IIFT 2019

It is given that by using the escalator, one can get out of the station in 80 sec.

Let us consider the number of steps =80

Speed of the escalator = 1 step/sec

When the escalator was not working, the time taken by a person = 50 sec

The speed of the person = 80/50=8/5 steps/sec

Time taken by a person walking down a moving escalator = $$\ \frac{\ 80}{\ \frac{\ 8}{5}-1}$$

=$$\ \frac{\ 80}{\ \frac{\ 3}{5}}$$

=2.22 min

B is the correct answer.

$$x=8-\sqrt{32}$$ and $$y=2+\sqrt{2}$$,

We have to find the value of $$\left(x+\frac{1}{y}\right)^2$$

$$\left(8-\sqrt{\ 32}+\ \frac{\ 1}{2+\sqrt{\ 2}}\right)^2$$

$$(\ \frac{\ 8\left(2+\sqrt{\ 2}\right)-\sqrt{\ 32}\left(2+\sqrt{\ 2}\right)+\ \ 1}{2+\sqrt{\ 2}})^2$$

$$\left(\ \frac{\ 9}{2+\sqrt{\ 2}}\right)^2$$

$$\ \frac{\ 81}{6+2\sqrt{\ 2}}$$

$$\left(\ \frac{\ 1}{y}\right)^2=\ \left(\frac{\ 1}{2+\sqrt{\ 2}}\right)^2$$ = $$6+2\sqrt{\ 2}$$

=$$\left(\ \frac{\ 2-\sqrt{\ 2}}{2}\right)^{^2}$$

=$$\ \frac{\ 6-4\sqrt{2\ }}{4}$$

=$$\ \frac{\ 3-2\sqrt{2\ }}{2}$$

$$x^2=64+32-64\sqrt{\ 2}$$

=$$96-64\sqrt{\ 2}$$

=32($$3-2\sqrt{\ 2}$$) = 32*$$2y^2$$

we get, $$x^{2\ }=\ 64y^2$$

$$\ \frac{\ 81}{y^2}=\ \frac{\ 81}{64}x^2$$

D is the correct answer.

Alternative solution,

$$xy=\left(8-\sqrt{\ 32}\right)\left(2+\sqrt{\ 2}\right)=4\sqrt{\ 2}\left(\sqrt{\ 2}-1\right)\times\ \sqrt{\ 2}\left(\sqrt{\ 2}+1\right)=8\left(2-1\right)=8$$

(As $$xy=8\ \longrightarrow\ y=\dfrac{x}{8}$$)

$$\left(x+\frac{1}{y}\right)^2=\left(\frac{\left(xy+1\right)}{y}\right)^2=\left(\frac{\left(xy+1\right)\times\ x}{8}\right)^2=\left(\frac{\left(8+1\right)\times\ x}{8}\right)^2=\frac{81}{64}\times\ x^2$$

Centroid divides orthocenter and circumcenter in the ratio of 2:1

The co-ordinates of orthocentre and the centroid of a triangle ABC are (—5, 7) and (5, 5)

Let the co-ordinate of circumcenter be (x,y)

5=$$\ \frac{\ 2x-5}{3}\ =\ 5$$ and $$\ \frac{\ 2y+7}{3}\ =\ 5$$

x=10 and y = 4

Hence the circumcenter = (10, 4)

Area of the tunnel to be filled with concrete = 16*8-$$\pi\ \left(32\right)$$

Length of the length along which the tunnel has to be filled with concrete = 500m

Volume = 16*8-$$\pi\ \left(32\right)$$*500

= $$32(4-\pi)$$*500

Cost of concrete per $$1 m^3$$ = 1000

Cost of $$32(4-\pi)$$*500 = $$32(4-\pi)$$*500*1000

=$$16(4-\pi)*10^{6}$$

C is the correct answer.

Let the value of MN be p

Since NP is the angular bisector

NP= $$\ \frac{\ 2\times\ MN\times\ NL}{MN+NL}\cos\theta\ \ $$

2x=$$\ \frac{\ 2\times\ p\times\ 3x}{p+3x}\cos60$$

p=6x

NP: NL: MN = 2:3:6

B is the correct answer.

Derivation for the length of angular bisector:

Let CP be the angular bisector of $$\angle\ ACB$$. And lengths of AB,BC,CA, CP,AP,PB be c,a,b,x,m,n respectively.

From triangles CPB and CPA using cosine rule we get:

$$n^2=a^2+x^2-2axCos\left(C\right)\ and\ m^2=b^2+x^2-2bxCos\left(C\right)$$

According to angle biscetor rule: $$\ \frac{b}{m}=\frac{a}{n}$$

.'.$$\frac{a^2}{b^2}=\frac{n^2}{m^2}$$= $$\frac{a^2+x^2−2axCos(C)}{\left(b^2+x^2−2bxCos(C\right)}$$.

$$\frac{a^2}{b^2}=\frac{a^2+x^2−2axCos(C)​}{b^2+x^2-2bxCos\left(C\right)}$$

Upon equating we get, $$(a^2−b^2)x−2ab(a−b)Cos(C)=0$$

since x$$\ne\ 0$$,

x=$$\frac{2ab(a−b)Cos(C)}{a^2−b^2}$$

.'.x=$$\frac{2abCos(C)}{a+b}$$.

Let x be the number of women who decided to execute the interior and exterior decoration of the society in a week’s time

Let the efficiency of each of them be 1

Total work = 7x

Number of women who worked on first day = x

Number of women who worked on second day = x-11

Number of women who worked on third day = x-22

....

On similar lines, the number of women who worked on $$n^{th}$$ day = x-(n-1)11

Number of women who worked on day 12=x-121

x+x-11+x-22+........x-(n-1)11 = 7x

12x-(11+22+....121)=7x

5x=$$\ \frac{\ 12}{2}\left(2\times\ 0+11\times\ 11\right)$$

x=145

C is the correct answer.

The sum accrued by Joseph if he had taken the job = (80000*3+120000*3+150000*4)*12

=144*$$10^5$$

If Joseph has taken the MBA program, tuition fee for the program = 1600000+2000000

Sum accrued post MBA = 200000*12*4+250000*12*4=21600000

Net amount = 21600000-3600000 = 180*$$10^5$$

Net gain if Joseph had taken MBA over job 10 years down the lane= 36*10^5

Percentage of gain = $$\ \frac{\ 36\times\ 10^5}{144\times\ 10^5}$$

=25%

B is the correct answer.

Let us consider angle CAB=2x and angle ACB=5x

From Alternate segment theorem,

angle AEC = 2x

In triangle AEC, 60+180-5x+2x=180

3x=60

x=20

In triangle ABC, angle ABC = 180-7*20=40

A is the correct answer.

Let the total customers be 100.

Number of customers who said that they will return = 76x

Number of customers who said that will not return = 24x

Number of customers among those who said that they will return rated the restaurant as good = 76x*0.36

Number of customers among those who said that they will not return rated the restaurant as good = 24x*0.32

Number of customers who rated the restaurant as good = 76x*0.36+24x*0.32 = 35.04x

Percentage of customers who rated the restaurant as good = approx 35%

B is the correct answer.

Two couples can be selected in $$^4C_2=6$$ ways.

The four couples can be seated in $$3!=6$$ ways.

Thus, the total ways of seating arrangement = $$6\times 6=36$$

Hence, the answer is option A.