XAT 2017 Question Paper

There are 3 parallel lines AB, CD, and EF, in that order. 
Let the distance between CD and AB be 2x.
It has been given that the distance between CD and EF is x.

A quadrilateral PQRS is formed such that P is on AB, Q and S are on CD, and R is on EF. 
Also, the length of SR is the least possible value it can take. Therefore, SR must be perpendicular to the parallel lines. 

Area of quadrilateral PQRS = Area of triangle PQS + Area of triangle SRQ = 30 square cm. 

Area of triangle PQS = 2*area of triangle SRQ (Since they rest on the same base and height of SRQ is half the height of PQS)
=> Area of triangle SRQ = 10 square cm. 
Let the length of SQ be $$b$$. We know that SR= $$x$$
$$0.5*x*s = 10$$
=> $$xs=20$$
$$s=20/x$$
We do not have any other detail to evaluate the value of the expression. But, we have been given that d1 and d2 are integers. Therefore, the least value that 'x' can take is 1. 
The least value that S can take is 1. 
By Pythagoras theorem, $$QR =\sqrt{s^2 + x^2}$$
$$QR=\sqrt{20^2+1}$$
$$QR=\sqrt{401}$$
Therefore, the value of QR will be slightly greater than 20. Therefore, option E is the right answer. 

First, let us construct the rectangle using the given information.

Let the length of the rectangle be 'l' and the breadth of the rectangle be 'b'.
Area of rectangle ABCD = lb. 
We have to find the area of the triangle APS.

We consider length as 4cm and breadth as 2cm. ( Answer is irrespective of length and breadth) 

In triangle QRD

RS/RQ=XS/DQ

1/4=XS/2cm

XS= 0.5 cm

SY= 4cm-0.5cm= 3.5cm

We can now find the areas of all the sections.

area (RDQ)= 1/2 x 1cm x 2cm= $$1cm^2$$

area (ABP) =1/2 x 1cm x 4cm = $$2cm^2$$

area (ARS) =1/2 x 1cm x 0.5cm = $$0.25cm^2$$

area (PSQC) = Area (PSZ)-Area(QCZ)= 1/2 x 2cm x 3.5cm  - 1/2 x 1cm x 2cm =

                       =   $$2.5cm^2$$

Area (ABCD)= 4cm x 2cm = $$8cm^2$$

Area (APS) = 8 -1 -2 - 0.25 - 2.5 = $$2.25cm^2$$ 

Area (APS) / Area (ABCD) = 9/32  =  36/128

Hence answer (A).

Since only the specialist bowlers complete their quota of 4 overs, the maximum over a non-specialist bowler can bowl is three.

Total number of overs bowled by non-specialist bowlers = 8

If one of the three bowlers bowls 1 over, the other two will bowl 3 and 4 overs. Since the maximum over a non-specialist bowler can bowl is three, this is an invalid case.

If one of the non-specialist bowlers bowls 2 overs, the other two will bowl 3 overs each.

Thus, the minimum a non-specialist bowler can bowl is two overs.

Since the above calculation did not need either of S1 and S1,

The correct answer is E.

As per  Statement given  paragraph the economy rates are 6,6,7,9 which exclude best bowler and worst bowler

And  Economy rates of specialist bowlers are lower than that of nonspecialist bowlers.

Let’s take economy rates of specialist bowlers as 5, 6, 6.

Hence their cumulative runs would be (5 × 4) + (6 × 4) + (6 × 4) = 68
As per Statement 2, Total runs for non specialist bowlers =69. And their
economy rates are 7, 9. Let’s take economy rate of worst bowler be x
 (7 × 3) + (9 × 2) + (x × 3) = 69        

=>   x =10
For the lowest values of the specialist bowler we take 4. Then value of the worst bowler will be less than 9 which is incorrect.

Hence only a single case is possible.
So we can find the economy rate of worst bowler using both the statements.
Hence answer is option D. 

Profit  = selling price - raw materials cost - selling cost - labor cost 

Hence, raw materials cost = selling price - profit - selling cost - labor cost 

A worker takes 2 hour to produce a kurta.

Labor cost for Kurta = 2 * (Labor cost/ hour)

Raw materials cost per Kurta

for factory 2,    5300-800-45-400*2 = 3655

for factory 3,    5800-900-60-550*2 = 3740

for factory 4,    5500-800-68-450*2 = 3732

for factory 5,    5400-600-75-600*2 = 3525

for factory 6,    6000-875-65-400*2 = 4260

for factory 7,    4900-500-85-350*2 = 3615

for factory 8,    5300-600-70-420*2 = 3790

Sales margin = Profit per kurta/ selling price per kurta

for factory 2, sales margin = 800/5300 = 0.151

for factory 4, sales margin = 800/5500 = 0.145

for factory 5, sales margin = 600/5400 = 0.111

for factory 6, sales margin = 875/6000 = 0.146

for factory 7, sales margin = 500/4900 = 0.102

Factory 7 has lowest sales margin.

Profit  = selling price - raw materials cost - selling cost - labor cost 

Since selling cost is zero.

New Profit  = Original Profit + Selling cost

Profit per Kurta

for factory 1,    775+60 = 835

for factory 2,    800+45 = 845

for factory 3,    900+60 = 960

for factory 4,    800+68 = 868

for factory 5,    600+75 = 675

for factory 6,    875+65 = 940

for factory 7,    500+85 = 585

for factory 8,    600+70 = 670

Arranging all factories in decreasing profits,

Factory (3>6>4>2>1>5>8>7)

Factories 3, 6 and 4 will ensure maximum profit.

Profit  = selling price - raw materials cost - selling cost - labor cost 

Hence, raw materials cost = selling price - profit - selling cost - labor cost 

Raw materials cost per Kurta 

for factory 2,    5300-800-45-400*2                      ....(1)

After technology is introduced, a worker takes 1.5 hour to produce a kurta.

Labor cost per Kurta = 1.5 * (Labor cost/ hour)

New Profit          = selling price - raw materials cost - selling cost - new labor cost 

Profit per kurta 

for factory 2       = 5300-(5300-800-45-400*2)-45-400*1.5 = 800 + 400*0.5   =  Original Profit per kurta + 0.5*Labour cost per hour

                                                                                     = 1000

Similarly,

for factory 3,    900+550*0.5 = 1175

for factory 4,    800+450*0.5 = 1025

for factory 5,    600+600*0.5 = 900

for factory 6,    875+400*0.5 = 1075

Factory 3 will yield highest profit.

Histogram is a graph which depicts the frequency of occurence of values in different ranges.
For example, lets consider 50 students attended an exam. The distribution of number of students in different score ranges is depicted by the following histogram

In this question we have histograms for different personality dimensions. 
However the x and y co-ordinates are not given. 
Since the y co-ordinate gives us frequency/number of occurences of a certain value, we can use a histogram to calculate mean median and mode even without details of x-y co ordinate axes

Option A : "Extraversion" has 2 modes.

We can calculate mean of a histogram table by averaging all the values

We can calculate the median of a histogram table by arranging the histograms in ascending order and the bar/value which is in the middle will get the median.

Median for "arousal_plac" = -3

Avergae for "arousal_plac" = (-4-3-2+1+0-3-4)/7 = -15/7 = -2.143

Median for "arousal_caff" = 0

 Average for "arousal_caff" = (-1.5-1.5+2+1.75+1.5+1.75+1+0-0.5-2-2)/11 = 0.0454

Median for "performance_plac" = 0.075

Average for  "performance_plac" = (0.025+0.075+0.05+0.1+0.2)/5 = 0.09

Median for "performance_caff" = 0.075

Average for "performance_caff" = (0.05+0.25+0.09+0.075+0.08+0.075+0.085+0.1+0.1+0.15+0.18)/11 = 0.1122

+1 being the extreme linear positive relation and -1 being the extreme linear negative relation. Relationship is weakest when the value is closest to 0.
Among the given options, A is the weakest.

OR

Among the given plots, scattered plot between "extraversion" and "performance_caff" is the most scattered. Therefore weakest relationship.