IIFT 2018 Question Paper

Options A and D have errors in parallelism. Option B mentions 'from...and to' which is not correct. Option C is correct, considering all the aspects. The answer is option C.

The idea conveyed by the whole sentence is that while the square root of a negative number has no real value, equations involving such numbers may have practical applications. Due to the usage of "not necessarily," the answer we are looking for should be phrased as "such equations have no practical applications" and the double negation will give the sentence a positive essence.

The only answer that conveys the required sense is B.

B is an opening statement in which it is suggested that that RBI has taken over several aspects of monetary policies from the government. A indicates that despite having certain powers, the final decision rests with the government. C elaborates A with an example. D poses a question for the reader to decide after reading all the above statements.

Hence the correct order is BACD. 

Sentence A gives the basic premise of what is being discussed in the passage i.e. that all demands are subject to change. Thus, sentence A is apt as the opening sentence of the paragraph. D further discusses the individual demands and C discusses aggregate demand. DC discuss the demands chronologically and hence ADC is the mandatory pair. B is driving the paragraph on a tangential nuanced idea apart from demand. Thus ADCB is the right choice.

Z's funds = 10000 x 100/5 = 2 lakhs

Y has funds which are 2 times the funds of Z => 4 lakhs

X has given loan to Y which is equal to 10% of funds of Y

y's funds + 10% of Y's funds = 4 lakhs

Y's funds = 4/1.1 lakhs

Y's donation = 5/100 (4/1.1lakhs) = 18181.81 

Let height of liquid level before the ball was dropped be h

The radius of the surface of the liquid level then is 4cm

$$\frac{h}{3}\pi\ \left(4^2+3^2+4\cdot3\right)\ =200cc$$

=> h = $$\frac{600}{37\pi\ }$$

The liquid level raises by 0.3h as a result of the solid metal ball being dropped.

By similarity, the radius of the surface of the liquid level after metal ball is dropped is 4.3cm

Volume of he solid metal ball = $$\frac{1.3h}{3}\left(4^2+3^2+4.3\left(3\right)\right)\ -\ 200$$ = 83.82cc

The base on any one of the lines can be formed in $$5_{C_2}=10\ ways$$

To form a triangle, the base has to be joined to any of the 15 points on the other lines

Number of triangles that can be drawn when the base is on any of the lines = 10 *15 =150

and number of triangles that can be drawn totally = 150*4 = 600

S is the midpoint of QR. So, area of PSR = $$\frac{1}{2}\left(Area\ of\ PQR\right)\ =\ \frac{5.8}{2}=2.9$$

PT||XS

So area of XSP = area of XST ( same height and same base)

Adding area of XSR on both sides, we have area of PSR = Area of RTX

=>  area of PSR = Area of RTX = 2.9 sq.cm

Q(x,y) = x^2- xy + y^2.$$

Q(9,4) = $$9^2-4\cdot9+4^2=81+16-36=61$$

P(7,61) = $$7^2+7\cdot61+61^2=\ 49+427+3721=4197$$

PA = QB and A=B = $$90^{\circ\ }$$

=> PQBA is a rectangle.

Area of the shaded region = (Area of PQBA - Area of the segment PRQ)

$$PQ=AB=25\sqrt{\ 5}$$

$$PO=QO=\frac{PQ}{\sqrt{\ 2}}=25$$ = radius of the circle.

The perpendicular from O to PQ is $$\frac{25}{\sqrt{\ 2}}$$

Distance from the foot of the perpendicular to AB = $$30-\frac{25}{\sqrt{\ 2}}$$

Area of PQBA = $$\left(30-\frac{25}{\sqrt{\ 2}}\right)\left(25\sqrt{\ 2}\right)=750\sqrt{\ 2}-625$$

Area of segment PRQ = Area of sector OPRQ - Area of triangle OPQ = $$\frac{90}{360}\pi\ 25^2-\frac{1}{2}25^2$$

Area of shaded region = $$\left(750\sqrt{\ 2}-625\right)-\left(\frac{90}{360}\pi\ 25^2-\frac{1}{2}25^2\right)$$

= option C