IIFT 2020 Question Paper

The slope of any line parallel to 2y - x - 5 = 0 would be 1/2
The slope of any line parallel to y + 2x - 7 = 0 would be -2

Product of slopes = -1 which shows that these 2 lines would be perpendicular i.e. we have a parallelogram whose diagonals are perpendicular to each other. This means that the quadrilateral is atleast a rhombus and it would be a square if we can get info that all sides are equal to each other. 

Let the original amount of money i.e. Principal taken by Suhani = 2P
The amount of money that she owes back to the loan company at the end of 2 years = 2P + 0.24*2P = 2.48P

Now, she loans amount P to Vishamber.
Amount returned by Vishamber to Suhani under given conditions = $$P\cdot\left(1.08\right)^4$$ = 1.36P

She loans amount P to Kalawati.
Amount returned by Kalawati to Suhani at end of 2 years under given conditions = $$P\cdot\left(1.12\right)^2=1.2544P$$

Amount received by Suhani at end of 2 years = 1.36P +1.2544P = 2.6144P
Earning of Suhani = 2.6144P - 2.48P = 0.1344P

As she paid 0.48P for this earning of 0.1344P.

The percentage earning can be calculated as (0.1344P/0.48P)*100 = 28%

Hence the answer is Option B 28%.

Probability that recession will happen = 0.3 and probability that value will increase in this case = 0.2
Expected probability of mutual fund value increase in this case = 0.3*0.2 = 0.06

Probability that recession will not happen = 1 - 0.3 = 0.7 and probability that value will increase in this case = 0.75 
Expected probability of mutual fund value increase in this case = 0.7*0.75 = 0.525

Hence, total expected probability that mutual fund value will increase = 0.585

As all 3 are mutually exclusive, we can assume (in one extreme case) that these 3 form the entire range of probability and hence,
0.25(1-x) + 0.5(1-2x) + 0.25 (1+4x) = 1 which gives us x = 0 
Hence, x is bound by 0 on one end. 

Also, none of the probabilities can be negative as any event probability is atleast 0. Hence, for the second event, we can see that x cnnot be greater than 0.5
hence, x is bounded from 0 to 0.5

Let the river flow from Mathura to Agra. Let the speed of boat in still water be S kmph

Hence, if journey is to be done in 6 hours, then:
$$\frac{48}{S+6}+\frac{48}{S-6}=6$$

Solving this equation, we get: (S-18)(S+2) = 0 which gives us S = 18 kmph

The situation given in the problem can be summarised as: 

By 30-60-90 theorem, we can say that P'M = 4 meters and MQ' = 3 meters
Also, triangle P'MQ' would be a right angled triangle. 
Hence, P'Q' would be the hypotenuse of length = root(16+9) = 5 meters

At exactly 10 AM, angle between the two hands = 60 degrees

Applying the concept of relative velocity, we can say that the minute hand has to travel 180 - 60 = 120 degrees more at a relative velocity of 5.5 degrees/minute 
Hence, time taken to make 180 degree angle between the 2 hands = 120/5.5 = 22 minutes 

Let the mixture so created to be sold was exactly one kg.
Hence, amount of first variety procured to contribute to this mixture = (2/5)*1000 = 400gm or 0.4 kg
Hence, amount of second variety procured to contribute to this mixture = (3/5)*1000 = 600gm or 0.6 kg

Let the cost price of second variety = Rs. a/kg and hence, cost price of first variety = Rs. 2a / kg
Hence, cost price of the entire mixture = 0.6a + 0.4*2a = 1.4a rupees / kg

Also, the mixture is sold at Rs. 28 per kg and it means a profit of 25% ==> CP per kg = Rs. (28/1.25) per kg

Equating both, we get a = 16, hence, price of the first variety = 32 rupees per kg

Using all the information in the question, the given figure becomes as shown below.

Since the perpendiculars from centre to chords of equal lengths are always equal, $$\triangle$$OMN becomes an equilateral triangle with MN = 6 m.

In $$\triangle$$MNP, PM = PN  [by symmetry]

Thus, using the cosine rule,

$$MN^2=MP^2+NP^2-2\times MP\times NP\times cos(120)$$

$$36=MP^2+MP^2+MP^2$$

$$MP^2=12$$

$$MP=2\sqrt 3$$

Thus, each of them has to travel a distance of $$2\sqrt 3$$ m in order to meet each other.

Hence, the answer is option C.

If ACB is a right angled isosceles triangle and AB is hypotenuse with length 7*(root 2) length, then sides AC and BC = 7 cm
Now, the circle inscribed is an incircle.
Radius of in-circle = {7+7-[7*(root 2)]}/2 = $$7-\frac{7\sqrt{\ 2}}{2}$$

Now, if center of circle is marked as O, then quadrilateral OZCY is a square. 
Hence, CY = CZ = $$7-\frac{7\sqrt{\ 2}}{2}$$

Hence, AZ = 7 - $$7-\frac{7\sqrt{\ 2}}{2}$$ $$=\frac{7\sqrt{\ 2}}{2}$$
Also, BX = BY and BY = BC - CY $$=\frac{7\sqrt{\ 2}}{2}$$
Hence, BX $$=\frac{7\sqrt{\ 2}}{2}$$

Hence, the given ratio becomes 1:1:[(root 2) - 1]