Option A is the correct choice. 

Let the number of hours needed by Surojit working alone be S hours and the number of hours required by Debjani when she works alone be D. 
The time required to finish the work when they work together = $$\frac{S\cdot D}{S+D}$$

Hence, we can frame the given 2 conditions as: 
$$S-\frac{S\cdot D}{S+D}=\frac{21}{5}$$ ............(1)
$$D-\frac{S\cdot D}{S+D}=\frac{141}{5}$$ ..........(2) 

Solving the two equations simultaneously, we get S = 15 hours approx

Consider the instance when Suresh just starts from the office. 
Distance that Pradeep has already covered in that time = 50*1.5 = 75 km 

Now, by concept of relative velocity, time taken by Suresh to just overtake Pradeep = 75/(60-50) = 7.5 hours
Hence, the point of overtaking on the route is = 60*7.5 = 450 km from the office 

Subodh also overtakes Pradeep at the same point. Pradeep covers this 450 km distance in 9 hours. 
Subodh can cover this 450 km in 6 hours. 
Hence, Subodh can leave 9 - 6 = 3 hours after Pradeep does.