IIFT 2018

$$\alpha\beta\ \ =\ 14$$

$$\alpha\ +\beta\ \ =\ 8$$

$$(1 + \alpha + \beta^2)(1 + \beta + \alpha^2)$$ = $$1+\beta\ +\alpha\ ^2+\alpha\ +\alpha\ \beta\ +\alpha\ ^3+\beta\ ^2+\beta\ ^3+\left(\alpha\ \beta\ \right)^2$$

 $$\alpha\ ^2+\beta\ ^2=\left(\alpha\ +\beta\ \right)^2-2\alpha\ \beta\ \ =\ 64-28\ =\ 36$$

$$\alpha\ ^3+\beta\ ^3=\left(\alpha+\beta\ \ \right)^3-3\alpha\ \beta\ \left(\alpha\ +\beta\ \right)=512-336=176$$

Hence $$1+\beta\ +\alpha\ ^2+\alpha\ +\alpha\ \beta\ +\alpha\ ^3+\beta\ ^2+\beta\ ^3+\left(\alpha\ \beta\ \right)^2$$ = 1 + 8 + 36+176+14 + 196 = 431

The equation becomes

==>$$\frac{1}{\log_x yz + log_x x} + \frac{1}{\log_y xz + log_y y} + \frac{1}{\log_z xy + log_z z}$$

$$\frac{1}{\log_x xyz} + \frac{1}{\log_y xyz} + \frac{1}{\log_z xyz}$$

$$log_{xyz} xyz$$

=1.

Part of the work  completed on Monday, Tuesday and Wednesday = 1/9 +1/12 + 1/15 = 47/180

Part of the work completed at the end of 9 days of work = 47(3)/180 = 47/60

remaining part of the work= 13/60 = 39/180

Part of the work completed by ram on 10th day = 1/9 = 20 /180

Part of the work completed by Rvi on 11th day = 1/12 = 15/180

remaining part of the work = 4/180

Ratan can do 12//180th of work on 12th day. Hence he needs 1/3rd of the 12th day.

Proportion of the completed work done by Ravi = 3(1/12) + 15/180 = 1/3

 Let the side of the square x be denoted by $$S_x$$

Let the radius of the circle y be denoted by $$r_y$$ and diameter by $$d_y$$

$$S_{s1}=4$$

$$r_{c1}=\frac{d_{c1}}{2}=\frac{4\sqrt{\ 2}}{2}=2\sqrt{\ 2}$$

$$S_{s2}=d_{c1}=4\sqrt{\ 2}$$

$$r_{c2}=4$$

..

.

.

A1 = $$\pi\ r_{c1}^2-S_{s1}^2\ =\ 8\pi\ -16$$

A2 = $$16\pi\ -32$$

A3 = $$32\pi\ -64$$

Sum = $$\frac{\left(8\pi\ -16\right)\left(2^n-1\right)}{2-1}\ =\ \left(8\pi\ -16\right)\left(2^n-1\right)$$

Ratio = $$\left(2^n-1\right)$$

Since the value of n is unknown, the answer can be C or D

Speed of joseph = 50/60 = $$\frac{5}{6}$$ m/s

Let the radius of circular playground be r

Speed = distance/ time

Time to diametrically cross a semi-circular playground t1= $$\frac{2r}{\frac{5}{6}}$$ = $$\frac{12r}{5}$$

Time to cross the playground along the semi-circular path t2 = $$\frac{\pi\ r}{\frac{5}{6}}=\frac{\left(6\pi\ r\right)}{5}$$

t1-t2 = 48

$$\frac{12r}{5}-\frac{\left(6\pi\ r\right)}{5}\ =\ 48$$

=> r = 35m

hence diameter = 70m

Garima had x notes of Rs 2000. She gave x/2 notes of Rs 2000. She is left with x-x/2 = x/2 notes

x/2  + 30 = 1.75x/2

=> x= 80

So, she remains 80/2 = 40 Rs 200 notes

So she must be having 50-40 = 10 Rs 200 notes.

H - volume of hole
R - volume of remaining solid
T - Total volume
It is given, $$H\ =\ \frac{1}{3}R$$
H + R = T
T = 4H
$$\frac{1}{3}\pi\ \left(25\right)\left(2x\right)+\pi\left(25\right)\left(3x\right)=4\times\pi\ r^2\left(\frac{10}{3}x\right)$$
$$r^2=\frac{275}{40}$$
$$r\ =\sqrt{\frac{\ 55}{8}}$$
The answer is option D.

Tap A alone can fill the tank in 12 hours
Tap B alone can fill the tank in 12*1.5 = 18 hours
Tap C alone can empty the tank in 36 hours
Let the total volume of the tank be 36 units
A fills 3 units, B fills 2 units and C empties 1 units in an hour.
Let the total time taken be 't'
t(3) + (t-2)2 - 6(1) = 36
5t = 46
t = 9.2 hrs = 9 hours 12 minutes
The answer is option C.

Let the point intersecting inside triangle ABC be 'E'.
It is given, $$\ \angle\ AED$$ = 60, $$\ \angle\ ABC$$ = 45, $$\ \angle\ EBC$$ = 30 and BE = 100
At point E, $$\ \angle\ BEF\ =60,\ \angle\ DEF=90\ and\ \angle\ AED\ =\ 60$$
$$\ \angle\ AEB\ =\ 360\ -\ 60\ -\ 60\ -\ 90\ =\ 150$$
In triangle AEB, $$\ \angle\ EAB\ =\ 180\ -\ 150\ -\ 15\ =\ 15$$
$$\ \angle\ EAB\ =\angle\ EBA$$
Therefore, EA = 100 (as BE = 100)
EF = DC = 50
In triangle AED, 
$$ED^2+AD^2=AE^2$$
ED = x, AD = $$\sqrt{\ 3}$$x
$$4x^2=\left(100\right)^2$$
2x = 100
x = 50
Height of summit = $$\sqrt{\ 3}x\ +\ 50$$ = $$50\left(\sqrt{\ 3}+1\right)$$
The answer is option A.

Total cost=Fuel cost+ Driver’s cost
Fuel cost=Total distance × Cost of Diesel per km
Fuel cost=1000×[1/400 {(1000/x)+x} ]×65
Fuel cost = 162500/x + 162.5x
Now, substituting the values from the given choices, we find that at x=36, the total cost is minimum.
Hence. Option B is correct.