XAT 2019

Let the sides of the smaller square be $$x$$ units and the sides of the larger square be $$y$$ units.

Construct diagonal HF on the larger square. Let the point of intersection of AD and diagonal be I and point of intersection of BC and diagonal be J.  

We know that since EFGH is a square, the diagonal will bisect the angle. Therefore $$\angle EHF=45^{o}$$

Observe $$\triangle AEH$$ and $$\triangle DHG$$ : 

AE=AH=DH=DG = $$x$$ units and EH=HG=$$y$$ units. 

Thus by SSS property, $$\triangle AEH \cong \triangle DHG$$

Thus, we know that $$\angle$$AHE=$$\angle$$DHG

Also, $$\angle$$AHE+$$\angle$$DHG+$$\angle$$AHD= $$90^{o}$$ (angle of a square)

$$ \Rightarrow (2 \times \angle AHE)+ 60^{o} = 90^{o}$$ (angle of an equilateral triangle)

$$ \Rightarrow (2 \times \angle AHE) = 30^{o}$$

$$ \Rightarrow \angle AHE = 15^{o}$$

Since $$\angle AHE = 15^{o}, \angle AHI = 45^{o}-15^{o} = 30^{o} $$

Since $$30^{o}= \dfrac{60^{o}}{2}$$ we can say that HF is the angle bisector of $$\angle$$AHD 

Since $$\triangle$$AHD is an equilateral triangle, we know that the angle bisector, median and altitude will all be the same line ie HI in this case. 

By symmetry, we know that JF will also be the angle bisector, median and altitude. 

Also, length of altitude of equilateral triangle ie HI and JF = $$\dfrac{\sqrt{3} \times x}{2} $$

Length of diagonal HF = $$\sqrt{2} \times y$$

From the figure, we can express the length of diagonal HF as given : 

$$ \sqrt{2} \times y = \dfrac{\sqrt{3} \times x}{2} + x + \dfrac{\sqrt{3} \times x}{2} $$

$$ \Rightarrow \sqrt{2} \times y = (\sqrt{3} \times x) + x $$

$$ \Rightarrow \sqrt{2} \times y = x \times (\sqrt{3} + 1) $$

$$ \Rightarrow y = x \times (\dfrac{\sqrt{3} + 1}{\sqrt{2}}) $$

Ratio of areas asked = $$(\dfrac{y}{x})^{2} = (\dfrac{\sqrt{3} + 1}{\sqrt{2}})^{2} = \dfrac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$$

Alternate Explanation: 

Let AB=BC=CD=DA=x, then the area the area of square ABCD will be $$x^2$$

Since AHD, DGC, CFB, and BAE are equilateral triangles with side x, the sum of area of these traingles will be $$4\times\ \frac{\sqrt{\ 3}}{4}\times\ x^2=\sqrt{\ 3}x^2$$

We know that angle HAD=60, angle DAB=90, and angle BAE=60, thus angle HAE=360-(60+90+60)=150

Area of triangles HAE+HDG+GCF+FBE=4* area of triangle HAE

In HAE, AH=AE=x, and angle HAE=150, thus area will be $$\frac{1}{2}\times\ x\times\ x\times\ \sin\ 150$$

sin 150=sin(180-150)=sin 30=0.5

Thus area of 4 triangles will be 4*0.5*x*x*0.5=$$x^2$$

Thus, total area= $$x^2+\sqrt{\ 3\ }\ x^2+x^2$$=$$x^2\left(2+\sqrt{\ 3}\right)$$

Thus area of EFGH/area of ABCD=$$2+\sqrt{\ 3}$$

In order to accommodate maximum number of points on the circular disk, all the points should be placed on the boundary of the disk as any point if placed inside the boundary would be less than 1 metre apart from the centre of the disk. Also, all the points have to be exactly 1 metre apart from each other to place maximum number of  points. Hence, a maximum of 6 points can be placed on the boundary of the disk (each adjacent points placed 60 degrees apart) and 1 point at the centre. 

Hence, a total of 7 points can be placed on the disk.

The figure can be redrawn as shown above. 

$$\because $$ AB and OP are parallel, POAC is a rectangle. 

In right $$\triangle$$QCB, QB = $$\sqrt{5}$$

$$\triangle$$QBO is an isosceles right angled triangle with $$\angle $$OQB = $$ 90°$$ (By Pythagoras theorem)

$$\therefore $$ $$\angle $$QOB =  $$45°$$

The critical points of the function are -4, -6, -8, ... , -98 ( 48 points).

For all integers less than -98  and greater than -4 f(x) > 0 always .

for x= -5, f(x) < 0 

Similarly, for x= -9, -13, ...., -97 (This is an AP with common difference -4)

Hence, in total there are 24 such integers satisfying f(x)< 0.

$$\triangle$$ ABC can be represented as shown above where AD $$\bot$$ BC

Given, Sin b = $$\frac{3}{5}$$

Hence, AD = $$\frac{3x}{5}$$ and BD = $$\frac{4x}{5}$$

Now, BC = 2 BD = $$\frac{8x}{5}$$

$$\therefore$$ area of $$\triangle$$ ABC = $$\frac{1}{2}$$ * BC * AD = $$\frac{1}{2}$$ * $$\frac{8x}{5}$$ * $$\frac{3x}{5}$$ = $$\frac{12x^{2}}{25}$$

Only option E satisfies.

Hence, option E.

We know that,

$$x^{3} - 1 = (x - 1)(x^{2} + x + 1)$$

Since, $$x^2 + x + 1 = 0$$

$$\therefore $$ $$x^{3} - 1$$ = 0 

=> $$x^{3}$$ = 1

Now, $$x^{2018} + x^{2019}$$ 

= $$(x^{3})^{672} * x^{2}$$ + $$(x^{3})^{673}$$

= $$1^{672} * x^{2}$$ + $$1^{673}$$

= $$x^{2}$$ + 1

= -x

Hence, option C.

Given, m and n are two positive integers having product less than 100

Statement 1:

mn is divisible by 6 consecutive integers. This means mn must be divisible by the LCM of these 6 integers its multiples.

Only numbers from 1 to 6 satisfy this with LCM 60. Any other set of 6 consecutive integers clearly exceeds 100 as its LCM.

Only such number satisfying the condition is 60.

60 can further be expressed as the product of two positive integers in the following ways:

1 x 60, 2 x 30, 3 x 20, 4 x 15, 5 x 12 and 6 x 10.

So, no unique values of m and n can be determined from the above.

Statement 2:

m + n is a perfect square.

From this statement alone too no unique set of solutions can be determined.

From Statements 1 and 2:

Only the pair 6 x 10 satisfies the condition.

Hence, both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.

From statement 1:

Red + Blue + Green = 17

Since, this ensures that there are at least 3 green marbles, so, remaining marbles must be red and blue. 

$$\therefore $$ Red + Blue = 17 - 3 = 14

Blue = 14 - Red = 14 - 8 = 6

But, this doesn't give us the total number of green marbles. So, statement 1 alone is not sufficient.

From statement 2:

Red + Blue + Green = 19

Since, this ensures that there are at least 2 red marbles, so, remaining marbles must be blue and green 

$$\therefore $$ Blue + Green = 19 - 2 = 17

Again, Green = 17 - Blue = 17 - 8 = 9.

Hence, statement 2 alone is sufficient to answer the question.

Total number of Physics major from all locations = 225

$$\therefore $$ Total number of students = $$\frac{225}{0.18}$$ = 1250

Number of students majoring in chemistry = 14 % * 1250 = 175

Hence, option B.

Total number of Physics major from all locations = 225

$$\therefore $$ Total number of students = $$\frac{225}{0.18}$$ = 1250

Number of engineering students = 20% * 1250 = 250

So, basis the given condition we have

$$\frac{18}{225}$$=$$\frac{x}{250}$$

=> x = 20

Hence, option D

(Note: This question has been presented as was in the actual paper and has been solved by the aid of visual inference)